Elements of Pure and applied mathematics

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International Series in Pure and Applied Mathematics

WILLIAM TED MARTIN, Consulting

Editor

ELEMENTS OF PURE AND APPLIED MATHEMATICS

International Series in Pure and Applied Mathematics

WILLIAM TED MARTIN, Consulting Editor

AHLFORS

Complex Analysis

BELLMAN

Stability Theory Advanced Calculus

BUCK

of Differentia]

Equations

CODDINGTON AND LEviNsoN

GOLOMB & SHANKS

Theory of Ordinary Differential Equations Elements of Ordinary Differential Equations

GRAVES

The Theory

GRIFFIN

Elementary Theory

HILDEBRAND

of Functions of Real Variables of

Numbers

Introduction to Numerical Analysis

HOUSEHOLDER

Principles of Numerical Analysis

LASS

of

*

LASS

Elements

Pure and Applied Mathematics

Vector and Tensor Analysis

An

LEIGHTON

NEHARI

Introduction to the Theory of Differential Equations

Conformal Mapping

NEWELL

Vector Analysis

ROSSER Logic for Mathematicians RUDIN Principles of Mathematical Analysis SNEDDON Elements of Partial Differential Equations

SNEDDON STOLL

Fourier Transforms

Linear Algebra and Matrix Theory

WEINSTOCK

Calculus of Variations

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES

19

"tinuing, (1.41) transforms into

1-1

2

-1

10

33

11 001 00-46 uuc otage we interchanged rows two and The equivalent system of equations is

t

From

10

9

(1-42)

three.

the above consideration the reader can prove by mathematical

iduction or otherwise that a system of n linear homogeneous equations i 771 unknowns always possesses a nontrivial solution if ra > n. \

We now

a]x>

,the

unique

|,mple

=

consider the case in

=

t,

trivial solution

The only

1.10.

[tion occurs for

or

1

The system

n.

j

=

(1.44)

1, 2,

x

=

2

x

n

=

if

|aj|

5^

from

possibility for the existence of a nontrivial

the case

|aj|

=

0.

Triagonalizing the matrix such that |fy| = 0. \\b]\\

o

ds a new equivalent triangular matrix [reader explain this. We thus obtain

Let

bl

-

000 ;h implies 6j =

0,

=

If (no summation) for at least one value of i. original system to one containing more

we have reduced our

liowns than equations, for which a nontrivial solution exists. 0, bZ~\

=

0,

then x n

=

If

and again we have more unknowns than

again exists. Continuing, we see the vanishing of at least one element along the main diagonal ies the existence of a nontrivial solution.

itions so that a nontrivial solution

ample 1.23.

The determinant

of the coefficient matrix of the

system

x+y+z+uQ 5x

-

y -f

(1.45)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

20

Triagonalizing the coefficient matrix yields

vanishes.

1111 0007 000-3 -3

System

(1.45) z

?>y

On

0,

becomes

7u

?>u

so that z

-1

-3

so that u

3A, y

z

A,

considering the system

x

-

2x x

+ +

=

2\,

-

0, o

and we have x

<

< +

X

-\-

y

= = -

y y y

=0,

z

-\-

oo.

(1.46)

exists. see immediately that only the trivial solution x = y = of linear a system Generally speaking, homogeneous equations in n unknowns, > n, does not possess a, nontnvial solution. The reader

we

m

m

is

referred to Ferrar's text on algebra for the complete discussion of this The rank of a matrix plays an important role in discussing solu-

case.

For a discussion

tions of linear equations.

of the

rank of a matrix see

the above-mentioned text.

Example

\

.24.

The column matrix X =

will

The number

he called a vector

JC"

thejth component of X. We call the number, n, the dimensionality of the The determinant of the matrix X r X is called the square of the magnitude of space. 7 the vector X. If the x', ? = 1,2, r?, are complex numbers, we define ]X X| as n = conjugate comthe square of the magnitude of X, where X 7 = Ha; ;? 2 x

}

is

called

.

.

.

,

'

1

*

||,

plex of

a?.

The system

of vectors

X =

r

r

is

said to be linearly

not

all

dependent

if

=

1,2,

(1.47)

there exist scalars X 1

equivalent definition of linear dependence

set of vectors (1.47) is a linearly

\i

=

X2

,

.

.

. ,

Xw

zero such that

XX a = An

,

X2

=

=

\m

=

0.

(1.48) is

independent system

the following: if

The

Eq. (1.48) implies

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES

We now

prove the following theorem:

sional space are linearly dependent The system of equations X a X a = linear

=

x la \ a

i

=

1, 2,

.

m

.

vectors in an n-dimen-

The proof

n.

as follows:

is

equivalent to the system of n unknowns X X 2 Xm

is

in the

homogeneous equations

m

Any

m>

if

21

1

.

,

=

n] a

. ,

Such a system always has a nontrivial solution

1, 2,

for

.

.

,

.

.

.

,

m >

n.

,

m Q.E.D.

Problems 1.

Solve the system

-

x

-

-f 3z

2y

=

4w

2x+y-z+u=Q -

&r 2.

x

+

-

2z

-

2/y

-

2.r

3.

+

y

Solve the system

-

z

3w = 4w =

+2+

;(/

=

u

Solve the system

=

u

o

<

u

x

-

+

5x 4.

solve

Determine

-

4// ?/

X so that the following

= =

2w 3w

2s

82

will

system

have nontrivial solutions, and

:

= 4r + y = -2x +

\x \y

y

1

6.

Show that

the vectors Xi

=

1

X

1 ,

are linearly independ-

1

2

2 erit.

Given

X =

1 ,

find soalars X],

X-2,

Xs

such that

X

* II

1.6.

P(x,

Quadratic Forms. The square of the distance between a point and the origin 0(0, 0, 0) in a Euclidean space is given by

y, z)

L =

x2

2

+

using a Euclidean coordinate system. an n-dimensional space yields

u-

y

2

+

The

from

2

(1.49)

generalization of (1.49) to

=

(1.50)

The linear transformation x = a a y a ,i,a = 1

z

l

(1.50), n

t

=l

1,2,

.

.

.

,

n, \a]

9* 0, yields,

ELEMENTS OF ITRE AND APPLIED MATHEMATICS

22 If

we

desire that the

?/'s

he the components of a Euclidean coordinate

system, we need

Comparing

(1.51), (1.52) yields

=

P

8

=

!

Ij* a if

t=

The system

of Eqs. (1.53) in

A = A r = A"

(1.54) implies in turn that

an orthogonal matrix. (1.54) column vectors of A, then Af A 2 = is

' (153) v

P

matrix form becomes

A^A = AA r = E Equation

^

^

1

called

0.

We

A

1 .

A

If AI,

(1.54)

||aj||

2

matrix

are

A

any two

satisfying different

say that the vectors are

perpendicular. n

For complex components

(1.50)

becomes L 2

=

x'x*,

y

and

for the

t=i y's to

be the components of an orthogonal coordinate system we find

that the matrix

A must

satisfy

the complex conjugate of matrix.

a*

We now

A T A = E or A T = A~\ where A = If A T = A" we say that A is a unitary ||a]||,

1

a].

,

consider the quadratic form

Q =

a a px a xP

a aj9

=

real; a,

1, 2,

.

.

.

,

n

(1.55)

and ask whether

X = BY

it is possible to find an orthogonal transformation such that (1.55) reduces to the canonical form

Q = Let the student note that (1.55)

2

X,(2/')

may

(1.56)

be written in matrix form as

Q = X TAX

(1.57)

r noting that X AX is a matrix of just one element and so is written as a scalar. Under the transformation X = BY, (1.57) becomes

Q = (BY) rA(BY) = Y r (B rAB)Y

(1.58)

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES so that (1.58) will have the form of (1.56)

if

and only

23

if

(1.59)

that

is,

B^AB must be

Our problem has been reduced

a diagonal matrix.

to that of finding a matrix

B

satisfying (1.59), arid hence satisfying

B-'AB = HA.SJ

AB =

or

since tion.

Q

s=

B r = B" is required if X = BY is to be an orthogonal transformaWe may consider A to be a symmetric matrix, A = A r since X r [^-(A + A T )]X + X r [-|(A - A r )]X, and X T [|(A - A r )]X = (see

Prob.

1

,

l|6 w

||,

=

1

A.

T )

is

a symmetric matrix.

B

to find the square matrix

attempt Eq. (1.60) becomes

aj

For j

+

while ^(A

9, Sec. 1.1),

We now B =

(1.60)

B||X,J|

=

/

i,j

=

satisfying (1.60).

1, 2,

.

.

.

,n

If

(1.61)

we have n

}

=

or

If

BI

is

the column matrix (vector),

i

or

=

(A

XBi

-

(1.62)

X,

XE)Bi

,

X

=

(1.62)

may

be written

= (1.63)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

24

Equation (1 .63) represents a system of n linear homogeneous equations n unknowns comprising the column matrix BI. From Sec. 1.4 a necessary and sufficient condition that nontrivial solutions exist is that in the

|A

a 22

-

=

XE|

a 2n

X

or

(1.64)

X

an

We

=

the characteristic equation of the matrix A. It is a so n in X of has X roots X n the 2 degree n, polynomial equation Xi, X t real or complex. The roots X], X 2 X n determine the column call

.64)

(1

.

.

,

.

.

,

vectors BI, 62,

.

.

Bn

. ,

B = The

.

,

,

.

,

which in turn comprise the matrix B, that

,

B

a square matrix

BJ<

2

is,

ABi = X]Bi

is called an eigenf unction, eigenvector, called the eigenvalue, or latent root, or charXi If B! is a solution acteristic root corresponding to the eigenfunction Bj. of (1 .63), so is Bi/length of BI, a vector of unit length. If the B r i = 1 2,

solution BI of

or characteristic vector.

is

,

.

.

. ,

n, are unit vectors, it is

,

easy to prove that the matrix

B

is

an

Let

orthogonal matrix.

AB = 2

^

X!

X 2B 2

X2

(1.65)

Then (ABO* s= BfA = BfA = XJBf so that BfAB 2 = XjBfB 2 and B[B 2 = 0. over Bf AB 2 = X 2BfB 2 so that (Xi - X 2 )B[B 2 = 1

71

.

,

X

t

i

=

1 ,

,

2,

.

.

.

,

the matrix

n, are all different,

B

is

MoreIf

the

an orthogonal

matrix.

Example

We now

1.25.

the quadratic form

Q

firxd

7x 2 +

the linear orthogonal transformation which transforms 2 We have -\- 7z* -f 6x/y -f- 8?/z into canonical form.

7?/

3

7

The

characteristic equation

is

7

-

3

X

3

7

-

\i

7 Eq. (1.63)

(7

-

X)(X

-

12) (X

becomes 36 2 =

-

2)

4

X

4

which reduces to

0|

374 047

Q-

7

=

0, 36i -f 46.

-

X

so that Xi

=

=

0.

=

0,

46 2

7,

X2

A

=

12, X 3

=

2.

For

unit vector BI whose

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES

25

4 5

components

b 1;

b 3 satisfy these

b>2 ,

equations

is

=

Bi

For X 2

_

-

12 Eq. (1.63)

3 5

5 \/2

becomes -56i

+

36 2

=

0,

-

36i

=

56 2 -f 46 3

0,

46 2

-

56 3

=

0,

so that

B

_1 2

V2 4 5 \/2

5 \/2

For A 3

=

2

we obtain B

3

5

V2

so that

The quadratic form Q = 7x 2

+

7y

z

Q =

-}-

7z 2

7it 2

+

+

under the linear orthogonal transformation

Each root X,

i

=

1, 2,

.

.

. ,

GJ-?/

12^ 2

+

+

Syz be(;omes

2w 2

(J.66).

n, of |A

XE|

=

determined a column

=0

If multiple roots of |A vector of the orthogonal matrix B. XE| occur, it appears at first glance that we cannot complete the full matrix B.

However, we can show that an orthogonal matrix F exists such that the Q = X T AX, A = A r is canonical in form for the trans-

quadratic form formation X =

FW.

X = BY, Y = CZ are = (BC)Z is an orthogonal transtransformations, then X We have (BC) T = C TB T = C^B- = (BC)" so that BC

First let us note the following pertinent facts: If

orthogonal formation.

1

1

,

an orthogonal matrix. In other words, the matrix product of orthogonal matrices is an orthogonal matrix. Next we note that, if B is an

is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

26

orthogonal matrix in a /^-dimensional space, k

<

n,

then

C =

an orthogonal matrix for the n-dimensional space. The reader can easily verify that C has the necessary properties for an orthogonal matrix. is

Finally the roots of |A

and conversely.

-

=

XE|

are the roots of |B- 1 AB

From B~ AB - XE 1

|B-

1

AB -

XE|

= B-^A -

-

=

XE|

0,

XE)B we have

= IB-KA - XE)B| = JB- |A - XE| |B| = IB- |B| |A - XE = |B-'B| |A - XE| = |A - XE| 1

1

)

which proves our statement. It is immaterial whether Xi let Xi be any root of |A XE| = 0. = for BI, BI a unit vector. a multiple root. We solve (A XiE)Bi We now obtain an orthogonal matrix B with BI as its first column. This can be done as follows (the method does not yield a unique answer) Let 62 = \\bt2\\ be the elements of the second column of B. In order that 62

Now

is

:

n

be orthogonal toBi, we need

Y 6A =

This

0.

2

is

a single homogeneous

1=1

equation in the unknowns 6 t 2,

i

=

1,

2,

.

.

n.

.

,

We know

that

we

n

can find a nontrivial solution which can be normalized so that } 6 22 t

1

obtain the third column,

=

1.

1

n

n

To

=

we need \

6 t i6 t3

=

*i

N

0,

6 t2 fr t3

=

0.

A

t=i

normalized nontrivial solution exists for n > process we construct an orthogonal matrix B.

2.

By

continuing this

The final column of B for which a nontrivial in n unknowns equations

1 involves a system of n solution exists. From the construction of

as the element of the

first

row and

first

B

it

follows that

B -1 AB

has Xi

column and has zeros elsewhere

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES in the first column. 1

(B~ AB)

we note

T

that

From

= B rA T (BB~ AB !

that

T

= B-'ACB 7 )- = B^ACB-

r

1

1

)

1

)-

= B^AB

1

We have used the facts a symmetric matrix. Hence under the orthogonal transformation

is

A = A B = B" X = BY, Q becomes T

27

1

.

,

Q = X rAX - Y T (B 'ABjY Xi

y

l

'

=

The

2

l

\\y

//"I;

//

characteristic roots of the matrix \i

-

B~ AB ]

\\c'a p\\

\

satisfy

-

~

6*22

X

X

c 32

=

Hence the remaining roots c 22

X

C nn

C n2

of |A

X

XE| *

r2a

=0 satisfy

*

'

c 2n

X

-

n

By

the same procedure

a,^

so that

Q becomes Q =

Y

we can reduce

2

1

\i(z

)

+

c a py"y ft

to the

form

=3

X 2 ^ 2) 2

+

with Xi

=

X2

if

Xi

ELEMENTS OF PURE AND APPLIED MATHEMATICS

28

Continuing this process reduces

a repeated root.

is

Q

to canonical form.

Q.E.D. 1.6.'

Positive -definite Quadratic Forms.

Let us

Inverse of a Matrix.

consider the quadratic form

Q = X r (S T S)X where S

S =

a triangular matrix,

is

(1.67)

= ||sj||,

>

for i

s]

we have Q = (X 7'S r )(SX) = (SX) r (SX) =

From

j.

It is

(1.67)

obvious that

1=1

Q > this

unless

Qx

tf

=

a

i

0,

=

2,

1,

.

.

If

n.

. ,

we know that

0,

|sj|

of equations has only the trivial solution

system

= Thus the special quadratic form

(1

=

0.

xn

=

.67) for |S| 9* r If

has the property

say that Q is a positive-definite quadratic form. be shown that

For such a form

unless x

|a

n

|

1

=

=

x2

=

an

>

x

ri

Q = X AX >

an

ai2

a-21

022

012

unless

Q >

X =

0, it

we can

> =

>

|A|

(1.68)

W. L. Ferrar, Oxford University Press). Conversely, implies that Q is positive-definite. The above considerations suggest that, if Q is a positive-definite form,

(see

"Algebra," by

(1 .68)

Q

X T AX, A = A r then there = A. We now show that ,

SrS

matrix S such that The equation S r S = A

exists a triangular

this is true.

implies

ij =

I For

i

=

=

j

we obtain

=

2, 3,

we have s^ = an

1

aij since

SnSij

.

.

.

$22

,

=

n.

For

(a 22

t

2i

=

*J 2 )*

jf

=

=

=

so that Su Ssi

=

"

*

= =

.

--a -Iy 2

22

=

,n

.

.

sn i

=

+

(1.69)

For

(an)*.

2 we have (i 2 ) 2

/ I

1,2,

i

(22)

2

=

/I

an

a i2

V

I

a2i

a 22

I

I

~

=

I,

Remember

that o#

o,-.

For

i

=

2, j

>

2

we have

>

1

a 22 so that

/

=

j

0, so that

s^Sij

+

S 22 s 2y

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES and

=

s<2j

(a 2j

29

Continuing this process, one obtains the

Si2Sij)/S22.

set of equations

n =

(flu)*

=

$22

(1-70)

+ =

Ski

?-,.*)]>

0H - -

Given the numbers a the elements s from (1.70). Let the reader show that tj

,

(1 1 1

(1

can be calculated step by step

tj

2

1

1

3

(1-71)

The above method

011

012

021

022

S can be used to

for determining

find the inverse

A provided X AX is positive-definite Since A = S^S, we = To find S- from S, we proceed as have AS^S 7 )" - S^S" = 0, i > j. From TS = E we obtain follows: Let T = S" 7>

matrix of

1

1

1

71

)

1

t i3

,

^

7t

Equation tn t\i

= =

(1.72)

For

l/'&'ii.

1.20.

= 5,

i,./

=

i

=

=

j

(1.70)

Thus

^13

sn

=

(1.72)

Thus t^Su =

t tj .

.

We

=

,n

.

1,

=

612 .

ti n ,

.

,

,

or

1,

so that fat,

^23,

invert the matrix

1

= -2,

.

2

1

From

.

we have nSi 2 + ^i2<s 22 = Continuing, we can compute / 13 1,

2

1

253 *12

1,2,

enables one to solve for the

tnSi2/s 2 2.

Example

1

-

s 12

-01lSi8

=

+

2,

1S

=

^12fi23)/33

1,

=

s 23

3

=

l/A/2,

J

4

1,

/22

s 33

=

A/2. 1, <23

From

= -

1

u -

(1.72)

/ \/2,

<33

=

1,

l/\/2.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

30

and

-5

11

A' - S-'(S1

r

]

3

)

-1

-1

1

has the property that Ki2B is a new matrix which can be obtained by multiplying the second row of B by k and adding these elements to the corresponding elements of the first

row

Thus

of B.

1

k

P

l

612

b\\

1

Let the reader show that placing k in the rth row and sth column of the unit matrix E produces a matrix E r , such that E r ,B is a matrix identical to B except that the elements b rct)

a =

C -

1, 2,

Now

r 7* s.

.

.

let

.

A

,

n, are

For the

||A, B||.

replaced

by

b ra

+

kb fa

,

r 7* s.

be a square matrix such that A|

A

in

Example

5^ 0.

Notice that |E r ,| = 1 for We consider the matrix

1.26

121100 253010 134001

(1.73)

We manipulate C by operations on the rows until the first three rows and columns of C These operations are equivalent to multiplying C on the discussed above. Let B be the product of all the E ra

become $. of the type

EM

BC Hence

BA -

with the

C

E, first

.

HE, B||

We obtain B from ||E, B||. For example, starting 2 and add to the second row, and multiply the first row by row from the third row. This yields

and B

of (1.73),

subtract the

= A"

HBA, BEII

by matrices Then

left .

1

.

we

121

100

011-210 013-101

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES

We can have

easily obtain zeros in the first

and third row

row

third

to the first

We

5-20 10 1-11

10-1

1-2

01 00 Adding ^ the row yields

of the second column.

31

now

2

row and subtracting

-3-

5

1

2

2

the third row from the second

1

-

1

-

1 2

002

1-1

Factoring 2 from the third row, multiplication by

1

100 010 oo

11-5

1.7. Differential

Equations.

We now

yields the inverse matrix

L

II

-1

1

1 2

1

I

consider the system of differ-

ential equations

-7-2-

=

l

a,

ax

a i,

a =

1, 2,

.

.

.

,

n\ a]

=

=

a{

constants

which can be written in the matrix form

^ = AX We

A = A^

(1.74)

look for a linear transformation which will simplify (1.74). so that (1.74) becomes

Let

X = BY,

(B~

1

AB)Y

(1.75)

From previous considerations we have shown that it is possible to find a matrix B such that B~ 1 AB is diagonal (nondiagonal terms are zero). For

this matrix

B we have no summation on

since

B-'AB =

i

(1.76)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

32

The

solution of (1.76)

is

+ From X = BY we can .

.

.

,

D*e -V^

solve for

=

i

=

x*(t), i

1, 2,

1, 2,

.

.

.

.

,

.

.

n.

,

n

(1.77)

The

y\

i

=

1, 2,

n, are called normal coordinates.

Let us consider two particles of masses m\ ra 2 respectively, moving continuum, coupled in such a way that equal and opposite forces proportional to their distance apart act on the particles. The differential equations of motion are d Xi , -

Example

1.27.

t

,

in a one-dimensional

=

(-JP

a(xi

(1.78)

.afo

,-Jj?

For convenience,

let y\

= \/m\

xi,

2/2

= \^m

~a: 2

2

**>

)

">\

-

a /mi,

a/ra 2

,

a k,

so that (1.78) becomes fl 2/i

-jgT

d*yi

2 - -?yi

_ "

i

H-

,

;

df 2

W~

r

The

characteristic equation

I

2/2

so that Xi

-

xi(t),

-|

{

(w? -f

col)

+

"2

I

1|

y\ |

I

AY

2/2

is

-co?

hence

k

*\

di*

[(i

-

-

X

w2) 2

+

4A; 2 ]i

|

.

Let the reader find

yi(t),

t/ 2

(0,

and

x z (t).

A matrix A may be subdivided into in turn each being thought of as a matrix. For array rectangular arrays, 1.8.

Subdivision of a Matrix.

example,

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES

The

linear

system y

may be

33

= a^

i

i,

a

i

=

=

1, 2,

.

.

.

1, 2, 3,

.

.

.

,

n

,

fc,

written n

k

a

x<x

+

a X"

+

*

.

1,

.

.

,

n

In matrix form y

1

-

a}

a\

2

2 2

al

a

'

-

-

al

y

k+l

y

a

Xl

or

Y2

AX

Hence

2

A

3

(1.79)

we

are not interested in solving for X obtain eliminate X 2 from (1.79). If

We X = A^CY, - A XO Xi = (A! - A A A )2

2

r<

3

M

1

(Y

1

,

,

|A 4

3 1

so that

2

Xi

is

stant,

We

x

=

a]

are the imped-

=

ax,

a

/7Y

xoc

at .

Can we

we can

|

dx In the calculus the solution of -r

Conclusion.

,

-

In a mesh circuit the y* are the impressed voltages, the l ances, and the x are the currents. 1.9.

xn

generalize this for the system -^-

=

con-

= AX?

would be led to consider matrices of the should we go about defining such a matrix? From the

see immediately that one

form

6 Ae

.

calculus e*

How = }

n-O define

% n /n\.

This suggests that

if

B

is

a square matrix

we

ELEMENTS OF PURE AND APPLIED MATHEMATICS

34

E

+B+

B

+

2

-

.

+

-

B*

+

This poses a new problem. What do we mean by the We define of matrices?

(1.80)

sum

of

an

infinite

number

B as the rth partial

^E + B+~B + 2

r

sum

of (1.80).

*

r

Br

element of

converges,

we

=

lim r

If

B

is

r

exists in the sense that

B

r

>

a square matrix of order n, whose terms

is,

|6j

< A =

each

so

define e*

that

B

lim

If

+ ~ B'

-

constant, for

i,

j

1,

are uniformly bounded,

65

2,

.

.

.

,

then

n,

lim r

B

r

oo

Its elements The proof is as follows: Consider the matrix B l Thus each element of B 2 in absolute value is less are of the form b j)f. 2

exists.

than

nA

nM

by

.

2 .

3

The elements of B 3 are of the form blfiffi which are bounded The elements of E k are bounded by n k ~ A k Hence every B r E is bounded by the series 1

.

element of

.

V

~

-

nk lAh

_ ~~

Li Thus each term of B r converges since each term of B r

E is a series bounded r

in absolute value term by

Y

term by the elements of the series 1/n

(n A ) /k k

!

L*l

*

which converges to sin B.

Is sin

2

B +

(\/ri)e

cos

2

nA as r

B = E?

=1

oo Let the reader define cos Let the reader also show that

>

.

B and

at

for

a constant matrix B. Problems

1.

AX 2.

Q

=*

Show that the

roots of (1.64) are real t\ 2 , X Xi Xi XX, assume X

+

if

a,

+ 1X2,

real and a, = a,.. and show that \2 =

is

Consider the quadratic form Q = a a & a x&, the s aj complex. We may write X rAX. If A =* A T show that Q is real by showing that Q = Q. A matrix A that A A r is called a Hermitian matrix. If A and B are Hermitian (see Prob. 2), show that AB -f BA and t(AB BA) are l

,

,

such 3.

Hermitian. 4.

Hint: Consider 0.

Show that the

roots of (1.64) are real

if

A is

Hermitian.

LINEAR EQUATIONS, DETERMINANTS, AND MATRICES 5.

35

Find an orthogonal transformation which reduces

to canonical form. 6.

Show

that the characteristic roots of the matrix

A

are the

same as those

of the

matrix B^AB. 7. Write the system

w -T^r +

in the matrix

form

X

where

Let

He-J7 = Ee

X -

CY,

|C|

^

0,

and

C

find

becomes

so that the system

d2Y

W +.Hell ^\0

Oi

v Y

-i

tl

-111

=2U

->|

t||j

Integrate this system of equations. 8. Solve the system xi -f x t

-

o-i

for xi Xz t

9.

by

first

eliminating

xi,

J-2

+ x* + + x, -

*4 jr 4

=

0-4.

Consider the system of differential equations

d2X dt*

^A

where A and B are constant matrices.

X

ss

rfX dt

+ BA

""

U

Let

C

e*C!

constant

and show that a; satisfies \w z d* + ^a] -f 6)| = if C is not the zero vector. 10. The characteristic equation of A is the determinant |A 0. This XE| polynomial in

See Dickson, istic

X,

"Modern

equation, that

Algebraic Theories," for a proof that

is

A

is,

An This

is

a

written

-f fciA"-

1

+ 6 A~ +

the HamiUon-Cayley theorem.

2

2

-

+

ZuE

-

satisfies its character-

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

36

REFERENCES Aitken, A. C.: "Determinants and Matrices," Oliver & Boyd, Ltd., London, 1942. Albert, A. A.: "Introduction to Algebraic Theories," University of Chicago Press, Chicago, 1941. S. MacLane: "A Survey of Modern Algebra," The Macmillan ComYork, 1941. Ferrar, W. L.: "Algebra," Oxford University Press, New York, 1941. Michal, A. D.: "Matrix and Tensor Calculus," John Wiley & Sons, Inc., New York,

Birkhoff, G.,

pany,

and

New

1947.

Veblen, 0.: "Invariants of Quadratic Differential Forms," Cambridge University Press,

New

York, 1933.

CHAPTER 2

VECTOR ANALYSIS

2.1. Introduction.

line segments.

The

Elementary vector analysis is a study of directed is well aware that displacements, velocities,

reader

accelerations, forces, etc., require for their description a direction as well as a magnitude. One cannot completely describe the motion of a particle

by simply

stating that a 2-lb force acts

upon

The

it.

direction of

the applied force must be stipulated with reference to a particular coordinate system. In much the same manner the knowledge that a particle has a speed of 3 fps relative to a given observer does not yield all the pertinent information as regards the motion of the particle with respect to the observer. One must know the direction of motion of the particle.

A vector, by definition, is a directed line segment. Any physical quantity which can be represented by a vector will also be designated as a vector. The length of a vector when compared with a unit of length The magnitude of a vector will be called the magnitude of the vector. is thus a scalar. A scalar differs from a vector in that no direction is associated with a scalar. Speed, temperature, _ volume, etc., including elements of the real- La a ?> >

number system,

are examples of scalars.

Vectors will be represented by arrows (see Fig. 2.1), and boldface type will be used to distinguish a vector from a scalar. The student

"""FIG

21

his own notation for describing a vector in writing. vector of length 1 is called a unit vector. There are an infinity of unit vectors since the direction of a unit vector is arbitrary. If a will

have to adopt

A

=

If |a| s= 0, we represents the length of a vector, we shall write a |a|. = a is zero that a a 0. say vector, 2.2. Equality of Vectors. Two vectors will be denned to be equal if, and only if, they are parallel, have the same sense of direction, and are of equal magnitude. The starting points of the vectors are immaterial. This does not imply that two forces which are equal will produce the same physical result. Our definition of equality is purely a mathematical definition. We write a = b if the vectors are equal. Moreover, we imply further that if a = b we may replace a in any vector equation by b

37

ELEMENTS OF PURE AND APPLIED MATHEMATICS

38

and, conversely,

we may

by a. Figure 2.2 shows two vectors b which are equal to each other. 2.3. Multiplication of a Vector by a Scalar. If we multiply a vector a by a real number x, we define the product xa. to be a new vector replace b

a,

parallel to a; the FIG. 2.2 is

parallel to

We note

magnitude

of xa. is

\x\

times

the magnitude of a. If x > 0, xa is parallel to and has the same direction as a. If x < 0, xa.

and has the reverse direction

of a (see Fig. 2.3).

that

= =

x(y&)

Oa

(xy)a.

= xya

It is immediately evident that two vectors are parallel if, and only one of them can be written as a scalar multiple of the other.

if,

(atb)-f-c

FIG. 2.4

FIG. 2.3

Let us suppose we have two vectors given, 2.4. Addition of Vectors. b, is defined as follows: A say a and b. The vector sum, written a triangle is constructed with a and b forming two sides of the triangle. The vector drawn from the starting point of a to the arrow of b is defined

+

sum a + b (see Fig. 2.4). From Euclidean geometry we note that

as the vector

+b = + b) + c = 3 (a + b) = a

(a

Furthermore, a c = b a d.

+

+

+

=

a, (x

+

y)a.

=

The reader should

b

+a

(2.1)

+ (b + xa + xb a

xa

+

?/a,

c)

and

(2.2) (2.3)

if

a

=

b, c

=

statements.

Subtraction of vectors can be reduced to addition by defining

a

-

b

d,

then

give geometric proofs of the above

(-b)

VECTOR ANALYSIS

An equivalent

We look for the vector

b is the following: c is defined as the vector a

definition of a

+

b

=

39

b (see Fig. 2.5). be vectors with a common Geometry. whose end points A, B, C, respectively, lie on a straight line.

c such that c

a.

Let

2.5. Applications to

origin

a, b, c,

A

A -b

FIG. 2.5

Let C divide in the ratio x:y, x y = 1 (see Fig. 2.6). propose to determine c as a linear combination of a and b. It is evident that

AB

=

c

a

+

AC.

Conversely,

=

But

let c

z)a

(1

+

=

ya.

+

xb, x

+

y

lies

on the

Q.E.D. Equation

=

1.

If a, b, c

(2.4)

have a common

end points lie on a straight line. xb so that c = a + x(b - a). Since x(b their

a vector parallel to the vector b of vector addition that the end point

C

We

AC = xAB = x(b - a). Hence c = (1 - z)a + xb = y* + xb

we show that

origin 0, c

+

line joining

(2.4)

is

A

= AB, we

We a)

write is

a

note from the definition

to B.

very useful in

solving geometric problems involv-

ing ratios of line segments.

Example 2.1. The diagonals of a parallelogram bisect each other. Let ABCD be be any point any parallelogram, and let in space (see Fig. 2.7). The statement a defines the parallelogram c d = b

ABCD.

+

The vector

(a

+

Eq.

The vector

BD.

_i_

\

Hence a

i/

A

o\

_

(2.4)].

There

the vector

p.

is

c

- b

+

d

-

I/K

_i_

A\

FIG. 2.7

AC

has its end point on the line joining c)/2 with origin at [see has its end point on the line joining (b d) /2 with origin at whose end point lies on these two lines, namely, only one vector from

Hence

+

P bisects AC

and BD.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

40

In Fig. 2.8, D divides CB in the ratio 3 1 does P divide CE AD? Imagine vectors a, b, c, d, e, p drawn from a fixed point 2.2.

Example 3:2.

:

How

E divides A B in the ratio

;

}

to the points A, B, C, D,

From Eq.

E, P, respectively.

(2.4)

we

have d

+3b __

c

e

_ -

2a

+ 3b 5

Since p depends linearly on a and d, and on c and e, we eliminate the vector

also

b from the above two equations.

This

yields

4d

-f

2a

=

5e 4- c

4

_L 2 a

=

5

or 2.8

[

+ fa with origin at + ^c has its end point

The vector -d Similarly -|e

been shown to be equal. p

Thus

P

divides

AD

d

e

_i_

l

c

must have its end point lying on the line AD. These two vectors have lying on the line EC.

Then

- |d

+ fa -

in the ratio 2

:

1

|e

+ -Jc

and divides

Why?

CE

in the ratio 5:1.

Problems 1.

Interpret

2.

a, b,

-,

c are consecutive vectors forming a triangle.

What

is

their vector

sum?

Generalize this result.

a and b are consecutive vectors of a parallelogram. Express the diagonal vectors terms of a and b. If xa. -f 2/b = la + wb, show that x 4. a and b are not parallel. m. I, y 3.

in

- b| ^ |a| - |b| |. graphically that |a| |b| ^ |a b|, |a that the midpoints of the lines which join the midpoints of the opposite The four sides of the quadrilateral are not necessides of a quadrilateral coincide. 5.

6.

+

Show Show

+

|

sarily coplanar. 7.

Show

that the medians of a triangle meet at a point

P which divides each median

in the ratio 1:2. 8. Vectors are drawn from the center of a regular polygon to its vertices. Show that the vector sum is zero. 9. a, b, c, d are vectors with a common origin. Find a necessary and sufficient "*

condition that their end points

lie in a plane. that, if two triangles in space are so situated that the three points of intersection of corresponding sides lie on a line, then the lines joining the corresponding vertices pass through a common point, and conversely. This is Desargues' theorem.

10.

Show

2.6. Coordinate Systems. The reader is already familiar with the Euclidean space of three dimensions encountered in the analytic geometry and the calculus. The cartesian coordinate system is frequently used for describing the position of a point in this space. The reader, no doubt, also is acquainted with other coordinate systems, e.g., cylindrical

coordinates, spherical coordinates.

VECTOR ANALYSIS

We let

i, j,

k be

41

the three unit vectors along the positive #, y, and z axes the vector from the origin to a point P(x y, z), then

If r is

respectively.

y

(see Fig. 2.9) r

The numbers

x, y, z

=

xi

+

+

yj

zk

(2.5) r.

Note

and

z axes,

are called the components of the vector

that they represent the projections of the vector r on the r is called the position vector of the point P.

x, y,

FIG. 2.9

By

translating the origin of

coordinate system

AI,

A

2,

As

A

to the origin of our cartesian

can be easily seen that

it

are the

any vector

A = AJ + A projections of A on

2j

+

the

Ask x, y,

(2.6)

and

z axes, respectively.

are called the components of A. Let us now consider the motion of a fluid covering

They

any point P(x, Thus u, v, w.

y, z)

V =

the fluid will have a velocity

u(x, y,

z, t)i

+

v(x, y,

w

The

+

z, t)j

w(x,

V

all of

space.

At

with components

y, z, t)k

(2.7)

depend on the point velocity components u, v, The most and the time t. vector encountered will be general P(x, y, z) describes of the form given by (2.7). a vector field. Equation (2.7) If we fix t, we obtain an instantaneous view of the vector field. Every point in space has a vector associated with it. As time goes on, the

A

vector field changes.

V =

will, in general,

special case of (2.7)

u(x, y, z)i

+

v(x, y, z)j

is

+

the vector

w(x,

y,

z)k

field

(2.8)

Equation (2.8) describes a steady-state vector field. The vector field independent of the time, but the components depend on the coordinates

is

of the point P(x, y, z).

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

42

The simplest vector throughout

all of

field

space.

occurs

when the components

of

V are constant

A vector field of this type is said to be uniform.

Example 2.3. A particle of mass ra is placed at the origin. The force of attraction which the mass would exert on a unit mass placed at the point P(x, y, z) is

_ _____

Gmr F . _ This

is

Newton's law

of attraction.

It is easily verified that

si

Note that F

Dot, Product.

We

(A*

yB 2 )j

l

2.7. Scalar, or

+

)j

l

,

field.

if

= BJ + B 2j = A+B (A + i)i + (A, + A + yE = (xA yBji + (xA 2

f tnen

a steady-state vector

is

+

+

3

(xA,

)k

+

define the scalar, or dot, product of

two vectors by the identity a

where

6 is

b

ss

a| |b|

cos 6

the angle between the two vectors when they are drawn from

common origin. Since cos how is chosen. From (2.9) it follows that

a

6

=

cos

there

0),

(

a-b = b-a za-t/b = xy&'b a a = |a| 2 = a a If |a|

a

is

7* 0,

(2.9)

=

== b, c

d

(2.10) 2

a

implies

no ambiguity as to

is

= 0. perpendicular to b, then a b |b| p* 0, then a is perpendicular to b.

c

=

b

d

Conversely,

if

a

b

=

0,

cos0-J FIG. 2.10

Now a of

b

b

is

equal to the projection of a onto b multiplied by the length

(see Fig. 2.10).

Thus

a-b = With

this in

(proj b a)|b|

=

(proj a b)|a|

mind we proceed to prove the a- (b

+

c)

distributive law,

=a-b + a-c

(2.11)

VECTOR ANALYSIS

From

Fig. 2.11

apparent that

it is

a

+

(b

= = =

+ c)]|a| + proj c)|a| (proja b)|a| + (proja c)|a| = b a + c -a

c)

[proj

(b

(proja

b

= a-b A

43

+ a-c

repeated application of (2.11) yields

+

(a

b)

+

(c

=

d)

a

+

(c

+

d)

b

(c

+

d)

c=b-a

a FIG. 2.11

Example

2.4.

Cosine

Law

FIG. 2.12 of Trigonometry (Fip 2.12) c c

*

c

r2

Example

2.5.

i

i

=

j

= k

j

= b - a = (b - a) (b - a) = b-b + a-a-2a'b 2 _ = ^2 2nb cos 6 _|_

k

A B =

#ii

-I-

B 2j

-h

==

1

Aii

i

=

j

+A

j

-f

2j

k

= k

i

=0.

Hence

if

Ak 3

^sk, then

A B = Formula

,

+ AB + 2

(2.12)

2

very useful. It should be memorized. Let Ox l x*x 3 and Ox x 2 x 3 be orthogonal rectangular cartesian coordinate systems with common origin 0. We now use the superscript convention of Chap. 1. The unit vectors along the x axes, j = 1, 2, 3, are designated by i,. Simij Let a be the cosine of the larly Iy, j = 1, 2, 3, is the set of unit vectors along the x axes. = 1, 2, 3. The projections of ii on the $*, 2 x* angle between the vectors i, 10, a, Let the axes are aj, a*, a?, respectively. Hence ii = a}ii -f o?i 2 + ajig afla (2.12)

Example

is

l

2.6.

j

,

.

reader show that (2.13)

We have

(Fig. 2.13) r

r,

so that x a i

- ^a. ft

-

From 1, 2,

3

(2.13)

x^lft

so that (2.14)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

44

(2.14) represents the coordinate transformation (linear)

Equation

In matrix form,

coordinate systems.

X^X - X 7'X, which

that

X =

in turn implies

X

2

l

A A = E

A 7 = A"

l

l

it

follows

"

77

or

U

1 .

U* are the components of a vector when referred to the x coordinate 2 system, and if t/ U U* are the components the same vector when referred to the x If

x

From } x x = ^ &x

AX.

U

between our two 3

3

3

1

,

2

,

1

-3 -ii -2 ^3 (x , x , x )

,

,

coordinate system, one obtains (2.15)

obtained in exactly the same

This result

is

manner

which Eq.

From

in

(2.14),

^

=

a

(2.14)

was derived.

so that (2.15)

maybe

written (2.16)

(2.16) will

Equation field (see

Chap.

be the starting point for the definition of a contravariant vector

3).

Problems 1. Add and subtract the vectors a = 2i 2k. Show -2i 3j 5k, b 2j that the vectors are perpendicular. 2. Find the angle between the vectors a = 2i - 3j -f k, b = 3i - j - 2k. 3. Let a and b be unit vectors in the xy plane making angles a and with the x axis.

+

Show

that a

=

cos a

i

-f-

sin

a

cos (a

j,

b

j8)

=

cos

=

cos a cos

ft i

+ sin /3

+

j,

-f sin

+

and prove that a sin

ft

Show

that the equation of a sphere with center at PQ(XQ, r/o, zo) and radius a is - y )2 (z- *o) 2 = a 2 (y 5. Show that the equation of the plane passing through the point PO(XO, 2/0, 20) normal to the vector Ai -f Bj Ck is 4.

-

(x

2

x<>)

+

+

.

+

A (x 6.

Show

B(y

-

-

2

)

-

that the equation of a straight line through the point PO(XQ,

to the vector

li

-f raj -f

nk

parallel

is

Prove that the sum of the squares of the diagonals of a parallelogram is equal sum of the squares of its sides. 8. Show that the shortest distance from the point PQ(XQ, yo, ZQ) to the plane 1.

the

+ By + Cz + D D Axo + Byo + Czp (A + B* + C*)i

Ax

-f-

2

to

VECTOR ANALYSIS

45

3 9.

For Eq.

(2.14),

Y

Y

=

3*3?

0=1

&&.

Y aXT

= =

For the a of Prob. 9 show that

if

L<

10.

Show

that this implies

0=1 a a

[

? 1

l if

* -*

J 7

U a = J^, F a =

agF^, then

3

3

V

V ^"^

Z

a

aya

l

2,

a=

1

1

2.8. Vector, or Cross, Product. One can construct a vector c from two given vectors a, b as follows Let a and b be translated so that they have a common origin, and let them form :

the sides of a parallelogram of area

A =

We

(see Fig. 2.14). |a| |b| sin define c to be perpendicular to the

this

of

plane

The

parallelogram.

with

parallelogram

magnitude equal to the area

of the

direction of c

obtained by rotating a into b (angle of rotation less than 180) and

is

FIG. 2.14

considering the motion of a righthand screw. The vector c thus obtained cross,

product of a and

=

c

with |E

=

1,

is

defined as the vector, or

written

b,

a

X b =

E-a = E-b =

|a[

|b|

sin 6

E

(2.17)

0.

The

vector product occurs frequently in mechanics and electricity, but for the present we discuss its algebraic behavior. It follows that

a

X b = -b X

a

is not commutative. If a and b are parallel, Conversely, if a X b = 0, then a and b are parallel proIn particular a X a = 0. 0, |b| 5* 0.

so that the vector product

a X b vided

=

0.

|a|

9*

The

distributive law, hold as follows: Let

aX(b + c)=aXb + aXc,

attempt to show that u = multiplication holds, we have

We

v u

=

v

a

X

(b

+

0.

c)

can be shown to

Since the distributive law for scalar

-

v

(a

X

b)

-

v

(a

X

c)

(2.18)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

46

In the next paragraph

for arbitrary v.

a

Thus

may be

(2.18)

u =

v

= =

(v (v

X X

X

(b

=

c)

be shown that

it will

x

(a

b)

c

written

a)

(b

a)

b

+ c) - (v X a) b - (v X a) c + (v x a) c - (v X a) b - (v X

a)

c

or v J_ u. Since v can be chosen arbitrarily, This implies that u = = so that to arid hence picked not perpendicular u, it follows that u

c)=axb

(2.19)

i, 0,k X k = 0,i X j - k, j X k Example2.7. Oneseesthati X i = 0, j X j X i * j- For the vectors a = aii + a^j -h #&k, b = 6ii + bzj -\- &ak we obtain a 3 ?>2)i -f- (a^bi a X b = (a 2 ba aib 3 )j -f- (ai6 2 a2^i)k from the distributive law.

k

Symbolically a

Equation

Example

(2.20)

is

to

X

b

-

i

j

k

cii

a2

as

b\

bz

bz

(2.20)

be expanded by the ordinary method

of determinants.

2.8

a

X

a (a

X

Example

b)

2.9.

b a

i

-

-

7

-

3j

+

2k

j

1

-3

2

4

1

-3

-

33

=

4i

+ -

7i

+

llj

+

13k

X

b)

-b

b

j

3k

k

i

+

26

(a

-

28

+

11

-

39

-

Assume that a particle is rotating about a L with angular speed w. We assume

Rotation of a Particle.

fixed line

that the shortest distance of the particle from remains constant. Let us define the angular velocity of the particle as the vector, <*>,

L

is along L and whose length choose the direction of c* in the usual sense of a right-hand-screw advance Let r be the position vector (see Fig. 2.15). of P with origin on the line L. It is a simple matter for the reader to show that the velocity of P, say V, is parallel to, and has the

whose direction is

w.

We

same magnitude as, <> X r. Thus V = X r. Example 2. 10. Motion of a Rigid Body with One Fixed Point. Let Oxyz be a fixed coordinate system, and OxyZ a coordinate system attached to the rigid body whose fixed point <*>

FIG. 2.15

is

the origin 0.

body. x

,

#, 2

Let

remain constant since the Oxijz coordinate system From (2.14) we have & Hence ajz'.

moving frame.

P be

a point of the rigid

As time progresses, the coordinates is

rigidly attached to the

VECTOR ANALYSIS d&

da!

.

-*---** so

-At~jx>,

that^~

-

AX

,*,

or

\\A}\\

47

dx'

+a *-S -

1

Ha;!!"

.

If

we define

*

*

--Aj^

we have

- -?"

(2 - 21) 8

3 v~^

However, y

^TA

xk xk represents the invariant distance from

to

P so that y

dx^ x* -TT-

=

3

and y can

to*J 2 T*

now be

=

From Example

0.

it

follows that w*

=

wj.

Equation

(2.21)

written as

dr

so that v

1.2

=

^j.1

rr

i

+

^j2 -TT- j

1

+

^3

=

~ir k

<*>

X

r,

o>

=

w$i -h wjj -f a>jk.

It follows

from

the result of Example 2.9 that the motion of a rigid body with one point fixed can be characterized as follows There exists an angular velocity vector co whose components, in general, change with time, such that at any instant the motion of the rigid body is :

one of pure rotation with angular velocity o>. This property is very important in the study of the motion of a gyroscope. It can easily be shown that the most general rigid-body motion consists of a translation plus a pure rotation. 2.9. Multiple Scalar and Vector Products. The triple scalar product a (b X c) has a simple geometric interpretation. This scalar represents the volume of the parallelepiped formed by the coterminous sides a, b, c,

since

a

where

A

is

(b

X

c)

= |a| |b| |c| sin 6 = hA = volume

cos a

the area of the parallelogram with sides b and c and h

altitude of the parallelepiped (see Fig. 2.16). Hence (a X b) c represents the same volume.

is

the

It is easy to see that

it is permissible to interscalar in cross the dot and the triple product. Since there can change be no confusion as to the meaning of a (b X c), it is usually written as The expression a X (b c) is meaningless. Why? We let the (abc).

reader prove that

a

b

X

c

=

(abc)

=

(2.22)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

48

From determinant

=

(abc)

theory, or otherwise,

=

(cab)

(bca)

it

follows that

= -(bac) = -(cba) = -(acb)

show that a necessary and sufficient condition that a, b, c same plane when they have a common origin is that (abc) = 0.

It is easy to lie

in the

=

In particular, (aac)

0.

bxc

FIG. 2.16

product a X (b X c) is an important vector. It is call a it V, since it is the vector product of a and b X c. certainly vector, We know that V is perpendicular to the vector b X c. However, b X c

The

is c.

triple vector

perpendicular to the plane of b and c so that V lies in the plane of b and If b and c are not parallel, then V = Xb If b and c are parallel, juc.

V =

0.

Since

V

=

a

we have X(b

0,

V = It

s

1

(b

x

can be shown that Xi a

The expansion

x

(2.23) of a

middle factor.

a)b

Xj[(c

X

+ + M(C

a)

-

(b

a)

=

so that

a)c]

so that

=

c)

X

(b

(a c) is

-

c)b

(a

b)c

(2.23)

often referred to as the rule of the

Similarly (a

X

b)

x

c

=

(a

-

c)b

(b

c)a

(2.24)

More complicated products can be simplified by use of the triple For example, we can expand (a X b) X (c X d) by considerproducts. ing a

X

b as a single vector and applying (a

X

b)

X

(c

X

d)

= =

(a

X

b

(abd)c

(2.23).

d)c

-

-

(a

(abc)d

X

b

c)d

VECTOR ANALYSIS Also

(a

x

b)

X

(c

= [(a x b) x c] = [(a c)b - (b = (a-c)(b-d) -

d)

a

a

c

Example

Consider the spherical triangle Let the sphere be of ra-

2. 11.

dius (a

d d

c)a]

(b.c)(a

d)

d

b-d

b-c

(see Fig. 2.17).

49

ABC

Now

1.

X b)

(a

X

= b

c)

c

-

b)(a

(a

c)

a = 1. The angle between a X b and a X C is the same as the dihedral angle A between the planes OAC and OAB, since a X b is perpendicular to tjie plane of OAB and since a X c is perpendicular to the plane of OAC. Hence since a

sin

cos

7 sin

A =

cos a

cos 7 cos

B

/3

Problems

Show by two methods that the vectors

1.

a

=

3i

-

j

+ 2k,

b

= -12i +

4j

-

8k are

parallel.

Fio. 2.17 Find a unit vector perpendicular to the = i - j -f 2k, b = 3i + j - k. * 0. 3. If a, b, c, d have a common origin, interpret the equation (a X b) (c X d) 4. Write a vector equation which specifies that the plane through a and b is parallel to the plane through c and d. 6. Show that d X (a X b) (a X c) = (abc)(a d). 2.

vectors a

6. 7.

Show that Show that

(a

a

X b) (b X c) X X (b X c) + b X

X X

(c (c

a)

=

(abc)

a)

+

c

X

2 .

(a

X

-

b)

0.

Find an expression for the shortest distance from the end point of the vector ri, to the plane passing through the end points of the vectors r 2 r,j, 14. All four vectors have a common origin 0. Let d = xa + j/b -f zc. Show that 9. Assume (abc) -^ 0. 8.

,

* (abc) 10. If (abc)

^

0,

"

(abd)

(abc)

(abc)

show that

c-d (abc)

11.

i(adc)

a

X ~

b "

b

-*

X

C

(abc)

(abc)

Consider the system of equations

-f Ciz

= (2.25)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

50 Let a

oil

+ azj -

Show that

and show that

-f ask, etc.,

^

az

+

^

0.

etc., (abc)

u = Ai(x,

At any point P(x

y, 2,

y, z)

y

Let us consider the vector

of Vectors.

2.10. Differentiation

Oi

d

-f cz

by

can be written as

(2.25)

+ ^ 2(3,

+

2, t)j

?/,

and at any time

A*(x, y,

z,

t)k

field

(2.26)

(2.26) defines a vector.

t,

If

we

keep P fixed, the vector u can still change because of the time dependence of its

+ dy, z +

y

find the

dz).

we keep the time

If

components A\, A%, A%.

the vector at P(z,

we note

fixed,

+

(or differential) of a single function of x, y,

change

that

y, z) will, in general, differ from the vector at Q(x dx, Now, in the calculus, the student has learned how to z,

What

t.

do we encounter in the case of a vector? Actually none, since we easily note that u will change (in magnitude and/or direction) if and only if the components of u change. The vectors i, j, k are assumed fixed difficulties

We

throughout the discussion.

in space

definition for the differential of a vector rfu

dA dAA = -r- dx ,

,

where

t

l

If x, y, z are

.

dx

= dA^i

+

+ dAi dy + -^ dy ,

functions of

t,

.

dA 2

dAt -r

are thus led to the following

:

+ dA

j

,

dA

+

dz

dz

s

k

(2.27)

,.

.

dt -^dt

i

%

=

t

rt

1, 2,

3

then

<

In particular,

let r == xi

particle P(x, y, z).

Equations

(2.30)

yj

+ zk

be the position vector of a moving

and

=

d 2r

d*x.

^=dp

(2.31) are,

by

1

+ ,

d*y

dh

/001X

^ + ^k .

.

.

(2 - 31)

J

definition, the velocity

and accelera-

tion of the particle. If

a vector u depends on a single variable

du dt

(see Fig. 2.18).

-

Then

dv as ^

,

and

+

2 29 >

_ Um

u(t

+

A*)

t,

-

we can u(Q

define

VECTOR ANALYSIS

51

u(f+At) ttff)

FIG. 2.18

It is easy to verify that (2.32) is equivalent to (2.28).

u =

u(x, y,

.

.

z,

If

.)

then

du

=

du

+

ufo

i;

lim

dAi.

^ = -aF

dA 2

+

1

Ax, y,z,

.

J

-dF

-

^ + AO

-

-

=

^

-j#*

lim

.

d -r-

Equation

(2.34) also

Thus

QQ\

(2 33) '

Au Av _+v _ + Au _

\

(

(u

v)

A

.

,

.

_

^-

w

,

.

w

.

at

ar

dv

= u

-j

.

h v

du

,

-j-

(2.34)

n

,

x

33

can be easily obtained by writing u and v in com3

ponent form.

.

z,

/

At

AJ-+O

or

u(x, y,

dAi. k + 17

= u Av

<p(t)

_J_J^

and

-

.)

= u(< + A<) v(< + A<) - u(0 v(<) = (u(<) + Au) (v(<) + Av) - u(0 = u Av + v Au + Au Av

v(t)

v(t

Hence

.

Consider

2.11. Differentiation Rules.

Afl

.

u // ^

<p

t

d<p __

vt ,

=

at

u% /7 j at

dv

+

// j

v l -rr>

at

du

Similarly (u

Example

This jg-

is

2.12.

-rr

v)

= u X

+

Let u be a vector of magnitude

a very useful result.

and u

X

0.

In particular,

if

u.

X

v

(2 35)

Then u u

w 8 so that

the magnitude of u remains constant,

This implies, in general, that

-T- is

perpendicular to u,

if

ELEMENTS OF PURE AND APPLIED MATHEMATICS

52 |u|

=

constant and

The reader should

0.

give a geometric proof of this

statement. position vector of a point

Example 2.13. Motion in a Plane. Let r be the moving in a plane having polar coordinates (r, 6) (see

Why r

is

= rR

P

R =

Since

unit vector.

cos 6

-f sin

i

R?

perpendicular to

we have P =

j,

Notice that

P

is

Now

Fig. 2.19).

=

=

r

sin 6

i

-^-

also a unit vector.

P

R

a

-f cos 6

j.

rR,

Differentiating

yields

dr

dv_dV ~

and

dt

dr

~*~

_

dR

,

drdR^ dt

dl*

dr

c/rc^

dR

.

~^~

do dt^~ dtdt

so that the acceleration of the particle

_

dedPdO

^0

r

d0

^~ r

dt 2

dt

dO dt

is

(2.38)

since -r-

=

dB

cos

central force field, sectoral area

sin 0j

i

f

=

=

/R, then

R.

^ (r

2

If

the particle

^=

swept out by the particle in time

so that dt is

moves under the action 2

~

dA = ^r 2 This

equal areas are swept out in equal periods of time. planetary motion.

=

h

-|-r

=

The

constant.

d0, so that is

of a

Kepler's

r-

=

first

^ and ;

law of

t

y

FIG. 2.20

FIG. 2.19

Example 2.14. Frenet-Serret Formulas. A three-dimensional curve, T, in a Euclidean space can be represented by the locus of the end point of the position vector given by (2.39) r(0 - x(t)i y(t)j -f- z()k

+

where

i

is

the space curve, then to define

-

a parameter ranging over a set of values

-T-

-7-

ds

-r

ds

^

to

1

t

^

fa.

If

is

arc length along

from the calculus.

as the unit tangent vector to the space curve

T

It is natural

(see Fig. 2.20).

Since

VECTOR ANALYSIS

=

53

to t. a unit vector, -y- is perpendicular tells us how fast the Moreover, ~r * * ds ds Hence we define the curvature, direction of t is changing with respect to arc length s. t

-r ds

is

'

r by

of the space curve

K,

*2

=

-7-

The

varies from point to point.

**,

-j--

in general,

a function of

is

*,

and hence F

principal normal vector to the space curve

defined to be the unit vector, n parallel to

-7-

=

*n

;

g

K is

Thus

(2.40)

The reciprocal of the curvature is called the radius of curvature, /> *= I/*. At any point P on r we now have two vectors t and n at right angles to each other. This enables us to set up a local coordinate system at P by defining a third vector at right We define as the btnormal the vector b = t X n. The three angles to t and n. fundamental vectors t, n, b form a trihedral at P; any vector associated with the space curve r can be written as a linear combination of t, n, b. Let us _ ds

.

t

now

=

since

ular to b (b -r-

=

^n,

evaluate

To

r

0.

and

-y->

as

*! = b ds

X

From b

-y-

as

Hence

-y- is

ds

we

definition

by

obtain

The famous

as

a unit vector),

is

where

curve T.

b n =

-y-

is

we note

+ ? ds

we

to perpendicular ^

see that -r-

must be

obtain

X

t

= b X

=

b

X

t

as

b

-=- is also

ds parallel to n.

-r-

=

or

as

perpendic* *

Consequently

r is called the torsion of the

so that

+m X

-en

t -f-

-7-

Since

t.

the magnitude of -7-

that n

ds

=

t

t

= -t -rb

Frenet-Serret formulas are dt

3-

=

Kll

ds

*!. -(rt+rb)

(2.41)

db

As an example

of (2.41)

we r

We have t

*

T-

a sin ti

(

consider the circular helix given

=

a cos

+ a cos

t i

-f-

t j

a sin

-f 6k)

t

j

-7--

by

4- bfk

From

t

t

=

1

we

obtain

so that t

(-a

sin ti

+ a cos

a sin

j)

t

j

+

2 6k)(a* -f 6 )-*

i

Thus *n -

j

- ( -a cos

f i

-

t

(a

2

-f-

b 2 )" 1 ,

and

K

-

a(a

2

+b

2

)-

1 ,

since *

-

i

yj*

?

ELEMENTS OF PURE AND APPLIED MATHEMATICS

54

From b

X

t

Q . rn Example

n

let

cos

(6

i

t

the reader show that b

+ 6 sin

t

j)(a

2

+

6')-

,

(b

am

r

-

6(a*

ti

b cos

+6

1

)"

+

t j

ak)(a

2

s -f & )~i,

1 .

M

Let us consider the problem of a missile pursuing a target T, the motion taking place in the xy plane. TT and VT are the position and velocity vectors of the target; FA/

Pursuit Problems.

2.15.

1

=

y

and VA/ are the position and velocity vectors of the missile; 0, <p, $ are defined by Fig. 2.21. Let

and

=

r

IT

~

r

=

= rFT

VT = rFr

Differentiating the identity r

If

t is

<p

(p

cos

VT - V T ^

r

the unit tangent vector to the curve r traversed

by the

(V r

-

Example cos

2.14).

^>

dt A i^ -n so that

Equation

- rV T -

sin

VA/)

^

(2.43)

^FT -

^Vy p-

(r

A

t

^

Tr

,

cos

?

-f

-

F r -57

<p)

^

For the special case of constant target speed,

-IT sin

^

6,

-

r

-^ ar

and

FA/ cos

(2.42)

cos

(2.43)

VT

target, then rft

n

since

rft

*

-r-

dt

dt

- FT

^ (r cos +

rVr

Thus

VA/.

*

Fyt

.

(see

becomes

V, -

-f

r

VA/

yields

cos

and

r

-

cos

rV M

- FT

^

FIG. 2.21

-

cos

at

VT -

r

eu

VT

so that -^

TA/

FA/FT cos

-f Fj,

(<p

0, (2.44)

FT

6)

-r.

r

y?)

cos ?

~

(2.44)

become

(r

cos

<

(2.45)

Let us apply (2.42), (2.45) to the dog-rabbit problem. At i the rabbit starts at the origin and runs along the positive y axis with constant speed VT- A dog starts at = and pursues the rabbit in such a manner (direct pursuit) that = (a, 0) at t throughout the motion.

The constant speed

j

of the

- FT

cos

dog

*

is

-

Fj/.

We

have

FA/ (2.46)

c*

V\i

snce

r/2.

Equations

Fr

FA/FT cos (2.46)

(r

<f>

VT

j,

cos

can be written as

cos

)

+ y*

-

Integration yields Frr cos <p 4- FA^ a, ^> Fiv)^ -f FA/C, since r (Fj, - Vj.)-. at e 0. The rabbit is caught when r - 0, or at t FA/a(Fj,

ir/2,

VECTOR ANALYSIS

55

Problems 1.

Prove

2.

Show

=

3. r

4.

^ (r X ^Q a cos wt + b sin

-

that

u><;

^

X

r

u are constants.

a, b,

R is a unit vector in the direction r, r

5. If

X

a

^

ae"'

6. If r 7.

(2.35), (2.36).

a,

+

~=wX

be~wf

a, b,

,

b,

|r|

show that

.

^

Show

that

X

-

(a

w constants, show

b)

r

X

-r.

X w

s

r

=

o>a

r

R X dR

d*r that -5%

X

b and

r -

f

(a

X

-

0.

b).

Consider the differential equation

g+2A*? + Bu A, B constants. constant scalar. are roots of

w; 2

4-

2Aw;

5

-f

0.

9.

For the space curve x

that

(2.47)

,

f

Find a vector u which

Show

=

Assume a solution of the form u(0 = Ce wt C a constant vector, w a Show that u(t) = Cie^i* + C2e 2 is a solution of (2.47) where toi, ?z

8.

10.

Prove that

- d

8

satisfies

3t

2

t

What

MI

if

-gj

-^

y

3J 2 , 2

<*,

+ d* n -

'

1^2?

2

-rr

3<

=

+

t*

such that u

i,

~

j,

show that

*crb.

11. Four particles on the corners of a square (sides -= 6) begin to move toward each other in a clockwise fashion in a direct-pursuit course. Each has constant speed V. Show that they move a distance 6 before contact takes place.

12.

In

navigational

pursuit

VM

= VT

sin

sin

<p.

Interpret

this

result

geo-

^ constants. 13. A target moves on the circumference of a circle with constant speed V. A missile starts at the center of this circle and pursues the target. The speed of the The pursuit is such that the center of the circle, the missile, and the missile is also V.

metrically, assuming

0,

Show that the target target are collinear. to the moment of capture. 2.12. tion.

The Gradient Let From the calculus

<p(z,

2/,

moves one-fourth

z)

of the circumference

up

be any differentiable space func-

*-& b+ 5* + S*

(2 48) -

The right-hand side of (2.48) suggests that the scalar product of two If r = xi vectors might be involved. zk is the position vector of yj the point P(x, y, z), then dr = dx i k. dz dy j Hence, to express d<?

+

+

+ +

ELEMENTS OF PURE AND APPLIED MATHEMATICS

56

as a scalar product, one need only define the vector with components -r^>

~

>

ox

del

<p

This vector

called the gradient of

is

dz

dy

We

3= V<p.

(?(x.

y,

z),

written

define

"K

+ 5 + S*

l

2 49 '

1

<

'

so that

=

d<f>

The in

dr

V<?

(2.50)

reader should recall from the calculus that

we move from P(x,

as

d<?

+ dx,

to Q(x

represents the change

+

+

dy, z dz), except higher order. Equation (2.50) states that this change in <p can be obtained by evaluating the gradient of <p at P and computing the scalar product of V<? and dr, dr being the vector from <p

y

of

for infinitesimals

Pto

y, z)

Q.

We now

From <p(x, y, z) we give a geometrical interpretation of V<p. can form a family of surfaces <p(x, y, z) = constant. The surface S given by <p(x, y, z) = <p(z 2/o, 20) contains the point P(z 2/o, ZQ). <p(x, y, 2) has the constant value <f>(x$, y^ ZQ) if we remain on this particular sur,

Now

face S.

from at P,

is

Q be any

let

(2.50), di

,

=

V<p

normal to

all

S near

point on

Hence

0.

P.

Since dy

0,

V<p is perpendicular to dr.

possible tangents to the surface at

we have, Thus V??,

P so that V<p neces-

must be normal to the surface <p(x, y, z) = ^(z 2/o, 20) at P(ZO, 2/0, 20). This is a highly important result and should be thoroughly understood

sarily

,

by the

reader.

Let us

now

return to (2.50).

If

ds

=

|dr|,

(2.50) states that

= J-V^ = u-V^ ^ ds ds where

|u|

u and

V<p.

The V<p,

=

Hence

1.

surface

is,

|V?>|

show that

<p(x, y, z)

at

the greatest change in

<p(x, y, z)

Example

=

cos

V(^>i

2.16.

=

+

^>(z

^2)

=

,

2/o,

zg).

V^i

+

P(x 0t

<p

where

0,

It is obvious that j- has its

greatest change in

that

-^ds

is

is

the angle between

maximum when

z/

occurs

This

7 (2.50') v

,

zo)

8

=

occurs in the direction of

when we move normal

to be expected.

-

xy

+ yz

1

to the

Let the reader

Let us find a unit vector perpendicular to the surface x9

0,

VECTOR ANALYSIS Here

at the point F(l, 1,1).

V* -

2.17.

By

obtain this result

by a

Example

-

=

1/r.

Example

2.18.

where

Vr

some

=

r

=

dt

ft

+#

(x*

constant *

is

2

2 4- z )*.

We

a sphere with

Now

ft.

/r dr

in

We

u(x, y, z).

have

f'(u)Vu.

The operator

keep

=

V = i-

del,

mind that V

hjr

da*

h

k

-

is

>

o2

a//

a useful concept.

and as a vector

acts both as a differential operator

sense.

ThusV, = V(Civ>i -F

(!+,+) =

C 2 ^2)

C 2V<p

CiV<^i -h

2

=

if

how

,J?

C2

Ci,

=

V(^i^ 2 ) Notice

=

dr

?*

Hence V/(w)

-

It is helpful to

in

+ yk

1, 1)

perpendicular to the sphere, Vr

is

Consider V/(w),

2.19.

Example

x)J

at P(l,

Q.E.D.

-

/'(w)

-

(*

and

computation Vr = r/r for method. The surface r

direct

dr so that/

k

-f

i

+

y)i

+ yz,

xy

different

Since Vr

center at the origin.

-

(2x

Vv>

z2

y, z)

<f>(x,

57

+

jg + kg-

Let the reader show that

are constants.

p2V^i

<piV<p2 -f

conforms to the rule of calculus

(2.51)

ItiBea.ytoshowthat

(2.51)

for the derivative of a product.

Problems 1.

Find the equation of the tangent plane to the surface xy

z

=

1

at the point

(2, 1, 1).

2.

Show

3. If r 4.

that V(a

=

(z

If *

2

(r

-f

X

Show

6. *

7.

a,

where a

is

a constant vector.

~ show that Vr n = nr n 2 r. = b that show X V*> b),

(r

X

r\

+

r2

=

c\

when they have the same

Find the change of <p = x*y -f- yz* X y 2 -f z*y = 3 at the point P(l, 1, Prove (2.51).

/

/(MI, W2,

-

X

(r

X

b)

when a

.

.

.

,

Wn), w*

*?(x, y, 2, 1),

show that

=

c 2 intersect at

xz in the direction normal to the surface

w*(x,

T/,

= gf

r2

ri

1).

2),

fc

-/-I >

a) -f a

and the hyperbola

n

9. If

X

foci.

+

8. If

(r

vectors.

that the ellipse

right angles

yx

=

-f z*)*,

a)

and b are constant 5.

r) 2

t/

^f

+^

-

1, 2,

.

.

.

,

n,

show that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

58 10.

The equation

V(ri

+ r) =

an

of

T is a unit tangent to

n

ellipse is

Vri -h Vr 2 Vri ,

=

>

Vr 2 *

+

V(ri

2.13.

-

r2

-f-

the ellipse at the point

The Divergence

Why

constant.

P?

V(ri -f r 2 )

is

is

V(n

+

T ^*0

r 2)

if

From

computed at P.

give a geometric interpretation to

T =

r2)

Let us consider the motion of a z, t), the veloc-

of a Vector.

fluid of density p(x, y,

ity of the fluid at

V =

as

f

A

= pv = Xi

+

+

Yj

Zk

concentrate on the flow of through a small parallelogram ABCDEFGH (Fig. 2.22) of dimensions At time t let us calcudx, dy, dz.

y

F

Let

t).

We now

E

d

any point being given

V(x, y,z,

fluid

amount

late the

2 22

of fluid entering the

box through the face

ABCD.

The

x and z components of the velocity contribute nothing to the flow through

ABCD.

Now

Y(x,

has the dimension

y, z, t)

j^, M

T =

=

mass,

L =

length,

time. Thus Y dx dz has the dimension MT~ and denotes the gain mass per unit time by the box because of flow through the face ABCD. dY \ f Similarly ( Y + dy 1 dx dz represents the loss of mass per unit time l

of

EFGH

because of flow through the face of

mass per unit time

is

thus

dY

dx dy

-r

If

we

t.

The

loss

also take into consider-

dy

we

ation the other faces of the box,

unit time

dz.

same time

at the

find that the total loss of

mass per

is

A7\ ~ + ^- + (AY f)V

dx

Hence

dX dx

dY h T

dy r\

The

d7 h -rdz

ILT

scalar -^

dx

\

h

is

\r

,

dy

)

dx dy dz

(2.52)

/

the loss of mass per unit time per unit volume. \

+

-fdz

dy

rr

-z-~ is

dz

called the divergence of the vector field f

,

written

divf

=

g+^+f v

Returning to the operator

V = 1^

*v

(2.53) \

f-k-r->we note that ox hJT; dz dy

VECTOR ANALYSIS

59

7

provided we interpret the reader show that

V

V Example

2.20.

V

f

For

-

r

- r~V

r

xi

+

Example 2.17). Example 2.21. What

and a

as both a vector

=

Of)

+

+

yj

Vr~ 3

V

r

3r~ 8

-

zk,

r

+

f

v?V

-

3.

f

V<f>

For

3r~ 4 Vr

Let

differential operator.

f

=

r

(2.55)

-

r-r,

3r~ 3

-

Sr" 5!

r

*

(see

This important scalar

is

is

the divergence of a gradient?

called the Laplacian of

<p(x,

t/,

2).

+f +

VV =

(2-56)

Let u,, i = 1, 2, 3, be the components of 2.14. The Curl of a Vector. Euclidean coordinate system. The the velocity of a fluid in an x differential change in the components of v is given by l

dv\

xV

1

,

dx

(dv,

-

dVj\

dx

dvt

-

.

,

dt

(2 57) '

-Bi

The terms

s j;

= T~

^

=

i,j

1, 2, 3,

now occupy

fy are the elements of a skew-symmetric matrix. three important elements, listed as

1

2

"

^ ~

** v *

dx 1

~dx*

3

The vector

^i

+

t$

+

tjs. is

As a

The

result there are

W ~ 5? _ " " dF w -

d?>2

8

our attention.

-

dvi

1

defined and called the curl of

v.

Using the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

60

V operator, we note that the

curl of v

be written

may

k

j

= V X

v

curl

o

=

v

o (2.59)

dz V*

Example

Let

2.22.

=

f

z

x yzi

2xyz

z

j

+

z

y zk.

i

Xf =

V

z y z

-2xyz*

Example

(2.7/2

v

We

X

Consider V

2.23.

-

-

x z t/j

-

(2yz

have, for

f

+

z

=

?/i

z 2 *)k

+

*>j

+

wk,

M

X

(2.60)

To obtain f

fixed,

(v>)

X

the curl of

<pf,

f.

Example

fixed,

<p

and

<p,

let

V operate on

yielding

X

The sum of these operations yields V The curl of a gradient is zero.

f,

yielding <?V

X

f

;

then keep

and complete the vector product

V<p,

(<pf).

2.24.

i

JL.

v

keep

and allow V to operate on

x

dx

.

j

k

A Adz dy

d<f>

d<f>

d<f>

dx

dy

dz

/

ay _

a 4.

i

J

dz dx

dx dz

k

I

dx dy

o

provided

^>

has continuous mixed second derivatives.

We

2.15. Further Properties of V Operator. to be the scalar differential operator

d

d_

**di

We

define the product

'dy

-

V

(2.61)

**~te

then have

-

u

dv

,-

VECTOR ANALYSIS

61

Let us now investigate u X (V X v). Assuming that we can expand term by the rule of the middle factor [remember that the expansion a X (b X c) = (a c)b (a b)c holds true only for vectors; V, strictly speaking, is not a vector], we have this

u X (V X

The

=

v)

subscript v in the term V,,(u

components

.

V,,(u

v)

-

v)

(

U V)v

means that

V,.

(2.62)

operates only on the

Thus

of v.

V.(u

= V ,(u x r T

v)

+

t

?VV

+

*/*>'*)

Interchanging the role of u and v yields

v

Adding

V M (u

(2.62)

v)

+

and

V v (u

X

(V

X

= V

u)

tt

(u

-

v)

V)u

(v

(2.(>3)

(2.63) results in

v)

= u X

X

(V

v)

+

v

X

(V

X

+

u)

(u

V)v

+

(v

V)u

and V(u

v)

= u X

(V

X

v)

+

v

X

(V

X

u)

+

(u

-

V)v

+

(v

V)ti

(2.64)

We

it

The above analysis in no way constitutes a proof of (2.64). leave The same remarks to the reader to verify (2.64) by direct expansion.

hold for the following examples:

V X

(u

= V M X (u X v) + V v X (u X v) = (v V)u - v(V u) + (u V)v - (V V (u X v) = V w (u X v) + V, (u X v) = (V X u) v - V v (v X u) = (V X u) v - (V X v) u

X

v)

V)WL (2.65)

-

We now

list

some important

identities

(2.66)

:

V(uv) = uVv + vVu V (v?u) = <pV u + (V?) u <3) V X (<pu) = <?V X u + (V?) X u <4) V X (V<f>) = V (V X u) = J5) - (V X v) u (6) V (u X v) = (V X u) v - v(V = (7) V X (u X v) (v V)u + (u V)v = u X (V X v) + v X (V X u) + (8) V(u v) = V(V u) - V 2u <9) V X (V X u) = u (10) (u V)r V-r = 3 '(1)

t2)

u) (u

-

u(V v) V)v + (v

V)u

ELEMENTS OF PURE AND APPLIED MATHEMATICS

62

=

(12)

V X

(13)

d? = dr

=

(14) dt

+ ^ dt

V?>

+ j dt

(dr- V)f

t

V

(15)

r

=

8

(r~ r)

Problems

Show

2.

that the divergence of a curl Find the divergence and curl of xi

3.

If

4.

Show Show

1.

6.

A

6.

If f

7.

Show

8.

If

A

9.

If

w

10.

11.

+

axi

+

/>//j

ttzk,

zero.

is

yj/x -f y, of x cos zi r) = 2 A.

V 2 (l/r) - 0, r - (a; 2 that V X l/(r)r] 0. wVt> show that f V X f

+

that

2

Show

V

that

that dt

X

(u

-

*

v)

(dr

z 2k.

2 4- * )*.

?/

0,

u not constant.

that (v V)v - -g-Vv 2 - v X (V X v). is a constant unit vector show that A [V(A v) xi -f yj -f zk, show that is a constant vector, r

Show

+ y\nxj

show that V(A

X

(V

V)f -f

^ dt

-

v

u)

if f

-

ol

-

(V

X

v)

V X (v V X (w

X X

A)] r)

*

=

V

v.

2w.

u.

f(x, y, z, t).

V u. u) - V(V u) Vv = 0. Let dr be the 13. Let u u(x, y, z), v * v(x, y, z), and assume Vu to Q; and Q are points on the surface u(x t y, z) constant. From vector from dv ** dr Vv, show that dv = and hence that v remains constant when u is constant. 12.

Show

V

that

X

(V

X

2

X

P

P

This implies that relationship

We

F(u

v)

t

assume that

f(u) or F(w, v)

v

dF du

= and

then

dF dv

Vu

Show

0.

X Vv =

0.

conversely that

Hiw<* vFCw,

do not both vanish

u and v satisfy a dF = dF T Vw -f T" ^

if

v)

r<; '

oV

oli

identically.

14. Prove that a necessary and sufficient condition that u, F (w, v, uO *" \s that Vu Vi; Vw 0, or

v,

w

satisfy

an equation

X

*, V,

This determinant 15. Let A - V 16.

Show

2.16.

that

is

"ch*

dw

dx

djy

dz

dr

dv

dv dz

dw

dx

dy

dw

dw

dw

dx

dy

Hz

v, w) with respect to X(x}Y(y)Z(z). Show that - [(v V)f] u. [(u V)f] v

called the Jacobian of (u,

X W), VV * (V X f) (u X

0,

*

v)

=

*

(a;,

y, z).

A V XA -

0.

Orthogonal Curvilinear Coordinates. Up to the present moment for the gradient, divergence, curl, and

we have expressed the formulas

Laplacian in the familiar rectangular cartesian coordinate system. It is often quite necessary to express the above quantities in other coordinate systems.

For example,

if

one were to solve V 2 F = constant on the sphere a: 2

F =

boundary condition that one would find it of great aid to express V 2 F

subject to the

+

y

2

+ z* =

a2

,

in spherical coordinates.

VECTOR ANALYSIS

63

The boundary conditions of a physical problem dictate to a great extent the coordinate system to be used. Let us now consider a spherical coordinate system (see Fig. 2.23). The

relationships between x, y, z

6

=

cos"

x

are

+

y

2

+

(2.67)

<

(2.68)

1L

x

r sin 6 cos r sin

y z

r, 0, <?

1

tanand

and

=

sin

r cos

Let us note the following pertinent facts: Through any point P(x, y 2), other than the origin, there pass the sphere r = constant, the cone = 6 = constant, and the plane }

<f>

These surfaces intersect in pairs which yield three curves through P. The intersection of the sphere and the cone is a circle. Along this curve only <p can change, This since r and are constant. constant.

curve <p

At

*)

appropriately the construct a unit

called

is

curve.

,

P we

vector, e f tangent to the <p curve. direction of e^ is chosen in the ,

The

direction of positive increase in <p. The Similarly one obtains e r and e. reader can easily verify that these unit tangents form an orthogonal trihedral at

The

FIG. 2.23

P

such that e r

X

=

e*

e^.

form a basis for spherical coordinates in exactly the same manner that i, j, k form a basis for rectangular coordinates. Any vector at P may be written f = /re r + fee 9 + /*&<?, where /r /*, f, are the projections of f on the vectors e r e*, e v respectively. Unlike vectors e r e*, e v change directions as we move from point to i, j, k the Thus we may expect to find more complicated formulas arising point. vectors e r

,

e,

e^,

,

,

,

,

when the gradient, Since Vr to c r

.

is

divergence, etc., are computed in spherical coordinates. perpendicular to the surface r = constant, Vr is parallel

Similarly V0

is parallel

to e,; V^>

is

er

hrVr

e*

heVB

parallel to e,.

Thus

(2.69)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

64

Then dr = dr Vr [see Let dr be a vector of length ds parallel to e r = = = dr h dr e ds. Vr Since so that h dr dr = ds, we have r r r (2.50)], Now let dr be a vector of length ds parallel to e$. We have h r = 1. d0 = dr V0, ft* d0 = dr heVO = dr e$ = ds. It is seen that he is that Thus factor which must be multiplied into dO to yield arc length. .

he

=

r,

and similarly h? ds

The

2

= =

In spherical coordinates

r sin 0.

dr

+

2

r

dO 2

2

+

r 2 sin 2 6 d<p z

2 2 2 positive square roots of the coefficients of dr d0 d<p yield h r ho, h?, ,

,

,

respectively.

We

now

can

write

er

e e<p

The

= = =

differential of

Vr = e* X e, = r 2 sin 0V0 rV0 = e^, X e r = r sin 0V^>

volume

Now we

Equation

from

(2.71) is the gradient

=

frr

(2.70).

z

sin 0V0

X

sin 0)

X

ee

=

rVr

(2.70)

V<p

hrhehp dr dO r 2 sin

=

dp

dr d0

(hrheh^}'

/(r, 0,

<p)

d<^

=

1

We

.

2

(r

sin 0)" 1

.

have

a scalar in spherical coordinates.

of-

the divergence of

= /re r

f

+ f r sin e

+ fe&e + f^^ we write 0V^ X Vr + f^rVr X V<^?

Then

r

2

Vv?

consider the gradient of

V f = V(/ r 2 sin 0) V0 X V^ since V (V0 X Vv?) = 0, etc. V(/rr

= =

dsg

show that Vr V0 X

To compute f

er

is

dV = di d2 It is easy to

=

r sin 0V^?

X V<p X Vr X V0

V0

X V^ =

ou

+ V(/*r sin 0) [see

(fr r

ra;

V^>

formula for 2

sin 0)V0

X

V

V0

(/'r2 sin e)

Vr

(u

X

+ V(./ v)].

Vr

X

But

X

"

^

8in

VECTOR ANALYSIS

We

65

thus obtain

V

= T 4-^ r sin B 2

~ (f

r r~

-

f

I

+

sin B)

Equation (2.72) is the divergence of If we apply (2.71) and (2.72) to f

f in

+

(./> sin B)

-^ 36

idr

-

(2.72)

(/,r)

d<p

spherical coordinates.

= VF, we

obtain

1

V2 F = r

2

sin B r \T Idr ( \

+ dO ( 8in ' 60 ) + ?) dr / \ /

*

d<p

(^ (^ \sm I-)] B d<p

/J

(2.73) is the Laplacian of F in spherical coordinates. obtain the curl of f let f = frh rVr feh e VB f^V^. It can be

Equation

To

+

,

easily

+

shown that hr e r

V X

f

A

A

6V

'dB

d (2.74)

krfr

For the general orthogonal curvilinear coordinate system where ds 2 = /if du\ + hi du\ + h\ du\ we list without proof

(HI,

u Zj

-

V-f =

(2.75)

V X

f

= h sfz 1

V/ = 2

It

1

I

I

:

1

|-

oiii

would be a good exercise

I

j-

du% \ hi

\ h\ dui/

du$ \ h$

du^/

du$

for the student to derive (2.75).

Problems 1.

For x

=

cos

r sin 6

^>,

2/

=

r sin

rfs

becomes d* 2. For x 3.

Express

dr 2 r

cos

V/,

V

+

r2

<f> t

f,

d0 2 -f

V

X

-

dx

2

<p t

=

2

+ di/

2

-f

r

cos

rfz

show that the form

2

r 2 sin 2

r sin

y

2

sin

f in

^>,

z

z

show that

(is

2

dr 2

-f-

r 2 d^> 2

cylindrical coordinates and show that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

66

6.

V'F Vs F = Show that V

7.

By making

4.

Solve

in spherical coordinates

6.

Solve

in cylindrical coordinates

X

V(r).

V

if

F(r). spherical coordinate.

= 0, r a V 2V - V(V

[/Mr]

use of

V

if

-V X

V)

X

(V

V 2V

V), find

for

V -

v(r)e r,

V

purely radial in spherical coordinates. 2 = /(r)e r -f v(z)e 2 , in cylindrical coordinates. 8. Express V V, for

V

0.

Consider the equation (X

\, p,

Assume

p constants.

+

s =* ^ tpt Si,

(x -f

Next show that

[V

+

2

M )v(v

+

(/i

/*](?

-

juV'S

p constant, s,)

+ PP )/X -f VV - 0, ^ -

+

S)

2

A - V X (XT), necessary that ^ 10. If

/z)V(V

*

i

(M

+

Si)

-

p

j

\/

X -f M

0,

an d show that

*

0.

-

PP*)SI

R(r)Q(9)*(<p),

1

show that A V

XA -

0.

Is it

J

3

11. If

<fc

2

-

da;

2

1

daf,

,

t/

8 ,

?/

),

-

i

1, 2, 3,

show that

where 12. 13.

Derive Derive

(2.74).

(2.75).

The Line

2.17.

Integral.

We start

with a vector f

=

X(x,

y, z)i

field

+

Y(x,

AS

+ Let

^

=

r(0

^

+

x(t)i

y, z)j

Z(x,

y(t)j

y,

+

be a space curve T jointhe ing points A and B with posia

t

b

y

tion vectors r(a), r(6).

One may

subdivide T into n parts by subdividing

Fir,.

(see Fig. 2.24).

that

tj-i

^

vector f(&).

,

t

into

2.24

Lot Ar,

t3

=

r(^)

,_i),

We

and

can compute #(,), ^ tj. forms the sum One then

let ;j)>

,

<

'

<

tn

=

b

be any number such

*(;)>

which yields the

(2.76)

If lira

S. exists independent of

how

the

& are chosen provided maximum

VECTOR ANALYSIS |

AT,

as n

>

As

we

oo

-

,

|

the curve T from

A

67

define this limit to be the line integral of

f

along

to B.

in the calculus the limit

is

written

continuous along T and

if

F has continuous turning tangents, that

If f is

is, -T- is

ds

continuous, then (2.77) exists from

Riemann

integration theory

and can be written *

b

-

f Ja

A

i

dx

v\

^

du

v\ '

dt

\

dt

It

We

use (2.77') as a means of evaluating the line integral. fields for which the line integral from A to

some vector

B

(2.77')

There will be will be inde-

pendent of the curve T joining A and B. Such vector fields are said If f is a force field, (2.77) defines the work done by to be conservative. the force field as one moves a unit particle (mass or charge) from A to J5. We now work out a few examples and then take up the case of conservative vector fields.

=

P(l,

1, 1),

and dr

-

+

xyi xyzk, and let the path of integration be the zj the integration performed from the origin to the point 4 the range of t being given by ( k, ^ t ^ 1. Along the curve, f = f*i tj

Example curve x

Let

2.25.

t,

y =

J

2

,

f

z

/,

+

-

dt

(i

+

2tj 4- k) dt so that f

-

rfr

+ 2/ 2

(<*

4

t )

dt.

Hence

rr>

-dl

If

we choose the

straight-line /"

P(l,

1, 1),

field f is

y

we obtain

P

/

-

F f

dr

=

path x

t,

y

ty

z

=

t

1

/

2 -f-

(<

<

<

3

from the origin to the point

V

=

) eft

T

It

is

seen that the vector

not conservative.

Example 2.26. Let f x* from (0, 0) to (1,

1

x*i 1).

f

+ y*j, Let x

and

-

t

let

the path of integration be the parabola

so that y

J

2 ,

^

J

^

1,

and

l

For the same f let us compute the line integral by moving along the x axis from x 1 from t/ to x J and then moving along the line x to y 1. Although the continuous curve does not have a continuous tangent at the point (1, 0), we need not be concerned since one point of discontinuity does not affec,t the Riemann integral

ELEMENTS OF PURE AND APPLIED MATHEMATICS

68 provided

is

-j-.

bounded

in the

mi)

f(i,o)

=

di

f

/

/(i.i)

t

/

\

y4

'-r

4-

constant

=

dr

f

dr

V<^

=

+

/

dx

=

dr

f

z

dt

+

d<? so

dt,

dy

dt

=

ri t

3

Along the second

~

Tg12

7o

Notice that

conservative.

f is

= 0.

and

= V>

I

/

(j-3

Hence

dt,

7o

become suspicious and guess that 4-

0,

=

dy

n

=

dr

t

\

y

t, t,

7(o,o)

34

.

dr

f

/

=

Along the first part of the curve x part of the curve x = 1, dx 0, y

-5-

have

7(o,o)

7(0,0;

We

We

neighborhood of the discontinuity.

,4

j3

34

=

^>

?

1

T-|- constant

-5--f-

that

- vU)

/B

dr depends only

f

upon the upper and lower

limits and is independent of the path of integration from A to B. have just seen from Example 2.26 that if Example 2.27.

f = V>, <p singleWe a conservative vector field. Conversely, let us assume that /f dr is independent of the path. We show that f is the gradient of a scalar. Define

valued, then

f is

<p(x,

.,

=

2)

?/,

dr

f /

JPo(xn,yn,zo)

The value

of

<f>

depends only on the upper limit (we keep Po

<p(x -f

Ax,

fQ(x + Ax,y,z)

=

y, z)

f

\

Then

fixed).

dr

J P(ro,yo,zo)

and

sp(x 4-

Ax,

i/,

'

=

<^(x, y, 2)

2)

X(x,

/

//,

2)

dx

4~ Y(x, y, 2)

dy

yp(x,y,2)

4- Z(x, y, z} dz

We dy

P

choose the straight line path from **

dz

*=*

0,

Q

to

as our curve of integration.

Then

and Ax

X(x,

/x-f

y, z)

dx

Applying the theorem of the mean for integrals yields <p(x -f

so that

assuming

Ax,

y, z)

-

<p(x, y, z) 1 -'-L-

T^

lim

X

continuous.

= X(, ^

'

y, z)

J

tL

Similarly

Y *

-^, d?y

,.

'

+y + J

7 =

Ax

==

lim

x

^

A'(.

^ ?/.

x -f Ax

z)

-^, so that dz

ft.i +j+k

Q.E.D.

=

X(x.

y. 2)

VECTOR ANALYSIS

A Note

69

quick test to determine whether f is conservative is the following: = V<p, then V X f = 0. Conversely, assume V X f = 0. that, if f

Then

= Xi

for f

+

+

Yj

Zk we have

dX = dF dx

dy

w-f

dZ = dX

dx

dz

Let

=

<p(x, y, z)

y

+

X(x, y,z) dx

/ Jxo

F(x

f Jyo

y, z)

,

dy Jzo

We now =

show that

X(x,

=

f

-

Y(x,

y, z)

Y(x,

y, z)

dX

= =

y, z)

Z(x,

y, z) f

.

-

=

V<p or

dx

y, 2)

+

'

+

Y(x

(2.79)

y, z)

,

dY

Yj

+

+

Z(x,

Zk =

y, z)

gi +

-

|?

Z(x

j

+

ye, z)

,

Z(x<>,

V X

y

,

z)

+ gk = Vv

f

=

is

0.

that

V X

We

f

=

Thus

0.

for f conservative

also say that such

an

f is

f

be the

we have

an irrotational

field.

Example

2.28.

It is easy to

show that

-

dx

f

=

2xye*i

+ jVj + x*ye*k is irrotational.

Then <p(x, y, z)

and

dz

,

gradient of a scalar

vector

-

z)

y^ ZQ can be chosen arbitrarily. proved that a necessary and sufficient condition that

We have f

z}

y

Z(x<>, y, z)

= Xi

The constants x

'

Y(x, v

,

Z(x,

Hence

the calculus

2/0,

y, z)

Y(X

= =

From

V<p.

Z(x,,

f

f* 2xye'

= x 2 2/e = V(x z ye*

-f

+ fj We* dy + f* O

constant)

2

-

e*

dz

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

70

Problems Given

1.

the point 2. f

that 3.

f

e*i

+ sin

(y

f

V*>.

Let

f

j/i

and radius

fi

Let

5.

x 8 from the origin to

dr along the curve y

+ zj + x

z)i

Evaluate

-f xj.

Show

cos zk.

that

f is

conservative, and find

so

<f>

di around the circle with center at the origin

Jf

a.

Show that

4.

written

evaluate /f

xyj,

(2, 8).

-

f

=

if f

(-j/i

<p

V<f>,

dr, vanishes.

xj)/x

-j-

single-valued, the line integral around a closed path,

Prove the converse. 2

Show

-f y*.

around any closed path surrounding the

ft

that

origin,

di

-

f

* V

tan~

!

and show that

(y/x) and integrate for this path

f

2ir

Why

does this integral not vanish? See Prob. 4. Notice that f is not defined at the The curve of integration contains the origin in its interior. This will be important in complex-variable theory. 6. If A is a constant vector, why is it true that dr = 0? origin.

^A

2.18. Stokes' s

Theorem.

We

begin by studying the locus of the end

points of the vector r

=

+

x(u, v)i

y(u, v)j

+

z(u, v}k

(2.80)

where u and v range over a continuous set of values and x, y, z are assumed to have continuous partial derivatives in u and v. For a fixed v = VQ the end points of r trace a space curve as we let u vary continuously.

For each

v

by by

exists, and if we let u vary, we obtain a locus which collectively form a surface. The curves obtained

a space curve

of space curves

= constant are called the u curves, and the curves obtained u = constant are called the v curves. We thus have a two-

setting v setting

parameter family of curves forming the surface. A simple example will illustrate what we have been talking about. Consider r

r

=

r sin 6 cos

We use

6

and

<p

<p i

+

r sin 9 sin

v

j

+

r

cos 6 k

= g g

constant <p

g

2*

>

(2.81)

TT

instead of u and

stant, so that the For a fixed 6 =

v. Let us notice that r r end points of the vector r lie on a sphere of radius

the z component of

r cos

r.

constant.

r, namely, 2* the end points of r trace out a circle of latitude. The curves are thus circles of latitude. It is easy to show that the 6 curves are the meridians of longitude. We can show that the 6 curves intersect

^

For

<?

,

is

^

<f>

\_

the

<p

curves orthogonally.

to a 6 curve, while

The expression

represents a vector tangent

represents a vector tangent to a

^

curve.

From

VECTOR ANALYSIS

71

\_

=

cos

cos

r

=

r sin

+

<p i

ou

sin

cos

r

+

^? i

sin

cos

r sin

k

r sin

^> j

^

j

d^>

we move from a point P(u,

If

* RD

-

we have

v)

_

to the point n-.

>

on the surface, then j

=

=

rfr

tfr

dr

.

9

dn*

-

where ds

is

arc length.

now

Let us

+

+ ,

Q(u

+ rfw, v + dv), P and

\-.

= PQ =

dr

-

ft

dr

Hence

cfo.

dv

dr

2

,

For the sphere ds 2

=

dr

+ .

rf?/

r2

dO 2

+

ar -r-

a?'

a?;

r 2 sin 2

consider a surface of the type given by (2.80) bounded lies on the surface (see Fig. 2.25).

by

a rectifiable curve F that

The

vector

X

du

is

dr

perpendicular to the surface since

As we are tangent to the surface. our head in the same direction as Tdu

X

we keep r

-r->

dv

to our

We now

consider a

on the surface formed by a

u

curves).

fine

The mesh

enough so

v curve

collec-

will

u and be taken

FIG. 2.25

are the

if (u, v)

that,

,

a

+

+

ABCD

The value

+ dv)

;

at

curve

mesh

+

coordinates of A, then (u du, v), (u du, v coordinates of B, C, D, respectively (Fig. 2.25).

v

dv

*S,

tion of parametric curves (the v

-r-

the curve F, keeping

with which we keep in touch. F will be called the boundary of >S. We neglect the rest of the surface r(u, v).

and

track of the area

It is this surface,

left.

move along

du

of f at

A

is f (u, v)

D it is t(u, v + f(u

+

du,

;

at

dv). v)

dv), (u, v

Now

+

dv) are the

consider

ftdT

B it is f(u + Now

=

f(u, v)

=

f(u, v)

du,

r)

;

at

+ dfu + du du TT

'

V

f

C

it is f (u

+ du,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

72

except for infinitesimals of higher order.

+ dv) =

f (ti, v

f (w,

v\

Similarly

+

dv

f

|

V

f

J

Hence, but for infinitesimals of higher order, .

ABCD

v

f.

dv

du ? dv .v)t\. / j du

=

Xdudy, dv

Xf)' du

(V v

\

(see Prob. 16, Sec. 2.15). v

The vector is

r\-

-r-

du

du

X

dv

A BCD,

the area of the sector

since

ABCD

is,

normal to the surface

-r- rfy is

so that

(> j

magnitude

except for infinitesimals of higher order, not a parallelogram. We define

ABCD

^X^dudv dv

&

out in pairs, leaving only

X

(V

dr

f

= V X

gets finer

and <

r

dr.

f

f)

dtf

-*

=fl ^

dd

f

Interior line integrals cancel

Also

We

finer.

f.rfr

(2.82)

du

except for infinitesimals of higher order. We now sum over the entire network.

mesh

Its

strictly speaking,

d6=

as the

S.

(V

X

f)

rfd

thus have Stokes's theorem (V

X

f)-d<J

(2.83)

Corwmente 1. Since r may not be a parametric curve, (2.82) may not hold for a mesh circuit containing r as part of its boundary. This is true, but fortunately we need not worry about the inequality. The line integrals cancel out in pairs no matter what sub-

division

we

use,

tribute little to

negligence.

and //

for a fine

V

X

f

network the contributions of those areas next to r con-

dd.

The

limiting process takes care of this apparent

VECTOR ANALYSIS 2.

73

Stokes's theorem has been proved for a surface of the type r(w, v) [see (2.80)]. is easily seen to be true if we have a finite number of these surfaces con-

The theorem

The

nected continuously (edges).

case of an infinite

number

of edges requires further

consideration, 3. Stokes's theorem is also true for a surface containing a conical point where no dti can be defined. We just neglect to integrate over a small area covering this point. Since the area can be made arbitrarily small, it cannot affect the integral. 4. The reader is referred to the text of Kellogg, "Foundations of Potential Theory," for a more rigorous proof of Stokes's theorem. It 5. The tremendous importance of Stokes's theorem cannot be overemphasized. relates a line integral to a surface integral, and conversely. 6. In order to apply Stokes's theorem it is necessary that V X f exist and be integrable over the surface S.

Examples of Example

Theorem

Stokes's Let

2.29.

=

f

curve in the xy plane. f

(f)

-

yi -f

and

-rj,

let

us evaluate

= //(V X

di

,

-

dy

x

ellipse

b cos

t

dt,

a cos

A

and

F any

rectifiable

-

S

Thus the area A bounded by

For the

dr,

k dy dx - //2k k dy dx - 2A

f)

8

=

f

(p

we have

Applying Stokes's theorem,

t,

^

I

F, is

A = \fx dy - y dx - b sin ^ ^ 2w, we y 2 = irab. sin + a6(cos 1) dt

have dx

t

/,

2

(2.84)

a sin

t

dt,

t

f = everywhere, it follows from Stokes's theorem that around every closed path. Conversely, assume $i dr = around every closed path, and assume V X f is continuous. If V X f ^ 0, then V X f T* at some in some neighborhood of P and V X f point P. From continuity we have V X f ^ Choose a small plane surface S nearly parallel to (V X f)p in this neighborhood. through P with boundary F in this neighborhood of P. The normal to the plane is

If

V

X

chosen parallel to (V

X

f)p.

Example

fi

di

2.30.

=

Then

dr

f

(p

=

//

V

X

f

>

dd

0,

a contradiction.

8

An

irrotational field f

(i)

characterized

= V^ f = i dr =

by any

of the three conditions

V X

(ii) (iii)

Any

is

(2.85) for every closed

path

of these conditions implies the other two.

Example

2.31.

Let

f

=

f(x, y, z)a,

where a

is

any constant vector.

Stokes's theorem yields

/a-dr - /Tv

X

(/a)-dd

S <f)fdT

-

(I vf

X

a

rfd

-

a

(f dd

X

V/

Applying

ELEMENTS OF PURE AND APPLIED MATHEMATICS

74 so that a

f

X

// dd

(Dfdr

Since a

0.

V/J

-

(f)fdi

// dd

X

(rfd

X

is

arbitrary, it follows that

(2.86)

Vf

can be shown that

It

=

r * f

The

//

V)

* f

(2.87)

asterisk can denote scalar, vector, or ordinary multiplication. f becomes the scalar /.

In

the latter case

Problems 1.

Show that

2.

Show that

X

a

(p

=

dr

(h r

by two methods. 2a

**

dr

r

dd

//

if

a

is

constant.

s 5. 4.

By

show

Stokes's theorem

Prove that (D uVv

*

dr

X

V

that

X

Vw

//

0.

VSP

Vv

dd.

5 6.

Prove that

fuVv

fvVu

dr

dr.

x(l -f- sin //)j, find the value of center at the origin and radius r. 6.

If f

7.

If

(f)

cos

E

// i

dr

4-

- -

C ot

J

8.

Show

that

dd for

all

surfaces

S show

dr around the circle with

that

V

XE - -

JJ S

V

X

-f

2-r^/j.

II

B

//

^f

dd

f

*

if

S

-

~

C ot

a closed surface.

is

S 9.

f

(0, 0),

-

(#

2

2 y/

)i

(1, 0), (1, 1), (0, 1).

Find

Do

around

dr

y*f

the

square

with

10. If a vector is normal to a surface at every point, show that its curl tangent to the surface at each point. 11. Let f = a X g, a any constant vector. Apply Stokes's theorem to

that <T)dT

X

g

-

X

(dd

JJ

V)

X

vertices

at

by two methods.

this

is

f,

zero or

is

and show

g.

S

Assume V

12.

X

f

&

0.

Show

that,

V

if

X

a scalar /(*,

W

)

y, *0

exists

such that

-

V X f - 0. dt X r| taken around a closed curve in the xy plane is twice the 13. Show that area enclosed by the curve. 14. Let f X(x, y}i -f Y(x, y)j, and let S be the area bounded by the closed curves Show that Fi and T* lying on the xy plane, Fi interior to F 2 then

f

|

.

// f-dd

Both

line integrals are

-

()t -dt -<

f-dr

taken in the counterclockwise sense.

VECTOR ANALYSIS 2.19.

75

The Divergence Theorem (Gauss).

F in

containing the volume

Let

S be

a closed surface

We assume S has a well-defined

its interior.

normal almost everywhere. We now subdivide the volume into many elementary volumes. From Sec. 2.13 we note that except for infinitesimals of higher order

II

dd

f

= V

f

AT

AS

where A$

is

we sum over we obtain

the entire surface bounding the elementary volume AT. all volumes and pass to the limit as the maximum AT

" f

-dd

=

/// Iff

V-fdT

If 0,

(2.88)

the divergence theorem of Gauss. It relates a surface integral to a volume integral. It has tremendous applications In the derivation of (2.88) use has been to the various fields of science.

Equation

made

(2.88)

is

of the fact that for each internal dd there is a

interior surface integrals cancel in pairs,

surface

S

dd, so that all

leaving only the boundary

as a contributing factor.

(2.88) may be interpreted as follows: Any vector field f may be looked upon as representing the flow of a fluid, f = pv. From Sec.. The 2.13 V f represents the loss of fluid per unit volume per unit time.

Equation

time throughout

total loss of fluid per unit

V

is

f/J

V

f dT.

Now

if

and V f are continuous in F, there cannot be any sources or sinks in F which would create or destroy matter. Consequently the total loss of fluid per unit time must be due to the fluid leaving the surface S. We might station a great many observers on the boundary S, let each observer measure the outward flow of fluid, and then sum up their recorded data. At a point on the surface with normal vector area dd the component of the velocity normal to the surface is v N, and pv dd = f dd represents the outward flow of mass per unit time. The total loss of mass f

per unit time

is

1 1

f

and thus

dd,

(2.88) is obtained.

For a more

and rigorous proof of Gauss's theorem, see Kellogg, " tions of Potential Theory.

detailed

Example boundary

2.32.

Let

surface S.

-

We

gr/r

s ,

and

let

F

wish to compute

be a region surrounding the origin with //

dd.

We

cannot apply the diver-

0. is discontinuous at r gence theorem to the region V since V by surrounding the origin by a small sphere S of radius

this difficulty

"Founda-

We

overcome

with center at

ELEMENTS OF PURE AND APPLIED MATHEMATICS

76

The divergence theorem can be applied S sphere) with boundaries S and S. Thus

the origin (see Fig. 2.26).

(V

less interior of

to the region

V

//*+//*-///*(?)* V S

We

have seen that V

S

3

(</r/r

)

=

so that

0, r 7* 0,

//"--//** z

,s

Now

on

S, r

**

e,

r/d

=

3

(-r/r) dS, so that

(<?r/r

dd

)

=*

(-g/e

2 )

d& and

Hence //

E

r/d

47rg

5 gr/r

8

is

the electrostatic

field

due to a point charge q at the

origin.

For a con-

FKJ 2.26 tinuous distribution of charge of density p in S

it

can be shown that

E'

P UT

Assuming that the divergence theorem can be applied Potential Theory" for proof of this fact), then

(see Kellogg's

p dr

"

Foundations

of

(2.89)

IJI

E = 4irp, provided V E 4-n-p is = 4irp, D = *E. For magnetism V B = comprise two of Maxwell's VV for an inverse-square force. In

Since (2.89) holds for all volumes we must have V In a uniform dielectric medium V D

continuous.

one has V

B *

equations.

It

empty space

p

0. The equations V D can easily be shown that

=

so that

V2 F

the solution of Laplace's equation,

0.

A

=

4irp,

E =

great deal of electrostatic theory deals with

V2 F *

0.

VECTOR ANALYSIS 2.33.

Example

We

Green's Theorem.

77

apply (2.88) to

fi

= uVv and

t>Vw

f2

and

obtain dr

-

dr

=

(uV*V

HI

+

Vw

V0) dr

+

Vt;

Vw) dr

=

/I uVv

dd (2.90)

/// V

Subtracting,

we

(Wu)

2

u

(t>V

JNhj

JJvVu-fa

obtain

=

f (2.90)

and

2.34.

A

//

(uVt>

-

dd

t;Vw)

(2.91)

(2.91) are Green's formulas. Uniqueness Theorem. Let ^> and ^ satisfy Laplace's equation We show that = ^. Let inside a region R, and let <p = ^ on the boundary S of #. - VV =" in #, and s on 5. Applying (2.90) = <p - $. Hence V 2 = = yields with w t;

Equations

Example

W

<f>

([ eve

so that

i

have ve

///

(ve

=

in

continuous as

R

=

V0) dr

-

and

B

=

we approach

Since V0

0.

continuous (V 2

is

Thus

constant.

<p

\j/

=

is

assumed

constant.

we must have

the boundary,

-

<?

=

to exist),

Assuming

^, since

^>

^>

= ^

we

^ is on the

boundary.

Example

Let

2.35.

f

=

/(x,

t/,

2) a,

where a

is

any constant vector.

Applying (2.88)

yields

fdd = Hence

We

HI

ff leave

it

to the reader to

vanishes

assume

f

is

A

2.36.

vector

fd" ~

dd

field f

* f

g,

solenoidal vector.

so that

V

Iff

f

-

whose

called a solenoidal field.

= V X

(/a) dr

-

a

JJJ

Vfdr

show that

II Example

V

= V

/77 (V

flux I

From

X

(V Is the converse true?

* f ) dr

up f

(2.88) g)

=

0.

If

V

f

it

dd

(2.92)

J

over every closed surface

Now follows that V f 0. Thus the curl of a vector is a =* V X g? 0, can we write f

"yes," and we call g the vector potential of f. Notice that g cannot be V X g. We now exhibit a method for determining g. unique for V X (g -f v>) Let f Xi -f Yj -f Zk, and assume g = ai -f 0j We wish to determine g so 7k. that V f provided V f 0. Thus a, ft 7 must satisfy g

The answer

is

+

X

ELEMENTS OF PURE AND APPLIED MATHEMATICS

78

dy dy

are to determine a,

-

I? dx

_ y

Assume a *

0, 7.

%

**

dz

|2 dz

We

_

__

Then

0.

_!

y

x

dz

dy

(2.93)

Of necessity,

-

0(x, y, z)

-

r(x, y, z)

dy

Hence

dx

x)

y,

(a:,

z)

-f r(y, z}

-a/J

/aF

/"*

dX

,

-r- dx ;o

dx

dZ\

.

,

dr

do

dy

dz

-\

dr

d<r

y. 2).

Let

,

.

Therefore

0.

f

I

/

x

V

7

J XO

/ since

+ <r(y,

dx

(x, y, z)

/J

We need only choose

and

<r

r so that

r

r(y, z)

A (xo,

"

dz

dy

- fv /

X(aj

,

<r

=

and

y, z) dt/

y^o

x*

and

j/o

Hence

are constants of integration.

g

-

j

1^ Z(x,

y, z)

dx

where r(y, z) is defined above and For example, let f * x 2 i xyj yields r(y, z)

- ["tidy yO

+ <f>

k

?/ ,

^)

-

[r( is

|J

F(x,

|/,

z)

dx J

+

V*

(2.94)

arbitrary.

xzk so that V

f

*

0.

Choosing xo

yo

*

*

0,

- ~(zxV2), f Z dx - -z 7o f *x dx

ya;o

2 and g (zW2)j -h (t/x /2)k + Vy>. It is to verify that f - V X g. Example 2.37. Integration of Laplace's Equation. Let S be the surface of a region R for which VV " 0- Let P be any point of R, and let r be the distance from P to any point Q in R or on 5. We make use of Green's formula

dr

VECTOR ANALYSIS

We

=

79

which yields a discontinuity inside R, namely, In order to overcome this difficulty, we proceed as Surround P by a sphere S of radius e. Using the in Example 2.32. 1 in R (R less the S sphere) yields fact that VV = VY = choose ^

at P, where r

We

leave

it

=

1/r,

0.

show

to the reader to

that, as

e

-

dd jf (l/r)Vp

0,

>

2

and

dt -> -4r^(P).

// ?V(l/r)

Hence

(2 ' 95)

This formula states that the value of

<f>

at

any point

P

in 72 is deter-

*\

dned by the value mined

of

a and

v?

N =

V^>

on

on the surface S, where

N is the

unit vector normal to S. Problems 1.

xi

If f

yj

+

-

(z*

l)k, find the value of

//

f

over the closed surface

dti

S

hounded by the planes

z

0, z

2.

Show

3.

Prove that

//

dd

=

4.

Prove that

1 1

dd

X

5.

Show

ifI

f

V*> dr

6.

If

v

* V X

that xi -f t/j/x 2

that

w =

-g-V

X

v,

-f-

^

-

-f

2

1.

?/

a closed surface.

is

III

-

and the cylinder y 2

solenoidal.

is

S

if

f

1, 2

V

X

// ^f

f dr.

dd

-

///

^>V

f dr.

show that

u,

rrr

V

7.

If

v

=

v>,

Iff

VV

*=

v * dr =-*

0,

ff

show that

u wdr

(uXv)'dt

for a closed surface

///

v 2 dr

//

^>v

rfd.

and f 2 are irrotational, show that f i X ft is solenoidal. Find a vector g such that yzi - zx] 4- (z* -f y 2 )k - V X g. 10. Find a vector g such that r/r s V X g.

8.

If fi

9.

11. If

12.

-

pv

Find a vector

dd

f

-

such that

[

dT for

a11

surfaces,

show that

V*f-2z+y-l,VXf-a.

+V

( P v)

-

0.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

80

REFERENCES Brand, L.: "Vector and Tensor Analysis," John Wiley & Sons, Inc., New York, 1947. " Kellogg, O. D.: Foundations of Potential Theory," John Murray, London, 1929. Lass, H.: "Vector and Tensor Analysis," McGraw-Hill Book Company, Inc., New York, 1950. H. B.: "Vector Analysis," John Wiley & Sons, Inc., New York, 1933. Rutherford, D. E.: "Vector Methods," Oliver & Boyd, Ltd., London, 1944 Weatherburn, C. E.: "Elementary Vector Analysis," George Bell & Sons, Ltd., Phillips,

London, 1921. :

"Advanced Vector Analysis," George

Bell

&

Sons, Ltd., London, 1944.

CHAPTER

3

TENSOR ANALYSIS

In this chapter we wish to generalize the notion In Chap. 2 the concept of a vector was highly geometric This spatial since we looked upon a vector as a directed line segment. is understood for a of vector of a easily space one, two, or three concept dimensions. To extend the idea of a vector to a space of dimension higher than 3 (whatever that may be) becomes rather difficult if we hold to the simple idea that a vector is to be a directed line segment. To avoid this difficulty, we look for an algebraic viewpoint of a vector. This can be done in the following manner: In Euclidean coordinates the vector A can be written A = Aii A&. We can represent the Azj vector A by the number triple (Ai, A%, As) and write A = (Ai, A 2 A*). The unit vectors i, j, k can be represented by the triples (1, 0, 0), (0, 1, 0), We define addition of number triples and multi(0, 0, 1), respectively. plication of a number triple by a real number a as follows: 3.1. Introduction.

of a vector.

+

+

,

(Ai, A,, A,)

Equations

+

(Bi, B 2 B,) a(Ai, A,, A,) ,

(3.1) define

(Ai, A,,

The elements

AI,

A

A

Ai(l,

As

2,

(A l

+

(aAi,

A2

Bi,

of

A

0, 0)

+

+

aA 2 aA

A,(0,

B,,

A*

+ B,)

.

( ^

'

3)

,

We

a linear vector space.

=

8)

= -

note that

+

0)

A,(0,

0, 1)

(3.2)

are called the components of the

number

1,

triple.

Throughout

we

this chapter

shall use the

summation convention, and

at times, for convenience, the superscript convention, of Chap. continue our discussion of vectors, let A = (Ai, A 2 A 8 ), B = (B 1 ,

be two vectors represented as product of A and B by

We

triples.

,

The square

of the

norm i

=

cosine of the angle between

If

L2 =

two vectors 81

is

,

8 )

(3.3)

(or length) of the vector 1, 2, 3.

2

define the scalar, or inner,

A B = A aB L 2 = A a A, Ai = A\

To

1.

B B

1,

A

defined

is

A

is

defined to be

a unit vector.

by

The

ELEMENTS OF PURE AND APPLIED MATHEMATICS

82

(3 4) '

not

It is

show that

to

difficult

|cos

^

0|

A^BjB* ^

We

1.

(A a

must show that

BY

(3.5)

)>

(3.6)

Let us consider

^

3

y for

x

=

(A ax

-

real.

Now y = ^l^z

+

2A a JS ao:

2

BjB represents a parabola. Since y ^ has no real roots or two equal real roots. Hence Let the reader show that, (3.5), the Schwarz-Cauchy inequality, holds. a = 1, 2, 3. If if the equality sign holds in (3.5), then A a = \B a J

=

for all real x, y

,

X

>

0,

cos 6

=

If

1.

X

<

0,

cos

=

If

1.

cos 6

=

0,

that

is,

A aB = we say

A

that

and B are orthogonal.

can define the vector, or cross, product of A and B algebraically as follows: Let A\ B\ i = 1, 2, 3, be the components of A and B, respecLet Ck, k = 1, 2, 3, be the components of the number triple tively. C = (Ci, C 2 C 8 ), where

We

,

Ck = The

t, 3k

A'B>

(3.7)

We

epsilons of (3.7) were defined in Sec. 1.2.

note that

= A*B C 8 = A B* - A*B> l

C\

1

Let us consider the scalar product of

A and

A-C = A k C k = But If

lJk

We

have

A*A k B>

= -;* so that A C = 0. Similarly B C = 0. A = (Ai(t), A 2 (0, -Ai(O), then is defined to be the kji

-rr> -rr at

<-

C.

at

)

If

2 8 ^(^S 3 ^ )

is

a scalar function of

.

T^ *

1J ;

/

TT

T^ * ;

y *

T' * ;

/

the gradient of

<f>

is

defined to be the triple l,A

A,* \

(3.9)

TENSOR ANALYSIS and

It is easy to define the divergence

number

V X

The divergence

A.

V'A = definitions

curl of a vector in terms of

Let the reader show that

triples.

yields the conventional

The

83

above

refer to

of

A

is

the scalar

AA

^

(3.11)

a

Euclidean coordinates.

In the above presentation geometry has been omitted. Everything depends on the rules for manipulating the number triples. One need

only define an n-dimensional vector as an n-tuple

A = The

(A l9 A,, A,,

.

.

.

A n)

,

(3.12)

definitions for manipulating triples are easily extended to the case There is some difficulty in connection with the vector

of n-tuples.

product and the curl. This will be discussed later. Let us jhope that the reader does not feel that it is absolutely necessary to visualize a vector in a four-dimensional space in order to speak of such a vector. He may feel that an abstract idea can have no place in the realm of science. This is not the case. No one can visualize a fourdimensional space. Yet the general theory of relativity is essentially a theory of a fbur-dimensional Riemannian geometry. One further generalization before we take up tensor formalism. Let S be a deformable body^tfhich is in a given state of rest. Let P(x 1 x z # 8 ) be any point of this body. Now assume that the body is displaced from Let 8i(x l z 2 x 8 ), i = 1, 2, 3, represent the its position to a new position. displacement vector of the point P. The displacement vector at a nearby 1 8 8 dx re 2 point Q(x dx\ z + rfx ) is 9

,

,

+

+

1

,

d*

*,*')+ except for infinitesimals of higher order.

We

M 5/ +

1 (*

2*.

d* The nine terms

of

,

5

= ~

I (*** 2

f

T~

\to .

+

+

T~ V

i,

j

=

2

(3.13)

can write

_

Vto

1, 2, 3,

M a?/

n (

i-n;

are highly important in

deformation theory. It is convenient to represent these nine terms as the elements of a 3 by 3 matrix. A simple generalization tells us that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

84

to speak of a large collection of elements represented

we may wish

r >>>*'/ =

k (rl r2 r*J'" db"-c\X

t,j,

.

.

.

,

a, 6,

fc,

.

.

by

n\

c

.

,

1, 2, 3,

.

.

. ,

n

/g

j\

Before we speak of tenis our first introduction to tensors. we must say a word about coordinate systems and coordinate

This sors,

transformations.

x n ) forms the arithmetic n-space, n. the x* real, i = 1, 2, By a space of n dimensions we mean any set of objects which can be put into one-to-one correspondence with the

The

2 1 totality of n-tuples (x x ,

.

.

.

.

.

,

,

.

,

arithmetic n-space.

Thus

P

<->

x2

1

(x

,

,

.

.

.

,

x)

(3.16)

The correspondence (3.16) attaches an n-tuple to each point P of our We look upon (3.16) as a coordinate system imposed on the space. elements P. We now consider the n equations of space y*

= yK*

*w )

1 ,

**,

,

and assume that we can solve

=

t

1, 2,

.

.

. ,

n

(3.17)

l

for the x so that (3.18)

We assume that (3J7) and (3.18) are single- valued. The reader may read that excellent text, "Mathematical Analysis/ by Goursat-Hedrick on the conditions imposed upon (3.17) in order that (3.18) exists. The 7

n-space of which P is a point can also be put into one-to-one correspondn ence with the set of n-tuples of the form (y l y 2 y \ so that a new coordinate system has been imposed on our n-space. The point P has not changed, but we have a new method for attaching coordinates to the ,

elements P of our n-space. It transformation of coordinates. 3.2. Contravariant Vectors. define a space curve, F, fri this ri x

__

p.

is

,

.

.

.

,

for this reason that (3.17) is called a

We

consider the arithmetic

n space and

V n by

r t(A x\i)

O

1

i

i, ^,

.

.

.

,

n n ,g

g.

In a 3-space the components of a vector tangent to the space curve are

~W ~HT' ~W

Generalizing,

(3.19) as the n-tuple f

we

define a tangent vector to the space curve

-,

-

-

-^-,

dxi -57

i

-

,

^Y

1, 2,

.

.

written

.

,

n

(3.20)

TENSOR ANALYSIS

85

The elements of (3.20) are the components of a tangent vector to the space curve (3.19). Now let us consider an allowable (one-to-one and singlevalued) coordinate transformation of the type given by (3.17). We immediately see that

=

y^

=

i y (x ^ x

.

,

.

.

= y^(t\

x")

,

x*(t\

.

a?(0]

.

.

,

=

(3.21)

tf(t)

n. Equation (3.21) represents the space curve F as the coordinate y by system. An observer using the y coordinate system will say that the components of the tangent vector to F are

i

1,

2,

.

.

.

,

described

given by

^

t

=

1, 2,

.

.

.

,

n

(3.22)

Needless to say, the x coordinate system describing F is no more important than the y coordinate system used to describe the same curve.

Remember

that the points of

P have not changed

We

these points has changed.

cannot say that

:

only the description of (

/ /77/1

is

the tangent vector any more than

we can say

that

-%-

>

j->

I

\

-^

J

(JlJ

(HlJ

r > -|-> at at

>

-

\ )

(it

/

If we were to consider all allowable coordinate we would obtain the whole class of tangent elements,

the tangent vector.

is

transformations,

each element claiming to be a tangent vector for that particular coordinate system. It is the abstract collection of all these elements which is said

We

now ask what relationship exists between to be the tangent vector. the components of the tangent vector in the x coordinate system and the components of the tangent vector answer this question since

in the y coordinate system.

dj_wdx^ ~ dx

dt

1,

4

note that, in general, on every -jr depends

summed from

1

We

to n.

,

n

dy a

dip

We is

-

*

dt

leave

it

=- since

to the reader to

We easily

(4-M)

the index a

show that '

dt

We now make .

.

,

x n ),

i

=

1,

dt

dy

the following generalization: The numbers A (x l n, which transform according to the law 2, l

.

.

.

,

x 2,

,

~

A'(x\

x*,

.

.

.

,

*)

i=l,2, ... ,n (3.25)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

86

under the coordinate transformation xl

= &(x x\ l

.

.

.

,

,

xn)

=

i

1, 2,

.

.

.

,

n

(3.26)

The contraare said to be the components of a contravariant vector. variant vector is not just the set of components in one coordinate system but is rather the abstract quantity which is represented in each coordinate system x by the set of components A'(x). One can manufacture as many contravariant vector fields as one desires. A n (x)) be any n-tuple in an Let (A (x), A 2 (x), l

.

.

.

,

=

x

2

(x*,

:r

.

.

.

,

,

rr)

In any x coordinate system related to the x coordicoordinate system. nate system by (3.26), define the A i = 1, 2, n, as in (3.25). We have constructed a contravariant vector field by this device. If the components of a contravariant vector are known in one coordinate system, 1

.

.

.

,

,

then the components are known in all other allowable coordinate systems, by (3.25). A coordinate transformation does not yield a new vector; it merely changes the components of the same vector. We say that a contravariant vector

is

an invariant under a coordinate transformation.

An

object of any sort which nates is called an invariant.

is

not changed by transformations of coordithe reader is confused, let him remember

If

The point does not change under a coordithat a point is an invariant. nate transformation; the description of the point changes. The law of transformation for a contravariant vector is transitive. Let

^-

Then

r

which proves our statement. Example 3.1. Let X, Y Z be the components of a contravariant vector in a The components of this vector Euclidean space for which ds 2 = dx 2 + dy 2 + dz 2 in a cylindrical coordinate system af e t

.

i

where

2

+

2

dx

dy

dz

r tan" (y/x), z (x y )*, 6 the same as the dimensions of X, F, and Z. 1

z.

Notice that the dimension of

The quantity r9 has the

is

not

correct dimen-

TENSOR ANALYSIS

87

sions. (Rj rO, Z) are the physical components of the vector as distinguished from the vector components (R, 0, Z). R, rO, and Z are the projections of the vector

A - Xi + on the unit vectors e r

,

e0, e t

=

+

Fj

Zk

k, respectively.

Problems

Show that, the components of a contravariant vector vanish in one coordinate system, they vanish in all coordinate systems. 2. If A 1 and B* are contravariant vector fields (the A* and B l i are 1, 2, , n, 1.

if

.

,

components of the vector

really the

variant vector

fields),

O

show that

A*

B*

.

.

a contra-

also

is

field.

What can be said of two contravariant vectors whose components are equal in one

3.

coordinate system? 4. If X, Y, Z are the components given in Example 3.1, find the components in a By what must 6 and 4 be multiplied to yield the spherical coordinate system. physical components?

B* be the components of two contravariant vector C(x) = A*(x)B'(x). Show that

1

-

3.3.

Covariant

x2

l ,

.

.

show that

,

We

Vectors.

x n ), which

.

,

5,

is

the

consider

scalar

i

1, 2,

.

.

. ,

function

point

assumed to be an absolute invariant

under the allowable coordinate transformation x

=

Define

fields.

A*(x)B>(x),

Referring to Prob.

6.

<p(x

A and

Let

5.

C*'(x)

= &(x

1

l ,

x2

,

.

.

in that .

,

x w ),

n,

(3.27)

where x

l

(x)

=

x l (x l

,

x2

.

.

.

,

,

transformation of (3.26).

=

1,

2,

form the n-tuple

d^

d^ a?'

'

is

t

We 1

which

x n ),

an obvious generalization

d <? '

'

'

J

.

.

.

,

n,

\

the inverse

/o ofi^ (3 28) '

a?y of the gradient of

is

<p.

Differentiating

(3.27) yields

Equation (3.29) relates the components of grad <p in the x coordinate system with the components of grad (# = <p) in the 5 coordinate system.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

88

More

numbers Ai(x l x 2

generally, the

,

.

.

.

,

,

x n ),

=

i

1, 2,

.

.

.

,

n

which transform according to the law A*(x\ V,

.

.

dx a

S) =

.

,

=

i

1, 2,

.

.

A a (x\ z 2 .

,

.

.

x)

.

,

,

(3 3Q)

n

are said to be the components of a covariant vector

The remarks

field.

of Sec. 3.2 concerning contravariant vectors apply here as well. One may ask what the difference is between a covariant and a contravariant vector.

law

It is the

of transformation.

reason that no such distinction was

Compare (3.25) with (3.30)! The made in Chap. 2 will be answered in

Sec. 3.7.

Scalar Product of

3.4.

Two

the

sum A aB a

The

letter x is

What

.

is

l

(x)

and

and a covariant

the form of

an abbreviation

A

Let

Vectors.

ponents of a contravariant vector

A aB a

in the

for the set Or 1

,

x2

,

B

t

(x)

.

be the com-

We

vector.

consider

x coordinate system? x n ). Now .

.

,

Ba

7*

A aB a

so that

dX a

i

= A"B a Hence the form

of

A aB a remains invariant under a coordinate transforma-

tion.

This scalar invariant

of the

two vectors

A

called the scalar, dot, or inner product

is

and B. Problems

7

rf 1. If

A =

2. If

<p

t

dX a , A A a -rr- show Athat dx ,

A

l

and ^ are

r

.

A

scalar invariants

grad (^) grad F(<?) 3.

If

C t; =

4.

t

show that

* =

<p

grad ^ -f ^ grad ^

F'(^) grad

<f>

B are the components of two covariant vector transforms according to the law

Show that

tively.

dX adx l r

At and

A Bj

A* and

Aa

t

B

CJ

=

A*5, transforms according to the law (?j()

fields,

==

show that

CJ(x)

r-^-

are the components of a contravariant vector and covariant vector, respec-

TENSOR ANALYSIS

6.

A

A (x)B t (x) for (x)B v (x) a covariant vector field.

Assume

6.

that

B

t

is

Let

3.5.

l

l

Show

Sa^pr

all

contravariant vector

fields A*.

covariant vectors defined above

are special cases of differential invariants called tensors. ponents of the tensor are of the form

where the indices 1, 2,

.

.

.

,

ai,

a2

Show

that

The contravariant and

Tensors.

89

. ,

.

. ,

ar

frl,

,

62,

.

.

.

,

b,

The com-

run through the values

n and the components transform according

to the rule

dx* (3.32)

The exponent If

N

=

0,

dx of the Jacobian

A/"

we say

that the tensor field

tensor of

of order

s.

called the weight of the tensor.

is

absolute; otherwise the tensor

For N = 1 we have a tensor density. weight is said to be contravariant of order r and covariant (3.32)

field is relative of

The

is

'dx

If s

AT.

=

(no subscripts), the tensor

is

purely contravariant,

and if r = (no superscripts), the tensor is purely covariant; otherwise we have a mixed tensor. The vectors of Sees. 3.2 and 3.3 are absolute If no indices occur, we are speaking of a scalar. tensors. At times we shall call T^l'.'.'.^(x) a tensor, although strictly speaking the various T'& are the components of the tensor in the x coordinate

system.

Two tensors are said to be of the same kind if the tensors have the same number of covariant indices, the same number of contravariant Let the reader show that the sum of two indices, and the same weight. tensors of the same kind is again a tensor of the same kind.

We

can construct further tensors as follows

:

(a)

The sum and

of the

difference of

two tensors

same kind

are again

a tensor of the same kind. (b)

The product

case.

so that

Let

of

two tensors

is

a tensor.

We show this for a special

ELEMENTS OF PURE AND APPLIED MATHEMATICS

90 (c)

Contraction.

Let us consider the absolute mixed tensor A}.

We

have

^) = ^(*)gg

(3-33)

The elements

of A] may be looked upon as the elements of a matrix. Let us assume that we are interested in the sum of the diagonal terms, In (3.33) let j = i, and sum. We obtain A\.

= A(x) = Hence A\(x)

is

a scalar invariant (invariant in both form and numerical obtained a scalar by equating a super-

From the mixed tensor we

alue).

a subscript bsolute tensor A^lm 3ript to

.

and summing. We have

Ivklm = A

Now

A\(x)

let

/

=

j,

and sum.

Let us extend this result to the

.

dx a dx* dx k dx dxm

pffT

We

l

obtain

^^ = ~

x

ak

A a* pffT

pffT

A

e pffr

= ftOT

so that write

A

B apr

T

==

Q_ k

dx~"

x ^

x

dx1

^!^!^!m dx dx k dx

dx" dx k dx m

are the elements of an absolute mixed tensor.

A

T

since the index

o-

is

summed from

1

to

We may

n and hence

B

a Notice that the contravariant and covariant orders of disappears. pr It is not difficult to see why this method are one less than those of Aj$ m of producing a new tensor from a given one is called the method of .

contraction.

One may wish to show that the elements B/(z) (d) Quotient Law. Assume that it is known that A 4Bjk is a are the elements of a tensor. tensor for arbitrary contravariant vectors

A\

Then dxk

TENSOR ANALYSIS SO that

A

This

is

the desired result.

is

a tensor.

fijb

.'.'.'

i

(

Bp y

Bjk

This

is

t >

If ^(a: 1 ,

a;

2 ,

.

.

.

,

xn )

=

Since A*

0.

is

arbitrary,

we must have

a tensor for arbitrary A*, then

If A*Bi..'. is

the quotient law.

The Kronecker

Excynple 3.2,

=

1 j-

91

delta

ax' ox**

__ ~

dx" Jx7 ^(ar)

is

6] (see

Sec. 1.1)

ox* ox^

ax*

"

^ "

d

dx*

~d

is

an absolute

scalar,

a mixed absolute tensor, for -

a

tf>(x)

=

$(x), then ^55 is

an

absolute mixed tensor whose components in the x coordinate system equal the corresponding components in the x coordinate system. Conversely, let Aj(x) be an = A](x). Theix isotropic tensor, that is, A](x(x}} .

,

.

_

dx

1

dx"

fix" dx>

Since

-

uX a

for all

can be chosen arbitrarily,

=

?'

occurring.

A*

=

^>5}.

follows that

a, it follows that A] Choosing i 5^ j and a a no summation intended. Then A <r, Hence the diagonal elements of A] are equal to Let the reader show that if A^ is isotropic then ^

a,

i, j,

choose

it

<r.

f.

Now

nmation

j,

^o

that

t

3.3.

Example

Let ^,(x) be the components of an abglmite covanant tensor so that

and If

gr a<r

=

g ffai

we

have, upon taking determinants, that

and

Now if A

T

are the components of an absolute contravariant vector, then

and da:

dx_

dx

A

1

**

Aa fix"

ELEMENTS OF PURE AND APPLIED MATHEMATICS

92

Thus B is a vector density. tensors into relative tensors.

This method affords a means of changing absolute

i

Assume

3.4.

Example

a g a & dx dx&

g a $ dx

Moreover we assume that g a p

Gap

for arbitrary

Hence the

It follows

da;'*.

0a/s(z)

an absolute scalar invariant, that

is

dx*

=

dX a

From dx a

g$a.

dx

1~ !^ dx

from Example

is,

a g a dx dx?

-

-

g

1.2,

v

d& we

have

dx dx*

that

are the components of an absolute co variant tensor of rank 2 (two

subscripts).

Let Ai and

3.5.

Example

#

t

Let the reader show

be absolute covariant vectors.

that

is

an absolute covariant tensor

of

rank 2 and that Ct j

=

C/.

For a three-dimen-

sional space

A iB'2 2

_

A 2 Bi)

The nonvanishing terms correspond

to the

components of the vector product.

Problems 1. If the components of a tensor are zero in one coordinate system, show that the components are zero in all coordinate systems.

-

gp a

g a&

=

2. If 0*0 3.

From

and g t) =

dx a dx& ga p

>

?(ga0 -f 00) H- ?(gafi a gap dx dxP

4. If 6.

If

-

show that $

-

gfa),

j(g a p

t,

=

g jt

.

show that

+ gpa

Ay is an absolute tensor, show that A\\ is A ap is an absolute tensor, and if A a^Apy t

)

dx dx*

an absolute 6",

scalar.

show that A"P

The two tensors are said to be reciprocal. 6. Show that the cof actors of the determinant |o,| are tensor of weight 2 if o u is an absolute covariant tensor.

is

an absolute

tensor.

the components of a relative

-

7. If 8.

9.

A*

is

an absolute contravariant vector, show that r-j

Assume Use

A]/t

(1.26)

is

and

an

isotropic tensor.

|^,l

-

I0.il

dx dx

to

Show that show that

is

not a mixed tensor.

(3.34) holds.

|0,/|t ci,k is

an absolute tensor.

TENSOR ANALYSIS 3.6. The Line Element. we have

For the Euclidean space of three dimensions

<fe

The simple

= =

i

(dx

1, 2,

.

.

.

,

+

2

1

(dx dx dx? )

2

2

)

a

b a ft

we apply a

=

=

2

+ dy* +

dx z

dz*

(3.35)

generalization to a Euclidean n-space yields

ds*

If

93

+

+ =

5 a/3

a

if

(dx

^

n

2 )

S^ =

0;

coordinate transformation, x n,

we have

=

dz<*

dx-a

d&, dx*

uX

=

i

=

1 if

x*(x

#

l ,

=

a

(3.36)

|8

n

2 ,

.

.

.

,

),

dx& -

uX

d$

v ,

so that (3.36)

takes the form

=

dx" dx"

(7 M ,

(3.37)

3

where ^Mv

= 60 5x

a

\ A Qx a dx a

dx^

^^ =

^ ^'

y a=

Riemann considered the general

l

quadratic form ds*

This quadratic form metric. ple 3.4).

The

A

=

gafi(x)

dx" dx*

(3.38)

element ds 2 )

called a Riemannian components of the metric tensor (see Examspace characterized by the metric (3.38) is called a Rie-

(the line

is

g a $(x) are the

mannian space. Theorems regarding Given the form etry.

this

Riemannian space

(3.38),

yield a Riemannian geomdoes not follow that a coordinate trans-

it

formation exists such that ds* = d ap dy a dy&. If there is a coordinate z n a transformation x % = x l (y l y 2 y ), such that ds = d apdy dyP, we the is The that Euclidean. Riemannian coordinate space y say system is .

.

,

.

,

,

said to be a Euclidean coordinate system. Any coordinate system for which the gt} are constants is called a cartesian coordinate system (after

Descartes).

We

can choose the metric tensor symmetric, for

and Tj(Qij ~ 9ji) dx dx = 0. The terms ^(gl} + gi% ) are symmetric. assume that the quadratic form is positive-definite (see Sec. 1.6). 1

Example

3.6.

3

In a tl^ree-dimensional Euclidean space using Euclidean coordinates

one has

so that

We

(dx

2

1

\\ga\\

)

-f (dx

-

2

2 )

4- (dx

100 010 001

8 2 )

ELEMENTS OF PURE AND APPLIED MATHEMATICS

94

For spherical coordinates xl a;

= =

2

xl

The

r sin

cos

r sin

sin

<p

=

cos 6

r

l y sin y* cos y* 2 1 l y sin g/ sin y

<?

y

l

cos

2 2/

^ t-, of the spherical coordinates system can be obtained from Ai*a

d(y

y*,

,

/

Hence

)

2/

a.

-

d\\

ain for

We

AI&

8

l

2

2 ?/

)

?

?^

obtain

-

2

This spherical coordinate system

Example Prob. of the

5,

We

3.7.

Sec. 3.5). lj

Example

3.8.

is

to

1

sin

2 7/

2 )

(%)

2

2 2 -f r sin

not cartesian since

constant.

#22 5^

as the reciprocal tensor to g %1 that is, g ll gjk 5j (see The g* } are the signed minors of the g n divided by the determinant are the elements of the inverse matrix of the matrix \\g %J ||. For the define

The g spherical coordinates of gtj.

^(dy^ +

4-

(di/i)

dr 2 -f r 2

We

$7*''

,

Example

we have

3.6

define the length

L

of a vector

A*

in a

Riemarmian space by the

quadratic form

L 2 = gc<0AA 8 The

associated vector of A*

It is easily seen that

A'

is

the covariant vector

0*0 A0 so

L2 The

(3.39)

cosine of the angle between

that

jAA*

(3.40)

a/

two vectors

A*, B*

is

defined by

(3.41)

Let the reader show that

|cos

0\

1.

Problems 1.

Prove

2.

Show

(3.40).

that |cos

9\

1,

cos e defined

by

(3.41).

TENSOR ANALYSIS

95

For paraboloidal coordinates

3.

Xl

..

ylyt C0 g yl yly* Sin y

X*

X8

Show

.

l.

[(y l),

_

(,)]

that

Consider the hypersurface x* &(u l w 2 ) embedded in a Riemannian 3-space. l l 2 u obtain the space curve x l we fixed, x(wj, w ), called the w 2 wj, i l curve. Similarly a;* = x (u ul) represents a u\ curve on this surface. These curves Show that the metric for the surface are called the coordinate curves of the surface. 4.

t

we keep u

If

1

ds z

is

h%t du*

du

where

j ',

intersect orthogonally 6.

if

The equation p(x l

that the

-~ oX

,

x*,

.

Show that the

g a & T~* ^~r* 1

hit

hu =

OU QU

coordinate curves

0. ,

.

,

a^)

=

Vn

determines a hypersurface of a

.

Show

are the components of a covariant vector normal to the surface. a g a p dx dxP.

6.

Show

that -r-

7.

Show

that the unit vectors tangent to the u l and w 2 curves of Prob. 4 are given

is

a unit tangent vector to the space curve x t (s) d )

i

,

,

An"*

by 8.

If ^ is

d*

,

^22~*

dw 1

2

dx* r

dW 2

the angle between the coordinate curves of Prob.

4,

show that

cos 6

If a space curve in a Rie3.7. Geodesies in a Riemannian Space. mannian space is given by x { = z*(), we can compute the distance between two points of the curve by the formula

The

geodesic

is

defined as that particular curve

x*(t)

The problem

which extremalizes

joining x^to)

and

(3.42). determining the to in the of We apply reduces a calculus variations. geodesic problem the Euler-Lagrange formula to (3.42) (see Sec. 7.6). The differential

x^ti)

equation of Euler-Lagrange

where /

=

(g^&&)*.

is

Now d^

~ i 2/ dx

of

ELEMENTS OF PURE AND APPLIED MATHEMATICS

96

instead of using

If

and/ =

t

as a parameter

we switch

to arc length

s,

then

t

=

s

Hence

1.

_ ~

df

1

2

dx*

=

so that (using the fact that gy

and summing on

Multiplying by g

dV + 4 d? ** 2

d x

r

_

\d? dx" da^

+ r ^ -rf7 d^ (Xo 060

or

,

where

rS

i yields

+ *&>da"

( d0a{

-I-

"

dF U'O

becomes

##) (3.43)

d9a/>

} to*/

_

=

T

,.^

,

dx "

W

~

dx<>

rt

1, 2,

H

441 (d>44j

dF .

.

.

,

n

_

+

(3.44')

Equations (3.44) are the second-order differential equations of the geodesies or paths. The functions F^ are called the Christoffel symbols of the second kind. Example

For a Euclidean space using cartesian coordinates we have g %J

3.9.

constant so that

=

-ry

or X T

0,

Example

-^ =

and

=

b T linear paths.

3.10.

a Ts

+

(3.44'),

Since g 21

r^ =

Hence the geodesies are given by

0.

,

Assume that we

Euclidean 3-space, ds*

Applying

(3.44') yields

=

(dx

2

1

)

on the surface

live

+

[(x

2

1

)

+

c 2 ](dz 2 ) 2

of a right helicoid

We

.

immersed

have

we have

unless

i

2,

we have

Similarly, nl 1

11

f\

Ui

rn fit

^

**

f\

v,

Til 12

r

=

r

-pi 2i

__

f\

o,

r 22 fil

-x_j

x ,

-p2 22

r

__

/\

u,

ip2 r 12

__ -p2 1 2l

, r

n2

,

in a

TENSOR ANALYSIS The

97

differential equations of the geodesies are

dxldx 2

2*1

^

ds z

(x

2

1

)

4-

Integrating the second of these equations yields

2

1

-j- [(x

Let the reader show that

r~

(

)

+

4- 7 1rv^2 (x )

) \ as /

-

c2]

+ ;

constant

= A

constant.

;

c2

Problems 1.

2.

Derive the Tg T of Example 3.10 Find the differential equations of the geodesies

Integrate these equations with 8

=

conditions x 1

initial

for the line

=

On,

element

dx l

x2

<po,

dx*

~T-

1,

-5

at

0.

3.

Show that Y ra8 =

4.

Obtain the Christoffel symbols and the

rj fl

.

differential equations of the geodesies for

the surface

x

1

x2 x3

The

surface

is

5.

From

6.

Let ds 2

(3.44')

-

7. If f},(5) 8.

the plane x 8

i7

=

=

0,

= u = u =

l

cos u z

l

sin

uz

and the coordinates are polar coordinates

show that

dw 2

+

rj7 (x)

2F

rfw

dv

+ G dv*. Calculate |^,|, ||^ r; + Jg-, g, show that rj7 - T ?, is a tensor. ||,

.

fc

gg^

Obtain the Christoffel symbols for a Euclidean space using cylindrical and

spherical coordinates. 9. The Christoffel symbols of the

first

kind are

(t,

jk]

=

g^jk'

Show that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

98 3.8.

Law

of Transformation for the Christoffel Symbols.

equations of the geodesies be given d*x %

_

,

^+

<**#

j and

i

r,

,

N

{x}

Ts dF

,-,d2'dx

!%(*)

-

~

=

.

k

Tg -^

/0 CN (3 45) .

'

,

....

(3.4(0

two coordinate systems x\ x in a Riemannian space. We now From x = x*(x) we have a relationship between the rj* and the T] k l

for the find

dx dxk 3

.

w + T*

Let the

by

.

~

= "

a^ a ds

ds

dx^ dx a ds

ds 2

ds

dx a ds 2

Substituting into (3.46) yields ^x* a aa:

dV ds

6 2 ff

2

dx^ dx* ds ds

ds* dx

dx

Multiply this equation by

Comparing with

dx

(3.45),

?fc

d| d^ dx^ dx^ d^ a dxf ds ds

=

ff

t

and one obtains

3k

+

dx<*

d#

+

(after

summing on

i)

= dx? dx" dx>) ds

ds

we have

(3.47) is the law of transformation for the Christoffel symbols. the Note that Tjk are not the components of a mixed tensor so that T) k (x) may be zero in one coordinate system but not in all coordinate systems.

Equation

Let the reader show that

Example

From Prob.

3.11.

From

Prob.

4,

Sec. 1.2,

and the

definition of the

tf*

we have

5, Sec. 3.7,

s TjjjT

so that

Hence

~^!

ft

(3.49)

TENSOR ANALYSIS 3.12.

Example

this coordinate

Let us consider a Euclidean space using Euclidean coordinates.

From

0.

system T] k (x)

P T

for the

99

x coordinate system.

k(x}

^

d **

.

,*x >

In

(3.48)

~

k

If the T] k also vanish,

then

^

of necessity, so

that

x

-

ff

-f 6"

aj

where a*, b* are constants of integration. Hence the coordinate transformation between two cartesian coordinate systems is linear. If the transformation is orthogonal, n

>? - &*

y a=

For orthogonal transformations

[see (1.53)].

.,

.

.

80that

Now

6"

let

vectors.

1

d' =

=

g* }

ga p

-r-r;

CrX

-rr-

oX

reduces to

^dx"

us compare the laws of transformation for oovanant and contravariant

We

Making use

have

of (3.50) yields n .a

\*\

ra (3.52)

cr-1

m

so that orthogonal transformations affect contravariant vectors exactly the way covariant vectors are affected. This is why there was no distinction

between these two types of vectors

in

same

made

the elementary treatment of vectors.

Problems 1.

Prove

2.

By

3.

Show

4.

(3.48).

tct differentiating the identity g g a j

=

8)

show that

that the law of transformation for the Christoffel symbols is transitive. Derive the law of transformation for the Christoffel symbols from

ELEMENTS OF PURE AND APPLIED MATHEMATICS

100 6.

Define g*f (x)

where

~

<pa

ft

3.9.

*e

and

uX

Show that

p(x)gi,(x).

r*J, TJ fc

are the Christoffel symbols for the

dAj

=

t

~>- ~'dF = We scalar t

_ dA

t.

dx

once apparent

obtain

aV

4

dx k

a

i

It is at

t.

respectively.

by the transformation

with respect to an absolute

since

#*,,

Let us differentiate the absolute covar-

Covariant Differentiation.

iant vector given

{/*,-,

tha,t P-^- Jis \jGfL I

not a covariant vector.

We

wish to determine a vector (covariant) which will reduce to the ordinary We accomderivative in a Euclidean space using Euclidean coordinates. plish this in the following manner by making use of the transformation k

-

law for the Christoffel symbols: Multiply (3.48) by

k

dx dx* dx A -^ = A M -^-jjl

to obtain

dx k

.

dx dx*

=

dx-*

d& d3 k

.

dxd&

d*x*

dx k

+

rjY A a -^-

Subtracting from (3.53) yields

Hence

dA

dx^

--^~

T%y A a -rr

is

a covariant vector.

For Euclidean coordinates rj7 the ordinary derivative derivative of

we choose

rr-

it

so that this covariant vector

We

call

"

TT

at

ac

A^ with respect

since

s=

to

involves ~rr-

t.

Its value

We

write

Since

dt

"

=7-

the intrinsic

at

depends on the direction

at

.

a rj^A MT

becomes

*

dt

TENSOR ANALYSIS is

a vector for arbitrary -j->

follows from the quotient law that

it

d^*M

is

of

a covariant tensor of rank

Ap and we y

101

We

2.

/o K

a A p l^A.

\

ld.Wfi

call (3.56)

the covariant derivative

write (3<57)

The comma in A M(T denotes covariant We now consider a scalar of weight

A = We

differentiation. AT,

ax dx

have ax ax ax* a

From

Prob.

9, Sec. 1.2,

we have a

ax dx dx" N dx" 3

=

so that

Multiplying r-,

"-

n

= TL

ax

+

y

(3.58)

dx

;

by

ax

A

and sub-

tracting from (3.58) yields

dx

- NAT'* = J

Hence the invariant weight N.

We

dx

\ idA _ NATff

dA (in

form),

NAT aa

ox

,

is

dx

a covariant vector of

write

A, a

SE

- NAT ao ff

(3.59)

We call A, a the covariant derivative of A. The comma in A, a denotes = and covariant differentiation. If A is an absolute scalar,

N

~ A * -the gradient of A.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

102

In general, it can be shown that weight N, then M

I

rpftixt

I"

^0101

meriaz 1

.

,40,

atr-p/t .

.

1

ft,

l

T0 $:::(x)

if

ar-pai 0.1 Mm

T

nrr

_t_

.

.

t

-T

a relative tensor of

is

I

motiaj 1 ft

im

:::zr:m

a relative tensor of weight

is

^'m

'''

r**

of covariant rank one greater than

AT",

n

the covariant derivative of

called

is

Since the covariant derivative

is

(3.60)

a tensor, successive covariant

2

.'.'-X-

differ-

entiations can be applied.

Example

We

3.13.

from Prob.

5, Sec. 3.7.

Example have

3.14.

apply (3.60) to the metric tensor

A "' * IF ~ ^ so that At,,

Let

Curl of a Vector.

=

A,, t

^ Ar

ft

have

We

be an absolute covariant vector.

t

S

A-

and

r?'

1

~ 4 r "'

A r

r

ox'

A

We

gr,.

is

a tensor.

It is called the curl of

ox*

A

and

is

a

covariant tensor of rank 2. Strictly speaking, the curl is not a vector but a tensor of rank 2. In a three-dimensional space, however, the curl may be looked upon as a

vector [see (3.10)].

A.

-

|^

where

<p(x\

x\

If

=

Ai

dto

T~> then curl At

0.

Conversely,

if

curl

.

.

.

x) - [* Ai(x\ yxo

,

x*,

.

.

.

,

1 a?) da:

rx* /

a:

Ai(sj,

2 ,

.

.

.

,

a;

w )

dx*

ajo*

1

A,,(zJ, xj,

xj, xj,

3.15.

variant vector

is

.

.

.

,

xj can

be chosen

(3.49),

ra<

.

.

.

,

xj-

,

x) dx n

(3.61)

arbitrarily.

Divergence of a Vector. The divergence of an absolute contradefined as the contraction of its covariant derivative. Hence

div A*

From

then

1

J

Example

0,

[see (2.79)]

+

The constants

At

-

-~ + ATJrt

A^

"gft, j. = dlvA ...

"A-a

-ZZT

+ i

i

i

i

\<?\~*

^

'

\9 AA "-7T^

(3.62)

TENSOR ANALYSIS

103

a we wish to obtain the divergence of A,, we consider the associated vector A* g* A The A" of (3.62) are the vector components of A. To keep div A dimensionally corFor rect, we replace the vector components of A* by the physical components of A*. If

.

1

spherical coordinates 1

Or

101*-

2

r 2 sin r 2 sin 2 6

*

2

so that div A* u \_oT _

(r

ing to physical components (see Prob.

div A<

-

2

+ -~ o<p

sin OA')

*

sm

1

J

and chang-

4, Sec. 3.2)

(rs sin

(r 8in

The Laplacian of a Scalar Invariant. The Laplacian of the scalar defined as the divergence of the gradient of <f>We consider the asso-

3.16.

Example invariant

+^ (r ou

sin $A')

<f>

is

ciated vector of the gradient of

<f>,

namely, g

a&

-^

(3.62) to g"&

Applying

^ yields

the Laplacian

Lap

<f>

VV =

3=

-

rr

(3.63)

j

In spherical coordinates 1

=

r 2 sin 1

r 2 sin 2 r

*

8m 'Problems

1.

Aa

Starting with A*

without recourse to

i

-

show that A*y

(A%),* - A*B

+

5.

Show Show Show Show

6.

Find the Laplacian of

V

in cylindrical coordinates.

7.

Show that

+

rj< AJ

2. 3.

4.

that

that (0rA),, that |gr<,U = 0. that 5*^

is

4.

a mixed tensor

(3.60).

-

?f *

^A ar

A; B,. fc

0.

f

-

for

an absolute mixed tensor AJ without

recourse to (3.60). 8.

Show

that

A? a -

Jj

^

(|^|Uf )

- A|r?

tt

.

9. Write out the form of A", (two covariant differentiations).

10. If

A0c.

t)

is

a covariant vector, show that

-^ ot

=

-

at

+A

*

is

dt

the acceleration vector

if

vi

is

the velocity vector.

tf

,

-37at

Show

that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

104

Since the Christoffel symbols Tjk are not

Geodesic Coordinates.

3.10.

the components of a tensor, it may be possible to find a coordinate system We now show that in the neighborhood of x i = q i such that T] k (q) = 0. Let this can be done. (3.64)

xu

x

so that

5}

neighborhood of x *'

=

0,

that

is,

nate system.

=

and

=

The

q.

the point x

=

Hence

1.

l point x

q

=

(3.64) l

q

now becomes

is

nonsingular in a

corresponds to the point the origin of the x coordi-

Differentiating (3.64) yields

(3.65)

since

T ^(q) = T^(

Thus

l

55

=

g

Differentiating

(3.65)

with

k respect to x yields

+ so that

dx k

d& 1

- r i

Now and evaluating at zl

dx? dx *

d&

dx*

0, yields

- r;() =

o.

Q.E.D.

Any system of coordinates for which (T}k) P = at a point P IB called a geodesic coordinate system. In such a system of coordinates the covariant derivative, when evaluated at P, becomes the ordinary derivative when evaluated at P. For example, -^7]

snce (r system.

=

0, if

the

a;

1

+

are the coordinates of a geodesic coordinate

TENSOR ANALYSIS

We now

105

show that

System (3.66) is true in geodesic coordinates at any point P. But if two tensors are equal to each other (components are equal) in one coordinate system, they are equal to each other in all coordinate systems. Hence (3.66) holds for all coordinate

Equation infinitely

systems at any point P.

yields one geodesic coordinate system. such systems since we could have added

(3.64)

many

There are

to the right-hand side of (3.64) and still have obtained 1^(0) = 0. A special type of geodesic coordinate system is the following: Let us consider the family of geodesies passing through the point P, x* = xj.

Each

=

*

dx l -jds

p

determines a unique geodesic passing through P.

This

follows since the differential equations of the geodesies are of second order. Suitable restrictions (say, analyticity) on the T*k (x) guarantee a unique solution of the second-order system of differential equations when the conditions are proposed. The & are the components of the tan-

initial

gent vector of the geodesic through P. We now move along the geodesic (determined by *) a distance s. This determines a unique point Q. Conversely, if Q is near P, there is a unique geodesic passing through P and Q which determines a unique * and s. We define

p8

x*

(3.67)

The x are called Riemannian coordinates. A simple example of Riemannian coordinates occurs in a two-dimensional Euclidean space. l

= tan 6. a unique geodesic through the origin with slope r coordinate of polar coordinates corresponds to the s of Riemannian

There

The

m

is

Thus

coordinates. 2

=

x

(

=

cos

2

6,

=

sin 6),

r

=

s,

and x l =

r cos

0,

In this case the Riemannian coordinate system (x 1 x 2 ) corresponds to the Euclidean coordinate system (x, y). The differential equations of the geodesies in Riemannian coordinates

x

r sin

6.

,

are

d 2x

But

= ds

ft

'

= ds 2

0,'

-

l ,

,_ N

dx dxk j

so that

= Since (3.68) holds at the point

P

(the origin of the

(3.68)

Riemannian coordinate

ELEMENTS OF PURE AND APPLIED MATHEMATICS

106 2*

system,

=

0) for arbitrary

Riemannian coordinate system 3.11.

=

Hence a 0. it follows that 1^(0) a geodesic coordinate system. We consider the absolute vector V\

',

is

The Curvature Tensor.

Its co variant derivative yields the

7.3 = '

On

_ ^ VT 4-

dx>

= dVk

>J

4'

Vr v

l

a3

a V* L Y ,a T jk

i

,j

l

we obtain

again differentiating covariantly l

V* Y ,1k

mixed tensor

,

,

k

Interchanging k arid j and subtracting yields

where

A

^ ^+^ -

=

R^, k

a]

T'ek

- T^T'9]

necessary and sufficient condition that V\ lk = V\ kl from (3.69) and the quotient law that R ajk l

It follows

(3.70)

that R\, k

is

is

a tensor.

=

0.

The

R^

tensor depends only on the metric tensor of the space. It is called the curvature tensor. Its name and importance will become apparent in the next

two paragraphs.

The contracted curvature

called the Ricci tensor

theory

The

(3.71)

u*/

v*>-

is

tensor

and plays a most important

role in the general

of relativity.

scalar invariant

R=

g

lJ

R^

is

called the scalar curvature.

The

tensor

~

Rh is

Qh R*'

(3 72)

called the Riemann-Christoffel, or covariant curvature, tensor.

Problems 1.

2. 3. 4.

Show that, for Riemannian coordinates, f k ()&& Show that (A] + B]), k - A] tk -f B} tk Show that JB, - #, Show that R akjt0 -f- #,* + /%*,, - 0. This is

0.

j

.

t

.

l

the Bianchi identity.

Use geodesic coordinates. 6.

Show that

6. If Rvj

R\\]k

kgij,

Rikjk "^

show that

R

Rh\ki>

nk,

-Rj*

m

0, K\jkk

** 0.

n the dimension of the

space.

Hint:

TENSOR ANALYSIS

107

If a space is Euclidean, there exists 3.12. Euclidean, or Flat, Space. a coordinate system for which the g^ are constants, so that r] (x) 2= 0, fc

dr*-

=

-T-J

0.

components are

its

(all

It immediately follows that the curvature tensor Rjkl vanishes

We now

zero).

show the converse.

the

If

curvature tensor is zero, the space is Euclidean. Let us first note that if an x coordinate system exists such that the g tj (x) are constants then If there is an x coordinate system for which r* (x) = 0, and conversely. fc

rjtCr)

=

then

0,

and conversely,

=

(3.73) holds, then T* k (x)

if

Our problem reduces

0.

to the following: Given the Christoffel symbols Tjk (y) in any y coordinate n, satisfying (3.73)? system, can we find a set of x\ i = 1, 2, 3, l

.

The system

We

.

.

,

of second-order differential equations (3.73) can be written

'

(3.74)

reduce (3.74) to a system of first-order differential equations.

Let

w" ^=

"

(3 75)

u\T] k (y)

(3.7(5)

-

ftn

so that

For each a we have the first-order system of differential equations given by (3.75) and (3.76). These equations are special cases of the more general system (3.77)

with z

*

=

=

fc

ul,

2,

1, .

.

.

,

.

.

.

+

n

,

= u^

z n+1

1;

j

=

1,

2,

.

.

. ,

n.

If

we

let

zl

= x,

(3.77) reduces to (3.75), (3.76).

a solution of (3.77) exists, then of necessity (assuming differentiability and continuity of the second mixed derivatives) If

=

,

dy

If

the

Fk are j

l

analytic,

+dz* dy

it

l

dy>

conditions z k

=

r

zj

dy

3

+*

can be shown that the integrability conditions unique solution satisfying the a ^ 2/ ^ 111- The reader is referred to the

(3.78) are also sufficient that (3.77) has a initial

dz

ELEMENTS OF PURE AND APPLIED MATHEMATICS

108

elegant proof found in Gaston Darboux, "Legons systemes orthogonaux et les coordonn6es curvilignes," pp. 325-336,

Gauthier-Villars, Paris,

1910.

The integrability conditions become

T?M =

RW The

when

(3.78)

=

referred to (3.75),

(3.76)

-

.

TfouZ

.

(3 79) '

o

from the symmetry of the T*k Hence a necessary and sufficient condition that a Riemaimian space be Euclidean is that the curvature first

equation of (3.79)

and the second

is

satisfied

is

satisfied

= Ufa

if

,

0.

tensor vanish.

Displacement of a Vector. Consider an absolute conA 1 in a Euclidean space. Using cartesian coordinates We assume further that the A % are constant. From 0.

3.13. Parallel

travariant vector yields T*k (x)

=

we have

__

T

A a

IV

*

t/t

L/ ,

._

Oft dx a dx y 6

Moreover Ty

=

since

ff

d*x l

dx& dx a

dx? dx dxt dx

from Example

1.4.

Hence

dA = - A'T^ i

dfr

(3.80)

Now the A\ being constant in a Euclidean space, can be looked upon as yielding a uniform or parallel vector field. Equation (3.80) describes how the components of this parallel vector field change in various coordinate systems. Since, generally, a Riemannian space use (3.80) to define parallelism of a vector field.

We Vn

is

not Euclidean, we can

i say that A is parallelly displaced with respect to the Riemannian i l along the curve x = x (s) if

TENSOR ANALYSIS

109

We

say that the vector suffers a parallel displacement along the curve. a vector suffers a parallel displacement along all curves, then from Ai A A i /7/T0 fj - ,, tt^l (IX* = u/1 -j -7T-R -T- it follows that as dx* ds If

,

.

=

~~

A

T

*

~dxP

A^ =

or

0.

Let us consider two unit vectors displacement along a curve. We have 3.17.

Example

_L_2

,

and

=

since g a p,y

=

r

0,

0,

OS

.

W _^D +WA

=

-

1 ,

B

l

which undergo parallel

=

cos 5 (cos B)

A

dA

0.

,

Hence,

uS

dBP

.

.

two vectors of constant magnitude

if

undergo parallel displacements along a curve, they are inclined at a constant angle.

Two

vectors at a point are said to be parallel if their corresponding If A 1 is a vector of constant magnitude,

components are proportional. the vector

B = l

<pA\

<p

=

scalar, is parallel to A*.

If

A*

is

also parallel

8A we have -r = l

with respect to the

V n along the curve x = l

ds

*

x { (s),

Now

ds

ds

ds

0.

OS

--%* as <p

We desire B

l

to be parallel with respect to the

a vector of variable magnitude

must

^= if it is

V n along the curve.

satisfy an equation

f(*)B*

of the

Hence type (3.82)

to be parallelly displaced along the curve.

The curvature

tensor arises under the following considerations: be an infinitesimal closed path. The change in the components of a contravariant vector on being parallelly displaced along this closed path is

Example

Let x

If

l

*

3.18.

^

x l (t),

we expand A",

higher order,

it

t

^

1,

r^ in a Taylor series about x\ can be shown that

=

x l (Q) and neglect infinitesimals of

(3.83)

where

R^ y

is

the curvature tensor of Sees. 3.11, 3.12.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

110

Problems 1.

Show that the Riemannian space

which

for

ds*

dO*

=

sin 2 9

-\-

d^

is

not Eu-

clidean. 2.

Derive

3.

Show that

(3.79).

the unit tangent vector to a geodesic suffers a parallel displacement

along the geodesic. 4.

that

B

If

^

l

satisfies (3.82),

-

show by

letting #*

= ^A*

that

it is

possible to find

^

so

0.

8s

6.

Derive

(3.83).

Lagrange's Equations of Motion.

3.14.

Let

invariant of the space coordinates (x .r 2 x n ), and the invariant time tives (i, #2,

L be an

,

,

.

L(x\ x\

.

.

.

.

,

.

.

.

.

,

x, x

1 ,

z

,

t,

absolute scalar

x n ), their time deriva-

1

t

=

Hence

I.

2 .

.

.

,

,

x*,

t)

under the transformations **

From

= ^(x\x = f

2 .

.

.

,

t

,x")

L

l

_

it is

is

dx a^ a 4

_

assumed that the x and l

concerned.

.

.

dx*

fr are

dx

_ J

,n

(3 84)

l

~dS

independent variables as far

Now aL ~ = ^Ldx^ dx a dx

d*

l

dLfrr*

dx a dx*

dL

d*x

X

^=~

Also ,,

.

'

6^~o^6^"~aS" as

1,2,

(3.84)

dx

when

=

,

SO that

d (dL\ ] \ox /

-jf [ -^r,

at

=

d (dL\ I TT; J \ox /

^7 at

dL di,

Subtracting (3.85) from (3.86) yields

~

*

TENSOR ANALYSIS Equation

(3.87) implies

111

immediately that

dt

a

dx a

\dx /

an absolute covariant tensor. Let us consider a system of n particles, the mass m,, i = 1, 2, n, We assume that the coordinates located at the point x", a = 1, 2, 3. Let are Euclidean and that Newtonian mechanics apply.

is

.

V(x

l

x 2 x* x 1 x 2 x 3

x

l

.

.

,

r8)

j2

be the potential function such that

=

*

"" <dx]

represents the rth

Newtonian mechanics F rs = kinetic energy of the

=

where g a $

5 a/9.

of the force applied to particle

component

system

d zxr

m -~ is

Euclidean

for

s

coordinates.

s.

In

The

defined as

The Lagrangian

of the

system

is

defined

by

L = T - V

- V

xfx ,

Thus

-

d

.

80 ., that

fdL\

a\wj

Newtonian mechanics.

=

m g a rX = a

m x + dV =

dL =

..

*

-dx]

r

,

n

*

a*.

T ) -r-:r is a vector, it vanishes dx 9 \dxj/ in all coordinate systems. We replace the x 9r by any system of coordinates n 1 2 (j g , g ) which completely specifies the configuration of the for

Since -^

a<

,

.

.

.

,

{

ELEMENTS OP PUBE AND APPLIED MATHEMATICS

112

Lagrange's equations of motion are

of particles.

system

dL ~ ^

#\**i Example vF.

We

3.19.

F =

We

"

r

-

.... n

1, 2,

consider the motion of a particle acted

(3.88)

upon by the

force field

use cylindrical coordinates.

L - T - V -

m(r 2

=

so that

2 -f r

2

+2 - F 2

)

m(rd*)

d fdL\

and one

of Lagrange's equations of

mr

motion mrd*

is

= + -rOT r0 2

r- represents the radial force, the term r

Since

must be the

radial accelera-

OT tion.

If no potential function exists, or if it is difficult to obtain the potential function, we can modify Lagrange's equations as follows: Since the kinetic energy is a scalar invariant, we have that

**

*-*()-

a covariant vector. Let the reader show that if the x l are Euclidean If /r is the force in a y coordinates, then Q r is the Newtonian force. then coordinate system, is

=

Qr so that

Q

r

dxr

is

ia

&

r Qr d *

=

/a

f?

dxr

=

fa dya

a scalar invariant and represents the differential of work

dW, and

We .

.

obtain Q+ "1 .

,

x

1

,

forces acting

"1 .

.

.

,

,

Example

2

l

,

,

on the system, and compute Qt

speed w. and bead.

x* to vary by an amount Ao:*, keeping x # x n fixed, calculate the work ATT done by the

by allowing 1

x*"

A

=

r lim

t

.

i *

.

not

,

summed

hoop rotates about a vertical diameter with constant angular to slide on the wire hoop, and there is no friction between hoop Gravity acts on the bead of mass m. We set up the equations of motion

3.20.

A bead

is free

TENSOR ANALYSIS The

of the bead, using spherical coordinates.

by F = Re r

+ $6^,

unknown, and the

3>

R,

G

mg

R mg Q Q e = mgr sin Qp = $r sin

cos

cos 0e r

mg

hoop on the bead

force of gravity

+

*B

113

force of the

is

given

is

sin 0e0

Hence r

From T = ^m(r 2

+

r2

2

f/y

if

+

r 2 sin 2

0<

v

u***

I

we have

2

),

if

j

("a

cos

sin

:

-

-

(r^

dt

-T

-

2

(r ^)

+ r sin

2

r 2 sin

4, (r at

^

2

2

v? )

cos B

g-rj

( -17

2

-g-rjto

J

gr sin

= -

m

(3.90)

r sin

constant

r

ro,

4>

r

=

r

co f

0,

integrating the second can be obtained from the other two

in finding 0(0

=

sin 2

cos

g *

m

=

tf>)

R and

equation of (3.90). Once this is done, Let the reader show that equations.

=

<p*

sin 2

2

The geometry of the configuration yields = 0. The solution of the problem consists <p

0r

Hamilton's Equations of Motion.

+

cos

From

by

constant

(3.91)

the Lagrangian L(x,

x,

t)

define Pl

We

;

l

(ft

(3.89) yields

-57; 2

3.15.

\dd /

d^ \c

d^>

we

) '

/dT fdT\ = d

rf

. 2

.

-

= mr

" 7

dT = mr 2

Equation

7

show that

vector.

Since

p?,

is

L =

=

i

|^

=

t

1, 2,

.

.

n

.

,

a co variant vector, called the generalized L,

(3.92)

momentum

we have dL _ dL

d:c

a

_ ^L

ax

tt

ax a

which proves our statement. sent the coordinate system.

H

We

shall

now

use

The Hamiltonian

- pa q" -

L(j,

(?,

instead of x i to repredefined by

q*

is

(3.93)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

114

From

=

i

pi

1,2,

= .

ss

.

.

,

F(g,

n, in

g,

we assume

t)

t,

=

H(p,

dL

01

T-TOff*

= p.

dq

~*T

From Lagrange's rv

T

equations of motion, /

7

%

dp,

,

so that

-j-

Also

I c

di\t

~dq

dH

=

^-^

=

g* *i~

~

=

PO ^PI

dpt

from

dJu

a

dL _ a ~~ .,

q, t)

-p _-_-_ ^ do

= since

,

of the p's, g's, arid time,

H

dH

Hence

solve for the g 1

Thus the Hamiltonian

terms of the p, g and time.

H now becomes a function __

we can

that

(3.92)

and

ql

d(f~

~di

(3.93).

Hamilton's equations of motion are d(f

= dH

dt

dp T

d& =

_dH

dt

dq

l

Whereas Lagrange's equations of motion are, in general, a system of n second-order differential equations, Hamilton's equations are a system of

2n

first-order differential equations.

Example

and

3.21.

#

Referring to Example 3.19,

Pr

- dL - mr

qr

-

Pe

= &L * mf

*'

=

pe

~-

- pg - L

W

d? 9L

=mz .

.

r

p - -

r

PQ mr"2 .

pt

**=*=n

TENSOK ANALYSIS

115

Hamilton's equations are

mr*

__

d*r

m

Hence

r//

=

2

m r

p}

dV

wr 3

dr

/

d0\

(I

I

[

r2

dt\

dV

]

rlie

~ dV

d zz _ ~

~dz

~dfi

These are Newton's equations of motion

dr

\dt /

dt) >n

dV

= mr Sde\z _ ~ ~

dr

for a particle using cylindrical coordinates.

Problems 1.

Find the components of the acceleration vector

in spherical coordinates

and

in

cylindrical coordinates for a particle. 2. A particle slides in a fnctionless tube

constant angular speed and the particle.

T =

3.

If

4.

Set

,

.

,

.

.

,

n a & q )q q

up Hamilton's equations

Hooke's law 5.

a a p(q l q z

which rotates in a horizontal plane with Neglecting gravity, find the reaction between the tube

03.

force,

F

,

show

2T =

that

for a particle

q".

moving

in

one dimension under a

k*x.

Integrate (3.91) under the assumption 6

ISO

.

Equations of Motion. We discuss the motion of a rigid The reader is urged to read Example with one fixed point. body The x coordinate system 2.10 the results of which we shall use. will be fixed in space, and the x coordinate system will be rigidly attached We shall use the S sign to represent an integration to the moving body. 3.16. Euler's

over the complete rigid body. In vector notation the angular-momentum vector

and

H

=

H

= 2me

Sr

is

defined to be

X mv

in tensor notation

From

k

t

t j k x'

x

3

(3.95)

(3.95)

m< V me ^xw xx + V m( 2_ ^\ - V xx'

dHk = dt

k

2^m^,

*x ""k xx

/

i-,

ik

^me,

aiyiPA c f** y*^ */ Blill'v/ ciiTc*" i_

/

/

~~ _'f*J ^* *' *it*

t

/

x

& yJ'ir* *t/b*' * /

/

^

>

i_'y*T 7

*ti**'

"

~* il

(3.96)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

116

The moment

or torque vector

X

r

defined as

is

d 2r

=

f

-^ ** (3.97)

T = Lk

Hence

(3.98)

at

Now since

/:/*

# =

>z

co/a

Since

[see (2.21)].

space coordinates,

we

H We

define

/*'

and

e ijfc

o>f(0 are

independent of the

write

=

k

= Sra#^ as the

e^fc

Since x l

inertia tensor.

= A\x a we ^

have

_

d&

dx

The components

of the inertia tensor relative to the

are constants since the frame

This

is

time.

Remember

time.

Thus

this

moving frame

rigidly attached to the rigid body. not the case for the I d since the o are, in general, functions of is

that the a^ are the direction cosines, which vary with be useful to deal with the x coordinate system. In

it will

frame

H

= e^wf

k

The # transform like the components

=

transformations since -

_

\a]\

dX a

= jj

[see (1.25)].

_

dHk _ Aa dH dt

From A?aa = l

6},

d}

x7

it

Ja

,

Aa rr

dA%

jj

dt

follows that

Ctt

^

so that

dt

= - A* da*

an absolute tensor for orthogonal

of 1

(3.99)

= A? AKZ

Moreover

TENSOR ANALYSIS Hence

117

[using (3.99)]

~

Lk

(3.100)

one form of Euler's equations of motion. Since I is a quadratic form, we can find an orthogonal transformation which Let us now use this diagonalizes this tensor or matrix (see Sec. 1.5). is also fixed in the new coordinate system (it moving body). We omit have We the bars of (3.100) for convenience. is

Equation (3.100) %1

7i

72

00/3 +

It is easy to see that I\ z axis,

The

etc.

72

= A

z

,

for k

Equation (3.100) becomes,

=

moment

the

xyz coordinate system

is

of inertia

fixed in the

about the

moving body.

1,

= Li UW Z

I

(/2

IY/22

dux

.

or

_

(/3

,

du,, A v~"Ti

,

.

Similarly

/ll)

+

(

AA X ~ A4

.

z

)u z u x

=

Ly

=

L,

j

(3.101)

j..

Problems 1.

For a

free

body, L

0,

show that l -f

and

l&l

+

+ Aw^ = = +

constant constant

follow from (3.101). 2.

Assume

motion 3.

4.

for

2

that, at

>

=

for

a free body,

co x

= w

,

ww

=

co

=

0.

Describe the

Solve for o^, co y o>* for the free body with A x = A v that the body of Prob. 3 precesses with constant angular speed about the .

,

Show

angular-momentum 5.

t

0.

vector.

Derive the second and third equations of (3.101).

3.17.

Let us

The Navier-Stokes Equations first

of

Motion

for a Viscous Fluid.

consider the motion of a fluid in the neighborhood of a point

ELEMENTS OF PURE AND APPLIED MATHEMATICS

118

P(x) of the

u

=

l

g va

ua

P

Let the velocity of the fluid at velocity at a nearby point Q(x

fluid.

The

.

+

be given by u\ or is, except for

dx)

infinitesimals of higher order,

u

l

(x

+

= u

dx)

-

(x)

l

*(*>

+ +

du

t

*** 1

/du, i

?i V

du a \ '"./"*'

_

2\a.r

The

1

/ da

<i

I

,

i^

d,,.\ '

X

I"

2\d.r

r).r'/

partial derivatives arc* evaluated at the point P.

i

/

r/.r

&r / i

Strictly speaking,

not a vector, so that we should be concerned with the intrinsic The above can be written differential du is

du,

t

_I_ fj \ |^ \bJU

r

r)

?y ( (I>^^1/

We now

.

~~

J_

r

\ ( f>7^.t /

?y

|^

_

1

(1 i ^\''*"i.,a

} f] Ta i*wl/

>ii

"'cr,i/

_L l/ ?y j^ ^\' *'t,ff ;

-4- ?/ ") f?T a 1^ ^OL,%) WJi'

analyze (3.102), which states that the velocity of

Q is

(3 0'2 ) VO. 1i\J4)

sum

the

of

three terms:

u

1.

t

(x),

The term

2.

Q

the velocity of P.

a translational

is

carried along with P, so that

is

u

t

(x)

effect.

u a>l ) dx" corresponds to a rotation with angular

2-(?i,, a

= ?V X u. A sphere in the neighborhood of with center velocity <o would be translated and rotated under the action of terms 1 and 2. at

P

P

terms 1 and 2 are rigid-body motions. Hence the term ^(u lta + u a>l ) dx a must be responsible if any deforma-

It follows that 3.

tions of the fluid take place. define

We

=

*tf

i(ttw

+

*M)

(3.103)

=

x i temporarily, that is, let P be the origin of our coordinate system, x the coordinates of the nearby point Q. We can write stj x l = ^V(s t; x x ) using Euclidean coordinates. However,

as the strain tensor.

Let dx

1

1

7

s tjX

l

x

J

is

a quadratic form~which can be reduced to

9 =

Sn^

2

1

)

+

M^ + 2

2

5

)

M (*

2

8 )

under an orthogonal coordinate transformation. In the system, Vp = Su&i + s 22 2j + Ss3^ 8k. Thus, along with the rigid-body motions of terms 1 and 2, occurs a velocity whose components in the 5 1 5 2 x* The sphere surdirections are proportional to 5 1 5 2 5 8 respectively. rounding P tends to grow into an ellipsoid whose principal axes are the ,

,

2

1

35

8 ,

axes.

Terms

1, 2,

,

,

and 3 characterize the motion

of

order to discuss the dynamics of a fluid, one tensor iff. The face refer to Fig. 2.22.

We

,

,

a

fluid completely.

must consider the

ABCD is in

In

stress

contact with a

TENSOR ANALYSIS

119

part of the fluid. This part exerts a force on the face. This force per unit area has components *", t w t* v The y refers to the fact that the normal to the face in the points y direction. By considering .

,

ABCD

the other two principal faces one is led to the stress tensor P',i,j The total force on a closed surface S is given by

=

1,2,3.

(3.104)

where

N

3

is

show that

the unit normal vector to the surface area dd.

(3.105)

t$dr

so that

t\\

The

9 if .

Let the reader

(3.104) can be written as

represents the force per unit volume due to the stress tensor equality of (3.104) and (3.105) is essentially the divergence

theorem of vector analysis. In order to obtain the Navier-Stokes equations of motion, we make the fundamental assumption that the components of the stress tensor be proportional to the components of the strain tensor. Thus t]

We

further assume that a*g

Example Combining

(see

is

=

ajg2

(3.106)

an isotropic tensor so that

3.2).

(3.106)

and

and

t\

(3.107) yields

= =

k(x)b]s Zks\

+

+ ls\

(3.108)

l(x)s]

=

(3fc

+

I)l

In hydrostatics

-pOO -p -p so that

t\

=

general case.

3p.

The

pressure p

is

defined to be

p =

Thus I

.

$t\ for

the

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

120

Equation (3.108) becomes

and

=

tj ti

or

= -pj -

j if

From

=

Sj ktt

iw,, fcl

g

+ JM*,* and = I

Fj

=

=

t<

+

~ s% tj

pj

+

J0*8jfcf

ak

Tsvg

u a ,k}

(3.109)

,

becomes

2v, (3.109)

+

p, 3

0-^db,,

ls} ti

4- vg

kl

u

(3.110)

jt ki

In vector form

F = -Vp where F

is

+ o~V(V-u) +

i>V

2

u

(3.111)

the force per unit volume due to the stress tensor

Let

t].

f

be the external force per unit mass so that (F + pf) dr is the total force Newton's second law of motion states that acting on the element dr.

j

(p dr u)

=

(F

+

pf ) dr

=

(F

+

pf ) dr

t

p dr -TJ

or

since p dr

is

p

where

v is

Hence

a constant during the motion.

=

*J (/r

_ vp +

pf

the viscosity of the

V(V ^ o

fluid.

-

u)

+

V'2 u

Equation (3.112)

Stokes equation of motion for a viscous

(3.1 12)

is

the Navier-

fluid.

Problems 1.

For an incompressible

show that

fluid

du _

2.

.

Consider the steady state of flow of an incompressible fluid through a cylindrical

tube of radius

a.

Let u

Then show that u

j-

(r

= 2

2 w(r)k, r

a 2 ), ~-

=

z* -f

A =

2 2/

.

Show that p =

p(z),

vV 2w

=

~

constant.

Solve for the steady-state motion of an incompressible viscous fluid between two one of the plates being fixed, the other moving at a constant velocity, the distance between the two plates remaining constant. 8.

parallel plates (infinite in extent), 4.

Find the steady-state motion of an incompressible viscous fluid surrounding a is rotating about a diameter with constant angular velocity. No

sphere which

external forces exist.

TENSOR ANALYSIS

121

REFERENCES Brand, L.: "Vector and Tensor Analysis," John Wiley & Sons, Inc., New York, 1947. Brillouin, L.: "Les Tenseurs," Dover Publications, New York, 1946. Lass, H.: "Vector and Tensor Analysis," McGraw-Hill Book Company, Inc., New York, 1950. McConnell, A. J.: "Applications of the Absolute Differential Calculus," Blackie & Son, Ltd., Glasgow, 1931. Michal, A. D.: "Matrix and Tensor Calculus," John Wiley & Sons, Inc., New York, 1947.

Thomas, T. Y.: "Differential Invariants

of Generalized Spaces,"

Cambridge Univer-

New

York, 1934. Veblen, 0.: "Invariants of Quadratic Differential Forms," Cambridge University sity Press,

Press, New York, 1933. Weatherburn, C. E.: "Riemannian Geometry," Cambridge University Press, York, 1942.

New

CHAPTER 4

COMPLEX-VARIABLE THEORY

The

4.1. Introduction.

of

reader

We

complex numbers.

enter

is

already familiar with some aspects into a discussion of some of the

now

In order to attach a solution simpler properties of complex numbers. 1 = 0, the to the equation x* mathematician^ forced to invent a new

+

2 \/ 1. We say that i is an number, i, such that z + 1 = 0, or i order number in to distinguish it from elements of the realimaginary number field (see Chap. 10 for a discussion of this field). The solution of the quadratic equation ax 2 + bx + c = 0, a 9* 0, requires a discussion The set of all such of complex numbers of the form a + pi, a and ft real. complex numbers subject to certain rules and operations listed below is called the complex-number field, an extension of the real-number field. We note that the complex numbers are to satisfy the following set of rules or postulates with respect to the operations of addition and multiplication 1. Addition is closed, that is, the sum of two complex numbers is a complex number. :

(a 2.

2s

=

+

4.

di)

=

(a

c)

+

If z\

=

+

+ d)i a\ + bit, 2 = (6

2

a2

+

bzi,

+

The unique

= that

+

(c

Addition obeys the associative law. b s i, then 3 21

3.

+

bi)

+

0-2

+

(22

+

2 8)

(zi

+

z*)

+

z*

zero element exists for addition.

and

2

+

+ 2=2

=

for

any complex number

Every complex number z has a unique negative, written + (-z) = (-z) +2 = 0. If 2 = a + fo, then

z,

z

such

z

-2 = (-a)

(-fr)i

= -a -

bi

commutative.

5.

Addition

6.

Multiplication

is

+

is

defined as follows: If z\

=

a\

=

+

a 2 bi)i

+

then ziz*

(aia 2

fci6 2 )

122

+

(aib 2

bii, z*

=

a*

+

bji,

COMPLEX-VARIABLE THEORY

123

The product is

of two complex numbers is again a complex number. the closure property for multiplication. 7. Multiplication obeys the associative law.

=

1(2223)

The unique element

8.

=

1

=

z

1

+

1

2

1

i

=

=

for all 2

62

a2

+

+

b2

distributive law holds with respect to addition.

2i(2 2

4.2.

+

If z

commutative.

10. Multiplication is

The

has a unique inverse, written 2 = a b- 7* 0, then bi, a

z

1.

+

a2

11.

=

z~ l z

2i2a23

exists for multiplication.

z

Every nonzero complex number

9.

z~ l or l/z, such that zz~ l

=

(2i2 2 )2 3

This

+

2 3)

=

+

2i2 2

2i2 3

(Z\

+

2 2 )2;

{

The complex number

The Argand Plane.

We may

= z

+ + iy admits of a

2i2 3

=

x

consider z as a vector

very simple geometric representation. whose origin is the origin of the Euclidean xy plane of analytic geometry and whose terminus is the point P with abscissa x and ordinate y. The of

mapping

complex numbers

manner

in this

yields the

Argand

z plane.

Addition of complex numbers obeys the parallelogram law of addition of vectors. We call x the real part of z, written x = III 2, and we call y the imaginary component of

y

= Im

The length ing z of

2,

=

2,

written

2 (see Fig. 4.1).

x

+

written

of the vector representiy is called the modulus

mod z =

The argument

of the

\z\

=

(x

+y

2

^/^

\

y =Imz

2

)*.

_

complex number

the angle between the real x axis and the vector 2, measured in the

2 is

counterclockwise

ment

of z is not single-valued, for argument of z for any integer n.

by the inequality

||

Example 4.1. - 1, Arg 2 Example 4.2.

If z T.

If

< Arg

TT

1

If 2

-f

-

we use

I0

The argu-

sense.

i,

g

z

then

|*|

z

+

iy

arg

2,

then

9

2rn

is

also the

define the principal value of arg z

*

and Arg 2 r/4. Arg z -*-/2.

-y/2, 1,

polar coordinates, 2

=

6

'

?r.

\z\

-2, then

if

We

'

we may

''(cos

+

write

i sin 0)

If z

1,

then

ELEMENTS OF PURE AND APPLIED MATHEMATICS

124 r,

\z\

arg z

If z\

0.

to the reader to

+

ri(cos 0i

sin 00, 2 2

i

r 2 (cos

2

+

i

sin

2 ),

we leave

it

show that

=

ZiZt

r2

22

+

rir 2 [cos (9\

- =

#2)

-

[cos (0i

2)

+ +

+

sin (0i sin (0i

t

2 )]

-

r 2 7*

2 )]

x-

We leave it to the reader to attach a simple geometric interpretation to multiplication and division of complex numbers. Example 4.3. The reader should verify that Rl

(Zi

Im

(21

+2 +2 =

ziZz

arg

r2

=

=

arg

Im

zi -f

|2i|

|2 2

=

arg

_

2l

-W ^Rl2 -|2|

=

If 2 z

=

ing

x i

x

+

iy.

by

^ Im

\z\

+

Zl,

Z2.

zz\

^

2irn

z2

n an integer

|2|

Obviously

=

2 Rl z 2i

Im

zz

= =

x*

+

22) (21

+

22)

z

+

a rg

z2

|

^2/> we define the complex conjugate of z by the equation The conjugate of a complex number is obtained by replac-

i.

|si

2irn

22

Im

|2|

^

2

+

z

z

z

From

2i

arg z z

21 -f

Rl

Rl

2)

\Z\Z*\

arg

+

2)

22

|^i

|

2

=

|

+

(^i

\Zz\y

+

and from

z

=

y*

we

let

this that

\zi

2 |z|

the reader deduce that zz

\

^

|

\Zi\

\z*\

\

for all

4.3. Simple Mappings. Henceforth the complex number z will stand always for the complex number x + iy> We now examine the complex number w = z 2 = (x + iy) z = x 2 It will be highly beney* + 2xyi. ficial to construct a new Argand plane, called the w plane, with

w

+

= u

iv

Euclidean coordinates. The equation w = z 2 may be looked upon as a mapping of the complex numbers of the z plane into complex numbers of the w plane. The transformation w = z 2 maps the point P with coordinates (x, y) of the z plane into the point Q of the w plane whose 2 v = 2xy. coordinates are u = x 2 A curve in the xy plane will, in i/ For example, the transformageneral, map into a curve in the uv plane. <x> tion w = z 2 maps the straight line x = t, y = mi, < t < oo , into u and

v

,

the straight-line segment u

The hyperbolas x y u c. The hyperbolas xy 2

2

= =

(1

m

2

constant

** c

map

), v

=

c

=

g t < . 2mt, into the straight lines

map

into the straight lines

v

=

2c.

COMPLEX-VARIABLE THEORY Example

We now examine the

4.4.

z =s r(cos

w =

sin 0),

-f- i

u

=

iv

-\-

u

-\-

iv,

and

?/

?'

From

+

cos 2

=

sin 2

i

-f

=

--

(r

2

_

=

a

7* 1

+

map

(r

f/r)

^

0.

We have

-

-f-

-

(cos 6

i

sin 0)

cos

J

sin

)

-

__

f

^

I//)

into the ellipses r/2

(a

+

,,2

I/a)

2

Ma

-

I/a)'

Into what curve does the circle

of the uv plane.

1/z, z

we have

1

(r

circles r

sin 0)

= (r

__ The

+

z

so that

+

r(cos

125

transformation w =

r

1

map?

Problems a

+

=

1.

If

2.

Show that

7>?

c -f di, a, 6, r, 2

=

2

=

d. real, show that a = c, 6 by use of the distributive law and the

</

for all z

definition of

the zero element. 3. 4.

From |z, -fShow that

formula of 2s

=

i f Z4

z2

(cos 6

De Moivrc. = 1? 2 6 i,

+

(zi -\-

i

2 2 )(2i

sin B) n

+ =

22)

show that

cos (n6)

+

i

|zi

-f 22!

sin (n0),

Obtain an identity involving cos

^

|i|

-f-

|22|.

n an

40.

This is a integer. Solve for the roots of

Examine the transformation w = z 1/z in the manner of Example 4.4. = z -f h, b complex, w = Examine the transformations w az, a complex, w Examine the important bilinear transformation w (az -f- b)/(cz + rf), a, 6, 6. 6.

1/z. c,

d

complex. 4.4. Definition of

a Complex Function.

plex numbers defined in or set of rules we can set

Now

some manner.

Z =

\z} be a set of comassume that by some rule

Let

up a correspondence such that

Z

for every point z

there corresponds a unique complex number w,j^nd be the totality of complex numbers obtained in this manner

of

let

W

=

\w\

by exhausting We thus have a mapping Z > which defines a complex the set Z. The correspondence between the element z function of z over the set Z. and the element w is usually written

W

w=f(z) It is

customary to consider

f(z) as the

complex function of

(4.1) z.

We

can

write f(z) = u(x, y) iv(x, y), since, given z, we are given x and t/, which in turn yield the real and imaginary parts of w, called u(x, y) and

+

v(x, y), respectively.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

120

z

correspondence

domain

Z be the set of complex numbers z such w such that w ** z/(\z\ 1) defines a complex

>

of definition of

/

\

U(X. U) y

x

=

/

t

Let

+

Z be

>

2 ?y

,

=

?

The

+

the finite set of complex numbers

for all z of Z.

m

Note that

y

-

2

i

1

2 correspond to ?r = 1 correspond to ir =* 5, 2 a complex function of z defined over the set Z. Example 4.7. Lot Z be the set of real integers, and

w

.:::....

vV +

1

constant

1.

y

=

1

x2 4.6.

\

V(X V)

-

Vx^TV

with

Example

>

\z\

function of z for the

In this case

z.

2

2

that

Let

4.5.

Example

this case

z

i.

= 1 and 2 = 2. Let This mapping defines

let z of

Z

correspond to the

more than one element

of

Z

corresponds to the same value w. A complex function can be a many-to-one mapping. Remember, however, that only one w corresponds to each z. This is what we mean by a single-valued function of z.

We define continuity of a single-valued 4.6. Continuous Functions. complex function in the following manner Let the points z of Z be mapped into the points w of W, and consider the two Argand planes, the z plane :

We say that /(z)

and the w plane.

continuous at

=

z z in Z, if the following holds: Consider any circle C of nonzero radius with center at WQ = /(zo). We must be able to determine a circle C' of nonzero radius

with center at

w

W

zo

is

such that every point

z of

Z

interior to C.

of

-/(Z

I/CO for all z of

Z

such that

<

z

|z

|

)|

maps

into a

means 6

that, given any (the radius of the

>

<

This

5.

,

interior to C"

Analytically this > (the radius of the circle C), there exists a circle CO such that

point

z

is

the usual

,

5

definition of

If /(z) is continuous at every point continuity of real-variable theory. z of Z, we say that /(z) is continuous over Z.

If

Zij

=

i

is an infinite sequence of Z tending to zo as a then the above definition of continuity implies that

1, 2, 3,

limit, Zo in Z,

.

.

.

,

lim /(z) e

Let the reader verify Example

4.8.

for all z since

=

/(z

)

*o

this.

Let f(z)

z for all z.

we need only

pick

6

We

easily note that this function is continuous for every choice of e 0. defined over the same set Z [z\, and assume

>

e

Example 4.9. Let/(z) and g(z) be We now show that/(z) that/(2) and g(z) are continuous at the point z in Z. is continuous at z Choose any e > 0, and consider e/2 > 0. Since f(z) tinuous at ZQ there exists a 61 > z such that \f(z) < f(z )\ < e/2 for \z .

\

A

similar statement holds for g(z), with 5 2 replacing able to show that \(f(z) g(z)) (/( ) 0(o))| < c for

Z.

+

smaller of

5i, 6j.

-

+

The

61. \z

-

2

|

4- g(z) is

con-

61, z

in

reader should be

<

a,

z in Z,

the

COMPLEX-VARIABLE THEORY

127

Example 4.10. Let the reader prove the following statements: Let/() and denned over Z, and assume /(z) and g(z) continuous at ZQ in Z. Then

~

continuous at ZQ. continuous at z f(z)/g(z) is continuous at z provided g(z Q )

!

/(z)

be

0( z ) is

2. f(z)g(z) is 3.

g(z)

.

^

0.

Problems 1.

Prove the statements of Example

2.

Show

for all

z,

z

^

Show

3.

*

that/(z) oo

aoz

w

+

aiz

n-1

4.10.

+

-f

o n n a positive integer, ,

is

continuoun

.

that the /(z) of Examples 4.6 and 4.7 are continuous over their domain of

definition.

Let

4.

z

-

/(z)

-

8 (

-

l)/(z

-

1), z 5* 1,

Show that

3.

/(I)

/(z) is

continuous at

1.

Show that /(z) is uni5. Let /(z) be continuous over a closed and bounded set. formly continuous (see Sec. 10.12 for the definition of uniform continuity). 1. Show that /(z) is discontinuous at z 5. 6. Let/(z) = 1/(1 z), z 7* !,/(!) x sin (1/x) -f ly, x 7* 0, /(O) = 0, is continuous at the origin 7. Show that /(z) (0, 0).

Let

4.6. Differentiability. z

/(z)

be defined for the set Z = {z}. Let be any infinite sequence ,z n

be a point of Z, and let Zi, Z2, 3, Z which tend to z as a limit. None of the z,, i = ], 2, We say that /(z) is differentiate at z if is to be equal to z .

.

.

,

.

.

.

of elements of .

.

.

Jin,

exists

3,

.

,

-

/fa)

/^

(4.2)

We can state independent of the sequential approach to z an equivalent manner, /(z) is differentiate at ZQ .

differentiability in if

a

there exists a constant, written/' (20), so that for > such that

any

>

there exists

6

/CO z

whenever

|z

In the cases

ZQ|

<

we

shall

-

5 for z in

-

/(go) zo

Z, z ?^

be interested

(see Sec. 10.7) for the definition of

<

ZQ.

in, ZQ will

an

be an interior point of Z In this case, (4.2)

interior point.

becomes /(2o)

=

lim

(4.3)

independent of the approach to zero of Az. f(z)

=

w(x, y)

and investigate the conditions that

+

will

Let us consider

iv(x, y)

be imposed on u(x,

y)

and

v(x, y)

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

128

in order that

shall exist

= Ax

Ag

+ Ag) = /(g + Ag)

u(x

f(z

so that

We

independent of the approach to zero of Ag.

+ +

have

Ay

i

+ Ay) + iv(x + + Ax, y + Ay) -

Ax, y ?/(x

/(g)

ax

Ag

^

v(x

.

-r- z

Ax, y

+

ay

+ Ay) Ax + Ay

+

Ay)

w(x, y)

Ax, y

v(x, y)

..

(4 ^

'

'

t

First

we

Then

(4.4)

/(g

+

Ag

let

Ag)

by keeping y constant, that

>

-

/(g)

_

+

u(x

-

Ax, y)

u(x, y)

is,

and

ox

exist.

Now we

v

let

Az

-

c/x

Ax =

0,

=

kz

i

+ A^).....-----

_ldu --

f(z)

m

Az

and Tay

exist.

Assuming

ay dtt

--

ax ^i_

Ax,

Ay =

0.

dy

.

1-

so that

=

^

ax &u

,

+

Ax, y)

-

y(x, y)

Ax

by keeping x constant,

=

dx dv

dx

dv -p

dy

i

dv

provided

=

Then

Ay. /(*

11 111

/<x

*-

hm /JiMlI/<E>

and

that

Ag

becomes

Ax

provided

is,

~7r~

dy .

du

differentiability of /(g) yields

a/

(

_

dy

.

du

?

dy

&v

/4

___

-N

(4.o)

dy

du dy

The Cauchy-Riemann

conditions (4.5) are necessary if /(g) is to be have, however, neglected the infinity of other methods by which Ag may approach zero. But it will turn out that the

differentiable.

We

Cauchy-Riemann equations (4.5) will be sufficient for differentiability we further assume that the partial derivatives of (4.5) are

provided

continuous.

Let us proceed to prove this statement.

The right-hand

COMPLEX-VARIABLE THEOltY

may

side of (4.4)

u(x

129

be written

+ A//) + u(x, y + Ay) - u(x, y) Ax + i A// + As, y + A?y) - r(j, y + A//) + r(x, + Ay) - v(x, y) Ax + 'Ay Ax (0tt/a.r + tO + Ay (du/dy + fa) s AJ + A?/ Ax (frj/to + fa) + Ay (dv/dij + Ax + i Ay

+

Ar, y

-L. -

-

*'(

+

-

A,y)

?/Q, y

?/

'

"

"

i

?

.

,

fa)

"*" l

where

fa, fa, fa, fa

of the

mean

tend to zero as Az

>

We have applied the theorem

0.

The Making use

of the differential calculus.

tinuity of the partial derivatives.

t

>

we assume con-

if

of the

Cauchy-Riemann

equations yields f(z

+

Az)

-

f(z)

= du

A

We

leave

it

show that -

+

-

;

-

f( z

Ax

fa

+

fa

Az)

/(z) --^ exists .

,

Az

Ay + f (fa Ax Ax + i Ay

-

Ax ^ Ax + ?a A// + 7(6 ---------7. Ax + i Ay

H,n in

r Hence lim

dv <3.r

to the reader to

11

TT

.

(9.r

+

du and equals .

,

dx

fa

Ay)

.

\-

.

+

fa

Ay)

u

dv

i -^-

dx

.

,

,

x

,

of independent ^

the manner of approach to zero of Az. We have proved Theorem 4.1. THEOREM 4.1. f(z) = u iv is differentiate if u and v satisfy the

+

Cauchy-Riemann equations and if, furthermore, the four first partial derivatives of u and with respect to x and y are continuous. The following example shows that we cannot, in general, discard the /'

Let

continuity of the partial derivatives.

Here

u(x, y)

(*,

Moreover

Similarly

dx

2/)

=

=

%

w(0, 0)

=

[-^77!

KO, 0)

=

^

*

0,0

= -1,

^

=

Riemann equations hold

1,

-

T =

at z

=

0.

1

>

at 2

=

0,

However,

so that the /'(O)

Cauchy-

depends on the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

130

approach to the

=

x(\

+

=

v

f/

f(z}

along y

=

rax,

- -

origin, for let y

mx, so that

xm

i)

3

i)

and

,./(*) -/(O)

= rm

^2-0

- m(l - t) i) +ro )(l+roi)

+

(1

(1

+ f) - m (l (1 + m)(l + mi) 3

x x

2

=

(1

Let the reader show that

which depends on m.

does not by showing that lim -.Q OX

the origin

(l

nr^

t)

discontinuous at

is

dx

exist.

-

DEFINITION 4.1. Let/Os) be differentiable at z = 20. If, furthermore, there exists a circle with center at ZQ such that /() is differentiable at every interior point of that circle, we say that f(z) is regular, or analytic, ZQ. Analyticity at a point is stronger than mere differentiability at If /(z) is not analytic at 20, we that point (see Prob. 10 of this section). is that a of ZQ say f(z). singular point

at

We

4.11.

Example (x 4-

f(z)

=

2 t'/y)

show that

x2

v

2

so that the

could have shown that

Let the reader show that

Example

We

X

Thus

The reader should

4.12.

(1)

~

(2)

~

assume that

\f(z)

+

x2

=*

_ ~

~~

#

-

y

\-

i

=

!

c/X

2 ,

=

t>

everywhere.

2xy.

it is

2x

obvious that the partial

-f 2yi

integer,

verify that

2(x

c/3J

and equals 2z by applying

n an

Since

Hence

_ _

rt

Moreover

hold.

f'(z)

zn ,

/(z)

differentiable

is

?/

j

f'(z) exists

if

u

r

_ ~

Cauchy-Riemann equations

derivatives are continuous.

We

_

_ ~

//

z*

/(z)

we have

-f 2tf?/i,

then/

;

(z)

if /' (z<>), g'(zo)

+

^2/)

" 2z.

(4.3).

nz!*~ l .

exist,

then

g(.

[/(s)(/(r)]i

/(z)

and

0(2)

have the same domain of

definition.

Problems u and v satisfy Also show that 1.

If

(4.5)

and

if

their second partials exist,

du du dx dy

.

dv dv

dx dy

*

~

show that V*u

V*t;

0.

COMPLEX-VARIABLE THEORY Give a geometric interpretation of

is

a vector normal to the curve u

The

u and

Remember

this last equation.

_ - du Vu

=

.

1

+ .

du j-

131 that

.

J

constant.

must

satisfy Laplace's equation proves to be useful in the application of complex-variable theory to electricity and hydrodynamics.

fact that

Let

2. ~.

Show

u

f(z)

.

,

x

that v(x, y)

+ =

v

be differentiate, and assume u is given such that V*u fy du(xQ. y) du(x. y) T dx + / . r Let ay. / dx dy Jyo Jxo fx

,

be differentiate.

Find

Show Show

= -

4.

that f(z) that/(z)

-

that/(2)

z*.

Let sin

5.

where.

+

lif(x

=

z

sin

x*

3xy iy)

x cosh

-

z

iv(x, y)

-

j/*) is

-f t(3x*y

u(x,

-f- ^

?/

Sxy 2 ) =

.v)

+

*v(x, y),

show that

sin

z -f cos

2

cos x sinh

Show

y.

2

=

?^(z,

that sin 2

is

0) -f

it;(2,

0).

analytic every-

Define cos z

s

^

and

,

dz

1.

7.

8.

If/00

-

Then show

analytic everywhere.

show that/() constant. Prove the statements in Example 4.12. for all

6. If f'(z)

0.

show that/(z)

Is this definition of sin z consistent for z real? 2

,

not differentiate.

is

iy

3xy* -f

Note that V 2 (x a

v(x, y).

x

,

,

x*

f(z)

3.

0.

iv

z,

a n z n -f an-ix"- 1 -f

-

-

-f

V

-

a

a*^,

show that f (2) -

Y

Ar-O

If

9.

g\z\ 10. 11.

a2

/()

tt

converges for

^

|z|

<

na n 2 n

R, show that/' (2)

~1

for

<R

Show that /(2) = x 2 2 is differentiable only at the Assume /'(a) and g'(a) exist, g'(a) ^ 0. If /(a) //

z-+a g(z)

origin. gr(a)

0,

show that

g'(a,y

4.7. The Definite Integral. The reader is urged to read first those sections of Chap. 10 concerning the uniform continuity of a continuous

function, rectifiable Jordan curves,

the

Riemann

integral,

and Cauchy

y

Let F be a rectifiable curve 1 joining the points

sequences.

Jordan z

=

a, z

=

ft.

Let

f(z)

be a continu-

ous complex function defined on F. We are not concerned with the definition of f(z) elsewhere.

Neither do

we introduce the differentiability of The Riemann integral of f(z) f(z). over F

is

FKJ. 4.2

defined as follows: Subdivide F into n parts in any Call the points of the subdivision a = zo, z\, z%,

whatsoever. 1

o

Called a simple curve.

.

manner .

.

,

Zk,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

132 .

.

,

I,

i

form the

lim

If

n

=

2n

On

(see Fig. 4.2).

2n-i n choose a point

{2,

.

.

.

&+i,

*,

,

zoZi, ZiZ 2 .

.

.

.

.

,

. ,

,

n,

and

sum

partial

Jn = J

1,

each of the paths

and

exists

unique independent of the choice of the

is

>

as

and independent of the method of subdividing F provided only that, n tends to infinity, the maximum of the arc lengths from z*_i to z*,

k

=

&

1

,

2,

.

.

.

we say that f(z)

,n, tends to zero,

is

Riemann-integrable

/(*)& = yr if(z)dz

(4.7)

over F and write

j =

P P

f( 2 )

*

over T

dz

Jot

=

y(D

This definition agrees with the definition of the Riemann integral of realvariable theory if F is a section of the real axis, F: a ^ x ^ 0, and if f(z) is a real function of x. We now show that, if f(z) is continuous on the rectifiable Jordan arc given by x = /(/), ?/ = ^(0> ^o ^ ^ ^ <i, then the Riemann integral of /(z) over F exists. The proof proceeds as follows: Let us first look at any b and consider arc of F joining z = a to z

where

is

any point on the

division of this arc into ZQ

=

arc joining z a, zi, Z2,

.

.

= .

a ,

and

zn

=

z

=

b.

A

further sub-

6 yields the partial

sum

defined as in (4.6), n

Now 5 =

/({)(&

-

a)

=

/(()

V

(^

2*-,)

_ ^,) =

k=l

k

-1

w

S - Sn =

so that

If

z

furthermore the 6 is o-, then

maximum

(/({)

-

/(&))(*

-

*-i)

variation of /(z) on the arc joining z

=

a,

=

n

n

|S

- 8n

\

y

\f(&

-

/(&)| |*

-

ft-il

^

<r

^

k*

-

-i|

^

<rL

(4.8)

COMPLEX-VARIABLE THEORY where L

=

the length of arc from z

is

a to

=

z

133

Why

6.

is

he important in what follows. Now we use the property uniformly continuous on F. Choose a subdivision of F so maximum variation of /(z) on any segmental arc of the suband obtain J\ [see (4.6)]. Now impose a finer subdivision is less than This result

that f(z) that the

will

is

,

on the previous subdivision such that the maximum variation of f(z) on any segmental arc of the new subdivision is less than 1/2 2 and form a J 2 for this subdivision. Continue this process. For </ the subdivisions are so fine that the maximum variation of /(z) on any segMoreover the maximum subdivision tends mental arc is less than l/2 n division

,

.

We

to zero in size.

obtain the sequence of complex numbers

J2 J8

,/!,

We now

,

show that lim J n n

.

.

Choose any

e

. ,

exists.

Now

for

m ^

.

(4.9)

>

We

0.

can find an

*

< e/L for n n ^ HQ we have

n integer no such that l/2

of F.

,/,

.

.

,

-

\J m

^

<^L L

Jn\

where L

no,

the length of arc

a

Cauchy sequence

<

Hence the sequence

using the result of (4.8).

is

(4.9) is

and

Jn = J

lim n

We &.

must now show that J

For the new choice

Tf

J

and by

exactly the

is

j,

the same limit for any other choice of the obtain the sequence

& we

of the Jf

J

2,

If .

.

same reasoning

.

,

t/

n,

.

as above

lim J'n

=

.

.

we have

J'

n>*>

But

\J n

implies

J'n\

J =

< (V2 n )L, and we

leave

it

to the reader to

show that

this

J'.

The final step is to show that the same limit J occurs for any other method of subdividing provided the maximum length of the subdivisions K n ... of partial tends to zero. For any other sequence K\, K*, sums of the form (4.6) we can superimpose the subdivisions which yield .

.

.

,

,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

134

K n on the subdivisions of J n of

from

We

(4.8).

leave

Then

(4.9).

show that lim

to the reader to

it

Kn

=

n

lim n

Jn

.

*

The reader is referred to that excellent text Dover Publications, 1945, for a much by Knopp, "Theory clearer expository of the Riemann integral. For the reader who has trouble in understanding what has been attempted let us note that if /(z) = u(x, y) + iv(x, y) and if T is given by This concludes the proof.

of Functions,"

x

=

y

x(i),

=

=

f(z) dz

so that

fr f(z)

it

dz

y(t), to

^

g

t

-

u(x, y) dx

then dz

fa,

would be

logical to define

=

y)

jr u(x,

dx

-

+

+

i

dy,

i[u(x, y)

dy

u(x, y) dy

+

+

v(x, y) dx]

dy

v(x, y)

+ where

dy

v(x, y)

= dx

i

[|r

v(x, y)

(4.10)

dx]

u(x, y) dx, etc., are the ordinary line integrals of real- variable

I

It is

theory.

not very

difficult to

show that

this definition

and existence

with that discussed above for a continuous and hence a continuous u(x, y), v(x, y). In particular if the Jordan

of the line integral agrees f(z)

curve

regular in the sense that

is

-rr

and -~ are continuous, then

(4.10)

becomes

^

'

dz

=

(*(<), y(t))

(x(0, y(t))

Example

We

4.13.

evaluate

/

f(z) dz,

-

v(x(t),

-

+

where

y

-

t

4J,

^

1.

Moreover the vector

-

(3

+

4)"<

dt j

dt

=

z

and the curve F

(3, 4).

is

(4.11)

given as

F may be written x

3,

Along F /()

/(*) dt

^

v(x(f), y(t))

/(z)

the straight line joining the origin and the point

y(0)

z to the

-

2

-

curve

dt.

r

i

x -f ty is z

-

(3 4- 4i)(

(3 -f 4i)i so

that dz

(3

-f-

4t)

rf<

and

COMPLEX-VARIABLE THEORY

We

could save ourselves

work

all this

we knew

if

135

that

rp

zdz

f any simple path F from z Hint: subdivide F into a = z

a to

for

=

z

z,

z\,

,

Let the reader show that this

/3.

.

.

.

,

=

zw

0,

is

true.

and consider

n

Show that J w

4- J'n

we

In this example

on r

Example from

The

itself.

(0,

=

a 2 and

2 /3

n

let

,

notice that

We

4.14.

0) to (1, 1).

Here x

=

Jr /(z)

dz

=

x

1, 2

-

+

O/

f(z) dz for /(z)

/

t,

.

become apparent

significance of this will

evaluate

o

>

Zi-l

dz depends only on the end points of V

z

/

-

Zt_.(Zt

)

y

^

t,

=

(1

JJ

t

-

?)(!

in the next section.

with F the straight

iy *

+

and not

fl,

^ =

-

dz

(1

+

t) ctt

line

so that

1

let us take F as the sum of the two straight-line segments, one from the second from (1, 0) to (1, 1). For the first path x = t, y * 0, ^ t

For the same/(z) (0, 0)

to

(1, 0),

^ l,sothatdz = d dx = 0, y ,

dt

and/(z)

=

t

and

f(z) dz

/

\.

yi'i

g

=* i dt,

t

I

^

1,

dz

-

and

- 1

/(z)

-

i<

Along the second path

a:

1,

so that

JTs

Hence r

r

/GO

-

i

Thus, for

/(z)

=

x

iy,

the line integral

is

x

guess that the nonanalyticity of /(z)

not independent of the path. One may may be the answer. The next section

ly

will verify this fact.

Example

4.15.

The

function /(z)

**

1/z

is

continuous and analytic everywhere circle of radius a with

except at the origin. Let us compute ff(z) dz, where F is a sin center at the origin. It is easy to see that z = a (cos 4Hence dz sin sents the circle. + i cos 0} d0, and a( i

f

7

-

sin

cos 8

9+ i cos + i sin 8

=

(

_

8in(>

+

0),

-

I

c08(, )(cosfl

^

:S

2ir,

repre-

_i 8ine)(W

'2r

We

shall see that the Even though /(z) is analytic on F, //(z) dz 5^ 0, in this case. accounts for this fact. Notice that the singularity of /(z) at the interior point z value of the integral is not dependent on the radius of F. Indeed, it will be shown

ELEMENTS OF PURE AND APPLIED MATHEMATICS

136

that for any closed rectifiable path F, with the origin in

its interior,

one has

r z

Problems ft

1.

Show

that f

z*dz

=

Ja

2.

Show

f(z -

that

z

)

-

3

i(/3

m dz

8

).

=

^ w = -

if ra

1

if

1

27rt

w

a positive or negative integer, if the integration is performed around the circle F with radius a and center z The integration is performed in the counterclockwise .

sense. 3.

Define t

|/(z)| \dz\,

and show that

\Jr f(z)dz 4.

6.

Show

ff&dz

that

f J&

(r)

(F)

f(z)dz.

is

the length of

f(z]

6.

Why

is it

that

7.

Why

is it

that

8.

Evaluate

the 9.

line

same

(1,

at the origin.

4.8.

r

/

LM

dz

/(z)

show that

f/2

for r constant?

2(

)

path discussed in Example

)]

to (1, 1).

,

(|z

|,

0)

and Fs

is

Do

4.14.

along the path F consisting of FI and F 2 where FI

/

0) to

=

F,

^ 2 dz = + / 2 d? Jr /iGO + f^f*(z) Jr [/i( (z + ||) dz, where F is the straight line from (0, 0)

for the other

Evaluate

from

/

cf(z) dz

/

\f(z)\\dz\

a

= -

J <x

M along F and L

^

If \f(z}\

fr

the arc of the circle from

(|z

|,

is

the straight

0) to z with center

Is the value of the integral single-valued?

Cauchy's Integral Theorem. The fundamental theorem of comis due to Cauehy. There are various forms of this theorem. We present now a proof y) C(x, y) of one form. Let S be a simply connected open region such that the partial

plex-variable theory

derivatives of u(x, y} and v(x, y) are continuous and satisfy the

Cauchy-Riemann equations, f(z)

u(x, y)

+

iv(x, y)

In Sec. 4.6 we saw that] these conditions were sufficient be analytic in S. We shall show that, if T is any closed simple

at every point of S. for f(z) to

=

COMPLEX-VARIABLE THEORY

137

curve inside S, then j) T

The statement concerning prove

first

=

that $J(z) dz

=

f(z)dz

(4.12)

We

Cauchy's integral theorem.

(4.12) is

C

for the rectangle

(see Fig. 4.3)

contained

entirely in S.

We

can obtain a single-valued complex function of the two

F( x F(XJ y)

real var-

and y by defining

iables x

is

the

sum

x

=

y)

,

f

+

f(t

ij/o)

+

dt

y

f

i

f(x

+

it)

dt

of the integrals of f(z) along the straight lines

AB

and

AD

obtained by integrating /(z) along and DC. The integration of /(z) around the rectangle C in the counterclockwise sense is

BC.

<f>

where If

is

Similarly G(x, y)

c f(z)

=

G(x, y)

=

'

I

i

Jyo

we can show that

dz

F(x, y)

f(x,

=

-

F(x, y)

+

it)

G(x,

?y),

dt

+

G(x, y)

I* Jxo

then

f(t

/(z) (p J c

+

iy) dt

dz

=

0.

Let the reader show that

U*

We

f(*

Jx

+ W)

=

dt

dx J yo

real.

/(x

+

iy

)

dx

J yo

in order to obtain (4.13).

These statements are proved in Chap. 10 for write / = u + iv and apply the theorems

The student need only

of real-variable theory to is

(4.13)

need

4-1 dx

/

dt V-^ OX

J yo I/O

u and

v

separately.

The continuity

used to perform the differentiation underneath the integral.

of

-

ox

Also

we have dF = jj dC

if(x

+

=

iy)

if(z)

1

=

dG Ty

f(x

=

+ iy)

/(*) *

...

** + *) .

.

.

.

^^ df(t

+

iy)

(4.14)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

138

As

yet,

we have not made use

Similarly at

df

du

dv

dv

dx

dx

dx

dy

dF =

Hence

f(jc

-

^

(#o, 2/o),

of the

+

Cauchy-Riemann equations. du

.

+

L

it)

=

dt

the constant c

We now consider

(ft

is

/()

+

+

G

dv

where Fo

is

iy)

=

Since

c.

This proves that

zero. cte,

.

f(x

dt

and hence dF = dG, F =

^,

du

\dy

dy

f v df(x

,

'

f

.

__

Thus

(4.15)

F = G =

f(z) dz

a)

any closed

=

rectifiable

0.

Jordon

curve lying inside the rectangle ABCD. Parts of the curve may be segments of the sides of the rectangle (see Fig. 4.4). Let P(x, y) be any point on F and define F(x, y) at P in exactly the same manner in which ,

F was

From

defined at C(x, y).

dF

Hence

<f>

jTo

F =

*

+ = f(z) (dx +

f(z)

dx

dz

=

i

(4.14)

(4.15)

dy

=

f(z)

dy}

=

f(z) dz

dF -

(f)

and

/ To

r/T

(f>

J

+

dx

+

i

if(z)

dV =

<f>

/

I'o

dy

To

Certainly /df = fdV = 0. The final part of the proof uses the following reasoning: Let F be any closed simple curve in S. F is a closed set of points. Now S was assumed

where

/

f

+

7

zT.

we adjoin to S we obtain S, a boundary points, closed set, consisting of S and the

to be an open set.

If

its

boundary a

of 8, say, F.

minimum

There

will

be

distance between F and

T not equal to zero. We can thereimpose a fine enough mesh on S

fore A

(the

v

/

0>

mesh

consists of rectangles) so

that the rectangles which contain the

points of F will be entirely in S. over all integration rectangles interior to F plus the integration of f(z) over those boundaries which include F vanish from the above

The

results.

However,

all

integrations over the rectangles interior to F

vanish in pairs, leaving f(z) dz

=

(4.16)

<j)

The

condition that the partial derivatives of

can actually be omitted.

The proof

u and

that <b f(z) dz

=

v

be continuous

for the rectangle

COMPLEX-VARIABLE THEORY

C under First

the condition that /(z) be analytic in

assume

new

into four

Now

(p

of the four

by halving the

regions (see Fig. 4.5)

/(z) dz equals the

new

R

sum

(C and

as follows

its interior)

:

,

sides of the rectangles.

of the integrals of /(z) over the boundaries

regions since the internal integrations cancel each other in least one of the four integrals does not vanish. We

Hence at

pairs.

choose that boundary Ci for which

(p Ci /(z)

.

.

,

.

,

.

.

.

.

. ,

/?,

.

.

.

,

with boundaries

.

,

The regions R n n =

f(z) dz 7* 0.

We

this process indefinitely.

,

,

has the largest value.

dz

J

Again we subdivide, choose C 2 and continue obtain a sequence of regions /?, Ri, ff 2 such that ,C W C, Ci, C 2 C 3 (D

S can be shown

Subdivide the region

/(z) dz 7* 0.

(p

139

I,

,

... are closed and bounded sets, and the diameter of R n tends to zero From the theorem of nested as n > oo

2,

-H--

.

Chap. 10) there exists a unique Fio 4.5 point P which belongs to every R n J = 7 n be the point z Let Obviously z is interior to 1, 2, 3, or zo lies on C. Hence /(z) is differentiate at z = zo so that sets (see

,

.

.

.

/(*)

where

t(z, zo)

.

=

/(*o)

-

+/'(2o)(z

tends to zero as z tends to

can pick a region on C n Now

Rn

+

*) z

e(z, z )(z

z

)

Hence, given any

.

C n such

with boundary

-

fl

that

/?,

|c(z, z )|

<

c

>

eo

for all z

0,

we

.

But

+

cfe

Cn

so that

<fc

y

Remember

that

dz c n /(z)

$ dz

=

sufficiently large so that

0,

=

= L/2 W where L ,

<

Cn \z

is

=

j>z dz

of the diagonal of the rectangle

since l n

/

yCn

|c(z, ZQ)|

/(z) dz

f

f'(z*)(z

-

dz

./Cn

,

-

e(z

o)

For any positive e we choose n If l n is the length on (7 n

0. eo

-

)(

,

for all z

.

then z |

|dz|

^

4

the length of the diagonal of C.

<

4n

4L

2

4

L2

Hence

ELEMENTS OF PURE AND APPLIED MATHEMATICS

140 Since

can be chosen arbitrarily small, the constant

eo

u)

/(z) dz

The proof of Cauchy's theorem follows then the same manner as demonstrated above. This proof is due zero.

Q.E.D.

must be

in exactly to Bliss in

the American Mathematical Society Colloquium Publications, vol. 16.

A very strong statement of Cauchy's theorem is as follows: Let F be a simple closed path such that /(z) is analytic in the interior R of F and such that /(z) is continuous on F. Then ff(z) dz

=

The proof is not trivial and is omitted here. Continuity this case means lim /(z) = /(f ), f on F, z on F or in R.

We now theorem

state

some immediate consequences

of

of /(z) on

=

in

Cauchy's integral

:

A. Let /(z) be analytic in a simply connected region R. z

F

0, in

For

z

=

a,

R,

independent of the simple path chosen from z = a to z = /3 provided the path lies entirely in R. Let the reader verify this statement. B. Let /(z) be analytic in a region R bounded by two simple closed is

V

paths Ci, C 2 (see Fig.

4.6).

both integrations are done

Fi. sense.

Then

/(z) dz

(p

/

C'i

(D

/(z) dz '

J Ci

provided

in a clockwise sense or a counterclockwise

4.6

The proof is as follows: Construct the paths From Cauchy's theorem

Fig. 4.7).

Adding yields

and

=

/(,) dz

/(,)

AB

and

EF

(see

COMPLEX-VARIABLE THEORY

141

Notice that nothing need be known about f(z) outside of Ci or inside of and C*. We have assumed f(z) continuous on C. For Fig. 4.8 let the reader show that

d

C%.

<j>

T%

To,

FI,

ment

+

dz

<f>

F

/(z) is analytic inside

F2

=

f(z) dz

r j(z)

dz

and outside FI and F 2 and

is

continuous on

Generalize this state-

.

for the curves

F

,

FI,

F2

.

,

.

. ,

D. Let /(z) be analytic inside a simply connected open region R,

and consider

= where

and

zo

f'f(t)dt Jzo

The path R. omitted since the

z are in

of integration is

integral

F(z

+

Az)

-

=

F(z)

+

z

f Jz

the straight-line path from z to

Then

=

z

+

F(z

Since f(z)

Hence

independent of the path (see A).

is

We choose tion.

FIG. 4.8

is

0^/x^l, + Az) - * "Jo

+

dt

Az as the path of integra= dp Az, and

along this path, dt

M Az,

analytic at

z

f(t) dt

f(z

+

n Az)

z,

/'(*

and

f(z

where

lei

as &z

+ n Az) = f(z) + >

0.

)

+

(4.16')

Hence

rd + r Azff(g) Jo

Jo It is

Az

d

+^ r Jo

obvious that

so that F(z) is analytic in R. E. Let f(z) be continuous inside a simply connected open region

and assume F(z)

=

f'f(t) dt

R

9

ELEMENTS OF PURE AND APPLIED MATHEMATICS

142

independent of the path from

is

the path lying entirely in R. t is simply the complex

ZQ to all 2,

We show that F(z) is analytic in R. The variable

The reader can prove this statement easily variable of integration. D. Continuity implies that as in enough by proceeding

+

f(z

where

as Az

> 17

=/(z) +r,

z)

(4.16")

In the proof of D it was not necessary to use We could have used (4.16") in place of

0.

the analyticity of f(z) twice. (4.16').

The fundamental theorem

F.

well to the theory of in

D

of the integral calculus applies equally

Assuming the conditions stated

variables.

complex

yields

F(z)

=

/(*) dt

=

Let G(z) be any function such that G'(z)

-

and F(z) = G(z)

-

(F(z)

j|

- C =

G( Zo )

/(a)

f(z).

Then

=

G'(*))

- C =

Hence G(z)

C.

=

F'(z)

f* f(l)

l'"f(0dt

dt,

=

Jzo

so that 0(z)

As a simple example,

/

z

-G(*o) = Jf'f(t)dt 20 2

=

dz

1(0

a 8 ) since

3

8

-7-

^2

(|0

)

=

z

2 .

Problems 1.

Show

that

(7)

/

Show that

u)

/

2.

Use

^ Z

---

Z

=

2iri for

ZQ

=

if

any simple closed path

r, z

in the interior of T.

ZQ is exterior to F.

Z

Evaluate u) ----- for any simple closed curve F enclosing the circle r

C

above, and write I/ (z* I/ (z z) Construct a function f(z) such that

=

1.

1/2.

1)

f(z) dz

=

for all simple closed paths, /(z) not analytic. Docs this contradict Cauchy's theorem? 4. Let/(z, /) be a complex function of the complex variable z and the real variable /. 8.

Assume J(z

t

t)

and

-^-^

analytic in z for

F(t)

=

U ^

f'f(z,

t

/)

^ dz

t\ t

and consider

COMPLEX-VARIABLE THEORY If,

(z, t) is

furthermore, -^ vt

continuous in

and

z

f*G(u)du -

i 6.

Let

and

/(z)

show that

-

F(t)

F(t

)

/>/.*&*

be analytic

#(z)

t,

143

in a

simply connected region R. *f(z)g'(z)

From

+ f'(z)g(z)

show that

-/()?() - [* Ja The path

from

of integration

z

a to

e

=

|8

g(z)f'(z)dz

lies in 72.

A truly fundamental consequence of 4.9. Cauchy's Integral Formula. Cauchy's integral theorem of Sec. 4.8 is the following formula due to Cauchy Let /(z) be analytic in the simply connected open region R, and Then let F be a simple closed curve in R. :

a is an interior point of F. The sense of integration around F z such that as we move around F the region containing z = a lies to our left. The proof is as follows: From Sec. 4.8B we can replace the curve of integration F by any circle F with center at z = a, F interior to F. if

is

Then 27rz

jYz

a

2iri

/TO

a

z

f(z)b(- sin

!_

b (cos 8

2*i Jo o

+ since 2

=

a

+

the radius of F

.

+

17

>

as

fe

is

+

6 ( cos ^

^

i sin 6),

Since /(z) /[a

where

+

fr(cos B

+ icos 6) dB + i sin 6)

^

^

27r,

continuous at z

6(cos ^

+

i

sin 0)]

=

i siri ^)1

is

=

rf<9

(

the ecjuation of

we have

a,

+

/(a)

i\

Hence

* 0.

f

sin B)] dB

=

2ir/(a)

+

** 17

>2r

and

1

lim ^*v

^o

/ I

Jo

f[a

+

6(cos

+ i sin

B)]

dB

=

/(a)

d0

4

-

F

17/ )

,

b

ELEMENTS OF PURE AND APPLIED MATHEMATICS

144

Since the left-hand side of (4.17') is independent of 6, (4.17) must result. We can now observe one important consequence of analyticity. To evaluate the right-hand side of (4.17), one needs only to know the value

performed, the value to be analytic inside F, if /(z) is continuous on F so that Cauchy's theorem holds, then the values of f(z) interior to F can be determined if we know only the values of /(z) on F. Analyticity is, indeed, a powerful condition. Since /'(a) is known to exist, we may hope that /'(a) <<an be obtained from (4.17) by differentiating underneath the integral. If this were of /(z)

on the boundary

After the integration

F.

of /(z) is

known

possible,

we would obtain

in the interior of F.

Thus,

if

is

f(z) is

known

() d* r (z

Let us prove that (4.18)

We

correct.

is

a)

' v (4.18)

2

have dz

f(a /(a

+ +

- /() = - /(a)

ft)

h)

=

h

Since /'(a)

=

lim

r J^ 2iriT

--

-

For

ft

a)(z

-

f(z) dz

a)(z

-

a

_ -

h)

a) (2

-

-

a

-

-

a)

h)

T

(z

T

(z

-

a)

2

Consider

to obtain (4.18).

'.

(z

^-^ we need only show that

^

Ji->0

_

-

a

-

ft)

r

(z

2

~

-

a)*(z

sufficiently small it is easy to see that a) (2

uniformly bounded for turn yields (4.18).

By

-

a

a

-

h)

IS

on F, so that as h > (4.19) holds, which in mathematical induction let the reader show that

all z

(4.20)

embraces Cauchy's integral formula for /(a) and all its is important to note that analyticity of f(z) is strong enough to guarantee the existence of all derivatives of /(z). In realvariable theory the existence of f'(x) in a neighborhood of x = a in no way yields any information about the existence of further derivatives of f (x) at x as a.

Equation

(4.20) derivatives. It

COMPLEX-VARIABLE THEORY

145

Problems 1.

C.

Let f(z) be analytic within a circle C of radius R and center be an upper bound of |/()| on C. Show that Let

M

a, f(z)

continuous on

be an entire function. Use the results of bounded function must be a constant, /(z) is said to be exists such that |/(z)| < J\f for all z. Hence prove the first

2. If /(z) is analytic for all z it is said to

Prob.

1

to

bounded

show that an

if

a constant

entire

M

theorem of Liouville that a nonconstant entire function assumes arbitrarily large values (in absolute value) outside every circle with center at z = 0. 3.

Show

Consider the polynomial /(z)

-

I/WI

S

aoz

+

n

aiz*"

+

1

a 2 z"- 2

+

+

-

an

that

M

there exists an 7? such that, for z\ > R, \f(z)\ > M. for all z, and consider be the polynomial of Prob. 3. Assume /(z) 5^ = l//(z). Why is g(z) an entire function? Use the result of Prob. 3 to show g(z) that g(z) is a bounded entire function. Since g(z] ^ constant, use the result of Prob. 2 to deduce that a z exists such that /(z ) = 0. This is the fundamental theorem of algebra (Gauss). Deduce that /(z) has n zeros. 6. Prove Morera's theorem: If /(z) is continuous in a simply connected region R and if y/(z) dz for every simple closed path in R, then /(z) is analytic in R. 6. Prove (4.20) for n = 2. be a sequence of functions analytic inside and on 7. Let n (z), n 1, 2, 3, the simple closed curve T which converges uniformly to <p(z}. Show that <p(z) is

Hence show that for any positive 4.

Let

/(z)

()

t

.

<f>

.

.

,

analytic inside T.

4.10. Taylor's Expansion. could be written as infinite

In real-variable theory certain functions series, called

Taylor's series or expansion.

For example,

=

1

I

+ I

x

+

**'

I

jr-.

2!

+.+

+ I

I

**'

= /\ Lj

I

n\

** ,

n\

00

""

V LI

In

(!+)-,

-++ T2

T3

+ LC

1V

1

~1

n-O J-"

n (-i) ^ 2n+i

(2n

+

+"n n-1

The

function /(x) /(O)

=

e~ l/xt x ,

^

0, /(O)

=/(0) -/"(O) =

=

0, is

such that

/<>(0)

=

-

-

1)!

ELEMENTS OP PURE AND APPLIED MATHEMATICS

146

However,

We

no Taylor-series expansion about x

this function has

shall find,

however, that

if

f(z) is analytic at z

=

a,

=

0.

a Taylor-series

expansion will exist for f(z) at z = a. Remember that analyticity of /CO at z = a means differentiability of /CO for all points in some neighborhood Before proving this result the student should review the of z = a.

and theorems involving infinite series (see Chap. 10). The and uniform convergence of series, and the theorems regarding term-by-term integration and differentiation of a series, apply equally well for an infinite series whose terms are complex. definitions

definitions of convergence

We now proceed to the development of Taylor's theorem. Let f(z) be a function analytic in an open region /?, and let zo be an interior point 00

of R.

There

exists a

unique power

H

} a n (z

series

n

such that

ZQ)

=

n=0

for all z in

some neighborhood

of z

(4.21)

=

z

The

.

series (4.21) for

verges to /CO for

surrounding z

C

ference of

larities of /(z)

=

z

A

ZQ.

all z

=

z

,

/CO con-

inside a circle

C

and the circum-

contains those singuwhich are closest to

point

zi is

singular point of /(z) analytic at z = z\.

said to be a if

/(z) is

not

We proceed to the proof. Let C be the circle mentioned above. C has the property that there one point on C at which

Moreover

analytic.

is

at least

/(z) is

not

/(z) is analytic

every interior point of C. Now be any interior point of C, and construct a circle F with center ZQ at

let z

FIG. 4.9

containing z in

F

its interior,

Let f be any point on

F.

interior to C (see Fig. 4.9). From Cauchy's integral formula

_df z

_ 2

7r

(f

-

so)

-

(z

(*

-

2

)

.)/(f

-

(4.22) *.)]

COMPLEX-VARIABLE THEORY Z

Since r

2o

-

<K <

I

we have

(see Fig. 4.9),

___ -

LZJ?V

1

Equation

~

(Z

2

2o)/(f

147

( >)

n-0

becomes

(4.22)

00

The uniform convergence

of the series

^

y

\n+i

>

^ ^^ e variable of

n=

integration, ^

summation.

and z Hence

enables us to interchange integration and

fixed,

oo

= ^

To show

-

a n (z

20)"

uniqueness, assume

=

/(*)

n

for all 2 inside f.

For

-

2

n

=

20

a(2 for all 2 inside C.

But

=

we have a =

2

n )

=

60,

>

6 n (2

(2

2o)

(4.23) implies

00

(Z

2o)

>

a n (2

2o)

n~ 1

=

-

so that

2 )"

(4.23)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

148

Y

for all z inside C

which in turn implies

,

-

a n (z

for all z inside

C

Hence

lim 2

so that

=

6n

=

i

Y

,

Example

.

.

.

function

easily seen to

so that

=7:nj"

ZQ.

=

-

6,,0

w

1

5-0)

1

induction the reader can show that

Q.E.D.

.

./(z)

defined

= K

cos

h

t

~

-

sm

-f /r*

We er

dx

by

//

be analytic everywhere. /'(z)

=

z

*-^ f)

By mathematical The

Y

liin

Z~2o

/(z) is

=

l

2o)"~

3= 1

0, 1, 2,

4.16.

-

a n (z

^*

Zll

?>i.

n =

=

1

with the possible exception of

H

on

so)-

?/

have

=

cos y 4- 2e* sin y

by mathemat)(;al induction / (u) (2) =

/(z)

and / lw) (0) =/(()) =

/(z)

Hence

1.

00

n

Z2 We define /(z)

to be

e* s= e x c iv .

=

e T (cos

+

y

We note that,

if

and

(

that

so

a

-jdz

=

f"~ z

e a ~* for

any constant

-

__

2

.

it

p*i

Then

a.

= =

0,

r

Q

e*e a

"*

is

independent of z. Let ~ e ze a z ~ e. Now let

and we have

2 -f Zi

result

so that

e*e*i

*

e*+*i.

From

this

we havpe*"

Let the reader

|

iy)

z is real, /(z)

show that

Z

sin

reduces to e x By direct can be shown that e z e*i e****. An easier way is the following: Let the reader ez

multiplication of the series representing

|

i

j

cos y -\- i sin y, y real. show that this result of

Euler's also holds for y complex. Example 4.17. Let us define Ln z as fol-

FIG. 4.10

Let z be any complex number other from Example 4.16, TT < ^ IT. We

lows.

than

z

evaluate

z

0,

Ln

z

r(cos

s

/

y

+

t

sin 0)

=

re*

9

along any curve T not passing through the origin or crossing

the negative x axis (see Fig. 4.10). We can replace T by the path from x followed by the arc of the circle with radius r and center at z = until

1

to

x

= r

we reach

2.

COMPLEX-VARIABLE THEORY

149

This yields T

Ln2 =

f

dx x

r

/

Ji

= = Ln

2 is single-valued

and

real

positive,

Why If

2

Ln r/2

re ttf>

iO

-f-

Arg

-f i

\z\

T~

/

Jo 2

becomes the ordinary log e z

_

dn

I '

z2

2

Ln

so that

In

h

and analytic everywhere except at 2 =0,

Ln

d*

In r

e ire"? d<f>

f

.

=

z

Lnz _

(

_

z

1)!

/

__

^T

'

1)3 ~

"

does the series expansion for Ln 2 converge only for \z ir < 6 ^ TT, then Ln the condition

z

1|

2

<

1?

would not have been

Let the reader show that for this case

single-valued.

In z

number

= f*Q = l n

\

z

-f t'(Arg z -f 27rn)

\

path of integration r encircles the origin. If the a clockwise fashion, n is negative; otherwise n is a positive Let the reader show that the

is

integration integer.

For

*

'

we had not imposed

where n

-.

j^+i

sn (z

1)2

~

1 )

=

-7

easy to show that

is

-

(n

dz n

'

fc

It

x.

since

is

of times the

performed

in

Ln(-l) -TI In 2122 = In 2i {- In 22 Ln xijr<> = Ln xi + Ln x>

zirifi

-f*

x\

2

>

0,

xz

>

Ln 2 is called the principal value of In z. We imagine a cut exists along the negative a x axis which forbids us from crossing the negative x axis. We call the point z branch point of In 2. If we wish to pass from Ln z to In 2, we need only imagine that as we approach the cut from the top half of the z plane we have the ability to slide under the cut into a new

z plane.

In this new Riemann surface, or sheet, we have

= Ln

ln (1) 2 If

we now swing around the

origin

and

ln (2 > 2

slide into

= Ln

2

+

On

This process can be extended indefinitely. ln<*> 2

+

2

- Ln

a single-valued function. 1/(1 Example 4.18. The function f(z) please, has for its only singularity the point

2iri

a new surface,

we have

4rt

each Riemann surface

2 -f

2*ki

is

/(z)

about

z

must converge

=

for

|z|

<

1.

2), z 7* 1,

2

=

1.

/(I) defined in

The

any way we

Taylor-series expansion of

Indeed

(4.24)

f(z)

n~0 converges for

|z|

<

1

and can be used to represent 1/(1

z) for

|z|

<

1.

Let the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

150

reader show that

The

d"

1

a

\l

1

f

-

nldz* \l1

series expansion of 1/(1

z)

n =

1

Uo

zj

about the point

=

z

...

0, 1, 2,

i/2 should converge for i

since the point

from

=

z

1

z

We

.

\/5/2 units

is

I

1

I

= ?/2 write

-z

-

1

nb ! The

circles of :

*

(

convergence,

_ ,

i/2

2]

=

and

1

overlap (see Fig. 411).

In this shaded region of overlapping, (4.24) and (4.25) converge to the same value of 1/(1 z). Equations (4.21) and (4.25) are said to be analytic continuations of

each other.

Problems Obtain the Taylor-series expansion of sinh 0. same for cosh z s (e* -f e~ f )/2 about z 1.

cosh

sinh z

~j-

z, -7-

2.

Show

that

Ln

B.

Show

that

N

(1

cosh z

+

sinh

z s= (e*

Show

-

e~*)/2 about

that

cosh 2

2=0. The

=

z

2 -f sinh

1

z,

z.

(-1)" 2)

w

00

n

(z /n\) converges for all

Multiply the series

z.

n (z /w')

^ n-0

and

00

to obtain

I 4. If

one defines tan"

can one make 5.

e*e*i

Show

tan"" 1 z

that

e 2rtt*

1

2

1 by tan"

z

= f

> ,

what

difficulties

How

occur?

a single-valued function? 1 if

n

is

an

integer.

If e +a

*,

show that a =

2irni,

n an

pe i<p

show

integer. 6. Define w \fz as that function w such that w* = z. If z = re**, w? r also. that p - \/7, ^> - r/2. Since z - re *+ 2T) show that ? 0/2 show that w is double-valued. Construct a Riemann surface so that

+

(

,

valued. 7.

If

Is the origin

w - Ln

z,

=

For

w

is

a branch point?

show that

z

-

e-.

Hint:

Ln

*

-

In

\z\

-f i0,

"

-

<aU!+*.

t

z

^

0,

single-

COMPLEX-VARIABLE THEORY Define

8. IP

w

*

sP,

single-valued?

a complex, by the equation

Show

that for this case

w

z"

as** 1

-T

151

How can one make

1

e" "'.

.

00

Let /(*) be analytic for

9.

<

|z|

Y

-

JR,/()

n.

a*

Show

that, for r

< R

t

n-0

n-0 be analytic in a simply connected region 7? bounded by a simple closed be an interior on r, so that Cauchy's theorem applies. Let continuous path T, f(z) 10.

Let

/(z)

00

R

point of

^ a n (z

so that f(z)

z

n ( >)

Let

.

C

be a

circle

with center at

z<>

and

n=

radius

r,

C

inside

Show that

r.

assumed that |/(zo)| *= 1/00 theorem, which states that |/(zo)|

f r

all

1

4.11.

that,

if

exists.

^

2

|o

z

|

+

m

\Qi\*r*

R>

|/()| for all

+

2

|a 2

|

r 4 -f

|a

|'

if

it

is

Hence prove the maximum-modulus z in /if implies 2 on T if/(z) 5^ constant.

An Identity Theorem. Analytic Continuation. We have seen a function f(z) is analytic? at a point 20, a Taylor-series expansion If one desires, then, one could define the class of analytic func00

tions as the totality of series of the

form

N a n (z

2

n )

with nonzero

n=

Some of the series would be analytic continuations Thus one could start with a particular analytic function

radii of convergence.

of each other.

f(z)

=

Now

H a n (z ZQ) which converges for all z such that \z 2o| < r 7* 0. n-O choose a point, z\ inside this circle. Since /(z) is analytic at z\, we t

can find a series expansion for/(z) 00

in the

form

Y

n

b n (z

Zi)

which

n=0 converges for \z

-

all z

zi\

<

such that ri

^

This new region of convergence may extend beyond the original circle of

convergence (see Fig. 4.12). This process can be continued.

p an analytic continuaAll of them represent the original tion of its predecessor, and conversely. One f(z), which has now been extended to other portions of the z plane. might naturally ask, if a point z = f is reached by two different paths of analytic continuation, do the two series representations thus obtained

Each

series is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

152

converge to the same value in their common region of convergence? We cannot answer this question until we prove Theorem 4.2. THEOREM 4.2. Let f(z) and g(z) be analytic in a simply connected open region R. Assume f(z) = g(z) for a sequence of points z\, zz, zn in R. having z as a limit point, z t i = 0, 1, 2, n, .

.

.

,

.

.

.

,

,

Then /(z) s g( z ) in R. The proof proceeds /(zo)

=

.

lim /(zn)

=

.

,

.

.

.

,

as follows: First note that f(z n ) = g(z n ) so that = 0(z ). Moreover, since /(z) and 0(3) are

lim g(z n ) n

n

analytic at

.

>

zo,

=

g(z)

b t (z

-

fc-0

and

=

/(ZQ)

(Z n

0(20) implies ao

-

a

y

Zo)

fr

(2 n

=

-

Hence

/>o.

1

So)*"

=

(Z n

~

So)

Ar-1

for

n =

1, 2, 3,

for

n =

1, 2, 3,

.

.

.

/ Ar-1

.

....

This implies

Hence

oo

lim

Y

a t (2 B

-

zo)*"

1

=

lim

b k (z n

which implies ai = 61. By mathematical induction the reader can show This shows that /(z) s= gf(z) for some that a n = 6 n n = 0, 1, 2; neighborhood of ZQ. This neighborhood (a circle) extends up to the .

.

.

.

,

nearest singularity of f(z) or g(z).

construct a simple path To from

boundary

of

R

(see Fig. 4.13).

Now

let f

be any point of J?, and Let F be the

to f lying entirely in /. Call the shortest distance

z$

from F to

F,

radius of convergence of f(z) about z = 20 is ^ p. Why? Now choose a point Zi on F interior to the circle of convergence of /(z) and p.

The

Since /(z) and g(z) are identical in a neighborhood of z\ ZQ. prove in exactly the same manner as above that /(z) = g(z) for some circle of convergence about z\ whose radius is greater than or equal to p. Let the reader show that z = f can be reached in a finite number of steps. When f is interior to one of the circles of convergence, g(z)

we

about

easily

y

COMPLEX-VARIABLE THEORY

153

It is important to note that at 0(f)> which proves the theorem. each point zo, Zi, 22, Zk the actual given function /(z) and its derivatives are used to obtain the Taylor series expansion of /(z). Analytic continuation is not used since we are not at all sure that the value of

/(f)

,

FIG. 4.13

would be equal to the value at z = f obtained by analytic That this is true for a simply connected region requires continuation. some proof. This is essentially the rnonodromy theorem, whose proof we omit. The formal statement of the monodromy theorem is this:

/(z) at z

=

f

00

Let

R

be a simply connected region, and

let f(z)

= ^ a n (z

n

Zo)

w0 have a nonzero radius of convergence. If f(z) can be continued analytically from z along every path in 7?, then this continuation gives rise to a function which is single-valued and analytic in R. We now state some consequences of the identity theorem (a) The identity theorem holds for a connected region. :

Let zi, z 2 z 3 be analytic at z = z zn sequence of points which tend to z as a limit such that/(z n ) = Then f(z) = in a neighborhood of z ZQ. 2, 3, .... (6)

Let

f(z)

.

,

,

,

,

.

0,

.

n

be a

=

1,

Let /(z) be analytic at z f(z) ^ constant. A neighborhood, N, of z zo can be found such that /(zi) ^ /(z ) for z\ in N, z\ 7* ZQ. (d) Let /(z) and g(z) be analytic in an open connected region R. z in #. It can be shown Assume / (n) (zo) = ^ (n) (^o), n = 0, 1, 2, . (c)

,

=

.

that

/(z) ss gr(z) in 72.

Problems 1.

2. 3.

4.

Prove Prove Prove Prove

(a). (6). (c).

(d).

.

,

ELEMENTS OP PURE AND APPLIED MATHEMATICS

154

Consider f(z)

5.

1

=

sin 1

-*

<

\z\

Show

1.

number

that f(z) has an infinite

of

Does this contradict (6)? zeros in the region |z| < b. Show that, 6. Let ?(x) be a real-valued function of the real variable x, a ^ x if it is at all possible to continue <p(x) analytically into the z plane, the continuation is 1.

Hint: Assume /(z) and g(z) analytic for z real, a results of Prob. 6 to show that

unique.

^

z

-

6,/(x)

^(3)

g(x).

Use the

7.

-

><

is

n-0 is

the only possible definition of e*. a ^ aLlw Define 8. Consider z

ft as the operator of continuation by starting at the path of continuation encircling the origin. Show that .

2

and returning

Hint'

Ln

z

In

to

|z|

z,

-f-

t'0.

After encircling the origin In 2

* In

+

|z|

^0

+

2iri.

00

Let

9.

/(z)

be analytic at

2

=

Zo

so that /(z)

=*

n

y a(2

converges for

2o)

n-O |z

2o|

<

0^5^

/2.

2r.

Show that/(z) has at least one singular point on the circle z = Zo -f Re* e Hint: Assume /(z) analytic at each point of the circle, obtain a circle of convergence at each point, apply the Heine-Borel theorem, and extend the radius of convergence, a contradiction. ,

4.12. Laurent's

Expansion.

A

generalization of Taylor series is due Let /(z) be analytic in to Laurent.

an annular ring bounded by the two circles KI, Kz with common center z

=

KI

a.

Nothing

or outside

is

K

said of /(z) inside (see Fig.

2

4.14).

Now let z be any point in the annular and construct

Ci and a such that z is interior to the annular ring FIG. 4.14 Moreover lying between Ci and C 2 C\ and C 2 lie in the annular ring between K\ and K*. From Cauchy's integral formula and theorem ring,

Cz with centers at z

circles

=

.

(4.26)

__

This result can be obtained easily by using the method found in Sec. 4.8B.

Now

-

2

(f

-

O)

-

(2

-

O)

r

(f

-

/(f)

0)[1

-

(2

-

o)/(f

-

O)]

-

Z

-

Since

COMPLEX-VARIABLE THEORY

~~~

CL

<

a

f

<

r

1

-

-

(z

and interchange the order

C^ we

f on

1 for

-

a)/(f

write

a)

~

f

n0

of integration

uniform convergence of the

155

<*

and summation because

of the

This yields

series.

(42?)

-B^8^n

-*',*...

For the second integral

c,

f

-

_

-

7c, (f

2

-

a)

-

(2

a)

(z

Since

<

a

s

<

1

-

1 f or

f

-

(f

on

t'i,

a)/( 2

-

we

-

o)[

-

1

(f

-

o)/(

-

a)]

write

Zv

a)

V

-

a

Interchanging integration and summation yields

=

o.

2.

(r

Ci

-

CO

Hence

f(z)

a n (z

n=0 Sj (z

-

a)

+ V

n

Si(z

a)

.

.

.

a_ n (2

-

a)~

n

n-1

-

+

a)

2 fc

00

where

1, 2, 3,

00

= Y

=

n =

a)-'/(f) df

a) 00

= Y

a n (z

-

a)

n

Y

a^ n (z

a)~

n

=

S*(z

a)

n-l Since

a)~

(2

(w-f !)

between KI and an n ,

=

0,

1,

/(2)

and

(z

a)

n~ l

f(z) are analytic in the

annular ring

K^ we can choose any path of integration to calculate the 2, provided the path F encircles the point z * a .

.

.

,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

156

Hence we can write

exactly once.

/(z)

=

-

a n (z

>

a)" (4.29)

n-O

/(f ) df (f

-

^ -

U,

+11, +2J,

.

.

.

a)

Equation (4.29) is the Laurent expansion of /(z) valid for all z in the annular ring between K\ and 2 We leave it as an exercise for the reader to show that &\(z) converges

K

and that

for all z inside 7f 2

common

region of

.

The

$2(2) converges for all z outside K\.

convergence

the above-mentioned annular ring.

is

4.19. Consider /(z) = l/(z l)(z 2). Certainly /(z) is analytic for such that 1 < |z| < 2. Let us find the Laurent expansion for/(z) in this region. We wish to write /(z) as the sum of two It is not necessary to find the a n of (4.29). We write series, one series converging for |z| < 2, the other for |z| > 1.

Example

all z

11

1

z-2

(-l)(s-2)

1111

z-l

21-Z/2

oo

, _

V ^L- V

n+l L, 2

first series

converges for

answer by using

where T |z|

-

2.

is

\z\

<

(4.29) to find the

2

1

Li n = 1

n^()

The

zl-l/z

ac

zn

and the second

for

>

|z|

Let us check this

1.

er r,.

any simple path enclosing the write, for n ^ 0,

origin

and lying between the

circles

=

|z|

1,

We

1

=

__^_+_A_++^+ ^^

.

.

.

2

_L.

IIence

^ ^^--^L,.-

. A

+

C

find A, multiply (4.30) by f - 1, and then let f -^ 1. ** To find C, multiply (4.30) l/2 n+1 easy to see that n+1 Hence have Cso that C - 1

To

B

.

-

0-A+JS + ^

-

l/2

+_!_ ^

MSO^ l

;

Why?

We obtain - 1 = by

f,

and

let r -^

A.

It .

is

We

.

53

Let the reader show that a n

=

1

for

n

<

0.

We had previously stated that z is a singular 4.13. Singular Points. point of f(z) if /(z) is not analytic at 20. If, moreover, f(z) is analytic for some neighborhood of 2 with the exception of 2 we say that 20 is an ,

COMPLEX-VARIABLE THEORY isolated singular point. Zo| verges for \z

<

<

157

In this case the Laurent expansion of f(z) conr, where r is the distance from ZQ to the nearest

We distinguish singular point of /(z) other than z itself. 3 types of functions which have isolated singular points.

now between

00

Case

The

1.

series f(z)

=

Z O ) M is

a n (z

)

such that a n

=

0, for all

L-i

w-0

n <

0.

at z

=

We need only redefine f(z) at z to be a and/(z) becomes analytic A singularity of this type is said to be a removable singularzo. 00

Thus

ity.

= Y

f(z)

n

(z

^

/n\) for z

0, /(O)

=

2 can be

made

analytic

n~0 00

=

at z

by

defining /(O)

=

lim *-+o

Case

All but a finite

2.

we say

case

'

w

that

(z

-

n

(z

number

a pole of

ZQ is

T ^ n=0n

We

/(z).

=

ZQ is said to

be a simple pole of

The reader can a pole.

Case case

An

3.

we say

,

n

<

0,

In this

vanish.

write

'

be a pole of order n.

zo is

1.

*)"

zo is said to

if

=

of the a n

and

z

/nl)

/(z)

easily verify that

A pole

If

.

|/(z)|

becomes unbounded as

likewise called a nonessential singular point. infinite number of the a n n < 0, do not vanish. In this

that zo

is

,

is

an essential singular point.

has an essential singularity at essential singularity is

Theorem

z

=

4.3,

For example,

An important property of an 0. due to Picard, which we state with-

out proof.

THEOREM 4.3. In any neighborhood of an essential singularity a single-valued function takes on every value, with one possible exception, an infinity of times. Let us consider

number answer

=

as an example.

any neighborhood of z = 2Tnt = 1, we have "yes"! Since e

of z in

is

e 1/z at z

Are there an

infinite

for which e l/z = e 60 ? 1/z +* rni = e e 50 so the ;

The

equality

ELEMENTS OF PURE AND APPLIED MATHEMATICS

158 holds

if

to e

=

2wni

=

The

i.

e 60

We shall see in Chap. First

development

\z\

=

0.

=

and investigate

1/t

Example

4.20.

/(I//) at

-

(a) /(z)

analytic at z

(6) /(z)

.

which

for

e 1/z

=

zero.

is

Let the

0.

t

-

/(I/O

1/z,

*

an isolated singular point

> H >

defined to be the nature of f(z) at z

by

=

of z

of infinity is

such that

all z

as a

5 that the point at infinity will play an important of differential equations in the complex domain.

we say that the point

analytic for

is

2,

1,

=

Let the reader show that the same applies

.

is

role in the

Let n = which tend to z

2irni).

exceptional value stated in Picard's theorem

no z in any neighborhood reader show this. There

1/(50

sequence of z n

infinite

=

and such that e 1/Zn

l/*

=

50, or z n

and we have an

3, etc.,

limit

+

l/z n

z

.

so that,

t,

-

1/z, /(I//)

f(z)

then replace z in f(z) is nature of f(\/t) at t =

The o>

if

We

0.

we

if

\/t

/() =0,

define

/(z) is

which has a simple pole at

t,

00

t

We

** 0.

say, therefore, that /(z) has a simple pole at infinity,

= N

(c) /(z)

>

7J-0 00

Zl n-O

Since /(I//) has an essential singularity at

-

?

an

lias

essential singularity at z

= w

we say

t

that f(z)

.

Problems

1

1.

Find the Laurent expansion of

2.

Find the Laurent expansion of

<

|z|

<

2; for

|z|

>

8.

Show that

4. 6.

Find all Find a function

6.

Define cosh

=

f(z)

l/(z

=

/(z)

2

-^

-

-f l)(z

^or

4. o\

^pry?

<

2) for 1

|z|

<

2.

< M ^

*

^ or

2.

the coefficients in the Laurent expansion (4.29) are unique. i. the roots of e* z

which has a simple pole at

/(z) (e*

+

cosh (z

+

=

z

=

0, z

=

cos

9)

1,

and

z

>

.

Show that

~*)/2.

fl

*)

+

Y

a

n rl

an =

rt~ >"*

for

|z|

7.

>

Let

cos

/

JO

0.

f(z)

have a simple pole at

z

ZQ,

=

/(z)

Z

^ + ZQ

a(z

>

Zo)

n .

We call

/

n-0 a_i the residue of /(z) at z

z

Show

.

that o_i

lira

(z

zo)f(z).

2-+ZQ

8.

Let/(z) have a pole of order

ffl

.

A-

at z

z

.

Show that the

-j

Find the residues of f(z)

-

*V(c

+ *)< at its poles.

residue a_i

is

given

by

COMPLEX-VARIABLE THEORY

~l+z+z*B,tz~

Find the residue of /(z) - e^'X'- 1/*). t)

10.

Let

12.

13. oo

z

-

f(z

+

2ir).

By

.

.

z 2,

zi, .

,

~

cos (w0

-

t

z

<

>

|*|

is

sin 0)

infinite strip given by a < use of Laurent's theorem show that

Im

a,

a

>

0.

<*>

.

,

Let

14.

for

Let /(z) be an entire function (analytic everywhere) with a pole of order n at and show that Obtain the Laurent expansion of /(z) for ^ |z| < /(z)

at

=

be analytic in the

/(z)

/(z)

t)

*

= (-!)./(-

,/

Assume

of f(z,

-

t)

n

where

.

Show that the Laurent expansion

11. Let/(z,

/(Z,

159

n.

=

a

-f Qiz

+

+

a n zn

be analytic everywhere with the exception of a finite number of poles >. The order of the pole at z is a,, z zt and a pole at z 1, 2, Consider

/(z)

.

.

.

,

-

<^(z)

and show that/(z)

is

(z

-

z,)i(z

-

z 2 )s

-

-

(z

z*)*/(z)

a rational function, the quotient of two polynomials.

The result

of Prob. 13 is useful. 15.

2

Let p(z) have simple poles at the finite points z oo Consider z*p(z) is analytic at z

,

17,

f.

Also assume that

oo.

Hence show that

.

and show that P(z) has a pole

at

most

A z

where A, B, 4.14.

C are

of the order 2 at z

C

_J*_ \

z

17

=

z

f

constants.

Contour Integration. Let z = z be an isoFor /(z) single-valued and analytic in the we have the Laurent expansion

Residue Theorem.

lated singular point of /(z).

region

<

\z

3o|

< R

Let F be a simple closed path encircling \z

-

*o|

<

ft.

Then

z

=

2

lying in the region

<

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

160

00

<I>/(z)

=

dz

Y

<

a n (z

n

z

dz

)

ao

V

n

since (n

=

dz

zo)

(z

y r

n

for

J!

?*

and

1,

(p Y

T

= z

The

2wi.

inter-

ZQ

change of integration and summation can be justified. We leave this as an exercise for the reader. The constant a_i is called the residue of /(z) at z

=

z

.

Theorem

Let the reader prove

4.4.

R

be a simply connected open set, and let T be a THEOREM in Let /(z) be single-valued and analytic in R R. closed path simple number of isolated singular points. Then finite of a the with exception

Let

4.4.

/(z) dz

(f)

The

=

residue theorem (4.31) discuss

(4.31)

highly useful in the evaluation of real

some examples. dx

Let us evaluate

4.21.

Example

is

We

definite integrals.

of residues of /(z) inside T)

(sum

2iri

We

r-

/

deal naturally with

= r4-.,

/

Now f(z) has a simple pole at z i. To apply the residue theorem, we look for a path containing z = i in its interior. At the same time we desire that part of the simWe ple closed path be part of the real axis. choose as T the straight-line segment exR to x ** R and the tending from x = upper semicircle Since /(z)

lim^1 z-+i

-r-

+

2

1/(1 -f

=

)

has a "simple pole at

^

lim

J

rfx

+

M* M

z

Prob.

(see

f*Rie**de Jo

i

+

I

since

f-

I

\R*e**

e

-f

infinite, (4.32)

7o

1

+

-

1|

1

Jo

.

-

fl

1,

dx since

lim R*/(R*

-

1)

-

0.

1

+a

=R >

I

(see Fig. 4.15).

>

1.

9 \

"

i

is

(4.32)

^ R* = R* -

If

=

Hence

2i

-f R*e** fl

1

.

Jrl

de \l

becomes

-

_ "

iw

</?/"' K -

R*e** 9

\z\

the residue of /(z) at z

7, Sec. 4.13)

e

Now

it

we

1

allow

#

to

become

COMPLEX-VARIABLE THEORY The method used above

is

known

to advantage to evaluate integrals of the type

Example sin z)/(z

(z

+a

we

2 )

/

consider /(z) e is

which introduces the term

sin x.

e

t

^

=

ze"/(z

=

cos x -f

lt

2

much

is

The

Instead of dealing with

dx, a positive.

a

-f i

Remember

2

).

f

R

xelt

T J-Kx +o where C

On

2

;

j dx

o 2

than sin

easier to handle

" f ze dz -T* 2 2 ./cz

.

-f

+a

/

C, z

=

1

be

/<we*

o

27T*

(a

I

2at

=

/te', dz

determine lim

difficult to

*

f -f

/

#0, (-R,

riz

xe

fl a

,

-R +

>

o2

+

r

Jo

see that

2

a2

(

72,

0),

0),

(/?,

^w}^^i_dy r +

(ft

2

?//)

+

>

dx

fl

xe tf dx

/ftft B

/

by equating the imaginary parts. Example 4.23. As an illustration

(2

-

x2

+

""

a2

T sin x ~~~jl

f

and

=

efo

=

sT

2o)"

A

-

Ot(z

2 "_,

z

(z)

We

see that z

2

/

is

x

g(z)

an (n

such that

-

n(z - z - Zo) (z

^>(z)

w )

+

Oo

2o)*

(z

/(z)

-

l)st-order root of /'(z)

(z

- ZoMz)

so that

theorem consider

^

zo)V(z)

* 0.

-f (z

-

j

2 )*VC0 9 that there

for all z in this neighborhood.

5^

-V(2)

fl

"

.

neighborhood of

i~

"

TC

of the residue

3"

We say that z = z is an nth-order zero of /(z). We can also write /(z) * (z Let the reader show ^(zo) ^ 0, tp(z) analytic in a neighborhood of z exists a

(/2,

Then

/?).

/(Z)

it

We abandon the semicircle and choose

.

JC z*

z

as our contour of integration the rectangle with vertices at

J-R x* +

-

tire

le We ^X? a /t + 4^'

= -f^rA 2 Z H- a 2

so that

flie'' de,

fl-oo

ft -f

As the con-

z.

since f(z) has a simple

\ = 1 /

.

\

a,

the upper semicircle of Fig. 4.15.

is

Z

will

=

i

that

sin x

tour of integration let us use that of Example 4.21 with R > = ai. We apply the residue theorem and obtain pole at z

/

choice of the closed

80

One does not always pick a semicircle

always apparent.

Let us evaluate

4.22. 2

This method can be used

f(x) dx.

/

J

contour of integration is not as part of the contour.

161

as contour integration.

z

-

Moreover

z

Then

ELEMENTS OP PURE AND APPLIED MATHEMATICS

162 so that z

Z Q is

a simple pole of g(z), since v(z Q )

path F surrounding

which

ZQ for

Similarly, let /(*) (z

*

w/(z

z

9* 0.

for all z inside

Hence for any simple r we have

-

a* (*

analytic in a neighborhood of

^>(z)

/'(z)//(z)

^

<?(&)

z

=

z

,

The reader should now be

^(z

m

-f v'(z)/<(>(z).

)

able to prove

a-m

)

5^ 0.

-

so)* so that

Let the reader show that

called the order of the pole at z

is

Theorem

closed

=

2,,.

4.5.

THEOREM 4.5. Let f(z) be analytic in a simply connected set R with the possible exception of isolated singular points. Let C be a simple closed path in R enclosing a finite number of these isolated singular points which are poles, f(z) j

Then

on C.

(4.33)

N

where is the number of zeros of /(z) inside C, the order of each zero counted in determining JV, and P is the number of poles, the order of each pole counted in determining P. Problems

Show

1.

x*dx

that

-oo (Z*

+

0*)

8

x*dx Evaluate - x 4 -f 5z 2 -f 4 Integrate e*/2 around the rectangle with vertices at

2.

f

3.

at the origin,

and show that

dx

/ oo

J

X

-r

TT,

z

R,

R

Ri indented

see Fig. 4.16.

Or

R

FIG. 4.16

Prove (4.33). Let /(*) - aozn

4. 5.

1/00

1

>

1 for

\z\

concerning f(z)

-

>

-f

r.

"' 1

-f

-f On,

and

Show that ~~.

dz j) f-jj-

0.

let

-

C

n.

be a

circle

From

\z\

-

r

such that

(4.33) state a

theorem

COMPLEX-VARIABLE THEORY Consider

6.

-

sin

Use

z

e*

-

(l/2i) (*

9

7.

e~**

Integrate

>

b

i

sin

1/2).

this result to evaluate

y

+

cos B

=.

/

Show

6.

-

163

-

that cos 9

?(z -f l/),

-

2 + cos Jo around the rectangle whose sides are x

ft,

x

#, y

0,

and show that

0,

1

cos 2bx dx

e-**

dx

Then show that *

e~ z dx

/

from 8.

9.

= (

db

/

e~** uos

I

2bx

=

dx)

(6.78).

For

<

Show

that

<

a

f

2 show that 111

10.

For a

>

~(1

~f- 3*) .

/

Jo

-h

1

f 2

~- dx #2

11.

>

0, 6

>

t

= v

In 2.

TT

^2

sin

~\

^

show that f

x4

Jo

For a

dx =

-^~~

+

a4

2\/2a

s

show that r(o -f 26) i& 3 (a -f 6) 2

12.

For a

>

6

>

show that

?>

4.15.

/^

. _^_JL^

_ 2

a2

)

b*

\

_ CL\

6

The Schwarz-Christoffel Transformation.

closed polygon

in the

w plane.

a /

Let us consider a

We

assume that the polygon does not intersect itself (see Fig. 4.17). wish to find a function

We

F(w) or w = f(z) which maps the polygon into the real axis of the Let PI be mapped into z plane. As we (xi, 0), Pz into (rr 2 0), etc. z

=

,

move along the polygon in the w plane, we move to the right along

Q

the x axis in the z plane. Now if such a transformation exists, then dw arg dz.

Along the x axis, arg (h

along the polygon.

=

FIG 4.17

=

arg

f'(z) <Lr

=

arg

dw - arg/' (2)

arg

dw = A

dz

0, so that

Hence

A

arg

and arg dw

f'(z)

=

arg/' (2)

+

ELEMENTS OF PURE AND APPLIED MATHEMATICS

164

along the polygon, where A represents an abrupt change in the value of arg dw. This occurs, for example, at PI. As we turn the corner at PI, Now con1 < a < +1. arg dw changes abruptly by an amount ai^r,

=

sider f'(z)

zi)-

(z

ai ,

i

real,

so that arg /'(z)

=

arg

i

zi).

(z

Thus

A

arg/'(z)

= aiA arg (z = ai[argz>Il (z = -ai(0 - TT) =

Zi)

zO

arg f<Zl

Zi)]

(z

iri

This suggests that the transformation we are looking for

dw

A(z-

The reader can

A It

-

-

(2

z n )- a

(4.34)

easily verify that

arg

dw = A arg/' (2)

k

=

1, 2,

.

.

n

. ,

(4.35)

can be shown rigorously that w(z)

is

xi)*(z

satisfies

= A

(z

-

xi)*(z

the required transformation.

-

-"

A and B

(*-

x n )-n dz

+B

(4.36)

are constants in the Schwarz-

Christoffel transformation given It

(4.36).

by

can also be shown that

the interior of the polygon

maps

into the upper half plane of the z n

Since

axis.

7 a, ffi

=

2,

the reader

can easily verify that w(<x>) and

w(oo)

exist.

By

integrating

around the closed path of Fig. 4.18, allowing the radii of the small semicircles to tend to zero, and allowing the radius of the large semicircle to become infinite, the reader can verify = tt?( oo ) not neglecting the fact that 1 < a t < + 1 i = 1 oo that w( ) (4.34)

,

2,

.

.

.

,

If

we

When

z

,

,

n.

desire that x n be the point at infinity, we define z = z n 1/f = x, f = . Moreover dz = (1/f 2 ) df so that (4.36) becomes -

(f

I

Jo

(z

-

Oi)-(f ai)-

a >(*

-

o,)

a 2 )-a '

(f

-

an-

(4.37)

COMPLEX-VARIABLE THEORY since

=

at

^

The a

2.

t

i

,

1, 2,

.

.

.

,

n

165

are constants.

1

Equa-

i-i

tion (4.37) is exactly of the

Example

Two into

form

omitted

consider the polygon of Fig. 4.19. *>i ir/2 ir/2 polygon are at into (a, 0) of the z plane. At P, a\ ~

and Q

+

+001.

t

,

map P

Let us

at Q,

,

so that

can be written

A

w(z)

(z

+

\

z*

a)-l(z

-

+B

a)~* dz

yo

The transformation

-

a2

=

z

a sin

w,

=

dz

a

rot-

oi cos w, yields

dit,

rt

B

,

sin" 1 i

From (z

>

We

4.24.

a, 0)

(

=

with the term x n

of the vertices of the

(4.36)

M

of (4.36)

the conditions a,

w

=*

7T/2)

a

=

(z

a,

we have

sin" 1 -

=

1

2

a sin

10

(4.38)

a

If we let 10 = ir/2 *#, then 2 = a ir/2 -f- iR, we see that 2 Similarly if we let w = tion (4.38) unfolds the polygon. For z = j 4- t//,

+

-f iy

x

= =

V

Hence x 2 /(a 2

a sin (u

+

t;

a cos

w

i/

sinh

2 y*/(a? cos w)

sin 2 u)

ti?

a sin u cosh

w)

a sin w cosh

oo

1

v

as

/^

w

-|-

-{-

.

The transforma-

.

w, (4.38) becomes

KI cos a sinh

so that the straight lines

t;

M

Uo

^ ,/

into hyperbolas in the z plane. ellipses in the z plane.

map

Example

4.25.

The

Similarly the straight lines v

Schwarz-Christoffel transformation

certain problems in two-dimensional electrostatic theory.

an analytic function,

w

w(z, y)

/(z)

-f-

iv(x, y),

is

v*

map

into

very useful in solving

First let us note that for

and V 2

we have V 2 u

t;

0,

the electrostatic potential, then V*v 0. The 1, lines of force, w(2, y) =* constant, are at right angles to the equipotential curves, u But for the analytic function, w constant. ^> we know that the v(x> y) (see Prob.

Sec. 4.6).

If f(z, y) is

+

constant at right angles curves w(x, y) * constant intersect the curves v(x, y) (see Prob. 1, Sec. 4.6). Thus, to solve a two-dimensional electrostatic problem, we need only find w = /(z) * u -h iv such that v(x y} satisfies the electrostatic boundary t

conditions.

Once

m Now origin.

let

It

this

is

done,

E

=

Vv yields the

electric field.

Moreover du?

- i^.

d2

us consider an infinite line charge q whose projection in the xy plane is the 2 2 is easy to show that V w - 0, u - u(r), r - (z y)i, has the solution

+

ELEMENTS OF PURE AND APPLIED MATHEMATICS

166 2q In

u(r)

If

r.

we

+

u so that v

2g In r

2#t In

u> *

(z

ZQ).

is

w

consider

2qi In

2gi In

it>

field

due to both charges

(ran

2<7?

2qi In

z

(z

consider a point on the x axis,

(4.3

))

x

will yield the electric field for

infinite line

the charge were placed at placed at ZQ yields

If q,

In (2

2

z

z 2<?i

then

*

z

an

ZQ

In

x,

(4.39)

then

is real,

In

x x

ZQ

-z Hence

for the x axis.

so that v

grounded plane (the x axis) due to an that the imaginary part of w of (4.39)

infinite

charge placed at ZQ. Remember equation and satisfies the boundary condition v

consider a

,

2o)

(z

satisfies Laplace's

Now we

z

)

2^i In

)

t

Let the reader show that 2<n In

2q$

be obtained from

= we

i2q In r

(re**)

similar line charge,

w *

If

we have

the required potential.

A

w The

z,

=

more complicated example where we

when y shall

0.

make

use of the Schwarz-Christoff el trans-

Consider

formation.

two

semi-

infinite grounded planes intersecting

at an angle

^>.

We find w(z) for an

charge q placed at ZQ First we find the (see Fig. 4.20). transformation which maps the polygon AOB into the line v(x, y) = 0, which is the u axis. We need only infinite line

*

FIG. 4.20

changed.

We map

2

=

into

w =

apply the Schwarz-Christoffel transformation (4.36) with z and w inter= 0, is easily seen to be 0. a, at z

= AoW'*" 1 dw and z = Aw*'*, w = Bz T/*. ^>)/T <?/*, so that dz (w The charge at ZQ maps into WQ = BZQ*'*. The complex function for an infinite grounded plane with charge at WQ has been worked out above. =

1

It is

w

W

so that

V(x, y)

is

- U

+ iV

2qi In

the required potential function.

% Z

** """"

(4.40) ZQ

COMPLEX-VARIABLE THEORY

As a

special case let

U

+ iV

-

t/

F =

g In

1

Equation

(4.40)

becomes

Z T" Tot

r

\z

g In

2

In (z

t)

t>o|

\z

2q[ln

=

r &-

LZ_

2<^'[ln

V =

0,

=

2gz[ln (z

= For

*, 20

2qi In

= = Hence

=

v>

167

+

ir

*

arg

In

\z

\

.

/

+

r t)] t>

(2

+

ir

)]

\]

V

2

.

0.

Problems 1.

What

2.

Map

3.

Find the

electrostatic problem is solved by the transformation (4.38)? the rectangle of Fig. 4.21 with two vertices at infinity into the real axis of 1 Show that w the z plane. (a/V) cosh" z. electric field

grounded planes

due to a charge

q placed

midway between two

infinite

(see Fig. 4.22).

(w) B(o, a) (2)

-1

1 7T>

o c FIG. 4.21

FIG. 4.22

B(o, h)

-1

C FIG. 4.23

Map

the polygon of Fig. 4.23 into the real axis of the z plane, and show that 1 (&/r)(Vc* - 1 + cosh" z). 5. Consider a closed curve C given parametrically by x - /(O, I/ 6. *>(0, a ^ * Consider the transformation z x -f iy Show that under this /(to) -f tV(to). 4.

w

transformation the closed curve

C of the z plane maps into the real

axis of the

w plane.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

168

The Gamma Function. The Beta Function. It was Euler who obtained a function of the real variable x which is continuous for x = n, a positive integer. We positive and which reduces to nl when x 4.16.

first

consider e-'t*-

The t

=

dt

(4.41)

existence of the integral depends on the behavior of the integrand at first write and at t = <x

We

.

>

For x

we have

known

that

l

=

T(x)

well

1

that e~ t

/

x~ l

+

1 e-'t*" dt

f l

t

x~ l

behaves

e-tt*- 1 dt

like

>

x

dt exists f or

I"

t

x~ l

for

near zero.

t

It is

since

Jo

lim

t

^o

To

x ~i

dt

=

lim

7

*-+o

investigate the second integral,

-

e -tl*-\

Hence Since

for

t

^

7

Ae~ ^ 1

/

grate (4.41)

T

-</ 2

+

1)

z

A

>

m

e-/2j*-i

A such e^t^

/

1

=

o

that le"^" 1

]

<

Ae~' /2

.

We now inte-

dt exists.

by parts and obtain '"* <-00

+ '

-if x Jo F(n

ij

dt exists, of necessity

<-o

+

"

*

we note that

(<r-'/2^-i)

t

=

=

x/

there exists a constant

Jx

Thus T(x

- -

-

V

I

1)

=

nF(n)

x

If

xT(x).

=

er'Fdt

n(n~

is

f" / /

t 1

x

e-iV- dt <r

Jo

x

=

r(x

+

x

> (4.42)

1)

a positive integer, x

l)T(n

1)

and,

=

by repeated

n,

we have

applications,

we obtain p(n since T(l)

+

1)

=

=

J* e~* dt The gamma function

n(n

=

-

l)(n

-

2)

l

is

of In

of a

By

eln

=

n!

(4.43)

complex variable

*'~ l

=

c ( *~ 1)Ln

z is defined

by (4.44)

where Ln t is the principal value ', the same reasoning as above one deduces that T(z) exists for

defined as ^~ t.

=

ir(l)

1.

e-'t'- 1 dt

t*~~

2

1

COMPLEX-VARIABLE THEORY

We

169

can write T(z)

=

/

e-'t*- 1 dt

+

Jo

V(_lUfc+-l ^ /C

j_^

e-'t*- 1 dt

Ji

converges uniformly on

'

(0, 1) for

Rl

z

>

we can

0,

=

t

interchange the order of integration and summation.

Hence

t/:

V^

/""

Now

c~

/

l

t*-

1

dt exists for all

and

z,

>

(_ rA

i)* ,v ,

converges for

all z

k~0

=

Hence T(z) defined by = 0, 1, 2, = at z n At analytic everywhere except w, (4.45) defined a is has the 2 = simple pole. r(s) by (4.45) analytic ft, r(z) continuation of T(z) as defined by (4.44). other than z

0,

2,

1,

.

.

.

,

n,

.

.

.

.

is

.

The function .

.

.

at z

and

=

is

0,

= T (z)T( -z)

<p(z)

analytic elsewhere. 1,

2,

.

.

.

and

is

has simple poles at z

=

0,

1,

.

.

.

3,

2,

The function

I/ (sin vz) has simple poles elsewhere. It can be shown that analytic

(4 46) '

Now the right-hand side of (4.46) never vanishes. Hence, if a zo exists such that r(z ) = 0, then, of necessity, F(l ZQ) could not be finite so ft and zo = 1 + n, Hence 1 z = that 1 ZQ is a pole of F(z). = = n! a Thus contradiction. ?^ n) T(z) has no zeros. F(ZO) T(l 0,

+

We state

without proof Legendre's duplication formula,

i)

(4.47)

ELEMENTS OP PUKE AND APPLIED MATHEMATICS

170

An important

result is Stirling's approximation to n\

n\

This result

defined

B(p, closely related to the

exists

if

>

Rl p

substitution u

=

=

q)

fc

>

q

- OA

(1

-

9)

z)

= I

Jo

To

real

pole of e"/(l

+

eO at

-*

C

"i

1

dt

(4.49)

The convergence

of (4.49)

choose

or

-

Z

=

up

/

+

1/(1

~

w) in (4.49) yields

du

l

(T

~du +u

< Rlz <

1

I

yields

/

closed rectangle

and S

e'

^

/-

evaluate

J

ft

=

substitution u

O'-

function.

Jo

r(*)r(l

-

V>~ l (l

We

0.

v

The

(4.48)

labor that

-

Thus

1

It is

by

gamma

and Rl

One can show with some

The

large.

be obtained in Sec. 10.25 by use of the Euler-Maclaurin

will

sum formula. The beta function

is

V2irn(~

torn

^t

dt

one applies the theorem of residues to the

with vertices at

/

=

#,

/

=

S,

t

= S

+

27ri,

and

+ <

The positive, / the complex variable of integration. in. The residue of e') inside this contour exists at t

=

=

id

is

By letting R and S tend to infinity and using the fact that < Rl z < 1 one can obtain T(z)T(l - z) = T/sin ** [see (4.46)]. By analytic con-

COMPLEX-VARIABLE THEORY tinuation of I/sin

it

171

domain

follows that (4.46) holds throughout the

of definition

irz.

Problems

2.

that r(i) - VTProve that (2n)\ = 2*n\T(n

3.

From

1.

Show

(4.45)

show that jr

4.

)*-*.

(z -f

n)V(z)

=*

l)"/n!,

(

n a positive

integer.

n

*

Show that /2

/T for

+

lim

>

Rl p

0,

Rl ?

>

.

sin 2 ""

cos 2 "

1

l

0dO = ?R(p,

q]

0.

l l Integrate t*~ e~ around the complete boundary of a quadrant of a circle indented at the origin, and show that

5.

for 6.

<

Rl

z

<

1.

Show that

REFERENCES Ahlfors, L. V.: 1953.

"

Complex Analysis," McGraw-Hill Book Company,

Churchill, R. V.: "Introduction to

Book Company,

Inc.,

New

York,

Complex Variables and Applications," McGraw-

New

York, 1948. Copson, E. T.: "Theory of Functions of a Complex Variable," Oxford University Press, New York, 1935. Knopp, K.: "Theory of Functions," Dover Publications, New York, 1945. MacRobert, T. M.: "Functions of a Complex Variable," St. Martin's Press, Inc., Hill

Inc.,

New

York, 1938. "Functions of a Complex Variable," Intel-science Publishers, Inc., New York, 1943. Titchmarsh, E. C.: "The Theory of Functions," Oxford University Press, New York, Phillips, E. G.:

1932.

Whittaker, E. T., and G. N. Watson:

Company, New York,

1944.

"

A

Course of Modern Analysis," The Macmillan

CHAPTER

5

DIFFERENTIAL EQUATIONS

General Remarks. A differential equation is any equation involvderivatives of a dependent variable with respect to one or more ing 6.1,

independent variables.

Thus

+ +

= &

x

(5.4)

are classified as differential equations. Equation (5.3) is called a partial differential equation for obvious reasons. The others are called ordinary differential equations. One may have a system of differential equations

involving more than one dependent variable [see

(5.5)],

*

*L

(5 5) '

_

In addition to their own mathematical interest differential equations are particularly important since the scientist attempts to describe the behavior of certain aspects of the universe in terms of differential equations.

We

list

a few of this type.

dx -

+

1

X

= Acoso)t

(5.7)

(58) (i} -* }

-0 a

=

172

(5-9)

(5.10)

DIFFERENTIAL EQUATIONS

The order

173

of the highest derivative occurring in a differential equation

called the order of the differential equation. Initial 5.2. Solution of a Differential Equation.

is

and Boundary Con-

Let a differential equation be given involving the dependent variable <p and the independent variables x and y. Any function <p(x, y) which satisfies the differential equation is called a solution of the differditions.

For example,

ential equation.

VV =

satisfies

VV =

-r

ax 1

+

=

-

easy to prove that

it is

We

0-

say that

e* sin

y

<?(x, y)

is

=

e* sin

y

a solution of

dy*

important to realize that a differential equation has, in For example, y" = admits any general, infinitely many solutions. function y = ax + b as a solution, where a and b are constants which can be chosen arbitrarily. To specify a particular solution, either initial It is

0.

conditions like

=

y

boundary conditions

or

=

y

must be given

=

x

is

y the solution. \

2,

when x

1

= 3

like

when x

2

=

T/'

and

3

y

=

4 when x

=

1

In the first case in addition to the differential equation. is the solution, and in the second case y = ?x A full study of a differential equation implies the deter-

+

mination of the most general solution of the equation (involving arbitrary may or may not be constants) and a discussion of how conditions must be imposed in order to fix uniquely the additional many

elements which

Without attempting arbitrary elements entering in the general solution. an exact statement or proof, at the moment, we state the fact that "in '

the most general solution of an ordinary differential equation n contains exactly n arbitrary constants to be uniquely determined by n initial conditions. general' of order

Problems 1.

Verify that the following functions arc solutions of the corresponding differential

equations: (a) ...

2.

y

= z2 -

e*

(6)

u-x*-y

(c)

+x +

+

1

d zu

9

y

Verify that y

1

=

+

y'"

w+*Txdj

=

(y

ae x 4- be**

-

is

3.

=

+|

x}y'

-

(y*

-

-

*

-*

i

x 2)

*(!

-

e*)

,, 1 )

=

a solution of the differential equation

-

3i/'

Determine the particular solution which x

y

du du - 0/ , 2(1

.

y"

for

-

xy"

+

2y

satisfies

the

initial

0, y'

3

condition y

1

conditions y

**

0.

Find the solution of

when x *

1.

y'

+

3xe*

which

satisfies

the

initial

ELEMENTS OP PURE AND APPLIED MATHEMATICS

174

The

5.3.

Differential Equation of a

The family of the by b, equation y = ax To fix these constants means to

Family of Curves.

+

straight lines in the plane is characterized

where a and 6 are arbitrary constants. select one member of the family, that

is,

to

fix

our attention on one

differential equation y" = OJs among called the differential equation of this family of straight lines because b satisfies this equation and, conevery function of the form y = ax

straight line

all

The

the others.

+

= is a member of the family y = ax b. versely, every solution of y" The differential equation y" = characterizes the family as a whole without specific reference to the particular members. More generally, a family of curves can be described by

+

y

=

/(x, ai, a 2

.

.

.

,

,

a)

a n ) = 0, in which n arbitrary conor implicitly by F(x, y, ai, a 2 The differential equation of the family is obtained by stants appear. successively differentiating n times and eliminating the constants .

.

.

,

between the resulting n

+

,

The

relations.

I

differential equation that

results is of order n.

Example

5.1.

To

find the differential equation of the family of parabolas

~

y

we

ax

+

bx*

a

+

2bx

differentiate twice to obtain y'

y"

The tion.

y

-

26

equation is solved for b, and the result is substituted into the previous equaThis equation is solved for a, and the expressions for a and b are substituted into ax + bx*. The result is the differential equation

last

y

The

-

xy'

-

\x*y"

elimination of the constants a and b can also be obtained

by considering the

equations

+ x*b + (-r/)l + 2x6 + (-y')l -

xa tt

2b

4-

(-!/")!

-

homogeneous linear equations m a, 6, 1. The solution and hence the determinant of the coefficients vanishes.

as a system of trivial,

x I

xz 2x

-y'

2

-y"

y

-

Expansion about the third column yields the result above.

Problems Find the 1.

2.

y y

-

differential equations

Cie*

+

C*r*

Ci cos 2x -f C* sin 2x

whose solutions are the families

(a, 6, 1) is

non-

DIFFERENTIAL EQUATIONS

175

-

*(Ci + C 2z) y Find the differential equation of all circles which have their centers on the y axis. 5. Find the differential equation of all circles in the plane. 6. Find the differential equation of all parabolas whose principal axes are parallel to the x axis. 7. Find the differential equation of all straight lines whose intercepts total 1. 3.

4.

Ordinary Differential Equations of the First Order and First Degree. Let y be the dependent variable, and let x be the independent The most general equation of the first order is any equation variable. 6.4.

If y' enters in the equation only linearly, that is, involving x, y, and y'. the first in the only equation is said to be of the first degree. Such power, an equation can be written in the form

We

under suitable restrictions on f(x, y) there always exists a unique solution y = <p(x), such that 2/0 = ^(#o) and shall see later that

The

discussion of (5.11) will be restricted to the simplest cases at

present

:

=

(a) f(x, y)

(fr) f(x, y) is

n

=

a(x)0(y). of degree zero, that

homogeneous

is,

f(tx ty) y

=

t

n

f(x, y),

0.

(c)

Exact equations.

(d)

Integrating factors.

= -p(x)y +

(e) f(x, y)

Case

(a):

q(x).

Separation of Variables.

dy

then M(x) dx

+

F(x

9

N(y) dy y)

=

=

0.

If

_

M(x)

. .

Consider

dx |* M(x)

+ f

v

N(y) dy

=

constant

(5.13)

We wish to integrals in (5.13) are indefinite (no lower limit). that y as a function of x given implicitly by (5.13) satisfies (5.12). The

show

We

have

^ + ^^ = dx

dx

But

I?

=

*&

dy dx

dy

=

"<*>

8

=

- dF/dX

that

dF /dy

5*0 dy

%--'

Q

-

RD

-

Con "

ELEMENTS OF PURE AND APPLIED MATHEMATICS

176

versely, let y

=

y(x) be

any solution

of (5.12), so that

_ _M(x)

dy(x)

dx

=

Consider F(x, y(x)}

y(x)

+ f

dx f* M(x)

N(y) dy.

We

have

Hence any solution of (5.12) from (5.14) so that F(x, y(x}} constant. can be obtained from (5.13) by solving implicitly for y in terms of x.

We

Example 5.2. ables, we have

solve y dx -f

-f

(1

x 2 ) tan"

dx (1 1 Integration yields In tan" x

+

+

=

In y

y

=

dy

,

tan- 1 x

x*)

C\

=

1

x dy

_ ~

=

0.

Separating the vari-

~

y

In

C2

1

Cjftan- x)'

so that y tan -1 x

- C

2

or

1

Problems

dy =_0 2

VxeW dy

y(x*

+

1)

d#

= -

6. For what curves is the portion of the tangent between the axes bisected at the point of contact? 6. Find the general equation of all curves for which the tangent makes a constant angle ^> with the radius vector. = 4, and 7. Find the function which is equal to zero when x 1, and to 1 when x

whose rate

of

change

is

inversely proportional to

-

its

value.

v^7 ^.

8.

Find

9.

Find the family of curves intersecting the family of parabolas y* ** 4px at right all p. These curves are called the orthogonal trajectories of the system of

r(x)

if

r

ax

angles for

parabolas. 10.

The

area bounded

variable ordinate

is

by the x axis, the arc of a curve, a fixed ordinate, and a proportional to the arc between these ordinates. Find the equa-

tion of the curve. 11.

Assume that a drop

area of surface.

(sphere) of liquid evaporates at a rate proportional to its

Find the radius of the drop as a function of the time.

Brine containing 2 Ib of salt per gallon runs at the rate of 3 gpm into a 10-gal with fresh water. The mixture is stirred uniformly and flows out Find the amount of the salt in the tank at the end of 1 hr. at the same rate. 13. It begins to snow some time before noon and continues to snow at a constant rate throughout the day. At noon a machine begins to shovel at a constant rate. By 1400 two blocks of snow have been cleared, and by 1600 one more block of snow is 12.

tank

initially filled

cleared.

constant.

What time

before noon did

it

begin to snow?

Assume width

of street

a

DIFFERENTIAL EQUATIONS Case

We

(6).

=

if f(tx, ty)

say that /(z, y) is homogeneous in x and y of degree n 2 For example, f(x, y) = x 2 f(x, y). y is homogeneous

+

n t

= tV

of degree 2 since /(to, ty)

now

+

=

2

t*y

= where

/(to,

=

ty)

Moreover /(x,

to)

<

=

n =

0,

that

We

n

/(x,

y).

n

x /(l,

<)

t

+

2

2

(z

2 i/

)

=

t

We

z

f(x, y).

let

a

Tx

^

to,

that -p &X

so

=

(1 '' )

homogeneous dx x

-t

t

+

x-j uX

= M(x,

dy dx

-

We

In particular,

N(x,

(

of degree zero,

/(I,

and the variables have been separated. xt(x) satisfies (5.15).

'

=

T/

=

X

dt t)

(5 15)

/( *> y)

so that (5.15) becomes

+

/(x, y) is

is, if

/(I,

y

/

consider

Tx

If

177

<

5 16 ) -

then

*

(5.17)

can solve for

t

=

t(x),

and

if

y) y)

and M(x, ?/), N(x, y) are homogeneous of the same degree, then M/N is homogeneous of degree zero, so that the substitution y = to yields an equation in t and x with variables separable. Example

Consider

5.3.

.

dx x> y)

=

(x 4- 2/)/(z

T/)

is

homogeneous ,

tan~ l tan- x 1

t

-

-?

,

In (1

\ In

(

\

1

-

x

y

d

_ x

+ + x^ ) t

2

)

/

Problems

^ 2

dy

=

yi_

6tX

X

^

y 4-

~~" ^

x*

x

d!x

S

rfy '

dx

_ ~

x

+ x

y

+

Let y

of degree zero.

+

fcr

_

In x -f In In

Cx

1

C

-f- /

te,

so that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

178 4.

5.

x

^

x

ax

+y + y T

Discuss (5.15)

Case

(c):

*

if

=

Hint: Let x

=

/(I,

* -f

=

y

a,

y

+

6,

and

find a

and

6 so that

in (5.17).

t

We

Exact Equations.

consider

_ d* M(x, y) dx

or

We say that Eq.

is

(5.18)

exact

From

(5.18).

then

+

if

d<p

=

N(x,

y)

dy

=

(5.18)

there exists a function

= M(x,

d<p(x, y)

If (5.18) is exact,

tf (*, y)

y)

and

+

dx

y)

<p(x,

N(x,

y)

<p(x, y)

such that

dy

constant

(5.19) is

a solution of

the calculus

so that for (5.18) to be exact

we must have

|f

= M(x,

y)

(5.21)

Further differentiation yields

If

we assume

ay

ay

dy dx

dx dy

dy

dx

(5.22)

continuity of the second mixed partials, of necessity

dx

Equation

(5.23)

is

a necessary condition that (5.18) be exact.

Conversely, assume

such that

d<p

=

= dy

We show sh that a function p(x, y)

-T

ox

M dx + N dy.

exists

Consider

(5.24)

o

and

t/o

are arbitrary constants**

Then

DIFFERENTIAL EQUATIONS

^ = M(x, **

y)

r -^

=

dy

+

dx

N(XQ, y)

Jxo

-r. = =

Hence

179

N(x,

y)

N(x,

ij)

-

+

y)

,

N(x,

y)

(5.24) yields the required function v?(j, y).

seems familiar, there is a good reason, since the same material was covered in Chap. 2, Vector Analysis. Let f = M(x, y)i + N(x, y)j. dx + N dy. If f = V<p, then dtp = dx + N dy. But Then f dr = f = V<p implies V X V<p = 0, which is statement (5.23). If this

M

M

Consider

5.4.

Example

(In

M

=

In

+

1

x

y

dM =

-,

,

In

x

1,

dy

+\ny + -I

,

X

=

T

TV

})dx

->

y

+

-dy

dN =

-

dx

y

dM = -r

]

-r

y

=0

(5.25)

.

._

A

so that

>

__ N

(5.25)

dy

.

is

exact

Applying (5.24) yields <f>(?,

=

y)

(In

I

x

+ln

y

+

I)

dx

+

-

I

dy = constant

Integration yields

*

<?(j, y)

x In

|a?

a:

In

a:

l+xiny

a;

y.y

+

In 1

\ny-\-x

\-}-\ny**c x In (xy)

We

chose the lower limits to be 1 rather than zero since In x easily that

is

=

c

not defined at x

0.

The reader can check

d[x In (xy)}

(In

a-

-f In

y

+

1)

d*

+

-

Problems 1.

2.

2 cos

(xe* -f

z

\

4-

o 2

/

(x

2/)

3. sin y e wn x

Case

i/)

**'

dx

v

dx

2x sin y dy

+

-f (ze

= o ;

x Bin w

2 cos y sin y) dy

cos y

It

(d): Integrating Factors.

not exact, that

is,

-r

dy

^

Q

readily turn out that (5.18)

Let us assume that this

ox

one can find a function n(x,

may

-

y)

is so.

is

Perhaps

such that

dx

+ uNJy =

(5.26)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

180 is

now

exact.

We call p(x,

an integrating factor

y)

The

of (5.18). flii

tion of (5.26) is also a solution of (5.18) since in both cases -~

soluii/f

=

*&

ax N In general it is very difficult to find an integrating factor. There are two Let us cases, however, for which an integrating factor can be found. determine the condition on such that \L = n(y) is an integrating and factor. Of necessity

M

dM = __

_M +

dp

,

and

,,

N

,

= " T/

"

t

-

so that

dN __

5=

nay If

(5.27)

a function of y only, then (5.27) can be solved for

is

Tf

"

'

M

?/,

for in this case

<M XP

M(?/)

Example

5.5.

_dM

dx

f /

dy

-

,

dy

Consider

-ydx + xdy =

We

=

have

-- =

so that (5.28)

1

1,

1,

dN/dx

is

not exact.

- dM/dy __ ~

M

(5.28)

__

However,

2

y

so that an integrating factor exists, namely,

Multiplying (5.28) by

1/V

yields

The reader can

- dx

-f

- d^ =0

(5.29)

2

easily verify that (5.29) is exact.

The

solution of (5.29)

is

x/y

con-

which is also a solution of (5.28). For an integrating factor of the type n(x), one needs of necessity

stant,

--

dM/dy - dN/dx Problems

+

(2/2

xy*)

dx

1.

2xy dx

2.

(x

-

-

3x 2 )

-f x*y

<fy

dy

-

ss <p(jc)

(5.30)

DIFFERENTIAL EQUATIONS 3.

Prove the statement concerning

Case

(e):

Linear Equations of

j|

181

(5.30).

The equation

the First Order.

+

=

P(x)y

q(x)

(5.31)

The expression called a linear differential equation of the first order. "linear" refers to the fact that both y and y occur linearly in (5.31). The functions p(x) and q(x) need not be linear in x. If q(x) = 0, (5.31) is

f

is

said to be

The

homogeneous; otherwise it is inhomogeneous. homogeneous equation

solution of the

j| is trivial

since the equation

y

is

+

=

p(x)y

Hence

separable.

= A

exp

/p(x) dx]

[

(5.32)

A is a constant of integration, fp(x) dx is an indefinite integral. Now we attempt to use (5.32) in order to find a solution of (5.31). We where

introduce a

new

variable z(x)

= =

y or

z

where y(x)

is

=

^

=

z

exp [-fp(x) dx]

y exp Jp(x) dx

a solution of (5.31).

Jx

But

by the equation

-

q(x)

df 6Xp j P

^ dx + yp exp J

p(

p(x)y from (5.31) so that

-J?

p2/)

Hence

Then

=

z(x)

exp

/

pdx

+

py exp

/

p(x) dx

exp

/

/

q(x)

(exp

/

+C

p(x) dx] dx

so that y(x)

=

exp

[

-

The reader can

J

p(x) dx]

easily

\J

show that

q(x)

(exp J

(5.33) is

p(x)

dxjdx

a solution

+C

of (5.31).

(5.33) J

ELEMENTS OF PURE AND APPLIED MATHEMATICS

182

Consider

6.6.

Example

dy ax

+

-f-

Here p(j)

*

cot x, ?(x)

= exp

y(x)

f

l/(x)

We

We now

5.7

Example

~

from

(5.33)

-f-

THEOREM tion

is

the

exp [-Jp(x)

dj] -f

j/i(x)

C exp ~/p(x) 0.

(

cot x dx

/

dx

J

-f

C

We

detail.

rfx][J(/(x)

have

(exp Jp(x) dx) dx]

the general solution of the homogeneous

is

dxj

j/a(x) is

a particular solution of (5.31) obtained

In other words

The most

5.1.

sum

Moreover

C

(exp

more

investigate (5 IW) in

p(x)y

by choosing

x* esc x

fe lnn- fa 4.

o

- C exp |-/p(jr) - i/i(*) -f

notice that

equation

/

dxj

f

/

becomes

j* cscz so that (5.33)

cot j

/

x l csc x

y cot x

we have Theorem

5.1.

general solution of the inhomogeneous equa-

of a particular solution plus the general solution of the

corresponding homogeneous equation. This important result is valid also for linear equations of higher order and is discussed later. In simple cases a particular solution can be found by inspection. For example, consider

te

We

+ y-*

(

look for a particular solution of the form y o

holds

if

a

+

6

0,

a =.

+

aj

+

so that

1

=

6 6

=

ax

+

5 34 ) -

Then

b.

x

=

1

.

Hence y

=

x

1

is

a

This method of obtaining a particular particular solution of (5.34). solution is called the method of undetermined coefficients. The general solution of the corresponding

y

*=

ce~~*,

so that

y is

homogeneous equation,

*

<#-

-f

x

the general solution of (5.34). Problems

1

~+y=

0,

is

DIFFERENTIAL EQUATIONS

n

8.

x -r ox

4.

Change

* c08 x

y

+ p(x)y

What

P* 1.

183

if

n

-

g(x)y" to a linear equation

by the substitution

z

y

1

"",

1?

Review Problems

2.

A body

falls

because of gravity.

There

tional to the speed of the body, say, kv.

a function of the time. y) dx + x* dy [sin (xy) + xj/ cos (xy)] dx

If

is a retarding force of 1 riot ion proporthe body starts from rest, find the distance

fallen as 8.

(x

+

x 1 cos (xy) dy - x*y) dy t-**(2xy 6. Brine containing 2 Ib of salt per gallon runs at the rate of 3 gpm into a 10-gal tank initially filled and containing 15 Ib of salt. The mixture is stirred uniformly and flows out at the same rate into another 10-gal tank initially filled with pure water. This mixture is also uniformly stirred and is emptied at the rate of 3 gpm Find the amount of brine in the second tank at any time (. 4.

- 2*V)

5.

6.6.

An

cte 4-

-f-

<T*'*(3x

We

Existence Theorem.

5-

We

wish to show that

consider the equation

f(x, y)

(5.35)

/(x, y) is suitably restricted in a neighborhood there exists a unique function y = y(x) which 2/o) satisfies (5.35) such that y = t/(xo). The restriction we impose on is the Assume a that constant ^ QO exists such that /(x, y) following: if

of the point PO(ZO,

M

for all points P(x, y), Q(x, z) in

some neighborhood

!/(*,) -f(x,y)\

of

P

O (XO,

<M\z-y\

A function /(x,

y) satisfying (5.36) is said to

An immediate

consequence of (5.36)

is

yo)

we have (5.36)

obey the Lipschitz condition.

the following:

^

If

exists in

a

dy

neighborhood of PO(X O

,

2/0),

then

~

is

uniformly bounded

in

that neigh-

borhood, for [from (5.36))

=

lim

;/(.) -f(x, I

Conversely,

if

^

is

2

-

v

y)

<M

continuous for a closed neighborhood of /V#o,

yo),

then /(x, y) satisfies the Lipschitz criterion for that neighborhood. Remember that a continuous function on a closed and bounded set is uniformly bounded over the set. We assume further that /(x, y) continuous so that the integrals of (5.37) exist.

is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

184

The proof of the existence of a solution of (5.35) is based upon Picard's method of successive approximations. Define the sequence of functions yi(x), y*(x),

.

.

.

,

...

i/nGr),

as follows: *

/ Jxo

2/200

f( x

= /,*/(*>

=

dx

+

2/0

yi(*})

dx

+

2/o)

,

2/0

+ .................

y z (x)

n -i(x)

=

(j))

dx

/

*/(r, 2/n-2(x))

/

f(x, ?/n-iOr))

</o

+

dx

Jxo *

=

J/nCr)

2

[*f(x, Jxo

^+

(5.37) ?/

2/0

Ja-o

The

y/,(a-),

manner. obtain

i

=

In /(j,

1, ?/)

2,

.

.

n,

. ,

we

.

.

.

are obtained in a very natural

,

replaco y by the

=

such that

initial

constant

?/

,

integrate,

and

We now

z/o. ?/i(x ) replace y inf(x, y) by */iU), and obtain i/z(x). This process is continued indefinitely. The next endeavor is to show that the sequence thus obtained converges to a Let us note that function t/(x) which we hope is the solution of (5.35).

t/i(j)

integrate,

y n (x)

S

//o

+

-

(?/,

Hence the investigation

|/

+

)

of the

(2/2

-

+

y\)

+

'

'

(//n

-

l/n-l)

convergence of the nequence hinges on the

convergence of the series 2/o

+

(2/1

-

2/o)

+

(2/2

-

yi)

+

+

'

'

-

(2/

//-i)

+

(5.38)

Let A' be an upper bound of f(x, y} in the neighborhood of PoCr which the Lipschitz condition holds. Then from (5.37)

,

2/o)

for

A> -

/

!

Also from (5.37) \y t (x)

-

y

t

(r)\

=

from the Lipschitz condition. |i(i)

-

J/I(JT)|

[f(x,

,)

Applying

<

-WA' j

-

(5.39) yields

f*

|*

y

<

J

/(ar, y.)] rfx

" tK TT

-

ioi rfx

x

|

(5.39)

DIFFERENTIAL EQUATIONS

185

Let the reader show that

and by mathematical induction that

~

1

Hence each term

of the series (5.38)

is

hounded

(5.40)

absolute value by the

in

terms of the converging series

Since the series for M]T rof converge?; uniformly in any closed and bounded set, the series (5.38) converges uniformly for those x in the region We have for which the Lipschitz condition holds. <>

=

lim y n (x)

//(x)

n

and the convergence

From

uniform.

is

T

=

y*(x)

/ Jxo

/(J",

2/-iGr))

dx

+

y<>

we have y(x)

=

lim y n (x)

=

lim Jr

=

jj

*f(Xi //n-i(:r)) dx

lim /(x, 2/_i(x))

c/x

+

y

+

i/o

because of uniform convergence. The Lipschitz condition (5.36) also guarantees continuity of /(x, y) with respect to y, for lim since 3/|s

y\

>

as z

*

|/(x, 2)

y.

Hence f(x, i/(x))

and

~

=

=

/(x, y)\

/(x, y(x))

dx

y(x

)

+ *

i/o

2/0

so that y(x) satisfies (5.35) and the initial condition. Q.E.D. must now show that y(x) is unique. Let z(x) be a solution of (5.35) Then such that Z(XQ) y*.

We

~

/(x, ^(x))

ELEMENTS OF PURE AND APPLIED MATHEMATICS

186

and integration

yields (

Hence

-

z(x)

=

y(x)

>

-

-

-

|z(ar)

a Since

B

.

-

j

J

.

is

;?

converge,

Let

of

xo|

y(x)\

y(x)\

< M"L

L

V/

4l 4 the nth term ofr 4l

""

u- u which

<>

-,

/^

!

is

known i

to

we have

fix( k d

-

,

=

w!

-*

any

L he an upper bound

we obtain

,

for

/(s, t/(*))]

< ML\x -

y(x)\

n

,

y*

(5.41)

and, applying (o 41) successively,

\z(x)

+

-y(z)\dx

\z(x)

by applying the Lipschitz condition. Then y(x)\. \z(x) \:(x)

-

[/(*, 2(2:))

f

<M

-y(x)\

\z(jc)

x *(*)) dx

*

Hence the difference between z(x) and y(x) can be made l"his means that z(x) s= y(x). Q.E.D.

.r.

f

as small an w< please.

Problems

8.

Show that Show that Show that

4.

Consider ~

1.

2.

/(x,

x?/ satisfies

!/)

the Lipschitz condition for

|x|

< A <

.

(5 40) holds.

/(J,

J"

//)

j/,

sin

*

y(0)

?/

1.

does not satisfy the Lipsdutz condition for

Obtain the sequence

hm

j/nU)

-

(5.37),

all x.

and show that

e*

-*

^

- x

f

6.

CoiiHider

6.

(Consider the system

-f

t/

,

j/(0)

=

Obtain y

t

(x),

y t (x), y s (x),

j/*(x)

from

(5.37).

g-.ru.,,*>

Impose suitable restrictions on /(x, y, 1} and ^(j, y, f ) in a neighborhood of the point and ^(ir, Vo, *), and obtain a sequence of functions j/i(x), Vi(x), y.(x), .

.

.

,

.

.

.

,

DIFFERENTIAL EQUATIONS a sequence of functions

Zi(x), z*(x),

.

.

.

,

*(),

.

.

.

,

187

such that

lira

y(x)

j/(x),

n

lim

z n (x)

-

z(x) t

where

and

t/(x)

Show

z(x) satisfy (5.42).

that y(x) and z(x) are

n

unique

if

y(x

yo,

)

z(x

)

~

2 o-

Linear Dependence. The Wronskian. A system of functions 2/n(#), o ^ ^ ^ 6, are said to be linearly dependent y\(x), 2/2(2;), cn over the interval (a, 6) if there exists a set of constants r jt r 2 5.6.

.

.

.

,

.

,

not

all zero,

.

.

,

,

such that n

c^(x)

m

(5.43)

x on (a, 6). Otherwise we say that the y(x) are linearly independent. Equation (5.43) implies that at least one of the functions can be written as a linear combination of the others. Thus, if c\ ^ 0, then for all

Linear dependence will be important in the study of linear differential Let us find a criterion for linear dependence. Assume the equations. y,(x) of (5.43) differentiate

n

I

-

y(x)

Then

times.

j

0,

1,

2,

.

.

.

,

n

-

1

(5.44)

by successive differentiations. The system (5.44) may be looked upon as a system of n linear homogeneous equations in the unknowns r,, i = 1,2, . n. Since a nontrivial solution exists (remember not all c vanish), .

.

,

we must have \y?(*)\

= o

or

W(y

/,(*) t

(5.45)

,y,,

Determinant (5.45) is a necessary condition that the ?y,(j) be linearly dependent on a <J x ^ 6. This important determinant is called the If W(yi, ?/ 2 Wronskian of y it y^ yn //) ^ 0, the # are .

.

.

.

.

,

,

.

.

t

,

linearly independent.

Let us now investigate the converse.

We take first an

easy cae.

Let

ELEMENTS OF PURE AND APPLIED MATHEMATICS

188

for a < x g b. The same b. Assume yi(x) ^ a g x Then 6. would be obtained if we assumed y*(x) -^ for a g x

for

2/12/2

-

2/22/1

result

o

y\

and i~ dx

~ =

(

so that

)

2/ 2

=

cy\.

which shows that

y\

and

are linearly

\2/i/

dependent. The proof for the general case is not so easy. We assume that (5.45) holds for a ^ x g 6. Furthermore we assume continuity of the derivatives, along with the assumption that at least one of the minors of the last

row

of (5.45) does

not vanish for a

^

^

x

6.

For convenience we assume 2/n(2)

1/2(3) 2/2(2)

2/3(2)

0(2)

.n 2/3

We now

show that under these conditions the y are ^ g over the range a ^ x ^ 6. First expand (5.45) about linearly dependent the first column to obtain for a

x

6.

l

= Hence

is

a solution of the linear differential equation *~ l

y

,

n~ l

Moreover,

column

if

we

pi

replace y\(x)

+ -r

,

~l

dx n

by y^x),

by

y[(x)

=

y

(5.46)

7/J(x), etc.,

in the first

determinant vanishes since two columns are the

of (5.46), the

Hence (5.46) is also satisfied find that y\(x), 2/2(2), 2/3(2),

same.

we

1

.

.

by

.

,

With the same reasoning The

y*(x).

y n (x) are solutions of (5.46).

reader can verify easily that any linear combination of y^ t/ 2 yn Now we fix our attention at a point 2 is also a solution of (5.46). ,

.

.

.

,

,

a

g

XQ

g

b.

We

have

2/1(^0)

2/2(20)

=

so that the system of

homogeneous equations

n /^ Ct^'(#o)

=0

j

=

0, 1, 2,

.

.

.

,

n

1

(5.47)

f-1

has a nontrivial solution in the

c,-.

Consider a set of c not

all

zero which

DIFFERENTIAL EQUATIONS satisfies (5.47),

189

and form n

=

y(x)

(n .

.

~ 2)

2/

,

=

(#o)

as well as y

(n

)

(

Moreover y(x Q ) =

(5.48) is a solution of (5.46).

Equation .

^ c*^x " 1)

=

(xo)

Now =

conditions y(xo)

initial

that y(x) of (5.48)

=

y(x)

y"(xo) identically zero for a

is

= ^

= x

^

b.

(n

=

0,

condi-

initial

and the

(5.46)

~ 2)

i/

)

-

be shown in

1

certainly satisfies

=

?/'(o)

will

when n

Sec. 5.7 that Eq. (5.46) has a unique solution

tions are imposed.

It

0.

0, y'(x

5 48 )

(x

)

=

0,

so

Hence

n

c#

J g

for a

^

x

b.

t

m

(x)

Q.E.D. Problems

Show that sin x and cos x Show that sin x, cos x, e

1.

2.

e tx a*

that

cos x

+

*

Assuming

e**/ 2

i,

show

sin x.

Consider yi = x 2 ^ 2 - x|x|, -1 ^ x ^ 1. Show that Tf (y lf y 2 ) for x ^ 1. Does this imply that y\ and ?/ 2 are linearly dependent? Show that 1 ^ x ^ 1. and yz are not linearly dependent on the range Does this contradict

3.

,

^

1 7/1

are linearly independent. are linearly dependent.

the theorem derived above?

Let

4.

i/i

(x)

and y 2 (x) be solutions y"

^

for a

x

^

b.

Show

+

of

p(x)y' -f q(x)y

+

1?

that

-^

W(y\,

A

y>t)

=

p(x)TT(?/i,

exp

W

/

[

|/ 2 )

=

and hence that

p(x) dx]

for a: = x then TF = for all x on a ^ x ^ Show that if determine the constant A ? Show that if for x = #0 then 7* a g x ^ 6. 5. Let 2/1 (x), yz(x) be linearly independent solutions of

W

y"

^

for a

x

^

6.

+

P(x)y

Let y z (x) be any solution of

b.

W

How does one ^ for all x on

-

(5.49).

(5.49)

Show

that

-f

(:

6.

a

g

Show

Let x ^

that W(yi, y* y t )

yi(x),

2/ 2 (x),

.

.

.

,

=

for

y(x) be

a

^

x

^

6.

linearly independent solutions (on the

range

6) of

(5.50)

ELEMENTS OP PURE AND APPLIED MATHEMATICS

190 If y(x) is

any solution

show that

of (5.50),

CkVk(x)

k^l for

a

^

6.7.

x

b.

an nth-order linear

called

is

about

The

Linear Differential Equations.

differential equation

What can we say = 2/0,

differential equation.

a solution of (5.51) subject to the initial conditions i/(xo)

=

=

- !)

=

n - 1)

? Assume y^x) a can written as a system of n We note that (5.51) be solution of (5.51). of 1 new first-order equations by the simple device introducing n follows: Define as variables. y z y*, yn tf'(zo)

2/J,

y"(zo)

.

2/'o',

.

,

.

(n

.

.

.

2/

,

(zo)

(

2/

,

(5.52)

dx

The

equation of the system (5.52) is (5.51) in terms of 2/1, 2/2, . . . y n (x)] is a solution of (5.52), then Conversely, if [i/i(x), y*(x), yn Now (5.52) is a special case be solution of seen to a is (5.51). y\(x) easily last

,

.

.

.

.

,

of the very general system 1

= 3J

P(x, y

n

l ,

y*,

>

2/

i

)

=

1, 2,

.

.

n

.

,

(5.53)

As in Sec. 5.5 it can be shown that, if the/* satisfy the suitable Lipschitz 1 y*(x) of conditions, there exists a unique solution y (x) y*(x), .

1

(5.53) satisfying the initial conditions y*(x The Lipschitz condition for the /* is

)

=

yl,

=

i

1,

.

.

,

2,

.

.

.

,

w.

n n

2

!/*(, 2/S

2/

>

>

2/

"" )

2 1 /^(^, 2 , ^ ,

.

^

.

,

z n )|

<

Af

V t-i

for all 1

x in a neighborhood of x

The exponents

=

Xo,

Af

are superscripts, not powers.

fixed.

|i/*

2*|

(5.54)

DIFFERENTIAL EQUATIONS

191

= 1, 2, n of (5.52) are continuous in a neighborHence (5.51) has a then is (5.54) XQ, easily seen to hold. the n above. to initial conditions stated solution subject unique The reader can show easily that if 2/1 Or), yz(x), n y (x) are linearly If p(x), Pi(x), f

hood

of x

.

.

.

,

=

.

.

independent solutions of the

and

y(x) is

if

.

,

homogeneous equation

any particular solution

of (5.51), then

n

y(x)

is

d n

=

+

<W,(*)

(5-56)

(*)

the most general solution of (5.51). Let the reader show that the are uniquely determined from the initial conditions on y and its first 1 derivatives at x XQ.

In general it is very difficult to find the solution to (5.51), and it is necessary at times to use infinite series in the attempt. This method There is one case, however, will be discussed in a later paragraph.

which

is

far. Naturally, one expects the difficulties in be alleviated to some extent if the p (x) of (5.51) are study now this case. The homogeneous equation

the simplest by

solving (5.51) to

We

constants.

t

pt = constant:

=

i

1, 2,

.

.

can be solved as follows: Assume a solution of the form y

.

=

,

n

emx .

(5.57)

Sub-

stituting into (5.57) yields e mx (m n

Hence

if

+

piin*-

1

+

p 2m

n -*

from

=

+

-

m is a root of the polynomial equation Z + Pl*"- + P22"- + + /(*) 2

1

then y

+

e mx is

(5.57).

a solution of (5.57). dk v

One

by

replaces -p[

roots of (5.58) are distinct, call solution of (5.57) is

The reader can show that

if

Equation zk , k

them

mi,

y

T

the

w* are

=

m

0, 2,

.

pn )

=

Pn

=

(5.58)

(5.58) is easily 2,

1,

.

.

,

.

.

.

,

mn

,

n.

obtained If

the n

then the general

(5.59)

distinct,

^

then

ELEMENTS OF PURE AND APPLIED MATHEMATICS

192

show that

First let the reader

W=

exp

1

Next

let

1

mn

*i

ml

(5.60)

/

the reader notice that

F(u)

-

a polynomial of degree n certainly vanishes for the n

is

If

1

ra 2

mk\

x }

\

1 t\

1

The polynomial equation F(u) =

in u.

1 distinct

roots u

=

ra 2

,

ra 3

.

.

.

,

,

mn

.

F(mi) were zero, then

=

Why?

Continuing with this line of reasoning, the reader can show that as a = one finally obtains consequence of assuming

W

a contradiction, since

The

case for which

ra n 5^ ra n _i.

some

of the roots of (5.58) are equal must be treated Before attacking this problem we shall find it beneficial to introduce the operator

separately.

d jdx

The notation

sf(x)

means 3- f(x) = ax */(*)

and for

s"

any

s ~. We scalar a.

-

define

s[sf(x)}

(s

(5.61)

Similarly,

f'(x).

=

[/'(*)]

+ a)f(x) =

=

sf(x)

f"(x)

+

af(x)

=

f'(x)

+

af(x)

Let the reader show that (5.62)

DIFFERENTIAL EQUATIONS a and b are constants.

if

+

(s*

or

pis"-

1

Equation

+

p 2s

n~ 2

m

m-i)(s

(s

using the result of (5.62). s(e

ax

If

=

y)

a

+

2)

is

can be written

(5.57)

+

Pn-is

+

m n )y

(

=

p n )y

(5.63)

=

(5.64)

we have

constant,

+ aeaxy =

e ax sy

193

+

e ax (s

a)y

Let the reader show by mathematical induction that

=

s n (e ax y)

Now w*

let

+ a) n y

x

($

(5.65)

m

us assume that (5.58) has the distinct roots mi, w 2 3 a* so that (5.64) can be written i, a 2 ,

of multiplicity

/()y =

We

e

note that,

if

.

,

(s

-

.

rai)*

.

1

. ,

,

^ - m

y(x) satisfies

.

.

,

2

)'

nik)

(s

ak

y(x)

(*- m k )**y = = 0, then 2/(#) also

(5.66) satisfies

Now

(5.66).

e -mk*( s

- mk

ak )

y(x)

=

s ak [e- mkx y(x)}

m

(5.65), so that any y(x) which satisfies (s Hence satisfies s ait (e~-mkx y} 0, and conversely.

from

k)

ak

y(x)

=

also

=

so that

=

y(x)

e m "*(d

+

C*x

+

+

1

C^-iz"-

It is easy to verify that (5.67) is a solution of (5.66).

=

(5.67)

)

If

ak

=

1,

then

mkx

Cie y(x) Equation (5.67) contains a k constants of integration. the same reasoning to the remaining roots yields n constants of Applying We omit the proof that the solutions thus obtained, integration. .

namely,

are linearly independent.

Example if

m

2

1

The

5.8.

or

differential

m

1.

equation y"

The most y

If

Ci

^,

C 2 = ~y, then

.

sinh x.

t/

y

=

admits the solution y

general solution

Cie* -f

C 2e~*

Ci

C2

If

?,

to verify that sinh x and cosh x are linearly independent. can also be written in the form

y

Example

We

5.9.

0, 2/'(0)

-

n,

Ai sinh x f

4

2

cosh

em *

is

then

j/

cosh

The most

x.

It is

easy

general solution

a:

wish to solve y" -f n 8 i/ subject to the initial condition if m* -f n* real. c w * is a solution of y" + n*y or

n

ELEMENTS OF PURE AND APPLIED MATHEMATICS

194

m

Hence the solution

in.

is

Aeni*

y

From

y

0,

B

A

or

=

(x

0)

A + B, and from n = (An Bn)i

we have

Hence

*

~

yl

Let the reader show that y

+

?/"

n*y

-

=

,

A

**

+B

nx

sin

and

5",

(x

*

^

j/(x)

cos nx

0,

is

n)

?/'

we have

e~ nt *)

(*>"*

sin nx.

also a general solution of

0.

Consider

5.10.

Example

+ Be-*'*

6 Jg+J*_ dx rfr

6

4

S <&r 8

+

4*-0 dj*

(5.68)

In operational form (5.68) becomes (s

or

or

-

(

are

m

2

l) (s* -f

+

2

y

.

y

- A + Ex +

e

(A

Bs)

-f

-

4)y

\/3

1

0, 0, 1, 1,

+ e-(C +

(C

+* -

6

4

Dx)

/>*)*

+

4* 2 )s/

-

The zeros of /(m) - m*(ra - l) 2 (w 2 Hence the general solution of (5.68) is

0. *

-f

6*

+ E exp

[(-1

+

+ 2m +

4)

\/Zi)x]

+ Fexp[(~l + c-(G cos V3 x + F sin V3 a

-v/3^>]

1

)

has been explained before that the most general solution of an inhomogeneous equation can be written provided It

:

A

(i) particular solution is known. (ii) The general solution of the homogeneous equation is known. The second part has just been considered. It remains to study meth-

ods by which a particular solution can be found. Here any guess can be made, and, if successful, no further justification is needed. Quite often a particular solution can be determined by inspection. The involves finding certain undetermined coefficients. For example: (a) If p(x) of (5.57) is a polynomial, try a polynomial of the

work

same

degree.

A

(6)

If p(x)

(c)

If p(x)

ae rx try a sin coz ,

Aerx

+

.

& cos

(*>#,

try

A

sin <ax

+B

cos

<#x.

few examples should make clear what we mean. (a)

y"

Then

y'

c

= =

2,

2c

+y a

-4-

solution of y" (b)

y"

""

2x 2 0,

+y **

y

We assume a particular solution of the form y

.

+ 2cz,

6

y"

-

60.

j/'

6

-f-

This yields 3c

-f-

bx

+ ex*.

** 2x*.

x

Assume a

2e**.

2r^ >r

,

2,

t/"

a

-

4ce lz

0, 6

,

particular solution of the form

a -h &z 4-

y

Then

a

2c so that of necessity 2c -f a -f bx -f ex* == 2z* and Hence t/ 4 is the general A sin x 4- 5 cos x -f 2x*

and 1.

cc s *

of necessity 4ee la a The general solution

bx is

cc 1 *

B

x

2e f *.

DIFFERENTIAL EQUATIONS (c) y" + form y = a

a sin x

2 sin x

2y

4j/'

sin

3e* 4-

x

6 cos

-f ce* -f

4a cos x 26 cos x

-

-

2ce*

+ 2d

4ce*

-

+

2fx

4/

=

3a

coefficients of sin x, cos x,

e*, etc., yields Equating - -3, 4f - 2d - 1, -2/ - -2, so that a - -^, The reader can verify that y sin x 1. /

3c

^

particular solution of

in the

Of necessity

46 sin x

-

We try a particular solution

2x.

1

+ 6 cos x + ce* + d + /-

x

195

2a sin x 2 sin x - 3c*

46 6

^

-36

2,

- -^, cos x

c

"~ e x

-

+

+

1

-f 4a -1, d ? +x

-

2x 0,

ls

|,

A

(c).

Although in many cases a particular solution is easily obtained by it is important to have a formula The problem is to find a particular solution of the valid in all cases. the method of undetermined coefficients,

equation

g+* 3+

+

"->

g+

-*

-

PW

(5-69)

Let g(x) be a solution of the homogeneous equation

+..*+ which

satisfies

0(0)

We

=

the

0'(0)

initial

=

g"(0)

-0

(5.70)

0<-(0) -

1

conditions

= f<-(0) =

=

prove that y(x)

is

+-2 +

a particular solution of y'(x)

=

g(x

-

= fc

(5.69).

x)p(x)

g(x

-

t)p(t) dt

Differentiating (5.71) twice yields

+

* g( *

~

j*

(x-t) since g'(0)

=

0.

Further differentiation yields

i}

/;

(5.71)

dx

P (t)

dt

ELEMENTS OP PURE AND APPLIED MATHEMATICS

196 (n

since

~ 1)

(0)

=

If

1-

these values are substituted into (5.69), one

obtains

yW(x)

=

+ aiy^

p(x}

The

l)

+

(x)

+

a n y(x)

*vo + ^po +

+

.

ai

.

.

expression (*

The

vanishes since g(x) satisfies (5.70). t, but since g(x) know that g(x) satisfies (5.70) at x

evaluated at x

5.11.

Example

Consider y"

yields

the solution

gr(0)

0, gr'(O)

Thus

=

=

e"*)

\(^ PX

5.12.

.

,

n, of (5.69)

A - B =

0,

1,

so that

y

x

~- p~

dt

t

2

=

= =

Ae*

sin

o Jo

3j/"

it/'"

- ^ sin x -f - | sin x

Be~*

-f

Aie 1

x

\e

-

\e~*

Bif" 7

-f

+

ty'

y'"

-

=

2y

+

3j/"

e x sec x.

-

4y'

2y

The

solution of

-

is 2/

If gr(0)

0, 0r'(0)

-

0,

=x

Ae x

^(x)

-

^"(0)

1,

4-

which yields

0,

**

[e*-*

j

-

e*

1,

C -

I

-

-f

C

+B

x

-f

-

e*

2B -

o 1

Thus

e*-* cos (x

sec tdt

e* In (sec

-1.

C *

A A A -

+ C cos x)

we must have

A B -

e*(B sin x

-f

I*

+ tan x)

-

t)]<*

sec

t

dt

(cos x -f sin

xc* cos x

a;

tan

+ c* sin

t)

which

0,

conditions

^, sin x

f /

are

B = -^.

A

e-t*-) sin tdt

sm

l

.

.

is

y

Example

^ -

e~

\ 2 Jo

solution

x

2,

1,

particular solution of y"

rx

~ The complete

A

=

i

x. We first solve y" y We now impose the initial

sin y Be~*.

sinh x.

x we certainly (5.71) is a

We have shown that

t.

A +B =

This yields

1.

^-(e*

gr(a:)

= Ae x +

g(x)

derivatives of g(x) in (5.72) are

satisfies (5.70) for all

particular solution of (5.69); the at , constants.

-

dt

x In cos

a;

is

DIFFERENTIAL EQUATIONS The

general solution

y

=

e*[A

+ B sin x +

C

+ tan z)

cos z -f In (sec x

+ pi(x} simplify things,

we

+

+

pt(x}

+ sin x In oos #1

x cos

Example 5.13. Method of Variation of Parameters. particular solution of (5.69) for constant coefficients. tion of

To

197

is

(5.71) is a valid look for a particular solu-

Equation

We

+

=

p " (x)y

p(x)

shall discuss the general third-order linear differential equation

S+

Pl(x)

S+

P2(X)

Tx

+

pi(x}y

=

(5 ' 73)

P(X)

Let 3/i(.c), 2/2(2*), 2/a(a;) be linearly independent solutions of the homogeneous equation derived from (5.73). 3

We

know

that y

^ i

tion.

The method

A

t

yl

is

the most general solution of the homogeneous equa-

=i

of variation of parameters consists in attempting to find a par-

ticular solution of the

inhomogeneous equation by varying the

by assuming that the

A,,

i

y

We

shall

+

uiyi

impose three conditions on

Differentiating (5.74)

are not constant.

1, 2, 3,

-5-^ i

and making use

uzyz

=

+

A

%

,

i

=

1, 2, 3,

u z yz

1, 2, 3.

that

is,

Assume (5.74)

Two

of these conditions will be

of (5.75) yields 3

u

Differentiating again

and making use of

t

%

(5.76)

(5.75) yields 3

-7

dx Finally

: z

=

>

Ui

Lf

-j^ dx*

we have 3

-~ = We

multiply (5.77)

(5.78).

by pi(x\

If y(x) is to

Y

(5.76)

u

l

-T-* -f

by p z (x),

be a solution of

Y -j1 -{

(5.74)

by

p*(x)

(5.73), of necessity

(5.78)

and add these

results to

ELEMENTS OP PURE AND APPLIED MATHEMATICS

198

.

Equations (5.75) and (5.79) are three linear equations in the unknowns Solving for

-~

~i i

=

1, 2, 3.

yields 2/2

/

2/3

2/2

y*

(580)

~dx

2/i

2/2

y(

2/2

2/3

2/2

2/3

yz, #3)

,

2/3

2/2,

n

n

//

vi

The Wronskian W(y\^ Hence

"

"

p(x)

du\

2/

1/2 \

does not vanish since

y\,

2/2, 2/a

are linearly independent.

.()

and

i,

2/2,

,,

2/2,

in general

"

r^ Ut(X)

where

2/4(2;)

?/iCr),

?/ B

(^)

=

^dx

Ja

2/2(x).

(5.81)

1/3)

The reader can

verify that

),

y.+2) (5.82)

i: a particular solution of (5.73). 2 x. (l/x }y Example 5.14. We consider y" -f (1/J")?/' 1 /x are linearly independent solutions of #" x, 2/2(0;) 2/i(x)

is

We

It is easy to verify that -f (l/x)y'

(1/x

2

0.

)?/

have

-2/a

A

particular solution

tion

is

y

* Ax

H

X

is

2

y

(x /4)(x)

+

(

x 4 /8)(l/a:)

x 8 /8.

The complete

solu-

h isO

Problems 1.

2.

3. 4.

Ly"

-f ^2/' -f

(I/O?/

y" - y * x y"" - 22/" + y - y' - * + 1 2/'"

-

0, L, 1?,

5 sin 2x

C

constants.

Discuss the cases R*

-

~|

0.

DIFFERENTIAL EQUATIONS 5.

y"'

+

-

-

r

y

199

2 sin x

- e* 4 sin x 5j/ a particular solution of y" -f p(x)y' -f ?(aOy ** nOc) and 1/2 is a particular r 2 (x), show that 1/1 + ?/ 2 is a particular solution of solution of y" -f p(x)y -\- q(x)y 6.

y"

7.

If

t/i

+

4j/'

+

is

r

+

y"

+

p(a;)y'

8.

y"

9.

]/"

10.

+y

(*)y

tan

**

2#'

-

y"

- n() cos 2 x

-1- i/

4y'

+ r*(x).

a:

+ 4y =

11. Solve y'"

2xe*

=

y"

1

+x

-f sin

z by the variation-of -parameter method.

We

6.8. Properties of Second -order Linear Differential Equations. consider

and

If p(x)

(a)

solution

=

y'(xo)

y'Q

a

,

^

x

6,

there exists a unique

cit/

2/o,

"ii

Moreover -^ ctx

6.

We now show that if y(x) infinite

^

x

subject to the initial conditions y(xoV

(5.83)

g

^

on a

g(x) are continuous

of

y(x)

is

continuous since ^-~ QfX

exists.

a solution of (5.83) then y(x) cannot have an ^ x ^ 6 unless y(x) = 0. The

is

number of zeros on the interval a

proof is as follows: Assume y(x) vanishes infinitely often on the interval a ^ x ^ 6. From the Weierstrass-Bolzano theorem there exists a limit We can pick out a subsequence xi, x$, xn point c, a ^ c ^ 6. which converges to c such that y(x n ) = 0, n = 1, 2, From .

.

.

.

.

mean

the theorem of the

=

y'(tn)

0,

Xn-i

^

^

{n

n,

y(x n -i)

y(x n )

n =

1, 2,

.

.

.

=

(x n

y'(c)

=

since the

x

(5.83),

y(x)

s

approach

Moreover

and

y(c)

=

= =

a:

y(c)

c as

a limit and since y'(x)

lim

f(x n )

n *w

=

0, j/'(c)

t/i(ci)

show that

=

Since y(x)

0.

But

0. is

y(x)

unique,

,

,

so that

=

*

is

continuous

=0

satisfies

we must have

2/1(^2)

1/2(0)

=

be linearly independent solutions of (5.83). and 2/1(3) 5^ for a ^ ci < x < c 2 ^ fe. We

y*(x)

= 0, Ci

separate each other.

<

c

<

Assume

c2

has no singularities for c\ ^ x 0? Moreover <p(d) = ^>'()

=

for Ci

^

{

g

,

that

is,

c2

.

^

c2

.

^>(c 2 )

But

the zeros of ^i(x) and y*(x)

for c\

y<t(x) j*

!/2(c) 5^

mean

=

.

0.

Q.E.D. Let 2/1 (x) and

(6)

Let

n also

c.

.

.

lim y'({ B ) n

at

.

a?n-i)y'(fm)

Hence

.

.

Why =

0.

is

<

x

it

From

<

c2.

Then

true that

2/2(^1)

^

0,

the theorem of the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

200

Since yi(x) and y*(x) are linearly independent, we know that W(yi, t/ 2 ) ^ ^ x ^ 6. Hence ^'(f) cannot be zero, a contradiction. y*(x) must

for a

be zero for some x, Ci < x < c 2 Q.E.D. (c) We can write (5.83) in the form .

<K*)V-0

We

multiply (5.83) by exp

& [exp (/; so that K(x)

=

^ An

p(t)

|]

+

*

exp f

Q(x)

p(t) dt,

q(x)

=

important differential equation

-

where X

is

^

K(x)

Qjjj

and write

^* p(t) dt}

dt)

CLX J

|_

is

+

exp (// p(0

exp

q(x)

/"*

=

Xi,

y

2 (x,

eft)

y

-

p(t) dt.

the Sturm-Liouville equation

Xg(x)i/

=

Assume K(a) = K(6) =

a parameter.

solution of (5.85) for X show that

(5.84)

(5.85)

0, arid let yi(x,

XO be a

=

X 2 ) a solution of (5.85) for X

X2

.

We

q(x)yi(x, Xi)y 2 (x, X 2 ) dx

We

=

Xi

^

X2

(5.86)

have

(5.87)

Multiply the first equation .of (5.87) by y 2 and the second by This yields

2/1,

and

subtract.

dx

dx

\

[_

(5.86) follows

CLX I

\

we

integrate (5.88) over the range a g x ^ 6. orthogonality property (5.86) will be discussed in greater detail in Chap. 6 dealing with orthogonal polynomials.

Equation

if

The

(d)

Let yi(x) be a solution of y"

tion of y" 2/i(a)

=

2/ 2

+ G*(x)y (a)

vanishes, x

>

= a.

ft

=

GI

0.

<

+ Gi(x)y =

0,

and

let y*(x)

be a solu-

Assume further that y^a) = ?/ 2 (a) = a > 0, G 2 We show that y*(x) vanishes before yi(x) .

The proof

is

as follows:

We

have

y{'

+ Gi(x)yi =

0,

DIFFERENTIAL EQUATIONS 2/2'

+G

=

2 (x)2/2

so that 2/22/1'

~

+ +

2/12/2'

"

(y*y(

jz

y&*)

-

(G\ (

~~

Gi

(72)2/12/2

#2)^12/2

= =

f

=

dx

2/12/D

(2/22/1

I

-

I

Gi)yi(x)y t (x) dx

Ja re

c

~

2/22/1

-

^

-

)

#1)2/12/2

+

a/3

=

a/3

- GJy^dx

(G,

/

Jo

Let

be the

c

first

(5.89)

g

for a

x

~ g

=

2/i(c)

2/ 2

Now(? 2 (x)

>

zero of y\(x), c

and

(?i(x)

>

a,

and assume

2/2(2;)

for a

?^

g

(5.89)

x

^

c.

we have

g

for a

C

=

(c)2/((c)

Hence

c.

dx

Ja

a

From

o

c

c

d T~ X Ja f

201

I

x

y<i(c)y((c)

(02

-

dx

G02/12/2

(5.90)

^ c, ?/i(x) > for a ^ > from (5.90), so

< c, y 2 (x).> that y[(c) > 0.

x

But a;

^

c *^

^

x<c

ar<c

> and x < c. This is a contradiction. Q.E.D. The same would have been obtained if 2/1(0) = 2/2(0) = a < 0. Let the reader show that, if y\(c\) = 0, y\(ci) = 0, y\(x) T for < x < c 2 then y^(x) = for some x on Ci g x ^ c 2

since

2/1(2;)

result

Ci

.

,

Problems 1.

Consider y"

+

p(x)2/'

+ g(x)2/

=

0.

Let

?/

=

uv,

and determine w(x) so that the

resulting second-order differential equation in v(x) does not contain T2.

Consider Legendre's equation

(1

Let yi(x,

a),

2/ 2 (x,

/3)

~

x2)

S^

be solutions of

/

2x

fx

+ n(n +

(5.91) for

n

t/i(a?, 0)2/2(0;, |3)

=*

l)y

=

a and n

(5 91) *

/3,

a

7^

/3.

Show that

dx

i

3.

4.

Prove the second statement of Give an example of (6).

(d).

Equations in the Complex Domain. Up to the present discussed differential equations from a real-variable view. Greater insight into the solutions of differential equations can be obtained 5.9. Differential

we have

ELEMENTS OF PURE AND APPLIED MATHEMATICS

202 if

we study the differential equations from a complex-variable point The reader may recall that the Taylor-series expansion of

of

view.

about x

+

1/(1

l

=

/(*)

1

X2

=

converges only for \x\ < 1. In a way this is puzzling since x z ) has no singularities on the real axis. However, the analytic

continuation of /(x), namely, /(z) i. The distance of z = at z =

= i

+

1/(1

and

+

the Taylor-series expansion of 1/(1 convergence equal to unity.

The

+

z

=

z

2

), is

known

z 2)

to have poles is 1, so that

=

from z about z = i

has a radius of

simplest first-order linear differential equation may be considered If p(z) is analytic at z = zo, the solution of difficulty.

without

^ + p(z)w = is

w(z)

The

= WQ exp first

non

and w(z)

I* p(t) dt

=

analytic at z

is

zo.

trivial case is the linear-second order differential

~+

ft 111

This equation

(5.92)

is

equation

fiitj

P(z}

^ + *W W =

(

most important to the physicist and engineer.

5 93 > -

More-

over, the methods used for solving it apply equally well to higher-order In attempting to solve (5.93) we must obviously consider equations.

the coefficients p(z) and g(z). = ZQ. q(z) are analytic at z are differentiate in

type

two z

Let

=

z

zo

be a point such that p(z) and

Remember

this

means that

some neighborhood

of z

=

said to be an ordinary point of (5.93).

is

radii of

convergence

ZQ.

A

p(z)

and

q(z)

of this

point z

R be the smaller of the

Let

of the series expansions of p(z)

= z We shall now prove Theorem 5.2. THEOREM 5.2. For any two complex numbers w

and

q(z)

about

.

unique function w(z) satisfying (5.93) such that w(zo) z Moreover w(z) is analytic for |z < R.

,

wj there

=

exists a

WQ, W'(ZQ)

=

w'

.

|

We Let

begin the proof by removing the first-order derivative in (5.93).

w =

uv,

u and

v

undefined as yet. ,

Then

-r-

dz

du\ dv

,

du dv

(d*u

dz

2

,

d u

Substituting into (5.93) yields .

= u -y-

.

du

+

v -T->

dz

203

DIFFERENTIAL EQUATIONS If

we

set

pu

+

2

-r =

we

0,

see that

u(z)

Thus the

substitution

w =

r

exp

^+

f* p(t) dt\ reduces (5.93) to

-%

\

=

,/(*),'

(5.94)

Let the reader show that J(z) is analytic for \z the reader can also show that if v(z) is a solution

w = is

equation.

exp

I"

-^

<

z \

Moreover

It.

of (5.94)

then

I* p(t) dt\

Next we attempt to reduce

a solution of (5.93).

Assume

v

-

such that

v(z) satisfies (5.94)

(5.94) to

V(ZQ)

=

VQ,

an integral V'(ZQ)

=

v'Q .

Then

or

-^ az

-

= -

V'Q

J(r)v(r) dr

/

J 2o

Integrating again yields

-

where

^>(f)

=

/

-

vo

v'9 (z

-

JMvM dr.

zo)

= -

*

Note that

f

F

I J

20

yo

<p(2

)

J(r)v(r) dr d{

=

0.

We now

Jzo

integrate

by parts and obtain

and

()

=

1*0

+

P,(

-

20)

+ Jzo f

(r

-

2)-/(f)''(D

df

(5.95)

The reader can show that

if v(z) satisfies (5.95) then v(z) also satisfies (5.95) is a Volterra integral equation of the first kind. It is a special case of the integral equation

(5.94).

Equation

v(z)

To find

= A (z)

a solution of (5.95),

imations due to Picard

(sfce

+

k(z, f)v(f)

we apply Sec. 5.5).

df

(5.96)

the method of successive approxWe define the sequence VQ(Z),

ELEMENTS OF PURE AND APPLIED MATHEMATICS

204

.

1/1(2),

.

.

...

v n (z),

,

as follows:

-

(r

(5.97)

-

S(

-

(f

20)

Now

so that the convergence of the sequence

\v n (z)

}

hinges on the convergence

00

of the series ^0(2)

+

/

-

v k (z)].

\Vk+i(z)

v k (z)

=

z

f

(f

We

can write

-

-

J Zo

w*_i(r)]

df

(5.98)

We choose as our path of integration the straight line joining z o to z so - 20), ^ t ^ 1, that f = ZQ t(z

+

with f

ZQ

=

df

Let

C

t(z (z

be the

ZQ)

and

ZQ) dt

circle of analyticity

of J(z) with center at

ZQ,

and

let z

be an interior point. We construct a circle S with center at ZQ interior to C and containing the given point z in its interior (see Fig. 5.1). z are analytic Since J(f) and f FIG. 5

is,

\J(f)(f

-

*)\

<

M for

and on /S, /(f)(f bounded by some constant and on S. Hence inside

1

all z, f in

'(r-

where M

is

any upper bound

of v$(z) in

and on

*S.

Similarly

M

z) ,

is

that

DIFFERENTIAL EQUATIONS

(' yo

ir

-

i*

-

2.1

205

i

r

2

^oi

**,

Jo

By

mathematical induction the reader can show that

Thus the

~ v k (z)\

bounded by the

series representing v n+ i(z) is n+l

yM

-

k \z

series

k

z,\

fc-O

which

in turn is

bounded by the

series of constant

terms

/z

}

^ 7 ^ K

m*J

since

I

*-0 that the latter series converges to e MR so test the series representing v n +i(z) converges that from the Weierstrass

R >

zo|.

\z

We know

M

Since each term of the series representing v n +i(z) is analytic, v n +i(z) converges to an analytic function t>(z) (see Prob. 7, Sec. 4.9). We now show that the limiting function, v(z\ satisfies (5.95). From (5.97) uniformly.

lim n

v n +i(z)

=

vQ

+

v'Q (z

ZQ)

+

lim n

*

v(z)

=

^o

+

^o(z

ZQ)

+

/

*

J/ *o

(f

lim (f

It is possible to take the limit process inside the integral because of the uniform convergence of v n (z) to v(z). Next we prove that v(z) is unique. Let u(z) also satisfy (5.95). It is u(z) satisfies easy to verify that r(z) = v(z)

r(z)

=

Z

I

yo

(r

~

*V(f)r(r)

*

(5.100)

Since w(s) and v(z) are analytic in and on S, the function r(z) by some constant K, |r(f)| < K. From (5.100) we have

|r(*)|

<

MK ^

\df\

= MK\z -

|

is

bounded

ELEMENTS OF PUBE AND APPLIED MATHEMATICS

206

this inequality to (5.100) yields

Applying

M

< M*K

|r(f)| |df|

|f

,|

\df\

2!

Continuing this process yields

< KM*

\r(z)\

Since lim n

as small as

we please.

This

*

is

n\ ""

*' g

'

for all integers n.

H

*

*

1

<go

=

'

we can make

0,

\r(z)\

!

if r(z) s=

possible only

so that u(z)

s

v(z).

Q.E.D.

An

point.

We

5.15.

Example

consider

analytic solution

is

w" zw known to

w

f

=* 0.

Certainly 2

The

exist.

easiest

way

is

an ordinary

to find this solution

00

to let

is

w

c nzn .

}

The

are to be determined

cn

by the condition that w

satisfy

n-O

w"

w

zw'

We

** 0.

have M

00

W " '

A

~1

nCn2!n

W"

n-l

n-2

* V* / n(n

so that

n l)c n z

~~

2

A power .

,

series in z

>

cnzn

-0

can be identically zero only

if

the coefficients of z n , n

+

2)(n

+

-

l)c n +2

+

(n

=0

l)c n

n -

...

0, 1, 2,

obtain the recursion formula

We

note that

c2

-

-

c /2, c 4

=^2 c 2 /4

C6

.

nCnZ*

n-l

cn+t

.

00

V*

Hence

vanish.

(n

We

00

V* /

n^2

.

"

n ^n

A

""

.

,

cjn

c

/2

nn!.

Also

""

C(

d -

ci/(2n

(2n

+

+

1)!!,

1)!!

-

Co

__ ~

c 8 /5

23!

- d/(l

-3-5),

C

"

"7

-

c /2 2 2!,

(2-4-6)

Ci/3,

*

-

4)

Co

4

6

C7

-

c /(2

*

ci

n =0,1,2, ...

(1

3

5

7)

where (2n

+

l)(2n

-

l)(2n

-

3)

5

-

3

1

-

0, 1, 2,

DIFFERENTIAL EQUATIONS The constants CQ and w zw' of w" =

207

are arbitrary constants of integration.

c\

The

general solution

tv

Also show that

is

+

(2n

n0

1)!!

Problems 1.

Show

if

that,

v(z) satisfies (5.95),

then

t>(z )

=

VQ

,

V'(ZQ)

v(z) satisfies (5.94) if v(z) satisfies (5.95). 2.

Prove

3.

Solve

4.

Solve

by mathematical

(5.99)

w" w"

z*w

+

f

-h 1/(1

zw

=*

z)w' -h

induction.

Taylor series about 2=0. Taylor series about 2=0.

for w(z) in a IP

=

in a

=

Let z a be an isolated singularity of either 6.10. Singular Points. =* a by a circle, C, of radius p, p(z) or q(z) in (5.93), and surround z such that p(z) and g(z) are analytic < an ordinary point of (5.93),

for

<

\z

\ZQ

a|

<

Sec. 5.9 there exists a solution of (5.93)

hood

of z

Wi(z)

= and

which

<

p.

Now

let zo

is

analytic in a neighbor-

ZQ).

about z = zo which may be the point

Wi(z, ZQ)

be

From

The Taylor-series expansion converges up to the nearest singularity

Call the solution Wi(z,

.

a|

p (see Fig. 5.2).

of of

=

a or a point on the circle C. In Fig. 5.2 C' is the circle of convergence. Now let F be any simple closed curve through ZQ lying inside C and surrounding z = a (see Fig. 5.2). We choose a point z\ on F, z\ inside C". Since w>i(z, ZQ) is defined at zi, we can compute Wi(z\, zo), w((zi, ZQ). Since z\ is an ordinary point of (5.93), there exists a unique function Wu(z) such that Wu(z) satisfies = M>I(ZI, ZQ), w(i(zi) = M((ZI, z ). w u (z) is an analytic (5.93) and Wn(zi) p(z)

q(z) 9

z

continuation of Wi(z ZQ). Its domain of definition is the interior of the circle 9

C" (see Fig. 5.2). This process of analytic continuation can be continued, and the reader can show that number of such continuawe can reach z Let w*(z) be

in a finite

tions

.

the analytic function which is the analytic continuation of Wi(z), both Wi(z)

and w*(z) analytic of z

ZQ.

in a

neighborhood

We

write w*(z) == Awi(z) A as an operator associ-

and look upon

ated with analytic continuation. We it as an exercise for the reader to

leave

show

that,

if

Wi(z)

and

w z (z)

FIG. 5.2

are lin-

early independent solutions of (5.93) for some neighborhood of z then Ate?i(z) and Aw 2 (z) are also linearly independent.

Since

z

=

a

is

=

o,

a singular point, we cannot expect necessarily that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

208

now attempt

Let us

SB Wi(z).

hw\(z)

to find a solution w(z) of (5.93)

such that

=

Aw(z)

\w(z)

(5.101)

Wi(z) and w^(z) are linearly independent solutions of (5.93) for a neighborhood of z = 20, we have

If

=

w(z)

The

last

In other words,

A

(5.101)

which

in turn implies

and

+

Cia 12

nontrivial solution in

ci, c 2

n

+

c 2 (a 22

exists

-

Case

is

two equal 1.

Let

=0 =

X)

and only

if

.-

if

(g

X

a quadratic equation in

so possesses

X,

two

distinct

roots.

<f>i(z),

the distinct roots

c 2a 2 i

-

o a 22

X

i2

Equation (5.104)

fact.

(5.102)

X)

roots or

c\Kwi and that

Let the reader deduce this

a linear operator.

is

From

A

=

equation of (5.102) implies that AciWi

Xi,

^> 2 (z) be the functions of (5.101) corresponding to X 2 that is, ,

A^r=

i

X,v?,

=

1,

2

(5.105)

an easy task to show that <f>\ and <p z are linearly independent. ie a = \z Let us note the following: Let z a\e and

It is

(2

The

We

-

analytic continuation of

choose

ay,

j

=

1, 2,

A(*

-

a) a

a)i

a)

(z

so that

=

=

|z

a

-

alV*'

around F increases

e** = Xy(

-

X,,

a)*

j

=

1, 2.

j

-

1,

6

by

2ir

so that

Hence 2

(5.106)

DIFFERENTIAL EQUATIONS

Combining

and

(5.105) A[(z

-

(5.106) yields

a)-^(z)] = A(z

= ^

-

(z

a)~ Fj(z) = (z with a possible singularity at

a

sion theorem

=

z

1,

2

(5.107)

= 1, 2, are single-valued From Laurent's expan-

j

><ft(z),

=

j

a)-><pj(z)

Thus the functions

C

Why?

a)-<A^(*)

- d)^\j9j (z)

= (zinside

209

a.

we have

-

(z

=

a

a)- ^(z)

-

b n (z

n=

a)"

ao 00

-

(z

d)-

=

a *<p<L(z)

-

c n (z

a)

n

or

=

a)

(2

Y

a"

n=

Thus both

(f>\(z)

singularity at z

Case

2.

and

=

^2(2) have, in general, a

is

=

X2

if

a)

n

branch point and an essential

a.

The reader

plex Variable,"

-

c n (2 oo

\i

<p(z)

-

(z

referred to .

-

MacRobert, "Functions

+

a)" [t0,(2)

^

In (2

a)]

where Wi(z) and w^(z) are analytic at z = a. An important case occurs when the functions <0i(z), <p*(z) have no essential singularities. In this case we can write

=

(z

$j(z) is analytic at z

=

*,(*)

where

of a

Com-

In this case

-

a)"fc(*)

j

=

1,

2

(5.109)

of (5.108)

(5.110)

Now

a; #/(a) 5^ 0.

= so that

Thus j

and

-

<f> 2 <f>{'

-T-

(^2^1 /

x

p(z)

^1^2'

+

^1^2)

p(^)(^2^{

=

~

--

= - ^7

=

<Pi^ 2 )

p(z)(<f>*<P\

^w/

i

1

r

2

dW(<pi, j dz

'

N

<pz)

(5.H2) ^ '

ELEMENTS OF PURE AND APPLIED MATHEMATICS

210

Making use

show that

of (5.110) to (5.112), let the reader

=

P

+ P*M

r

(5.113)

with p*(z), g*(z) analytic at

=

=

The same

a.

result occurs for

<p(z)

of

We

(5.109). z

z

notice that p(z) of (5.113) has at most a simple pole at = a. This suggests the q(z) has at most a double pole at z

a and

following definition:

DEFINITION

2

5.1.

=

z

said to be a regular singular point of

is

+

w"

+

f

p(z)w

q(z)w

=

(5.114)

if:

not an ordinary point.

ZQ is

(i)

(ii) (iii)

zo)p(z) is analytic at z

(z

(2

We now

w =

z*u

=

z a u(z),

=

z'

w =

+

[2az

NOW

z

a

z u'(z)

+

az a

2az-

u(z) undefined as yet.

~l

p(z)z*]u'

u,

1

u'

+

a(a

-

-

+ azp(z) + z*q(z)]u = = PQ + piZ + P2Z + = qo + q\z + q*z +

+

[a(a

is

'

P(z)u'(z)

+

Q(z)u(z)

=

(5.116)

chosen as a root of the indicial equation

P(z) Q(z) if

'

^

+

a(a

Conversely,

'

2

Equation (5.115) becomes zu"(z)

(5.115)

1)

2

Zp(z)

provided a

For

then u(z) must satisfy

z 2 q(z)

with

+

u"

z

If w(z) satisfies (5.114),

+

.

a an unknown constant,

we have w' = "

z*u"

2

simplify matters, we assume that z = 0. ZQ transfers the singularity to the origin.

To

regular singular point.

The transformation Let w(z)

=

analytic at 2 = 20find a solution of (5.114) in the neighborhood of a to attempt zo)*q(z) is

-

1)

+

= 2a + zp(z) = 2a = a(pi + P& +

ap Q + q + p + p& + p& 2 ) + 0! + q& +

u(z) satisfies (5.116), then w(z)

=

zu(z)

(5.117)

+

'

'

'

will satisfy (5.114)

provided a satisfies (5.117). The existence of a solution of (5.114) hinges on the existence of a solution of (5.116). If u(z) satisfies (5.116), u(0)

= w

DIFFERENTIAL EQUATIONS

211

then a double integration of (5.116) yields

zu(z)

=

-

[P(0)

and

=

u(z)

With

-

[P(0)

=

B(Z, r)

We leave

+

l]u*

[* Jo

-

+

P(r)

+\j B

I]u

2

-

[2

(z >

+(r -

P(r)

(T

r

-

z)(Q(r)

MO dr

P'(T))]I*(T) dr

*

*

-

*)[$(T)

-

(

P'(r)]

show that, if u(z) satisfies (5.118), then u(z) The reader can also show that if we define u(Q) = UQ

to the reader to

it

satisfies (5.116).

so that (5.118) holds for then u(z) of (5.118) is analytic at z = = w this, the reader must show that lim u(z)

To do

z

=

0.

.

z->0

Now

=

P(0)

of (5.117).

+

2a

If

i

po

=

2,

+

a2

=

then P(0)

s

1

and a\

=

1*00

i -

B(z, r)u(r) dr

/

^ z

1 po if <*\ and a 2 are the roots so that (5.118) becomes

(5.119)

Jo

If we attempt to prove the existence of a solution of (5.119) by Picard's method of successive approximations, we might expect trouble at z = 0. Let our first approximation be UQ(Z) = uo, and define 1

Let the reader show that lim Ui(z) = i/ so that, if we define = 0. Continuing Ui(z) becomes continuous and, indeed, analytic at z ,

this process yields

u n (z) with Mn(O)

= w

n =

,

0,

reader can show that u n (z) write P(r)

We

=

1

+ rf(r)

= 1,

is

-

2,

\B(z, T)|

< R <

2 for

UQ(Z)\

\Ul(z)

.

.

and B(z,

all 2, r

.

=

r)

we

let r

=

te,

I

<

g

1,

-

define 0.

r/(r)

u n (0)

=

Since P(0)

+

(r

-

for all z inside C.

=

z)[Q(r)

-

P(r)

/

+

+ (r

(r

-

-

z)[Q(r)

z)[Q(r)

-

-

and the

,

we can

1,

-

P'(r)].

P'(r)]\ dr

P'(r)]) dr

the variable of integration,

see that \Ui(z)

w

with center at z = such that inside and on the circle. Moreover

=

^

1

=

(5.120)

circle, C,

-rf(r) If

We

.

analytic at z

can certainly pick a small

"

B(z, r)Wn-i(r) dr

\

UQ(Z)\

<

A\z\

A =

constant

it is

easy to

ELEMENTS OF PURE AND APPLIED MATHEMATICS

212 Since

f

Ui(z)\

\u?,(z)

-

T)[UI(T)

\ z Jo

UO(T)] dr

we have

-

AR

ui(z)\

|r|

/

\dr

we obtain

Continuing,

-

\u n (z)

Y

Since the series A\z\

< A

u n -\(z)\

(R/2)

n~ 1

(ir

converges uniformly inside and on C,

n=l

remember R < 2, the sequence {w n (z) converges uniformly to an analytic function u(z). As was done in Sec. 5.9 for z = o an ordinary point, it can be shown that u(z) satisfies (5.116) and (5.115). It is then trivial to show that w(z) = z*u(z) satisfies (5.114). )

1/z,

p(z)

satisfies

l/z and

q(z)

We know

singular point.

w =

+ w'

zw"

5.16.

Example

zp(z)

+

w"

or

=

f

(l/z)w

z 2 q(z)

1,

(1/2)10

so that z

z

that a solution exists in the form

a quadratic equation and u(z)

is

analytic at z

=0.

of determining u(z)

is

to

assume

c r,z n ,

w(z]

is

substituted into zw"

More

-j-

w =

w'

specifically,

-

Y

and the

0,

The

series for

coefficient of

each power of z

is

we have oo

c n (n

+

a)z+

a- 1

Y

w"(z]

+ a)(n -fa-

c n (n

l)z

n +a -*

n=0

n=0 00

and hence w(z)

ways

n=0

no

w'(z)

where a

of the simplest

c n z n+a .

}

w(z)

n=0 equated to zero

a regular

00

}

u(z)

In this case is

z a u(z),

w

One

00

= 0. =

y

^ c n z n+a satisfies

zw"

-f w'

=

10

and only

if

if

n= 00

00

Y

c n (n

+

)(n

- D^^- +

+

The lowest power 1)

-1-

c

a

of z

s

which occurs

c [a(a

1)

+

+

a)z

-

is

z

a~ l

If (5.121) is to

-

1)

+

]

a(a

1) -f

-

The

.

be

c n z n+a

(5.121)

coefficient of z

satisfied,

(indicial)

a~ l

we must have

-

be arbitrary so that a must satisfy the quadratic

-

Y n=0

in (5.121)

].

C [a(a Co is to

c n (n

n+a ~ l

n=0

n=0

coot(a

00

Y

1

equation

is

213

DIFFERENTIAL EQUATIONS or a = 2

m

The

0.

roots of the indicial equation are

~ coefficient of zn+a l

The

0, 0.

is

(5.121)

+-!)+

c n (n

+

a)(n

If (5.121) is to

be

satisfied, c n (n

+

or

+

Cn (n

-

a)

+-!)+

n =

-

(5.122)

is

,

Ci

_ -

-

a)

1, 2, 3,

.

.

.

1, 2,

,

Co

C2

_ -

^

=

n -

_ -

Ci

Co

-

1, 2,

_ -

c3

.

C

2

-

c n -i

...

1, 2, 3,

the recursion formula for the c n n

cn

so that

+

c n (n

(^qf^yi Equation have

-

n

C H -I

we must have

a)(n

cn

.,

a =

.

(5.122)

.

.

.

Since a

.

0,

we

.

_

_

Co

Co

7

(37)2

and by induction

cn

=

Hence a solution

c /(n!) 2 .

= A

w(z)

zw"

+

10

10'

=

is

= A

Co (

n=

We

of

postpone the discussion for finding a second independent solution of

zw"

-\-

w

w'

00

Zz ,

n=

n ,

NOz

converges for

1,

2 2 g(2;)

^

<

|z|

.

This

is

exactly the

(n!)

region of analyticity of zp(z)

=

z.

The reader can find a proof in Copson, " Theory of Functions of a Complex Variable," that, if zp(z) and z q(z) are analytic for \z\ < R, 2

00

then

+

w"

p(z)w'

+

q(z)w

=

has a solution w(z)

= ^

c n z n+a

and the

n= 00

series

c n2 w

N

converges at least for

|z|

<

0.

The reader can

/J.

The

roots of the indicial

n=

eguation need not be equal. Example

5.17.

zw"

+

l)w'

(z

+

w = 00

is

Let w(z)

a regular singular point.

/

1

" 1

Cn^* "",

Cn (n -h

so that

a)^"^- 1

n-0 00

w"(z)

V n-0

c n (n -f

a)(n

+

a

-

l)z

w +a ~ 2 .

Substituting into

easily note that z

ELEMENTS OF PURE AND APPLIED MATHEMATICS

214

zw"

+

-

(z

+w -

l)w'

yields oo

+

c n (n

The

)( n

Y

- 2)z+-i +

+

smallest power of z occurring in (5.123)

c n (n

+

<*

+

l)*

n +a

z*" 1 whose coefficient

is

,

-

(5.123)

must be equated

Thus

to zero.

-

ro(a

-

2)

satisfies the indicial equation a(a We shall see 2) =0, so that a 0, 2. that the larger of these two roots produces no difficulty, the smaller of these two roots doGH produce a difficulty. ~ in (5.123) must he sot equal to zero. This The coefficient of z n+a n = 1, 2,

and a

l

.

t

.

.

,

leads to the recursion formula

+

c(n C

If

try a = 0, we have 2 and obtain r n =

Ct/4

=

Co/4!,

+

a

-

2)

+

=-n^~2

we

a

)(n

cn

2)

that

and by induction

c\

rw

=

(

and

C2

c

r2

,

+

a solution of zw" It is

not very

difficult to see

becomes meaningless.

=

n/2 =

We

try

r /2,

Hence

Az*e~*

0.

I)t0' H- to

(z

-

)

n I) (c /w!).

(-!). = is

+

"-1,2,3,...

c n ~\/(n

r w _i/n, so

r n -i(n

that a second solution cannot be obtained

by the method of series solution used in Examples 5.16 and 5.17 when the roots of the indicial equation differ by an integer. When the roots do differ by an integer the larger of the two roots presents no difficulty. ao

Let w(z)

= y

c n z n+a

be a solution of (5.114).

Then

n=0 00

w'(z)

=

00

c

(

n

+

<x)*

n+ ~ l

w"(z)

= y

c n (n

+ a)(n +

a

-

l)z

n+a ~*

n=0

and substituting into c n (n

n^O

+

a)(n

+

a

(5.114) yields

l)*+*~

2

_|_

pfy n^O

+

q(z)

cn

2^ =o

2+ =

(5.124)

DIFFERENTIAL EQUATIONS

=

Since zp(z)

+ piz +

po

+

p&*

=

2

-

z q(z)

,

215

+ q& + q& + 2

q

-

,

we have

+

[c n (n

-!)+ cn (n +

+

a)(n

+

Pi

Cn(?0

+

a)(p

+

+

)

O^-

$!*+

=

2

Of necessity a(a

The

1)

+-

coefficient of 2 n

+

c n [(n

a)(n

+

-

a

+ 2

<?o

must be

+

1)

+

ap

(n

+

=

indicial

which implies

zero,

+

a)p

equation

go]

n-l

+

c.[(

1, 2,

.

.

.

= some

(n

+

=

i

+ a)(n + + a)(n +

If

integer n.

we have a

1

i

QJ,

Since a ai

,

will

1

(5.125)

determine

cw

,

a a.

-

1) 1

+ (n + a)p<> + + p + ^o =

=

q

)

^

are the roots of the indicial equation, a ?o, so that

i,

ai =

po,

= =

F(a, n)

a

cn

n ^

in terms of Co unless

,

F(a, n) zz (n

for

formula for the

(5.125), the recursion

Equation

n =

+ g-J

a)p n -.

(n

+ a) +a

<*i)

(ft

n(n

i)

^ ai, F(a, n) 5^ f or n ^ 1. The only difficulty would occur if = negative integer. To obtain a second solution when the roots

of the indicial equation differ

by an

integer,

we turn

method

to the

of

Frobenius.

Method of Frobenius. in terms of Co and

...

We

use (5.125) to determine ci, 02, specific value of a is not inserted. .

.

.

,

The

a.

cn ,

The

series

10(2,

=

)

c

(

a ) 2"+

(

5 126 ) -

n-0

determined from (5.125) does not satisfy (5.114) but d*w(z, a)

^

+

,

,

dw(z. a) dz

x

P(z)

+ ,

/

x

/

?(*)(,

x

)

= where a\ and

2

satisfies

C (a

-

i)(a

-

a 2 )2 a

are the roots of the indicial equation.

(5.126) are well defined for the larger of the

two

~2

The

roots, say, a\.

(5.127) c n (a) of

Now let

ELEMENTS OF PURE AND APPLIED MATHEMATICS

216

= A (a

Co

z,

a)

so that (5.127) becomes

<* 2 )

,

,

N

to

dw(z, a) dz

= A(a The change

of c

to

A (a

since these coefficients

by A (a

replace Co

Since a and

z

=

Evaluating at a f

_JfM

2

yields

_|_

p(%)

(

Equation (5.129) shows that

Cl

'

(1 -f a)

2

i

ci, c 2 ,

.

.

.

it is

2,

we have

+

(

(

) J

is

=

q(z) (

(5.129)

)

a solution of (5.114).

a^a-i

Returning to Example 5.16, we have

5.18.

Co

~

If

)

\OQL /

Example

=

(5.128)

-

*

_

atfz~*

not necessary to differentiate (5.128) with respect to a.

We

2 ).

-

also to the coefficients

are independent variables,

_

il

a 2 ) applies depend on Co.

ax)(a

C2

~

Ci

= (2

+ a)

Co

2

(1 -f

2

a) (2

+ a)

'

'

'

2>

'

Co

Cn

'

2

a) (2 "+)""

(1 -f

.

."

2 (n -f a)

.

so that *,

a)

+ ^ To compute (T

J

we need

AT da 1(1 Let y

-

(1

+

a)-*(2 -f a)-*

In y

- -2

Y

j

to

?! (1 H-

2 )

+

(2

compute i

+ a) -

In (*

2 )

2

(2

+

a)

2

(n

+

1 2 a) J

2 (n -f <*)~ so that

+

)

and

-

&=

y

dot

-5

(n

+ + a) ^ 2

.

.

.1

J

DIFFERENTIAL EQUATIONS

217

1

where F(n)

n=0

The second

solution of

Cl

* ~

_ ~ ~

Co C2

=

"n

We

replace TO

by Aa and ^

a

Z H

r

)

\OCK /

a0

>

we need

AT -

-

(a

(a

+

n

-

+

-"!)(

+

1)-K

l)a(a

+

1)'

2)

y3 -__-__ 7

.

l)'

1

.

.

-|-

1)

+2) ...

(^=T)Ti "+"!)(

to

+

l)(a

(a

(a"

+ ^^2) +

compute l

da L( Let y

-

...

1)

T 1

a

1

(

(

C2

obtain

+ ~ 1)n

To compute

we have

_

3

'

[y2 a

5.17,

Co

-

- !)( +

(a

is

Example

_ ~

Ci

w =

w'

Referring to

5.19.

Example

+

zw"

!)(

+

^+

2)"

1

-

-

2)

+

(a

...(+ n

-

n

-

w , 2

1

2)J a .

2)-

1 ,

so that

n-2

-

In y

In

-

(

1)

-

Y

In (a

+ k)

n-2

n-2

dy

_ If

n =

3,

(

^

=0.

)

kj

(n

-2)1

Hence

n-2 w =4

The second

solution of zw" -f

w(z)

where F(n)

-(-!)"

- 5

2

w + w = f

(z

6-* In 2

1)

-

1

-

z

+

is

z*

-

z*

'

'

J

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

218

Problems 1.

Derive (5.113).

2.

Show

8.

Derive (5.125). w Solve zw"

4.

that

u(z) satisfies (5.118) then u(z) satisfies (5.116).

if

0.

S N-0

zn+l

+

nl(n

1)!

00

z n+l

[V_ ln Z

2

+

2, nl(n

n-O

1

1!2!

2

. JL

( 2!3!\1

5.

Solve

2 2 (1

z)w"

w

Zz)w'

2(1

-f-

-

for

in

10(2)

+

2 2

+n ^ + ^3/

.

.

1 ]

the neighborhood of

2=0. 6. 7.

Solve z(l Solve 4 8 ip

/'

z)w" + 4zw

+ z)w H- w + l)w f

(1 f

for 10(2) in the

J

for

(

in the

if;(2)

neighborhood of neighborhood of 2

2

0. 0.

To 5.11. The Point at Infinity and the Hypergeometric Function, determine whether the point at infinity is an ordinary or regular singular = l/t and investigate the point t = 0. We have point, we let z

dw _ ~ dwdt^ __ ~ ~

=

/4

az*

so that

w"

+

p(z)w'

Hence

=

oo

is

z

+

.

llidz

Ih

q(z)w

_^ dt 2

=

2

dw ~di

+

2p dt

becomes

an ordinary "point

if

2i (i)

is

analytic alytic

at

t

=

is

analytic

at

/

=

(5.131) (ii) (i

For

-4

q

this case a solution

Again z = point and if

oo

is

(

J

can be found in the form

a regular singular point

if

z

oo

is

not an ordinary

DIFFERENTIAL EQUATIONS

219

~

2t

is

(iii)

at

analytic

/

(5 132) '

1 (iv) 72 *

l\ # \7 I

is

v/

at

analytic

t

=

This implies

/A

1

MO-

+

Pit

+

+

pzt*

'

'

othat

moreover, we desire that the origin also be a regular singular point

If,

=

we must have

p(z)

5.20.

We

Example

po/z,

<?(z)

=

2 .

<?o/z

look for the differential equation (of second order) which has In this case z * is an ordinary point so z = 0.

but one regular singular point at that

n(*> - 2

and

-

_

g

+

.

.

.

VW-S+S+--In order that PO

Hence

g(z)

=

2

Pi

s

0,

= =

be a regular singular point, we must have '

and

=

'

p(z)

=

P

=

-

00

-

01

=

'

-

tfn

--

2/2, so that

now the following problem: We look for the differential which has exactly three regular singular points at z = 0, 1, equation all other points are to be ordinary points. Since p(z) has at most simple = = z we know that at z z(l poles 0, z)p(z) is an entire function. 1,

We

consider

;

Hence 2(1

-

2) P (2)

-

"

(5.133)

n-0 for all

z.

Since z

must be upheld.

= From

is

to be a regular singular point, condition

(5.133)

(iii)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

220

(0 x

y

must be analytic

at

MO; This

0.

t

Thus

=

p(z)

With the same type

is

if

possible

=

2)p(z)

z(l

and

-

t

-

#

+

a

+

and only

if

an

=

0,

>

1.

ai-s

r-A_ l

n

(5.134)

<s

of reasoning the reader can

show that

Let the roots of the indicial equation at z = <*> be a = a, a == 6. and also impose the condition that the indicial equations at z = 2 = 1 have at least one root equal to zero. Now at z = we have = lim zp(z) = c from (5.134). Also q<> = lim z 2 q(z) = pvB. The 7>o

We

*-o

s-o

=

indicial equation at 2

is

-

a(a

a

If

=

is

we pick

\i

if

z

= 2t

oo

-

+ \ivB =

we must have iivB so that desire q(z) The other root is a = 1 0. c. Let = 1 that, if one of the roots of the indicial equation at z we

zero, the other is

At

ca

to be a root of this quadratic,

=

the reader show is

+

1)

^

=

a

+A

1

and

M - i^i)

we have

-

2t

p(l/t)

ct

-

tA/(t

-

(5 - 136)

2

1)

-

c

-

t*

and

po

=

2t

lim

=

~

If

=

2

-

-

1)

a

2

+ +

(2 (1

-

+ c + c

the roots of this equation are a and

a

c

+A

I

B, and the indicial equation at

a(a or

t

t

t-^Q

Similarly g

A/(t

+

b

-1

6,

A)a A)a

z

=

oo

is

+B = +B =

we must have

+c-

A

-

1)

DIFFERENTIAL EQUATIONS

=

ab

A=a + b +

B, so that

l

l-c + a +

d?w (c

dz*

z

\z

Equation (5.137)

Our

c.

221

differential equation is

b\ dw

ab

/ dz

1

_

z(z

v

1)

the famous hypergeometric differential equation.

is

form

Its solution is written in the

(after Paperitz) oo

( I

= P

w(z)

\

1

a 6

c

\1

z\

a

c

b

(5.138)

}

The top row denotes the

regular singular points, and the other rows contain the roots of the indicial equations at each singular point. Note that the sum of the roots is unity.

The operator 6 = by

The

methods.

series

due to Boole

z-r-

is

very useful in solving (5.137)

9 has the

reader can show that

following

properties,

e n (z<f>(z)) =

+

z a (e

a)v

n

where 6V =

B(O^),

etc.,

)V = /^

+

and (0

r

The

n~ r

BV.

=

reader can show that (5.137) can be written in the form

6(6

We

{) a

+c-

solve (5.137) or (5.140)

l)w

by

=

Qw =

>

c*6z*+

Cfc*+

+

V)w

= 7

a) (8

+

b)w

(5.140)

a

=

*=o (6

+

Let

series.

W = Then

z(6

c*(fc

+

a)z

k+

fc=o

+

(fc

+

a

+

ck (k

+

a

+ b)(k +

c*(fc

+ a + b)(k +

c

fc

b)z

k

ffo 00

9(0

+

6)w

=

Y

a)z

k

+

*% 00

z(B

+

a) (6

+

b)w

= y

a

+

a)2^

+1

(5.141)

ELEMENTS OP PURE AND APPLIED MATHEMATICS

222 Also

+c-

9(9

=

l)w

c*(k

+

+

<**)(k

+c-

a

l)z

k+a

(5.142)

ib-O

Equating (5.141) and c*+i(/c

+

+

a

we obtain the

(5.142),

+

l)(fc

+

a

c)

=

recursion formula

+

c k (k

a

+

a)(k

+

a

+

b)

along with the indicial equation

+c-

a(a

For a =

+

(k

^i

1)

=

we have

x

so that

Ci

=

a)(k

+

b)

,

Co

^

1

c

(a

+

+

1)(6

C2

2(c

+

(a

1)

+1)

l)q(6

(c

+

+

1)6 Co 2!

l)c

and, in general, (

+^~

q

*

(c

_

4

r(o ~

+

gamma,

We

*

n

+

+

n)

a(b

+

n

- TT 7 7 " 7 ^ "

b C

1 )

n) _^ w!

or factorial, function of Chap.

- A ~

n)F(6 + V r(a +T(fTnT~

6; c; z)

r(c

+

by the equation

+ )

Problems

+

=

for 10(2) in the

2.

Solve Derive (5.135).

8*

Let w(z) be a solution of Legendre's

1.

formal

T (

^

n)T(b

,

2w'

A

n) 2n

Li n-O

cannot be negative integers. Why? define the hypergeometric function F(a,

w?"

4.

is

w(z) W(Z}

a, 6, c

+

n)r(b

T(c

where F(a) is the solution of (5.137)

*

1)

Cn

neighborhood of differential

z

~

equation

0.

n) i"

^

,

K

.

AA .

(5J44)

223

DIFFERENTIAL EQUATIONS Show

that

-1

oo

- P (1 {0

w(z)

Verify (5.139).

6.

If c is

-f

z\

1

-n

(O 4.

I

n

|

not an integer, show that a second solution of (5.137)

/

~ ,-c

x

w(z)

+

T(n

\

1

+

n-0 6.

What

is

7.

Show

that

a

-

T(n

c)r(w 42 - c)

1

+

b; 1; ,) In

n

V /

Li

/ [

=

_-- _--,

4-

7

n

-f

7

b

1

+ ;

is

7

n

l

2\

1

1

;

\a

of (5.137)

,

r

sr

c) z

the interval of convergence of (5.143)?

r(a)r(b)F(a,

,

-

n\

by expanding the left-hand sides in Taylor-series expansions. 8. Show that when c = 1 the second independent solution

where

6

+

is

I

n)

1

n=*l

5.12. The Confluence of Singularities. Laguerre Polynomials. In the discussion of the hypergeometric function and its associated differwere distinct. It ential equation the regular singular points 0, 1, when consider be useful to two or more what happens turns out to

regular singular points approach coincidence, a process usually referred " " The reader is referred to Chap. to as the confluence of singularities.

XX of Ince's text on differential equations for more details and for a very useful classification of linear second-order differential equations according to the number, nature, and genesis of their singularities by confluence.

We consider Example 5.21 Example

5.21.

Rummer's

by way

of illustration.

w(z)

-

*Fi(a, b; c; z)

+

1)

\z\

<

1,

which

'

'

(o c(c

n0 for

-

Consider the hyper-

Confluent Hypergeometric Function.

geometric function

+ * - 1)W + 1) + 1) (c 4- n -

-

-

(6

1)

+n-

1) af

n\

,

fi

(5 '

satisfies

2(1

-

z)w"

+

[c

-

(a -f b -f l)z]w'

- abw -

(5.146)

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

224

Now

let z'

fa, b;

=

c;

bz,

and then replace

n

+

= for

<

\z\

-

*

(l let 6

>

oo

+

(q

1)

-

n

obtain

,

we

+ 1) +n -

-

l)b(b

c(c -f 1)

which converges

we

We

z.

g fc

If

by

z'

(c

+

(6

n

-

,.-.-

1) g

(6 147) *

SI

1)6

2^1 satisfies

6.

+ +

a

-

Q w" + (c

1

tf

-

aw -

(5.148)

g)

fr

obtain (formally) the series

V

F ,. A _ fi(a,c,*) .

c(c

+ 1) + 1}

1.

Moreover

q(q

n=

which certainly converges

for

|z|

<

+ n - 1) 2" + w . 1} -,

(fl .

.

.

(c

,. (

can be shown that

it

,

14Q ) 5 149 -

iFi(a; c; 2)

satisfies

zw"

+

-

(c

- aw -

z)w'

(5.150)

oo obtained formally by letting fr Equation (5.150) is Kummer's confluent hyperis an geometric equation. It has a regular singular point at z = 0, but z = irregular point, that is, z = oo is neither an ordinary point nor a regular singular point. If c is not an integer, the other solution of (5.150) is .

zl

~f

iFi(a

-

c

+

I

;

-

2

(5.151)

r; z)

If r is a negative integer or zero, (5.149) becomes meaningless, while if c is a positive If c = 1, both solutions coininteger greater than 1, (5.151) becomes meaningless. The second solution can then be obtained by the method of Frobenius. cide.

We

+

now consider the function iFi(a; c tion of iFi(a; c 1; z) will terminate if a the coefficient of z m+1 is

+

(c

+

l)(c

+ 2)

1

is

;

z).

a polynomial of degree n co

(c

+

D(c

+

if

2)

n

series representa-

w, for

-

(c

Similarly higher powers of z have zero coefficients. 2) is

The

a negative integer,

is

Hence

a positive integer. (c

+ n) iFi{

_n

.

c

n, c

iFi(

We

+

+

1

;

define

1; g)

/C" I

as the associated Laguerre polynomial, can show that

n a

positive integer.

p + fr.M-oSa+.vw..

The reader

(6 153 ) .

DIFFERENTIAL EQUATIONS

We show now

we can

that

225

write n

(5.154)

(e-*z+<)

Let

v(z)

=

e-*z n+c so

that

~

dv j-

z

+

n

Differentiating

1

=

+ c)zn+c~ -

=

/

z n+c ]

l

<r\(n

and

\

i

(n -f c

z)v

times with respect to

z yields

n

If

we

let

d v u = T~ we have >

l

Now it is

let

w(z)

= e^~ w(g) = c

easy to see that w(z)

e

z

"

dn z~

-^

(n

+

z n+c ). (e~ z

1}l"

If

n

=

is

(5 155) *

a positive integer,

From

a polynomial of degree n.

is

=

u(z)

and the

+

C)

w(z)e~

z

z

c

fact that u(z) satisfies (5.155) let the reader

show that

satisfies

+ Since (5.156)

is

exactly the

Lf(z) Let

z

=

0.

From

+

(c

l

same

"

+

0)

as (5.153),

= Kw(z) =

Ke*z~ c

nw =

we must have

~

(e~*z

(5.152) the reader can easily

norm -

+

(c

1)(c

+

2) -

+

D(c

(5>156)

n+c )

show that

-

(*

+ n) ~

Let the reader show also that

=

e lz~ 2

so that

K

=

=

(c

1

n\

This proves (5.154).

+

2)

(c

+

n)

ELEMENTS OP PURE AND APPLIED MATHEMATICS

226

The

associated Laguerre polynomials have the interesting and impor-

tant property

=

e-*x cL%(x)U*(x) dx

To prove

m^

n

(5.157)

m<

n

(5.158)

we need only show that

(5.157),

Dn

er xx cx m

f*

c)

(x)

dx

=

f or

then (5.157)

is valid, for L*(x) is a polynomial of degree a (5.157) simply linear sum of integrals of the type (5.158). be obtained by integrating by parts m times after can Equation (5.158)

If (5.158) holds,

m<

n and

replacing

is

U(x)

in (5.158)

by

(5.154).

Problems

2.

Show Show

3.

Show that 14-

4.

Prove

6.

Show

1.

that iFi(a; that

c; z) satisfies (5.150).

L nc) (*) (

satisfies (5.153).

(0)

-

(5.158).

that

dx

r(c)

_

T(c

+n +

1)

/. Hint: Evaluate 6.

Show

n

f

e~*x x L% /./_

)

(x) dx.

that }

(2n

n>L^ (z)

+

c

1

-

zJL^^z)

(n

-\-

c

n ^ 2

W 0)

^

7.

is

dx

-

WW -LSIM

called the ordinary Laguerre polynomial.

Show

that

L(z) r-O

5.13. Laplace's

Equation.

We

discuss

This equation

Laplace's equation. matical physics. We consider

V2 F =

is

of

now a method for solving some importance in mathe-

in rectangular coordinates,

(5.159) V/.C-

VXfc>

^^i/

Let us look for a solution in the form V(x,

y, z)

=

X(x)Y(y)Z(z)

(5.160)

DIFFERENTIAL EQUATIONS

227

This attempt at finding a solution of (5.159) is called the method of separation of variables. Applying (5.160) to (5.159) yields

^=

x

For

V

we have upon

7*

1

d*X

X

dx*

1

+ Y

ld*X

by V

d2 F dy*

dz*

ld*Y

==

ld*Z (

equality in (5.161) cannot hold unless both sides of (5.161) are con-

starit, for if

u(x)

=

flu

v(y, z),

=

then

~ V-J-T = X ax 2

or

and u

=

.

.

constant

SE

,

constant.

Hence

M

A-

k*X =

jj^

The

d*Z

1

+Z

-J^ YW + Zd^

Thus

The

division

(5.162)

solutions of (5.162) are

X X X

or or

= A k cos kx + B k sin = A k ek * + B k e~kx = Ax + B if k =

kx (5.163)

Similarly

Y = Ci cos ly + DI sin Y = Cic^ + jDic-^ = F * Cy + D if

and or or

/?/

(5.164)

Z

Finally

^f + +

or

(

fc

2

T

/

2

)Z

=

2 solution of (5.165) depends on the magnitudes of k 2 2 signs preceding k and Z

The

(5,165)

and

I

2

and the

.

The

choice of the sign preceding the constants depends on the nature

of the physical

Example

5.22.

problem.

We

illustrate

with Example 5.22.

We consider a two0. For steady-state heat flow we have V*T The edge given by x 0, Q y < ois is the edge given by a: a, # < co.

dimensional semi-infinite slab of width o. 0, as kept at constant temperature T

ELEMENTS OF PURE AND APPLIED MATHEMATICS

228

The base of the slab given by y ^ x ^ a is kept 0, For the steady-state case we have by T /(x). d*T

,d*T_ ~

a? If

we

-

let 2T

^"(a:)F(j/),

we obtain

as above

ld*Y we have X T when

we choose our constant

to be negative, ary condition necessitates, however, that If

X

This cannot be achieved for we choose

n

"ay*

d*X

1

at a steady temperature given

Ake kx

Ake kr + Bke~ kx x = and when x .

BkC~ kx unless Ak

-j-

Bk

0.

Our bound-

=

a for

all y.

However,

if

_-= X dx*

X

= A k cos for -f #* sin kx. Since sin (rnrx/a) vanishes for x then provided n is an integer, it seems proper to consider

=

and x

-

v = Bn n sin nirx X a .

where

=

fc

rwr/a.

It follows that

d*Y

_

nV *

Y = Cn e (nir/a)v + D n e~ (nv/a}v

so that

infinite as

t/

becomes

infinite

Hence

;

so

~

U

a*

dy*

We

.

do not expect the temperature to become

we choose Cn

*

0.

= B n D*e-<*'**

r(or, y)

~

sin

(5.166)

The final boundary condition is T(x, 0) =/(z). Equation is an integer. (5.166) cannot, in general, satisfy this boundary condition, unless, by choice, we pick V*T is a linear partial differential equation, and it is sin (nirx /a). (x)

where n

A

Now

an easy thing

to

show that 00

T(x, y)

is

also a solution of

twice term

by

term.

V 2 !T To

T A ^ n=0

=

n a)v sin n e-< */

(5.167)

a

provided the series converges and can be differentiated boundary condition T(x, 0) f(x), we need

satisfy the

00

f(x)

Y

=

^sin

(5.168)

a

L*

n=0

Can the infinite set of constants This

is

An

,

n -

0, 1, 2,

.

.

.

,

be found so that (5.168) holds?

the subject of Fourier series and will be discussed in Chap.

6.

DIFFERENTIAL EQUATIONS Problem

=

for y

0,

y

Problem

-

Find a solution of V 2 F

1.

6,

F the

By

2.

for z

-f

method

Problem

We the

Do

3.

method

0,

x

-

o,

V -

67

a*F "*"

ay

2

a*

X(x)Y(y)T(t).

the same for

now

find

equation

-

-

x

of separation of variables find a solution of

dx 2 y,

for

.

d*V

Assume 7(s,

229

V -

such that

^

+ ~^ as 2

ay

- 42 ^?c a* 2

2

V2 F =

a solution of

in spherical coordinates

by use of In spherical coordinates Laplace's

of separation of variables.

is

'

3V I

Assuming

F(r,

=

0, <p)

sin 2

d (

p

TT~ \ \

2 '

_

.4-4-

JB(r)9(0)^(^),

d#\ "^7~ )

sin

.

_

R

ait, ,

I

/I

-

J. -L-

I

___

we obtain upon / sm

rf

c?6\

_

.

~l

^

I

twv

\

/

F

by

1

=

)

division

*tr

J

Let the reader show that of necessity

=

T-J

Now

physically

we expect and V(r,

constant

desire that

B, <p)

=

7(r,

0,

?

+

2r)

Let the reader show that this implies that the constant be chosen as the square of an integer, that is,

Thus

$>

= An 1

fi

and

cos n<p

+ B n sin n<p.

d ( 2 dfl\ V*

5

^J

= "

Equation (5.169) now becomes 1

eSO

d { sm 3 V

.

.

'

de\ 3 j

+ ,

n2

/c (

SETS

this implies that 1

dR\ = co^tant = ^TJ

d ( 2, r

.

RTr(

.

.

fc

r^ + 2r^-*.0 To

5

solve (5.171),

m yields r [m(m

-

we assume 1)

+

2m -

(5.171)

a solution in the form

fc]

=

0.

Hence rm

is

=

rm

This a solution of (5,171) JB(r)

.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

230

+

m(m

provided k = m(m

+

=

1)

(m+1) is a solution of (5.171) for Similarly rmost general solution of (5.171) for k = m(m 1)

+

The

1).

fc.

is

R(r)

= Cm r-

+ D m r-<+

(5.172)

We

need not have guessed at a solution of (5.171), for it is immediately The series r = is a regular singular point of (5.171). method of Sec. 5.10 could be used to obtain (5.172). Equation (5.170) obvious that

now becomes

~

sin 6

6

^J +

=

sin

(sin If

we

let

=

fji

cos

0,

dp

~ -

(1

M

-

2

)

M2)

+

[m(m

1) sin

d6, (5.173)

+

:

2^0 +

2

- n

6

2

]9

=

(5.173)

can be written

[m(m

[m(m

+

Equation (5.174) is called the associated Its solution in Paperitz's notation is

-

1)

9 -

(5.174)

J-5L-J

Legendre

oo

differential equation.

\

1

(5.175)

Problem Problem

4.

Deduce

6.

If

V

is

(5.175).

independent of

(1

-

-

)

<?,

that

is,

=

n

+ m(w

2M

0, (5.174)

H-

becomes

=

1)P

(5.176)

with 6 replaced by P. For m an integer show that the solution of (5.176) in the Also show that 0, written P,(M), is a polynomial of degree m. neighborhood of n

^-0*)P(M) ^M -

The

Pm (/*), m

great deal

V -

more

F(r, 0)

0, 1, 2,

.

.

.

,

for

m

5*

n

are called Legendre polynomials. We shall have a A solution of V 2 for

F

to say about such polynomials in Chap. 6.

is 00

V(r, B)

-

y

(A mr

+

B,,r-<-+'>)P,

(cos 9)

(5.177)

We have seen that the solution of Laplace's equation in spherical coordinates led to Legendre polynomials. The solution of Laplace's equation in cylindrical coordinates yields the Bessel function. In cylindrical coordinates

V2 F

becomes

DIFFERENTIAL EQUATIONS

231

3*F = . Q

W Assuming

V(r,

6, z)

=

(5 178) '

R(r)Q(B)Z(z) yields

_

Rdr\ If

we

desire F(r,

0, 2;)

=

Z

dr

+

F(r,

2

we need an integer

y

d02-

= A,

B(^)

^+B

cos

(5179) *' l

9

dz*

2ir, z),

y

9 so that

,_

,_

sin

v

^

(5.180)

Similarly

d *z

1

TT ~

7

^2 /

7 ^ rea * or a P ure imaginary i

so that

If

is real,

fc

we

This

let z

is

=

+

and

if

exponential,

Equation

trigonometric.

If

is

Z(^)

Z() = C**-

/?

fcr,

=

k

is

(5.181)

a pure imaginary, Z(z)

is

now becomes

(5.179)

w, (5.182)

D*e-*'

becomes

Bessel's differential equation.

We

see that z

=

is

a regular

singular point.

Problem Problem z

1

-

4s,

6.

Solve (5.182) for k

7.

Solve

6

x

-T-,

for

(5.183)

and write

and

v

the cases

an integer. v

=

integer,

(5.183) as

(02

-J-,,2)

w + xw a

oe

Let

10

=

x*

^

o rxr

,

and show that

Wi(x)

Ax v/

(~x)

'

r-O If p is

not an integer, a second solution

is

r

v

^

integer.

Hint: Let

ELEMENTS OF PURE AND APPLIED MATHEMATICS

232

We define

An

the Bessel function of the

kind of order

first

important recurrence formula for the J r (z) ,

To prove

(5.185),

=

+l (z)

v

to be

is

(5.185)

J,(z)

we note that

/A"

V (-l)-(s/2)*

1

^

LI r!l>

r-O

+

r)

rvLI

(-: r\T(v

+r+

2)

oo

\

Z / 4j

\

r-O

r=0

V

+

r=l

r"+

\)

+r+

1)

1

'

^2/

Ll

IT(v)

r\T(i>

4

\

However, ao

_ -

?! j (g) Jr(Z) z

2j; /*v

z

W

r

i

lT(v

+

+ 1)

^ r

2J Equation (5.185) Problem

8.

is

Show

L

V Li

(-2 r\T(v

2

/4)--

+r+

1)

v

4.

r-1

seen to hold since v/T(v

+

1)

=

that /,_,(*)

- J,(g) -

2J't (z)

(5.186)

DIFFERENTIAL EQUATIONS Problem

From

9.

(5.185)

and

(5.186)

-J'v (z) =

\J v (z) Problem

10.

Show

that

\i

(2 j Problem

Show

11.

(' - 0)

that,

if

a

7*

JoV,(rf)J,(0)

ft,

dl

2a 2 f*t[J,(at)]*dt

Problem

From

12.

v

= -

> X

sinz

1,

[j

v (ax)

(* -

^

2

J(ft)

0,

a

5^

^

j8,

>

)[/,(ax)]

2

-f

The

Hermite Polynomials.

-

Also show that for this case

1.

f*t[J,(at)]*dt

5.14.

^^1 _ /

the results of Prob. 11 show that

tf,(at)J,(0t) dt

provided J(a)

233

show that

=[/v+i()P differential equation

$ -,* + arises in

quantum mechanics

harmonic

oscillator.

We

(5.187)

in connection with the one-dimensional

shall see that the

Hermite polynomials defined

by

Hn satisfy (5.187) for

verify that

Hn

(t] is

=

2

a polynomial of degree n.

= n^B _i(<) 1.

Show

(5 188) .

(~l)VV2^g!_

any nonnegative integer

so that

Problem

(t)

n.

The reader can easily (5.188) we have

From

(5.189)

that (5.190)

ELEMENTS OP PURE AND APPLIED MATHEMATICS

234 Problem

From

2.

From

(5.189)

(5.189), (5.190)

show that

~? -

t

f + nH n

0.

we have

and, in general,

dr

Hn

(f)

=

dP

-

f

n(n

1N 1)

,

(w

-

=_H_

and

From

+ ,

r

(0

(5.191)

Taylor's expansion theorem arid using (5.191)

we

obtain

^n- 2 C)"~We now

look for a generating function

<p(z, i)

in the sense that

00

v(z,

If v?(^,

t)

=

//(/)

(5.193)

can be found, then the Taylor-series expansion of <p(z, Hermite polynomials. If <p(z, f) does

of z will yield the

powers then formally we have

assuming that one can differentiate term by term.

n=l

From

(5.189)

t)

in

exist,

we have

n-1

Integration yields ^>(z, t)

where

M(Z) is

=

an arbitrary function of integration. I

_|_

Thus

p'(z)]

(5.194)

DIFFERENTIAL EQUATIONS

From

(5.193)

235

we have formally

n=0

n=l

V

~

Li

"

z m JH

if //B+2(0

~

so that

=

~

tJ/ " +2(

5

tHn+l(i)]

w-0 00

-

+ !)#) rr

<n

n-0 eta

= Making use

Now

/i(z)

-

(p

of (5.194) yields

+

M "(z) Since z and

2

t

=

2

[M'(*)l

are independent,

+ zt =

+

(^

we

find that p(z)

+

The

2 2 /2 satisfies (5.195).

^(2,

/)

f

f)fjL

s

(z)

-1

must

(5.195)

satisfy //

+

=

z

0.

function

<-/*+*

(5.196)

H

There is no can be shown to be the generating function for the n (t). trouble in differentiating the series expansion of <p(z, t) term by term since <p(z, t) is analytic for all 2, t in the complex domain. Finally,

we show

that JS

Jjt

t}dt

We

if

= n\

m

consider

1=1J

-*>

e-^H^HnW dt /7mp-V

^C-D-H-W^r/* Integration

by parts

jdelds

f^oo

(-l)-i n

/

J

tO

Hn -i(t)~^^-dt wW

7*

n

ifm = n

rf<

Why?

\

(

5 197 ) '

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

236

Further integration by parts yields

If

m>

n, let

the reader deduce that 7

/

=

=

0.

/*^ dt If m = n,

n!

REFERENCES "

Differential Equations," McGraw-Hill Book Company, Inc., New York, 1942. " Partial Differential Equations of Mathematical Physics," Dover Bateman, H.:

Agnew, R.

P.:

Publications, New York, 1944. " Cohen, A.: Differential Equations," D. C. Heath and Company, Boston, 1933. Goursat, E.: "Differential Equations," Gmn & Company, Boston, 1917. Ince, E. L.: "Ordinary Differential Equations," Dover Publications, New York, 1944. Kells, L. Inc.,

M.: "Elementary Differential Equations," McGraw-Hill Book Company,

New

York, 1947.

Miller, F. H.: "Partial Differential Equations,"

John Wiley

&

Sons, Inc.,

New

York,

1941.

Petrovsky,

I.

G.: "Lectures on Partial Differential Equations," Interscience PubNew York, 1954.

lishers, Inc.,

CHAPTER 6

ORTHOGONAL POLYNOMIALS, FOURIER AND FOURIER INTEGRALS

SERIES,

A function f(x) defined over the be thought of as an infinite dimensional vector. = x Let g(x) be defined also over f(xo) is the component of f(x) at x In vector analysis one obtained the scalar, dot, or the interval (a, 6). inner product of two vectors in a cartesian coordinate system by multiplying corresponding components and summing. Obviously we cannot sumf(x)g(x) for all x, a ^ x ^ b. The next best thing is to define 6.1.

Orthogonality of Functions.

interval

(a,

b)

may

.

S(f,g) as the scalar product of /

orthogonal on the range

and

=

=

If S(f, g)

g.

(a, b).

0,

Generalizing,

r&

/ Ja

dx

f' f(x)g(x)

(6.1)

we say that / and

g are

if

=

p(x)f(x)g(x} dx

(6.2)

we say that / and g are orthogonal relative to the weight, or density, function, p(x), on the interval (a, b). In Chap. 5 we noticed that 6.2. Generating Orthogonal Polynomials. important classes of polynomials such as the Legendre, Laguerre, and Hermite polynomials arose in connection with the solutions of various In this section we shall show how to generate differential equations. orthogonal polynomials. The various theorems proved here will apply to every type of polynomial generated. Let (a, b) be any closed interval, p(x)

proceed as follows: a ^ x ^ b ^ >, and

^

oo

let

be any real-valued function satisfying the following conditions: (i) \ / (ii) (iii)

f\ P(X)/

^

>

The moments I

Ja

and are

for a

[' p(x) dx Joe b

exist

We

finite for

for a

n =

x p(x) dx 0,

x

a

< <

b ft

g

b

of p(x), say, ^n, defined

n

n =

< g

1,

2,

density, function. 237

....

0, 1, 2,

We

by

... call p(x)

a weight, or

ELEMENTS OF PURE AND APPLIED MATHEMATICS

238

We now

proceed to construct a sequence, or family, of polynomials,

P O (X) s

/MZ),

i,

/>,(*),

.

.

.

p n ( X ),

,

.

.

.

such that

P(X) =

+

Z"

(TnX"-

1

+

'

(6.4)

and such that b

f yo

Condition

P (x)P m (x}P n (x) dx

that

(6.5) states

P m (x)

and

m

=

P n (x)

n

3*

(6.5)

are orthogonal to each other

P n (x) has its leading mathematical induction to generate have already defined Po(x) = 1. Let Note that

relative to the weight factor p(x). coefficient equal to unity. apply

We

the family

{Pn (x)\.

We

=

Pl(z)

To

fulfill

condition (6.5),

X

+

ffi

we need

+

fp(x)(x

=

<n)dx

(6.6)

The

limits of integration are omitted with the understanding that they remain fixed throughout the discussion. Equation (6.6) yields

[see

(ii)

and

This choice of

of (6.3)].

(iii)

o-i

yields PI(X) orthogonal to

Now assume that we have constructed the sequence of orthogonal Po(x). Pk(x) satisfying (6.5). We know polynomials P (x), PI(X), P (x), 2

that k

is

at least

1

.

.

We now

.

.

,

show that we can extend our constructed

set of polynomials to include the polynomial

P k +\(x)

while preserving

Let

the important orthogonality condition (6.5). k

Pk+1 (x) where

cr

,

r

=

0, 1, 2,

.

.

.

,

=

fc,

x*+'

are

fc

+ +

cr

1

P

(6.7)

r (x)

undetermined constants.

We

desire

Jp(aOP.(x)P*+i(x) dx

With the

aid of (6.7), (6.8)

I However, for

x k +*p(x)

s

^

r

=

a

=

0, 1, 2,

.

.

.

,

k

(6.8)

becomes

p(x)P,(x)P (x) dx dx+^Cr r0 f r

we have fp(x)P(x)P r (x) dx =

=

Q

provided

(6.9)

5

g

fc,

SERIES, FOURIER INTEGRALS

ORTHOGONAL POLYNOMIALS, FOURIER r

g

by our assumption on the

fc,

set Po, PI, Pi,

.

,

P*(#).

239

Hence

of

necessity

~ _

C

fs*

+1

p(x)

dx ) dx

_ - <U>

.

2,

*

.

.

.

,

( (6.10)

The numerator

of c s exists [see (iii) of (6.3)], and the denominator of c from zero and is a linear combination of various moments of We leave this as an exercise for the reader. p(x), each of which exists. It is a trivial procedure to reverse the steps taken to derive the c a s = 0, of (6.10) are substituted into (6.7), one fc; that is, if the c 1, 2, P is shows that k+i (x) orthogonal to P r (#), r = 0, 1, 2, k, easily The principle of finite mathematical induction states relative to p(x).

is

different

,

.

.

.

,

.

that a sequence of polynomials Po(#), Pi(aO, Pz(x)

.

f

.

.

,

.

Pn(%),

.

,

-

-

-

exists satisfying (6.5).

Problem \P n (x}

\

is

1. Prove by mathematical induction that the sequence of polynomials unique relative to the interval (a, b) and the density function p(x).

For the interval

and

(0, 1)

for p(x)

1

the orthogonal polynomials are

We

the well-known Legendre polynomials. For PI(X) = x v^ we have

generate

now

PI(#), P*(x).

+

""5 """T"^" Jo so that l\(x)

=

x

-

T.et

|.

l\(x)

P we have

x 2 dx

/

+

do

!

we have Thus

2

f

P,(a:)

(x

=

x

^)jc 2

-

(x

dx

-

+

i)

x*

Pi dx

dx

/

=

+

a^Pi

=

1

-

i

I

=

2

(x 2 a-

)

-

A P

.

From

=

so that ao

a\

+

x

dx

-J.

=

0,

From

so that ai

=

1.

+I

Problem 2. For the interval (0, o) and for p(x] == e~* generate Li(x) and These are the Laguerre polynomials. Problem 3. Construct the Hermite polynomials HI(X), Hi(x) for the interval (-00,

oo),

An interesting result concerning orthogonal polynomials is the followThere exists a unique set ing: Let Q(z) be any polynomial of degree n.

ELEMENTS OF PUBE AND APPLIED MATHEMATICS

240

of constants

.

CD, Ci,

.

.

,

such that

cn

n

=

Q(X)

J C P (x) r

*

Cn

r

The constants depend, of course, on the family of polynomials under The proof is by induction. If Q(x) is of degree zero, say = then Q(x) = aPo(x). Assume the theorem true for all polyQ(x) a,

discussion.

nomials of degree g k. Now let Q(x) be any polynomial of degree k k+1 a ^ 0. Then Q(x) aix Ofc+i, 1, Q(x) = aox k. our is of a By assumption degree ^ a<>Pk+i(x) polynomial k

+

+

+

+

k

so that

Problem

-

Q(x)

= Y

P k+ i(x) =

a

cr

P

+

00

that

5.

Show

6.

If

for that $ P (x)x nP m (x) dx = > n. is a polynomial of degree < n, show that

r ri r

0, 1, 2, 3,

.

.

.

7.

Show

k

,

],

in the preceding line are

m

m

Q(x)

=

{p(x)Q(x)Pn(x) dx

Problem

=

c k +\

T (x)

Show

4.

unique.

Problem Problem

Q(x)

that the conditions

fp(x)x

k

P n (x)

yield a unique polynomial

dx

-

P n (aO =

k

xn

-

;

1

-f o*,,^"

)

1, 2,

.

.

.

n

,

-

1

(6.11)

and that

H-

=

rfx

m 7* n if P m (x) is generated^ in the same manner. Equation (6.11) could have been taken as the starting point for developing orthogonal polynomials. & F* +< find a linear transformation x = rx -\- s such Problem 8. For a 7* 6=s -f> find a linear when x = a, x = 1 when x = 6. For a? that into f transformation which maps x = a into x = 0, x = + + oo Problem 9. Show that JxP B _iP n p dx JP^p dx. for

<

,

5=0

>

;

.

6.3.

Normalizing

Factors.

Examples

of

Orthogonal

Polynomials.

The constants 7*

JPJ()p(a:) dx

n =

are called the normalizing factors of the

0, 1, 2,

Pn (x).

...

It follows that

(6.12)

SERIES, FOURIER INTEGRALS

ORTHOGONAL POLYNOMIALS, FOURIER

We

define the orthonormal set of functions

=

<p n (x)

The

Tn

\/pW

are not, in general, polynomials.

<p n (x)

by

{<p n (x)\

n =

Pn(x)

241

0, 1, 2,

.

.

(6.13)

.

possess the attractive

They

property ^ dx

if i 7*

j

=

.

.

f

^.

The {<p n (x)} form what is known as a normalized orthogonal system of functions associated with the family of orthogonal polynomials {P n (x)}.

We now

Example

list

a few families of orthogonal polynomials.

Let a

6.1.

=

=

0, &

P n (x) From

=

x k P n (x) dx

/

1

for k

1

with

4- a,jr

p(ar)

+

=

We

1.

choose

+

a 2x 2

-f

< n we have

yo

fc

We

+

1

^Ib

+2

^

r

k 4

j-

T =0

for k

can solve these n linear equations for the a t

,

i

=

-

1, 2,

0, 1, 2,

.

.

.

,

.

.

.

,

-

n

1

n, as follows:

We

have

k

+

1

frhere Q(k) 2,

.

.

.

,

^ is

<"+...+ +2

k

a

n

" -f

1

Lemma

+

a polynomial in k of degree at most n.

n

1,

1,

l)(fc

+

+2)

n

Since Q(fc) vanishes for

+ A;

1)

0, 1,

of necessity

(see

(k

Sec. 6.4).

2)

-

(k

-n +

1)

Thus

- n + 1) + (k 'k+n + l^(k+2)---(k+n + l) fe

If

we

set k

l,

we

obtain

C(-D(-2) 1-2-3

ELEMENTS OF PURE AND APPLIED MATHEMATICS

242 so that

t~\ (0

*

C - (-l) n

Thus

.

l + ai 4. + l^fc + 2^

a + ^ +r+

...

'

...

4. ^

1

A;

4. +

+n+

fc

n . (-l) fe(fe

1

1)

(A;

To

solve for o r

,

we multiply the above by

+r+

k

1

and then

set

-n

(fc+n

.

-

(* -h 1)(* -f 2)

=

A;

(r

+

1).

This yields

(-l)(r + D(r+2) (-r)(-r -f- l)(-r + 2)

-

Pn (x)

Hence

-

The Legendre polynomial with

+

n)(-l) 2

(-1)1

(n

-

r)

r)!

n

W

^ r0

(r

.

-

-

r!r!(n n

.

.

-

^ r-0

(

W

" 1)r

^

(r) (

leading coefficient unity

is

U;We now

determine

/

c/x.

F^(x)p(x)

Pn (x) 2

dx

-

We

have

a H x n f\(x) dx

I*

n

aw

/

ar

J '^ r

/

yo

^w

7=0

_ ~

Qr

r0 from

(a) above.

n+r +

n (~l) n!n! (2n + D!

Thus 2

^ /"^ (2n

n!yt!

n)\n) 6.4.

n

1

(2")!

.

(2n

-f 1)!

+

1

.

1)!

2n

H- 1

of the Orthogonal Polynomials. We wish to show that P n (#) of Sec. 6.2 are real, distinct, and lie in the fundamental We first state and prove two lemmas.

The Zeros

the zeros of interval.

LEMMA

If

1.

=

x

r is

P(x)j P(x) a polynomial.

= x

T

or If

a root of P(#)

The proof

=

0,

then

R =

x

-

=

0,

then x

as follows:

--

Q(#) H

P(x) P(r)

is

J?

=

a

r is

a factor of

We have upon division

=

constant

r

Q(x}(x

so that P(x)

-r)

=

+R

Q(x)(x

-

r).

Q.E.D.

ORTHOGONAL POLYNOMIALS, FOURIER

LEMMA real,

If

2.

then a

=

x

+i

a

+

ib)

243

b is a zero of a real polynomial, P(x), a, 6 It is easy to see that

b is also a zero of P(x).

i

+

P(a P(a Since P(a

SERIES, FOURIER INTEGRALS

=

-

= =

ib) ib)

b)

+

iV(a, b)

-

U = V =

we have

0,

U(a, b)

U(a,

iV(a, b) so that

-

P(a

=

ib)

0.

We now prove

the general theorem stated above. Let P n (x) be a real = If a b x is a of P n (x), then zero j* 0, ib, orthogonal polynomial. 2 2 s P n (x). This 6 is a factor of (a a) ib)] (x [x (a ib)][x

+

+

+

positive for all x in our for all complex zeros of n (x).

factor

is

fundamental interval

Now

P

P n (x)

N

^

n.

=

-

(x

-

xi)(x

x,)

since (x

= Ja

p(x)(x *

x\)(x

-

for all x

x2

.

.

This is true x = XN .

,

,

(a, b).

=

b

d^

.

=

from the nature of (?(#), a contradiction to the theorem is proved.

.

N<

p(x)Q(x) dx

or f Ja

Hence

(6.14).

N

xi,

,XN.

(6.14)

b

But f p(x)Q(x) dx > ya

(see Prob. 6, Sec. 6.2).

i, 2:2,

.

polynomial of degree

is

=

except for x

(a, 6)

- a*)PCc)

(x

XN)

(x

on

Ztf)Pn(z)

except for x

xi)

*

0^2)

-

(x

(a, b)

,

p(x)Q(x) dx

x\,

(a, b).

=

x

of

Let the reader deduce that Q(x) > ZArorQ(z) < for all x on 2, We have, however, .

=

y

Q(x)

.

x

odd multiplicity lying on the interval Assume N < n and let

be the real zeros of Certainly

let

=

w,

n

< and

We obtain 6.5. The Difference Equation for Orthogonal Polynomials. now an equation involving P n _i(#), P n (x), and P n +i(z). We have P n+ l(x) = X n+l + n +lX n + P n (x) = X w + (Tn^- + xP n (x) = n +l - n }x n + Pn+l(x) ~ Pn+i(z) xPn(x) '

<T

'

'

1

SO that

and since x n

= P n (x)

(r n

P n+ l(x) - xP n (x) is

ff

(<T

xn

~l .

(*+,

-

(T n

)Pn(x)

a polynomial of degree at most n

'

'

'

Hence

= P n+l(x) 1.

From

(x

'

a previous theorem

n-l

P-M(Z)

-

(X

+

<r n

+!

-

(T n

)P n (z) =

Y

Cr

P r (z)

r0

n-2 'n^!

+ Y

C rPr(x)

(6.15)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

244

We

multiply (6.15) by p(x)P n -i(x) and integrate over the interval Using the orthogonality property and the fact that

fxP^l(x)P n (x)p(x) dx = SP*(X)P(X) dx = (see Prob. 9, Sec. 6.2),

we obtain

=

y*

(a, 6).

7*

Thus

c n _i7_i-

n-2 Cr

P

r (x)

(6.16)

r-0 for To show that c r = 2. a polynomial of degree at most n need n only multiply (6.16) by p(x)P 9 (x) 2, one 1, 2, Hence 2, and perform an integration over (a, fe).

is

r s

= 0, g n

.

P n+1 (x) -

+

(x

Equation

.

.

er

9

,

n+1

-

<r n

+ -- Pn-i(x) B

)P n (z)

Tn

n ^

1

(6.17)

1

by the orthogonal show that the zeros of P0r),

(6.17) is the difference equation satisfied

Using

polynomials.

possible to

(6.17), it is

< x n are the zeros Pn+i(x) interlace in the sense that, if x\ < x < of P w (x) and if < y* < < y n < yn+i are the zeros of P M +i(a;), < y n < x n < n +i < b. We omit < X* < then a < yi < xi < 2

t/i

z/

y<t

this proof. 6.6.

The Christoffel-Darboux

P

tt

+!(0

-

(t

+

<Tn+l

From

Identity.

-

<Tn)Pn(t)

+

^

(6.17)

we have

P-l(0 =

(6.18)

Tn-1

so that

Pn (OPn+i(z) -(x -

(t

l(0

upon multiplying (6.17) by by 7^ yields

P n (t)

and

(6.18)

by

P n (a;).

-

Subtracting and

dividing

..

nil

Tn

We

(6>20)

have also

Pi(x)Po(Q

To

Pi(QPo(g)

=

(s

-

<rj

To

-

en)

^ *

(x

-

QPo(g)P,(Q To (6.21)

SERIES, FOURIER INTEGRALS

ORTHOGONAL POLYNOMIALS, FOURIER If

we

n

let

=

1, 2, 3,

.

.

.

make use

k iu (6.20), add, and

,

245

of (6.21),

we

obtain

P k+l (x)P k (t) " P

W (QP(X)

V

_

-0 Equation kernel,

K

(6.22) k (x; t),

is

What

if

c

11.

=

Problem

the Christoffel-Darboux identity.

VA r p

-

;

We

X^

(

r \p

'Wf' in

}

define the

W (f\

(6-23)

Evaluate Kk(x; x) by L'Hospital's rule. d are distinct zeros of Pjb(), show that If x = c, x

K

k (c',

=

d)

0.

dl

12.

Prove that

K

f* 6.7.

.

n-0

n-0 10.

,

(

by the equation

Kk ( X

Problem Problem

Pn(x)P n (t)

k (x-, t) P (t)

dt

-

1

(6.24)

We may

Fourier Coefficients and Partial Sums.

redefine the

orthogonal polynomials so that

=

p(x)pm(x)p n (x) dx

* 6 mn

J

n

HI

if

(6 ' 25) 1

= (l/7n)Pn(x), n = 0, 1, 2, ... (see Sec. 6.3). letting p n (x) Now let us consider a real- valued function of x whose Taylor-series expanWe have sion about x = exists.

by simply

(

We

6 26 ) -

see that f(x) can be expanded as a linear combination of terms from

the sequence of polynomials 1, x,

x2 x8 ,

f(x)

.

.

.

,

,

xn

,

.

.

.

= n-0

(6.27)

It seems logical to ask whether a given function f(x) can be expanded in terms of the infinite sequence of polynomials po(aO, Pi(a0, p(*),

.

.

.

,

Pn(x),

.

.

.

(6.28)

ELEMENTS OP PUKE AND APPLIED MATHEMATICS

246 If this is

the case,

we may

write

=

/(*)

n-0

To determine

the coefficients c n n

=

,

0, 1, 2,

.

.

.

,

we multiply

(6.29)

by

This yields

p(x)pk (x) and integrate.

=

p(x)p k (x)f(x) dx

Assuming further that the

ja

series

dx

c n p(x)pk(x)p n (x)

(6.30)

n-0

expansion converges uniformly on

(a, 6)

yields 00

p(x)p k (x)f(x) dx

= y

p(x)p k (x)p n (x) dx

cn f

=

ck

(6.31)

n-0

The

ck , k == 0, 1, 2,

cients of f(x).

If

.

I

.

.

,

of (6.31) are defined as the

p(x)f*(x)

inequality for integrals

dx

exists,

"Fourier"

coeffi-

then from the Schwarz-Cauchy

we have b

2 |c*|

Thus

c*,

g

I" P (x)f*(x) dx l

J&

Ja

P (x)pl(x) dx

(6.32)

given by

=

ck

exists provided

/

Ja

b

I p(x)pk(x)f(x) dx 2 p(x)f (x) dx exists.

k

=

.

(6.33)

Thus one can speak

of the Fourier

0, 1, 2,

.

.

coefficients of f(x) regardless of the existence or nonexistence of the expansion of f(x) in the form given by (6.29). If the ck k = 0, 1, 2, . . n, of (6.33) exist, we can form the series .

,

s n (x) is called

f(x).

,

the nth partial sum of the Fourier series associated with aid of (6.23) and (6.33) we have

With the

(6.35)

ORTHOGONAL POLYNOMIALS, FOURIER

The

difference

between f(x) and We have

SERIES, FOURIER INTEGRALS

247

the nth remainder in the

s n (x) is called

Fourier series of /(x). .(*)

- /(x) - (*) = b f(x)K n (x; t)p(t) dt K n (x; fa f' = f' K n (x;t)p(t)[f(x) -f(t)] dt

t)f(t)p(t) dt

>

(6.36)

JO,

b

since f

K n (x; t)p(t) dt

=

1

Now

[see (6.24)].

Jo.

so that

fn

The

~

(^ (pn+l(x} Pn (0 ja

p n +l(t) Pn (x)]

f(X}

~

P (t) dt

(6.37)

C

reader should explain the apparent difficulty at

t

=

x.

If

R n (x) =

lim n

f(t)

X

> oo

00

the Fourier

series,

Y

c n p n (^),

converges to/(x).

n-O Bessel's Inequality. orthogonal to Pk(x), fc = 0,

We

6.8.

b

f

Ja

[/(*)

-

s n (x)]p k (x)p(x)

1, 2,

dx

=

show .

.

. ,

first

that

n.

We

R n (x) =

f(x)

s n (x) is

have

b

[ Ja

f(x)p k (x)p(x) dx n Ci

t

pt(x)p k (x)p(x) dx

i-O

=

We

c fc

-

t-0 c*

=

dx

fe

(6.38)

show next that n

f'f*(x)

The proof

is

J

(6.39)

t|

t-o

as follows: Clearly

[/(x)

so that

P (x)

l*j*(x)p(x)

dx-2

-

s(x)]V(x) dx

*

f(x)Sn(x)p(x) dx

+

f*

sl(x)p(x)

dx (6.40)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

248

However,

n

=

f(x)s n (x)p(x) dx

-I f

s*(x)p(x) dx

fc

fc^O n

n

and

f(x)p k (x)p(x) dx

ck

= Y Y

Pj(x)p k (x)p(x) dx

f

c^c*

Since (6.39) holds for

so that (6.39) follows as a consequence.

all n,

the

n

sequence

tn

sequence

is

=

Y

n =

ef,

0,

2,

1,

.

.

is

.

,

monotonic nondecreasing.

bounded.

Moreover, this

This implies that lim n

tn

exists.

> oo

Let the reader deduce that 80

f*(x)p(x)

Formula

dx^

(6.41) is BessePs inequality.

Problem

13.

Why

Problem

14.

If f(x) is a

is it

=

true that lim ck

0?

polynomial of degree

dx

Problem

(6.41)

cl

15.

For

(a, 0)

such that a

^n

=

f

^ a ^

n,

show that

-

/8

^

6

assume that

g(x)pn(x)p(x) dx

/

Jot

exists.

Let f(x)

=

g(x) for

a ^ x ^

/3,

/(r)

=

otherwise.

Show

6.9. A Minimal Property of the Partial Fourier Sums. polynomial Q(x) of degree g n such that

/= f [/(*) -

Q(x)]*p(x)dx

that lim d n

We

0.

look for a

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS be a minimum.

shall

Let

249

be the nth partial Fourier sum of

s n (x)

f(x).

Then

=

-

P [/(*)

-

[/(*)

The second s n (x)

+

(*)

Ja

2 s(z)] p(*) dx

integral vanishes is

Q(x)

Q(x)]*p(x) dx

+

2 [" ya

from the

results of Sec. 6.8 [see (6.38)] since

^

a polynomial of degree

Hence /

fixed value.

-

(*)

will

The

n.

minimum when

be a

is

a minimum. This obviously occurs if a highly important property of s n (x). Problem

/b

Let

16.

Ql(x)p(x) dx

Problem to unity,

is

Q n (x) be to

we choose

\

Q(x)

s

s n (x).

This

a polynomial of degree n with leading coefficient unity.

=

be a minimum, show that Q(x)

Let {P n (x)

17.

Q(x)]*p(x) dx

[s n (x)

/

Ja

is

integral has a

first

P(x).

be a family of polynomials with leading coefficients equal rb

P n (x)

of degree

n such that

/

Ja

P*(x)p(x) dx

is

a

minimum.

Without using

the result of Prob. 16 show that b

Problem

Show

18.

that

7* for the

R n (x)

is

defined

if i

of

<

said to be complete

lim

dx = I* R*(x)p(x)

(6.36).

(

1,

1).

Pl(x) P (x) dx

Orthogonal Polynomials. A set of if for every continuous

is

by

*j

fundamental interval

xP;(a!)p(x) dx

g

Complete and Closed Sets

orthogonal polynomials function f(x) we have

where

<

-yj +1

Pl+i(x)p(x) dx

6.10.

-

P.(x)P,(x)p(x) dx

f

Ja

From

Sec. 6.8

(6.42)

we have n

b

[ Rl(x) P (x) dx Ja so that (6.42)

is

=

ff (x) P (x) dx - Lj y Ja 2

c|

equivalent to the statement 00

b

f Ja

f(x) P (x) dx

=

y d

Lj

(6.43)

We shall prove in Sec. 6.11 that the orthogonal polynomials generated above are complete for any finite range (a, b).

ELEMENTS OF PURE AND APPLIED MATHEMATICS

250

DEFINITION implies f(x)

A set of orthogonal polynomials is said to be closed if the Fourier coefficients of a continuous function f(x)

6.1.

the vanishing of

=

all

In other words,

0. b

I f(x)p n (x)p(x) dx

if

f(x)

=

is

continuous and

n -

0, 1, 2,

.

.

if

(6.44)

.

Jo.

s

then f(x) Problem

19.

0.

Show

that a complete set of orthogonal polynomials

/b

x nf(x)p(x) dx

is

-

a closed

forn

set.

0, 1, 2,

*

.

.

,

and conversely.

Completeness of the Orthogonal Polynomials for a Finite Range. (a, 6) can be reduced to the interval l^zglifa and b are finite. The famous Weierstrass approximation theorem states that a continuous function on a closed interval can be approximated arbitrarily closely by a polynomial (see " Courant-Hilbert, Mathematische Physik," vol. 1). In our case we wish 1 g x g 1 to within to approximate a continuous function f(x) on VV^o, where * is an assigned positive number and MO is the first moment, 6.11.

By a linear transformation the fundamental interval

/i

=

f

there

is

The Weierstrass approximation theorem

P(X) dx.

a polynomial Q(x) such that

!/(*)

We

~

denote the degree of

Q by

-1 ^

<

Q(x)\

m so that,

[f(x)

and

-

[/(*)

-l

[f(x)

Expanding

^

dx

-

2

Sn(x)] p(^)

s n (x)]*p(x)

[/(x)

dx<

dx as

x

g

-

P (x) dx

in Sec. 6.8,

y

c|

<

f or

n

sufficiently large so that

t-o

t-o

-

we have

Jb-O

c|

(6.45)

^

Q(x)]* P (x) dx

n

Y

1

from Sec. 6.9 for n

1

I J-l

But

states that

m,

ORTHOGONAL POLYNOMIALS, FOURIER

SERIES, FOURIER INTEGRALS

251

00

Since

J ~l /

f*(x)p(x) dx

Y

and

cl

are fixed constants whose difference can

o

be made as small as we please, we must have 00

1=

f^f*(x)p(x)dx=

cl

*

This completes the proof

[see

As an immediate

(6.43)].

continuous the relations

follows that for f(x)

corollary

it

=

0,

x nf(x)p(x) dx

[

imply f(x) = 0. 6.12. The Local Character of Convergence of the Fourier Series off(x). We are now in a position to examine the Fourier remainder R n (x) [see Let us assume that the p n (x) are 1 g x ^ 1. (6.36)] for the range < <x>, uniformly bounded on this range. This means that |p n (z)| < a constant. Let 6 be any small positive num1 g x g 1, for all n,

n =

0, 1, 2,

Then

ber.

.

.

.

,

-1 g

for

/"so-

f

M

M

x*

1

3

U-1

[xo+8

&+

Mn(,

\

I

fl *>(*,

/

7xo-5

Odt+

*

/info

/

yxo-h

J

(6.46)

where XQ [see

For the intervals

(6.37)].

function

o

" is

z

-1 g

g

5,

x

+

5

g ^g

1

the

Hence

well behaved.

n (x,()dt = r*~\ ~

lim >

oo

y

1

l

lim f

n_

(see Prob. 15, Sec. 6.8).

lim

-

2

n

n

z

y*o+

p n (x,

t)

dt

=

Hence

# n (zo)

>

I! We

<

need not worry about lim y n+i/y n since n

18, Sec. 6.9).

Whether lim n-n

the behavior of

f(t) in

y n +i/Vn

<

1

(see Prob.

*

R n (x<>)

is

zero or not depends entirely on

the neighborhood of

t

=

x

.

If,

for example,

ELEMENTS OP PURE AND APPLIED MATHEMATICS

252

A =

constant, for

all

t

near

then

XQ,

\R n (xo)\ for

n

sufficiently large.

that lim

R n (xo) =

Since

6

$ 6M 2 A5 M o

can be chosen arbitrarily small,

it

follows

0.

n

For a complete discussion of orthogonal polynomials the reader is urged to read G. Szego, "Orthogonal Polynomials," American Mathematical Society Colloquium Publications, vol. 23. 6.13. Fourier Series of Trigonometric Functions.

In the early part of the nineteenth century the French mathematician Fourier made a tremendous contribution to the field of mathematical physics. A study of heat

motion

Example

(see

Fourier to the idea of

5) led

Chap.

22,

expanding a real valued function f(x) in a trigonometric

=

f(x)

ao

+ +

cos x

fli

61 sin

x

+ a cos 2x + + bi sin 2x + 2

to

=

is-ao

+ X

series.

-

oo

a n cos nx

+ Y

n-l

6 n sin

nx

(6.47)

n-l

At the moment

let us not concern ourselves with the validity of wish to determine a, n = 0, 1, 2, and 6 n n = 1, 2, From were valid. we have (6.47) trigonometry

We

.

nx sin mx = [cos (n sin nx cos mx = ^[sin (n cos nx cos mx = ^[cos (n sin

.

.

m)x m)x m)x

,

.

,

(6.47). .

.

,

if

cos (n + m)x] + sin (n + m)x] + cos (n + m)x]

so that

-*

/: r

/; provided

m and

sin

nx

sin

mx

if

dx

.TT

nx cos

mx dx =

cos nx cos

mx ax =

sin

n are

if

n n

7*

m

= m (6.48)

if

7

T

.ii

n n

5^ ==

m m

integers.

multiply (6.47) by cos mx and integrate term by term (we are not concerned at present with the validity of term-by-term integration), If

we

we obtain by use

of (6.48)

cos mxf(x) dx

m=

0, 1, 2,

...

Similarly

(6.49)

bm

=

- /r* sin mxf(x) dx T y ~T i

m=

1, 2, 3,

.

.

.

ORTHOGONAL POLYNOMIALS, FOURIER The

a's

These

and

&'s of (6.49)

can exist

SERIES, FOURIER INTEGRALS

"

are called the Fourier

253

coefficients" of f(x).

the integrals of (6.49X.exist] regardless of the We can possibility of expanding f(x) in the series given by (6.47). The replace the interval ( TT, v) by the interval (0, 2x) if we so desire. coefficients

and

results of (6.48)

Example series

We

6.2.

-

1

[

-

(6.49) hold for the interval (0, 27r).

calculate the Fourier coefficients given

*

mx dx

-

*

irJ-T TT

m

5 2

(x

1

,

x cos

/

v ^ x

x t for

shall see later that f(x)

We

development. am

[if

[os mir

cos

mir)]

(

,

)

my-r

-*

m

/

5^

xdx -

a

-;/.: x

sin

2

mx =

cos

m

for

m even

for

m odd

4x

+

2

m Hence

has a Fourier-

(6.49).

I

m =0

\

ir,

If*.sm mx dx \

sin rax *

I

ir

^ by

f(x)

-

2(sin

x

-|

sin

Since the right-hand side of (6.50)

is

2x

+

-

sin

periodic, its

graph

sin

is

given by Fig.

)

6.1.

(6.50)

We note

O

FIG. 6.1

that /(ir/2)

x/2

-

2(1

|

+J -

y

+

This checks the result obtained

).

from

provided the series can be integrated term by term. Example 6.3. The function f(x) given by

will

f(x)

-

/Or)

-a:

x

-TT

^

or

be seen to have a Fourier-series expansion.

^T The Fourier

coefficientl of f(x)

i

ELEMENTS OF PURE AND APPLIED MATHEMATICS

254

-

fln

i

nx dx

cos

-

I

iry-ir -

fen

a

/.

=

a;

sin

nx dx

a;

d#

=

Thus

4

+ T./0 f

*

cos (2n

r H

-

1]

00

00

4

n -i[(~l) 7&V

cos

^ w2

Z/ L

-

x cos nx dx

^=

l)x

-

(2n

I)

V

_

(-!)"

sin (6.51)

2

l

the right-hand side of (6.51) becomes

For x

n-1 ao

Is

V/

7?:

/.^

(^7i

1

n-l

?T

TT4

"

2

In

"S"?

O

I)

complex-variable theory

it

can be shown that

(n

VI 00

If

we

allow z to tend to zero,

we

obtain

)

l^

TT

r~rro (Zn -r I)*

-^

2

~5~>

o

^ ne desired result.

The

71=0 limit process can

For x

?r

be

justified.

and x *

4

TT

the right-hand side of (6.51) becomes

T Z/ n-1

(2n

-

I)

2

4 7T

Notice that^

(

r)

'

,

T

^=

(2w

27T 2

"^/W . The right-hand ?L^Z . = T, yields the mean of f(x) point, "#

Z

-

I)

2

1

T

side of (6.51)

when

evalu-

JL

ated at either end graph of the right-hand side of (6.51)

is

given

by

at these end points.

Fig. 6.2.

2r FIG. 6.2

3r

a;

The

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS Example

Let }(z) be analytic for

6.4.

>

\z\

-

/(*)

255

Then

0.

a nz n (6.52)

C

where

can be chosen as the

circle

On C we have

** 1.

\z\

f

= e%

d

ie** d<p, so

that 00

/(*)

-

y

a nz

a"

J

27rjo

For

2

on the unit

circle

|z|

=

1

we have

= c* and

z

(6.53)

If

we

define

g'(^)

=s f(e t9 ), (6.53)

becomes

(6.54) r 2*

1

2ir

Example

We can

6.5.

A

cients of f(x).

employ the

results of Sec. 5.7 to evaluate the Fourier coeffi-

particular solution of

y" is

Jo

4- n*y

- /()

y(x)

where 0(x)

is

a solution of y"

-f

-

-

di

f*f(t)g(x

n2y

0, 0(0)

0, 0'(0)

sin n(x

is

(6.55)

given by

a solution of (6.55) with y(Q)

so that

An y(0)

6.

-

-

0, y'(Q)

*

/(0 sin nt

0.

dt

-

t)

=

1.

Thus

dt

Evaluating (6.56) at x

0, y'(0)

-

0.

*

2ir

yields

~ -

analogue computer can be used to solve (6.55) subject to the

-

(6.56)

$|57) initial

ELEMENTS OF PURE AND APPLIED MATHEMATICS

256

The height of the ordinate &tx = 2ir graphical solution can be obtained for y(x). Differentiating (6.56) and evaluating yields y(2ir), which in turn yields b n of (6.57). at x 2ir yields

A

cos nt dt

f(t)

=

~

(6.58)

y'(2ir)

Problems Find the Fourier

series for the following functions defined in the interval

TT

<

x <ir: 1. f( x )

2. f(x) 3. f(x) 4. /(x)

= = = =

5. /(:r)

6.

< x ^ 0, f(x) - 1 for < x < <x < e* for -TT < z < cos x for <x < cos a for ^ a 5^ 0, /(x) = cos x otherwise for -TT

|s

for

ir

--n-

ir

ir

2

ir

TT

|a:|

Let f(x) be an even function, that bn

~

-

7.

8.

=

is,

-

1

/"*

7T

J -TT

/

Show that any function can be

= f(x).

Show

that

sm nx dx =

TT

Let/Or) be an odd function, that

an

f(x)

f(x)

I

J

ir

is,

f(x)

/(x) cos

=

Show

f(x).

=

nx dx

written as the

that

sum

of

an odd function and an even

function. 9.

Find the Fourier

6.14.

series of /(x)

Convergence

=

IT

|sin x\,

^ x ^

TT.

of the Trigonometric Fourier Series.

We wish now

to investigate the convergence to f(x) of the Fourier series

+ y

i^o

a n cos nx

+ V

b n sin

nx

(6.59)

l

an

= -1 /f * J

w

cos

?ii

d<

n =

0, 1, 2,

...

-ir -

bn

= _1 7T

Some preliminary

/"*

/() sin

/

J -TT

discussions are necessary.

lim

r~

First let us consider

F(t) sinktdt

(6.60)

we feel that for very large k the function F(x) sin kx will be about as often as it is negative (with the same absolute value) positive since sin kx will oscillate very rapidly for large fc. Since the integral Intuitively

~* /F(f)

sin kt dt represents

an

area,

we

shall not

be surprised

if

,*

lim &

f* F(t) sin ktdt """"

^

=

(6.61)

ORTHOGONAL POLYNOMIALS, FOURIER

SERIES, FOURIER INTEGRALS

257

It is easy to see that Certainly much will depend on the nature of F(x). w ^ x ^ TT. (6.61) holds true if F(x) has a continuous first derivative for

Integration

f*

EI/

by parts \

j

7

F(x) sin kx dx

/

=

-

the fact that

*

F(x) ^~ cos kx #

J -IT

From

-

yields

jF'(z) is

+ .

T

I

y /C

f* F'(z) cos y _.

/

bounded on the closed

.

,

fcx

dx

TT^X^TT

interval

follows immediately that (6.61) holds true. Actually we can weaken or make less stringent the conditions on F(x) in order that (6.61) hold true. Nothing need be said about F'(x). Let 2 be on the interval TT ^ x ^ TT. now and F(x) [F(x)] integrable it

We

define

s n (x)

by fc

cos

fcx

+

6 A sin

(6.62)

fco:

A-l '

s n (x) is

finite

ak

=

-1 I/"* cos * J -* F(t)

=

-1

/"*

bk

^"

J ~T

F(t) sin

sum

called the nth partial Fourier

trigonometric

-

(F(x)

Moreover

series.

(*)]

=

[*(*)]*

kt dt

fc<

d

Note that

of F(x).

s n (x) is

continuous.

-

2F(x) Sn (x)

-

2

+

we have sn

One can

easily

)P<fc

=

show

(see Sec. 6.8) that

/* y-ir

[F)]

J

ctt

n

r" F(o.(o ^

;

From

"

the fact that f ~ 7T J

[F(t)

-

=

^+ y ,

(oj

+

+T

-(aj

+

s n (t)] 2 dt

^

s n (x) is

From

we deduce that

a

ELEMENTS OF PURE AND APPLIED MATHEMATICS

258

The

left-hand side 1

the constant*"

This

is

is

monotonic increasing with n and

/"* / J -r

is

bounded by

Hence

[F(i)]*dt.

Bessel's inequality [see (6.41)].

Since the series of (6.64) con-

verges, of necessity,

lim a*

=

lim 6*

fc-oo

lim bk

=

is

=

fc-*

precisely the statement (6.61).

*-*<>

A

simple class of functions which are both integrable and integrable square is the set of functions which have a finite number of bounded discontinuities in the sense that

function then lim

/(a;)

if

a:

and lim

JB-+C

X

X<C

X>C

=

a point of discontinuity of such a both exist but are not equal. We

c is

/(a;)

C

write

lim /(*) x

lim /Or) X

= /(c-0)

e

=/(c

+ 0)

*C

x>c

A function which has a finite number of bounded discontinuities of the type described above is said to be sectionally, or piecewise, continuous. Now let/(z) be sectionally continuous for TT ^ x ^ ?r, and assume that = f(x). We make use of the 2?r) f(x) has the period 2v, that is, f(x It is not necessary, however, that periodicity of f(x) in (6.67) below. = /( TT), since the value of f(x) at one point does not affect the value /(IT) In most cases /(x) is defined only for the of the integral [see (6.67)].

+

interval

IT

^

x

^

at other values of x

The nth Fourier

s n (x)

=

-^

/

IT. We*easily make f(x) periodic by by the condition f(x + 2ir) = f(x).

partial

f(t) dt

sum

of f(x)

+

f(t)

defining f(x)

can be written as

cos kt cos kx dt

sin

j /_'/(0 [j

+

^

cos

Jb

-

<

x)]

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS It is

259

s n (x) is a finite trigonometric series and that As long as the Fourier coefficients exist, it is always construct s n (x). Thus s n (x) exists independent of the possible

important to note that

s n (x) is

continuous.

possible to

development of f(x) as a Fourier series. We desire to show that, under suitable restrictions concerning the

first

derivative of /(#),

lim 8n(x)

=

+ 0) + f(x -

i[/(s

(6.65)

0)]

n

From

-

e wci-) x)

_

cos k(t cos

m

x)

__

-j-

_

x)

i s i n fry i

sin ^(1

_ _

x) x)

the reader can show that

/(x + if JIT

*LfrL+ 2 sm (w/2)

.)

r

(6.67) v '

The last integral results from the fact that the integrand has period From (6.66) the reader can deduce that :

v j _ T z sm (u/z) so that

and

/ (x)

.(x)

/

1

=

T

(in

- /(*) =

_ /"* /

v J -r

Now consider x fixed,

JL

[/(*

+

)

- /(x)]

-+-u-- -^ -M7/n\ sm

/(x

!

w)-

/(x)

7T~-

2

sin (w/2)

/ ( v

n

and define

~

M

be sectionally continuous

2 sin (w/2) if

lim F(u) *

Hence lim f F(u) /-*

du

2

-1

will

Gin

2*-.

sin (n

/'(x) exists, for in this case

=

+ ?)u du =

/'(x)

(see Prob.

2 of this section).

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

260

We have shown that, if f(x)

TT g x g ?r and is sectionally continuous for /(x) has a period 2w, then, at any point x such that /'(x) exists, the Fourier series of }(x) given by (6.59) converges to/(x). It may be that/'(x) does not exist at the point x but that

if

lim u-0 ;i

+

/(*

-

u)

f( X

+

0)

u

= f(f

+ Q)

= l/(.+u)-/(x-o) W

u-,0

W<0

do exist separately. We call f'(x + 0) and f'(x 0) the right- and lefthand derivatives of /(x), respectively. Now let the reader show that

f(x

i

+ u)~ f(x +

u

0)

- -

o 7 7o\ Sln 2 sin (u/2) !

w

T Joo

- + )-/(-0)-

It follows that

if

w

j-^

f'(x

+

0)

and/'(z

lim 8n (x)

-

= Hf(x

7T-'

sm

2

0) exist

+

,

T/ w

i

.

T~AV; s (w/2)

n(n

then

+ f(x -

0)

n

u

/(

TT

(

0)]

(6.68)

n

We formulate THEOREM and

let /(x)

6.1.

the above results in the following statement: Let f(x) be sectionally continuous for TT

g

x

At any point x such that

be periodic with period 2w.

g

?r,

/(x)

has both a right- and left-hand derivative the Fourier series of /(x) will 0) converge to the mean value of /(x) at x defined by |[/(x /(x 0)]. It should be emphasized that a sufficient condition for /(x) to be

+

+

written as a Fourier series has been given. There are no known necessary and sufficient conditions for /(x) to be developed in terms of a Fourier

Study of the Lebesgue integral yields greater insight into the development of Fourier series. References to this line of study are given at the end of this chapter. series.

.

1.

From

Problems

the identity

2 sin

cos ku ^ z

sin (k -f

sin (k

?)u

deduce that n

V/

,

cos ku

- -

fc-i

Hint: Let k

1, 2,

.

.

.

,

n,

1

-

2

Lj and add.

+ ,

sin (n -f i)u *

n 2 sin (w/2) .

.

i)w

ORTHOGONAL POLYNOMIALS, FOURIER Under the

2.

lim

-

F(t) sin nt

/ /-

n-*oo

261

[F(x)]* be integrable we have shown that

and

restriction that F(x)

SERIES, FOURIER INTEGRALS

I* F(0 cosn* -

lim

n _>,c ./-*

For the same F(x) why

is it

true that lim

f* F(0 -

lim n _eo

Show

that

0.

./

Let f(x) have a continuous derivative for v ^ x ir, so that x ^ w. Show that the Fourier coefficients of /(x) satisfy

3. IT

+ ?)t dt -

sin (n

0?

F(t) cos (t/2) sin nl d*

/ y-T

n _>oo

|/'(x)|

<

Af for

^

< ?M

|o*|

What

further restriction can be imposed

Let/(x) and

4.

2

-

*

1, 2, 3,

upon

.

f(x) so that

be integrable for the range

[/(x)]

.

.

^

TT

< 2M/kt

|6*|

^

x

Consider the

T.

finite

trigonometric series n

n

Sn (x) =

^

-|ao -f

dk cos

+

fcx

Show

J

that

I

S n (x)] 2

[/(x)

j -if

= ~

<*fc

7T

-

b*

*

sin (1/x),

6.

Let/(x)

6.

If /'(x) is sectionally

for 7.

TT

<

<

x

If /(x)

is

minimum

a

/(^) cos

fcx

rfa:

A:

=

k

=

for

0, 1, 2,

J -7T

TT

rfx is

^ /

/"* y -TT

sin kx

/()

kx

6* sin

2^

k=l

k=l

x ^ 0, /(O) = continuous for

dx

1. TT

1, 2, 3,

... .

.

.

Is/(x) sectionally continuous? ^ x ^ IT, show that /(x) is continuous

TT.

a continuous function such that /(x

+ T/3) = /(x),

show that

its

Fourier series has the form 00

-jr

+

00

y an

cos 6nx -f

nl 8. 41

0(x).

sin 6nx

Let (a n b n ) be the Fourier coefficients of /(x), ja n 0n! the Fourier coefficients of Find the Fourier coefficients of ,

,

MX) 9.

b

y n n=l

/(^ -t)g(t)dt

Let/(x) be continuous on the closed interval a = x x\, Xa, x n = 6. Now write

into a

.

,

.

.

^ x ^

6.

Subdivide this interval

,

n

//

(x) sin

ax dx

/ ^W

i

/(x) sin

/

ax dx

Jxi-i

i-i n

n

"

**

/ Lt t-1

/

Jxi-i

[/(*)

""

/fo-0]

sin

ax dx

+ Y L~t

t-1

*'

/

Jxt-i

/(x t .i) sin ax dx

ELEMENTS OF PURE AND APPLIED MATHEMATICS

262

From

the fact that f(x)

is

uniformly continuous deduce that

..

Extend

.

/6

~

f(x) sin

ax dx

~

l

this result for f(x) sectionally continuous

on a

x

6.

and Integration of Trigonometric Fourier Series. have seen that under suitable restrictions the trigonometric Fourier

6.16. Differentiation

We

series of /(x) will

Now we

converge to /(#). of a Fourier

term differentiation

series.

are concerned with term-byLet us return to the Taylor-

00

series

expansion of f(z).

If

a n z n converges to/(z) for

>

LJ

^

\z\

<

r.

we

n-0 00

know

a n nz n

V

that n

=

~

l

will

^

converge to/' (z) for

<

\z\

In the case

r.

l

of Fourier series, however, a simple example illustrates that the new Fourier series obtained by term-by-term differentiation may not converge The Fourier series of /(x) = x, IT < to/'(x) even though f'(x) exists.

x

^

?r,

is

given by 2(sin x

Term-by-term

+

^ sin 2x

i

3x

sin

-j

4x

sin

+

)

differentiation yields

2(cos x

cos 2x

+

cos 3x

cos 4x

+

)

This series cannot converge since cos nx does not tend to zero as n

becomes

infinite.

Certainly term-by-term differentiation of a Fourier series leads to a new Fourier series. If this new series converges to/'(x), we shall wish to

make certain that /'(a;) has a convergent Fourier we have seen that a sufficient condition for this is

series.

From

Sec. 6.14

that/'(x) be sectionally It is not necessary that

continuous and that /'(x) have periodicity 2ir. Let the reader show that, if f(x) is sectionally con?r) =/'(TT). Furthermore if /'(x) has periodicity tinuous, then f(x) is continuous.

/'(

2?r,

the reader can

show that/(

/(x) has periodicity

THEOREM

6.2.

2?r.

TT)

=

/(TT) is sufficient

to guarantee that

We state

Let f(x)

and prove now the following theorem be continuous in the interval TT g x ^ TT,

:

/(ir) = /(TT), and let/'(x) be sectionally continuous with periodicity 2ir. At each point x for which f"(x) exists the Fourier series of /(x) can be differentiated term by term, and the resulting Fourier series will converge tof(x).

The proof

is

as follows:

From

Sec. 6.14, f'(x) can be written

00

/'(x)

=

aj

+ Y nl

00

a'n cos

nx

+ Y n-l

b'n

sin

nx

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS with

f'

= -1

a'n

I

J

fl"

1 -

=

&n

=

/'(x) cos

nxdx

n

/'(*) s in

nx dx

n =

263

1,2,..

0,

IT

/"* /

1, 2, 3,

.

.

.

IT./-*

Integration by parts yields

a'n

=

/(x) cos nx ~ 7T

H

/

* J -v /(x)

|

sin

nx dx

f'IT

=

T

=

f

-

since /(

TT)

of/(x).

Similarly

/(?r)

nx

sin f(x) sir

rfx

= nb n

J -7T

b n is the nth Fourier sine coefficient

by assumption.

= -na n

b'n oo

so that

oo

=

/'(x)

~~

2,

na n

sin

nx

+ Y

The Fourier

nfr n

cos nx

(6.69)

nl

n-l series of (6.69),

however,

is

the Fourier series obtained by

term-by-term differentiation of 00

00

/(x)

=

-^a

+ X

fln

cos nx "^

nx

L,

This proves the theorem stated above. We turn now to the question of term-by-term integration of a Fourier series. The reader knows from real-variable theory that integration tends to smooth out discontinuities, whereas differentiation tends to introduce discontinuities. As an example, consider the function /(x) defined as follows

:

/(x)

=0

for-oo<z^0

/(x)

=

x

for

/(x)

=2

for

1

^ <

x x

g <

1 oo

and at x = 1. Moreover /(x) is disconf(x) does not exist at x = = tinuous at x 1. However, the Riemann integral of /(x) defined by

-

is

continuous for

oo

discontinuity of /(x)] /'(O)

does not

exist.

x < At any point x y 1 [x = 1 is the only we have F'(x) = /(x). Thus F'(Q) exists, whereas Let the reader show that F'(l) does not exist.

<

.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

264

We find that very little is required to integrate a Fourier series term by We state and prove the following theorem:

term.

THEOREM If s-a

Let /(x) be sectionally continuous for

6.3.

+ Y

+

a n cos nx

n-l

2&

sin

n

nx

is

/(x)

dx

g

x

g

TT.

the Fourier series of /(x), then

n-l

oo

oo

/

?r

=

~

+

(x

+ y

TT)

sin

nx

(cos

y

nx

cos

/ITT)

nl

nl

(6.70)

whether the Fourier series corresponding to /(x) or not. converges Equation (6.70) is not a Fourier series unless a = 0. The proof is as follows: Since /(x) is sectionally continuous and hence

Equation

(6.70) holds

we note

Riemann-integrable,

=

F(x)

is

that F(x) defined

-

/Cc) dx

by

|aox

(6.71)

Moreover F'(x) =

continuous.

continuity of /(x).

If

=

x

c is

/(x) ^ao except at points of disa point of discontinuity of /(x), the reader

From (6.71) it can easily verify that F'(c - 0) and F'(c 0) exist. follows that F(TT) = F(w) = ^aw. F(x) can be made periodic by

+

defining F(x)

for

>

|x|

TT

by the equation F(x

Sec. 6.14, F(x) has a Fourier-series

0)

or

since

+ F(x -

- $A +,

F(x)

F(x

+

0)

- ^A Q

0)]

= F(x

0)

An

+

2w)

+

An

cos nx

+

cos nx

-

J5 n sin

from the continuity of F(x).

F(x) cos nx dx

I

sm nx 1

_1

1

f*

mr J/ _ T

~

/"* /

WK J

,

-

[j"(^)

^ ao

]

F-"(x) sin

sin

TXJJ

dx

v

[' f(x) nwj.. r

3

F(x).

mnxdx= -in

#

+

parts yields

An =

=

From

development

nx

nx

sin

nx

(6.72)

Integration

by

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS Similarly

Bn =

(l/n)a w

We now

.

write

00

(%)

= i^o

265

00

- 6

/

+

cos nx

n

L~i

- an

/

si sn nx

n

L-4

00

From

= ^a Gir we deduce

F(ir)

=

that AQ

a

- 6 n cos

+

7r

/ ^ ^V =

n

these results along with (6.71) yields (6.70).

TITT.

Using

l

Q.E.D.

Problems 1.

Differentiate

and integrate the Fourier series which admit these processes

for those functions of Probs.

1, 2,

3, 4, 5, 9, Sec. 6.13, 2.

Sum

the series 1

3. It /' (x) is

-I +1_I+ 5 33 73 -T

+_T+11_ 1

.

1-

3

t-

+ f

'

'

'

j),

(2/t

= /( TT), and if can be shown that if f(x) is continuous for TT ^ x ^ TT, f(ir) TT ^ x ^ TT, then the Fourier series of f(x) consectionally continuous for

verges uniformly and absolutely for

TT

^

^

x

TT.

oo

fW

iao

+ n

/ =1

/(^)> integrate term by term (this uniformly), and show that

by

dx This identity

Multiply oo

an cos rue

-f-

is

6 n sin

y n=

nx

1

permissible since the

new

series

converges

=

due to Parseval.

is

The Fourier

If f(x) is defined for x on the range not periodic, it is obvious that we cannot f(x) a trigonometric Fourier series since such series, of represent f(x) by We shall show, however, that under certain be must periodic. necessity, conditions it will be possible to obtain a Fourier integral representation

6.16. oo

<

x

<

and

oo

Integral.

if

is

of/(x).

THEOREM interval oo

oo

is

<

<

(a,

x x

< <

6.4. fr),

\\f(x

that

oo, oo

represented

Let f(x) be sectionally continuous in every

and let/(#) be absolutely integrable

,

is,

/

J ~oo

|/0*OI

dx

= A <

oo.

for the infinite

finite

range

At every point

x,

such that f(x) has a right- and left-hand derivative, f(x)

by

its

Fourier integral as follows:

+ 0) + f(x -

0)]

=

-

da

f(t)

cos a(t

-

x) dt

(6.73)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

266

We

proceed to the proof in the following manner: The result of Prob. 9, methods of that section, enables us to state that

Sec. 6.14, as well as the

i[/(x

+

+ /((x -

0)

=

())]

lim - I a-*

at

a

any point x

<

<

x

b.

which

for

Now we

<

7T

dt

and left-hand derivative,

investigate the existence of

(6.75)

^T^hr

x we observe that sin a(t

The convergence such that f or a

of

/ 7oo

<

dt

\f(t)\dt

shows that

|/(OI dt

for

-

x)

x

t

,

dl

f(i) sin

f

>

a number

exists

a.

B

a(

rr

A

Thus the right-hand

side of (6.74j can be

made

to

from lim a

>

dt

7T

oo

e > provided This implies immediately that

a and b are chosen sufficiently large.

by any arbitrary

*[/(*

+

0)

+ f(x -

0)]

=

Urn ax 7T /""oo

sm

cos p(t

/

x)

dp

^

=

Jo

i[/(

+

t

0)

+ f(x -

0)]

=

1

lim ->

Before

we pass

to the limit as a

TT

9

-

x

f

/

_, ?

f(t) dt

/

~

*)

X

I

so that (6.76)

/""

y

<*(*

f(t)

J

Now

an

<5

independent of a. Similarly one shows that for any e/2 exists such that f or b ^ B

independent of

A

>

any e/2

^ A f(t) sin a(t

differ

(6.74)

Ja

f(x) has a right-

/_/<'> For a

oo

may

dt

(6J6)

be written

a

cos

yo

we wonder

if

/*(*

-

x)

dp

(6.77)

we can interchange

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS Let us

the order of the double integration.

^

x a parameter and

For

=

fj.

^

M

x) dt

We show that G (M) Y

a

a,

consider

first

cos n(t

f(t)

fixed.

267

is

continuous.

we have

MO

(t

-

cos M o(^

r)]

cos MO(

x)

*

x)] dt

cos no(t

x)

f(t)[co8 n(t

I

-

x)

/(J)lcos M(

/y

+

-

a;)]

dt

ip

|/00

than 2

we

Since

\f(t)\ dt.

j

1

and the

dt,

are given that

|/(/)| dt exists,

I

be found such that these integrals can each be made

>

T

0.

is

obtained after

cos n(t p.

between

Since f

/i

and

\f(t)\

middle integral

any

c

>

cos

x) M,

= A

dt is

we note

is

a

number, we can

finite

also less than e/3 for \p

there exists a

5

\fjL

/x

|

<

continuous at

This

5. /x

=

>

H(a)

exists,

and

dH = -7 ttOf

po\

<

-GG*o)|

I

dp

I

all p.

f(t)

is less

a

find

<

<

6

x)

ji(t

d

than

>

so that the

Hence

e/3A.

J

f(t)

for

e

Hence

cos p(t

x) dt

/** /

cos n(t

x)

dt.

By

considering

00

/(<)

let

than ^e for any

the definition of continuity, so that G(fj)

is

=

T can

such that

and, in fact, for

JJLQ

sin

/xo)

(fj,

that the middle integral

|G(M)

for

=

x)

/IQ(

less

a

From

chosen.

e is

last integral is less

the reader show that -r-

=

-3

cos

/.(

Since X(0)

-

x)

=

dn

H(0), of necessity

COS M

~

is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

268

Allowing a to tend to infinity enables us to rewrite (6.77) in the form given

by

(6.73).

If f(x) is

continuous at

=

f( x )

.

may

x, (6.73)

da I / * JO J-

be written

cos a(t

f(t)

-

x) dt

(6.78)

From cos a(t

-

= Ke^'-o

x)

+

e-*<*-'>)

one readily obtains

}(x)

=

^

^

= _L ZTT

Equation

(6.79)

/

da

da

e

j _

dt

j^ /(Oe*<-'>

oo

f(t)e-*< dt

7 -

(6.79)

can be written in the symmetric form

=

*

g(a}

We say that g(a)

is

1

v

"75=

=

the Fourier transform of the real function f(x), and

f(x) is called the inverse Fourier transform of g(a). more rigorous treatment of the Fourier integral requires the use of the Lebesgue integral.

A

Example and/(x)

=

6.6.

We

e~ 0x for x

consider the function f(x) defined as follows: f(x) = for x < Fourier's integral certainly applies to this function, 0, ft > 0.

^

|/(x)|

/ g(a)

= -4=

-"-* dt = -4=

I

ft

'

-

i 7T

From

r

cos

ax

cos ax

dx

TTT^-

Q sin ax

-f

+

sin

ax

.

JO

the definition of /(x) one has for x o

^

1

~

>

* /

I

aX

L

TT

;o

1

[

COS ax _

4-

.

a

sin

ax

a_ sin ax

,

a

exists.

From

(6.79)

ORTHOGJNAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS ,1

v JL

a -Bx

_

cos ax aa; j

(

f 20 s^-yD "Jo FT**

j.

sothat

(

p* -f-*

T ~** e

1

2

,

681) ' (6 81)

"

=

-asinax,

/"

2

269

Jo

Let the reader obtain (6.81) directly by contour integration. Example 6.7. Suppose /(x) is an even function. Equation (6.78) can be written o

= If

we wish

/*

-

/(x)

da

O -

/*

TT

;o

oo

I

ax dx

cos

f(t)

\

cos at dt

= f*

cos a^/(0

rf/

and obtain

(6.82)

=

/(x)

2

/"

7T

JO

e~ a cos

/

+

TT!

x

rfj

X2

Notice that /(x) is an even function and that this function for the application of the Fourier integral formula.

Let p

6.8.

simple example

We

(6.82)

Jo

to solve the integral equation

we apply

Example

eft

oo

/

/

COS aX

COS

f(t)

oo

e~

for /(x),

r

oo

I

is

Z(p)

=

=

has the necessary properties

and assume Z(p) such that Z(p)e

-^

n

aop

-f a\p

++

n~l

+

a n -\p

lfjlt

=

Z(iu>)e^

a n where p 2 ,

=

t .

A

d2 -,-->

etc.

consider the equation

Z(p)0

(t)

= 0(0

(6.83)

We assume that the input O (0 is given, and we output that (6.83) will hold. Let tf(w) and (w) be the Fourier transforms of desire to find the

t

H

(0, respectively.

(0

and

Then i

ff

V

= 27T

! () = ~7^:

-

/

/=

Q,(Z)

eo(0

=

,

(6.84)

r

V VJ2r J i

and

/

6-*'e (0 d

-

r e

/

^/-.

e

li

"H

t

(u)

du

" //oMd

'

Substituting into (6.83) yields

Assuming that ^(p)

(0 such t

H *^

dw

*

tto

^.() d"

Z(p)

)_

may

be placed inside the integral yields

elt

Z(iu)e

ttu>

HoM du

]_

**

I

e ttu

H

l

M

d<*

ELEMENTS OF PURE AND APPLIED MATHEMATICS

270

Since the two integrals are equal for the

Hence

is

called the operational impedance,

transfer function.

We now 8

Making use

fl

it

appears logical that

and

its reciprocal,

Y(p),

is

called the

have -

-4^ J-" V2r f

(0

c"jSr,(

of (6.84) yields

^

Too

1

JTT

= where

We

t,

-

0.(l)

Z(p)

range of

full

y(t

Too

-*

y

Bi(T)F(?w)6UU

-

y

,

(t

_

r)drrfco

oc

*

f y-oo

-

-r)</r

e,(r)j/(/

-

r)

-

f

ZTT

F^)^'-^ dco

oo

y

(6.85)

have assumed that the order of integration could be interchanged. can be written

Equation

(6.85)

9

-

(0

9.0

/

~

r)y(r) dr

(6.86)

Let t (< T). y(r) is appropriately called the memory function because of the term the reader compare (6.86) with the particular solution of a linear differential equation with constant coefficients given by (5.71).

No attempt

at rigor has been

made

in obtaining the results of this

example.

Problems 1.

Suppose that f(x) of (6.74)

?[f(x -f 0) -f f(x

If /(.r) is

2.

x

3.

/(x)

0)]

-

da f

r I

f(t)

Show that

cos at cos ax dt

x

odd, show that

?lf(x -f 0) -f f(x

for

-

an even function.

is

-

0)]

-

? f

"

f" jf(0

rfa

sin a* sin

ax

z

dt

Find the Fourier transform g(a) of the function /(x) defined as follows :/(x) for x > a. < 0, /(x) = I/a for ^ x ^ a, /(x) Find lim g(a).

Find the Fourier transform <r* for x ^ 0. Show that

for /(x)

denned as follows:

-

for x

* cos ax

-f-

sin

ax

,

-w

ve~ x

.

for

x

< _ >

/(x)

=*

for

x

<

0,

ORTHOGONAL POLYNOMIALS, FOURIER 4.

-

Consider g(a)

e~ tat f(t) dt such that

/ J-

lim

Show

0.

f(x)

271

that

ar-

=

g(a)

6.17.

SERIES, FOURIER INTEGRALS

la

We

Nonlinear Differential Equations.

discuss the nonlinear

differential equation

=

^)

M

l

(6-87)

which can be written as the system dx di di/

= =yy = -x -

-j.

The

solution of (6.87)

f or

M

=

y

=

x

is

^=

(6.88) rf(x, y)

= A

cos

-Asin(t

(t

+

+

B),

B)

A

The method

and J5 in the hope that an of Kryloff-Bogoliuboff is to vary be found. This is essentially the to solution (6.88) may approximate variation-of-parameter method discussed in Chap. We write

x y

= =

5.

r cos 8 r sin B

and obtain dx

=

-7-

dr -j-

.

.

cos

r

sm

dt

dt

-dB 6 -rr dt

dB dr dy = __ _ sm<,_ rcos 0_ .

.

.

Equations (6.88) can now be written -77

cos B

r sin

-77

-r-

=

r sin B

=

r cos

at

dt

r cos

sin B

-77-

dr

so that

-77

-r For the range

M/(^ cos

dv

(Zt

<W j-

=

n sin 0/(r cos

0,

r sin 0)

0,

r sin 0)

(6.89)

= -r -

of values

n cos 0/(r cos

(r, 0)

fj,f(r

0,

-r sin

such that cos

0,

r sin 0) <3C 1

0)

^

:

1

ELEMENTS OF PURE AND APPLIED MATHEMATICS

272

one has to a

first

approximation

(6

=

Integration yields 6

~ Since

-^-

=

<C

M

+

(0o

We

t.

of (6.89)

replace

t)f(r cos (0

+

-r

t),

+

*

r(t+2ar)-r(t)

+

sin (0

27r). Integrating (6.91) on this basis (t, t stant on the right-hand side of (6.91)] yields

interval

+

by

t

M)

so that

0)

(6.91)

the value of r will not change appreciably over the

1,

en

sm

+

do

-

=M

&& (0o+0/fr

cos (0

[r is

assumed con-

+0,-rsin

t

I"

H

= where

The

sin

f(r cos ^,

27r)

-

r(Q

-rr

<K

-^r itself,

=

r(t) for

r(t)

-

=

1

-f

=

9

G(r)

cos

/

ZiTTT

and with

-

= I

I

ranging

can be replaced

~ F(r)

(6.94)

The integration of (6.94) yields the first approximation for same method applied to the second equation of (6.89) yields

(It

t

This yields

at least to a first approximation.

^

limits of

M

dr

by

new

"S^W

the average change of

1,

(6.93)

introduced the

represents the average change of

Since

27T.

+

r sin ^) <i^

r sin ^)

a?

UK

+

f(r cos ^,

\l/

--

\fr

Thus

- -t

d^ (6.92)

sin

I

r(t

from t to

r sin ^)

r 2v

=

periodicity of sin

^

^,

MF(r) F(r)

integration.

^ f(r cos

\f/f(r

cos

\I/.

r sin ^)

r(t).

The

d\I/

o

+

I

1

+ ^;<?(r) 1

*cos^/(r cos

^-,

<

rsin^)d^

(6.95)

(6.96)

OETHOGONAL POLYNOMIALS, FOURIER

SERIES, FOURIER INTEGRALS

273

Before applying Fourier-series methods to obtain improved approximawe discuss an example.

tions,

Example

Here f(x,

Van

6.9.

y) =

der Pol's equation

x z )y, and

(1

f(r cos t,

so that

F(r)

Equation

(6.94)

whose solution

=

r

r sin

P* sin

t

=

*

^(1

^)

-

=

(1

r2

cos 2

r

For

.

r

-

plane

(

From

x,

=

y

(6.95)

To

d^

nr

- (l

r==-

x 2 -f 37

we

j

we note

2 2/

>

(6.97)

we have

5^

lim r(0

r2

tf)

^)r sin ^

is

0, r(0)

Since

r 2 cos 2

becomes

r(0

At

is

that to a

first

=2 approximation the motion in the phase

resembles a spiral motion into the limit cycle circle x 2

obtain G(r)

0,

so that

0o -h

t,

-f

y

2 *

r2

=

4.

which yields nothing new.

obtain an improved approximation to (6.87)

we proceed

as follows:

Let x (t)

=

r(t)

cos ^(0

+

fux(r)

+

M

a n (r) cos n

sn *and let us attempt to make (6.98) a solution of (6.87). r(t) and 0(i) will be taken as the first approximation given by the solution of (6.94) along

with 4,

We

(6.95).

....

shall

n (r), n = 2, 3, attempt to find (r), a n (r), 2 From (6.98) we obtain will be ignored. /*

Terms involving dx

dr r

sm

d6 -TT

dt

\^ M

7

L^

dd

sm nOn -j.

ttn

dt

Zjn 00

n$ n cos n0 j-

(6.99)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

274

da da The term M TT = M -rdr dt mi

dr -TT

F(r) da M -7^ T"

=

at

The same reasoning

neglected.

da n M

__ ~~

dp n

__

*

,

r

,

,

tne order of M

s

j and so

is

applies to the terms involving 2

da n dr _ ~~

da n nrj Zx dr

dp n dr

M dp n

"Wdt

dt

-

.

1S

dr

2ir

,

ju

.

2

__

<

Differentiating again yields

j

dt 2

= -2 ~ -.7

-

sin

r cos

dt dt

n

-2

sin n^?

Terms involving

>

-

have been omitted since

=

Moreover

+

1

at

d<

when we

~

(6.100)

*

irr

G(r)

27T

2 neglect terms of the order of M

Substituting x(t) of (6.98) and

-

d r 2

of (6.100) into (6.87) yields 00

M(r)

-

-F(r)

sin

-

-

-C?(r) cos

ju

7T

7T

(n

/

2

-

l)

n

cos nd

~~q

n2 l)/3 n

sin M0

=

nf(r cos

0,

-r

sin 0)

(6.101)

n-2 In the term p,f(x, x) we have replaced x by r cos 0, x by 2 neglecting terms of the order of M Since the left-hand side of (6.101) is a Fourier series,

r sin 0,

again

-

right-hand side of (6.101) in a Fourier 00

/(r cos

0,

-r

sin 0)

=

a

(r)

+ V n-l

we expand

series, 00

a n (r) cos nB

+ V n-1

6 n (r) sin

n0

the

ORTHOGONAL POLYNOMIALS, FOURIER SERIES, FOURIER INTEGRALS with

Oo(r)

=

a n (r)

=

1

P*

ndf(r cos

/ * Jo

= /p*

^

0,

r sin 0)

dd

0,

r sin 6)

dd

cos nOf(r cos

/

coefficients of cos

Equating

0>

n =

...

1, 2, 3,

Jo

= -i p* sin

&n(r)

r 8 ^ n 0)

~~~

f( r cos

/

-1 TT

F(r)

{*

^ Jo

H~

275

nO and sin nO in (6.101) yields

-r

sin B f(r cos

6,

cos 6f(r cos

0,

sin 6) dd

Jo

G( r )

= P*

a (r)

= -

a n (r)

/

JZTT

=

=

The

*/( r cos

/

0,

-r

sin 0) d0

(6.102)

Jo

-

r~2

rr

L)

ir(n

0n(r)

d0

r sin 0)

yo

-7-5

TT

TT(n*

1)

cos

/(r cos

710

/

dB

r sin 0)

0,

JQ

n - 9 J,

r 2ir

sin

/

nOf(r cos

-r

0,

~

.

tJ,

4,

.

.

.

sin 0) dB

JQ

values of F(r), G(r) given by (6.102) agree with their previous It can be shown that x(t) given by (6.98) satisfies (6.87)

definitions.

with accuracy of order Example

6.10.

2 /u

In Example

^

for

<

t

-

6.9, f(x, x)

<*>

(x

.

2

-

l)i, so

that

(r

s

\

.

r

sin &

J

r8

sin

36

Applying (6.102) yields (r) (r)

Thus the improved

*(r)

- -

/3(r)

=0

first

x(t)

with r(0 given by

-

r(0 cos

2, 3, 4,

.

n -

2, 4, 5,

.

.

g

(tfo

.

.

is

approximation

-

n -

-M

+

g

sin 3(d

+

(6.97).

Problems 1.

sin a

The

differential equation of

^

amplitude. 2. Solve x 3.

Solve

s tf

/6,

+ x + iub\\ 4-

a simple pendulum

and show that the period

H- ^(sign

0, a;)a:

M

=

0,

is

of oscillation

1,

for the

A(

1,

&

-f-

(g/l) sin

=

0.

Take

depends on the square of the

improved first approximation. improved first approximation.

for the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

276 4.

Show 6.

2 = 0, M Solve x -h x -h M( /3x )x 1, for the improved first approximation. that a limit cycle occurs and that the radius of this limit cycle is r = \/4a/30.

+

Consider the differential equation 'x

+

x

-f

=

Mx 2

1

M

(6.103)

written as the system

dx

=y

-dt

(6.104)

---, such that, at

t

= ,

x

= x,

i/

=

i/o,

= A

x(t)

y

=

=

/f

gin

(

-f-

0),

2

=

=

z

-77

Tin solution of x 1

ZD.

B

-f

=

B

=

is

y

cos

cos

4- x

(/

(/

-f

-f

C)

O

Tins

the

suggests

trans-

formation

w

x y z

in

a

= =

r

7

sin

/

cos 6

cos

an attempt to find an approximate solution of approximation

Let the reader slum that to

(6 104)

first

e (h

=

t

+

= WT

dt

and

iu(0

=

~~

~

"

o

Wltil

r

^ +

2

^o)

=

-

~

^

REFERENCES Carslaw, H. S.: "Introduction to the-Fouiier's Series and Integrals/' Dover Publications, Inc.,

New

York, 1930.

Chin chill, R. V.: "Fourier Scries and Boundary Value Problems," McGraw-Hill Book Company, Inc New York, 1941. Conrant, R., and D. Hilbert: "Methoden dcr Mathematischen Physik," Springer,

Veilag OHG, Berlin, 1931. Franklin, P.: "Fourier Methods," McGraw-Hill 1949.

Book Company,

Minorski, N.: "Introduction to Nonlinear Mechanics," J Inc.,

Inc.,

New

W. Edwards,

York,

Publisher,

Ann

Arbor, Mich., 1947. N.: "Fourier Transforms/' McGraw-Hill Book

Sneddon, I. Company, Inc., New York, 1951. Stoker, J J.; "Nonlinear Vibrations," Interscience Publishers, Inc New York, 1950. Szego, G.: Orthogonal Polynomials, American Mathematical Society Colloquium, ,

vol. 23, 1939.

CHAPTER

7

THE STIELTJES INTEGRAL, LAPLACE TRANSFORM, AND CALCULUS OF VARIATIONS

Functions of Bounded Variation. In an attempt to define arc length of a curve one is led to consider functions of bounded variation. Let us consider a simple curve, F, given parametrically by x = f(t), Two distinct values of i are assumed to yield two <p(), a ^ I ^ /3. y distinct points on F, so that as t varies from a to 0, the point P with coordinates x = f(t), y <p(i) moves continuously from one end of the curve to the other without retracing its motion. We now subdivide the 7.1.

interval (a, 0) into

=

a

For t Pit on

Pk

is

to

tk

F,

A*

<

ti

<

tz

<

<

'

we have the point

=

0, 1, 2,

.

.

.

,

Pk

n.

tk

<

t

<

k+ i

<

'

with coordinates x k

The

=

in

=

f(tk), yk

=

<p(4),

straight-line distance from 7V_i to

given by

The sum

total of these straight-line arcs

is

Now

<

Jn-i

if

for

all

manner

of subdivisions of

a

^

t

2

?(fc-i)]

^

ft

(7.1)

}*

a constant

A

exists

such that 71

I

*"' (7.2)

< \/2 A. Since the set of numbers {S n is bounded, of necesa least upper bound (supremum), L, will exist for this set. We define L as the length of arc of F. For a discussion of the supremum see Chap. 10. Conversely, one easily shows that if the {S n are bounded for then

<S W

}

sity,

\

ELEMENTS OF PURE AND APPLIED MATHEMATICS

278 all

A exists satisfying (7.2) for all subinequality of (7.2) leads to the following definition: Let f(x) be defined on the bounded interval a ^ x ^ I). If a constant exists such that for all possible finite subdivisions of (a, b) into subdivisions then a constant

The

divisions.

A

XQ

=

<

a

<

Xi

x2

<

'

xk

'

<

<

<

x n -i

xn

=

b

we have n

T W

/(.r*)

-/(A

,)|

< A

(7.3)

fc-1

we say that /(x) is of bounded variation on a ^ x ^ b. A bounded monotonic nondecreasing or nonincreasing function is always of bounded variation. If f(x) is a monotonic nondecreasing function,

then

!/(**)

Thus

-

bounded variation

derivative for a

^

^

x

[/(**)

A >

(7.3) is satisfied for

tion of

=

/(**-i)|

is

<

/(**_,)]

=

/(&)

-

/(a)

Another example

/(a).

f(b)

Assume

the following:

f(jr)

of a func-

has a continuous

Then

/>.

~ -/to-i)| = |to since |f(x)|

-

Jf for a

x

A-i)/'(fe)|

< M(x k -

.r,

Thus

b.

n

I/to)

-

/to_0| <

M

- ^-i = M(b -

k

a)

which proves our statement. Continuity of f(x) variation.

^x g

is

not sufficient to guarantee that /(a*) is of bounded = x sin (l/:r), x 5^ 0, /(O) = 0,

For example, consider f(x)

2/7T.

Since lim /(x)

=

x->0 /(x) is continuous at x

=

lim x sin ^ x-0 0.

x

7^ 0.

Let us subdivide

(

\

Then

lim

0.

|x|

=

=

/(O)

a

Moreover /(x) /

for

g

2\ -

1

'

7T/

is

/ into

(

\

easily seen to be continuous

2

0,

^ -----, (2n + l)r

~2 2n?r'

.

2\ -

.

'

).

TT/

THE STIELTJES INTEGRAL

279

00

V/ ^

Since

Lj =

i

2fc

diverges, no constant r-y 1

A

exists satisfying (7.3) for all

~T~

fc

modes

of subdivision.

A

fundamental result concerning functions of bounded variation is the following theorem A necessary and sufficient condition that /(x) be of bounded variation on a ^ x ^ b is that f(x) be written as the difference of two positive That the condition is sufficient moiiotonic nondecr easing functions. follows almost immediately from previous considerations concerning Now let us assume that f(x) is of moiiotonic nondecreasing functions. bounded variation on (a, 6). Let x be any number on the interval (a, 6). Let us subdivide (a, x) into :

a

=

Xo

<

Xi

<

X2

<

'

<

'

'

<

Xk-l

Xk

<

'

'

'

<

Xn

= X

Then n )l

Some

of the

terms of 8 n are such that/(x r )

<

are such that f(x 8 )

We

/(x_i).

Kn = where

sum

P n is

of

^

< A

(7.4)

/(x r _i), whereas other terms

write

Pn +

N

n

sum of terms of 8 n for which f(x r ^ /(x,_i) and A M is the terms of S n for which /(x ) < /(x __i). One easily shows that

pn - Nn =

T

the

)

g

fc

f(x)

-

/(a) so that

Sn =

/(x)

=

Sn

-

/(a)

+ 2N +2P n

-/(X) +/(a)

U

'

J

n

Since S n < A for all methods of subdividing the interval (a, x), we know from real-variable theory (see Chap. 10) that a least upper bound exists We call this least upper bound, F(a, x), the total for the set [S n \. The suprema (least upper variation of f(x) on the interval (a, x). bounds) of {N n and {P n are written as N(a, x), P(a, x), respectively. From (7.5) we have }

}

V(a,x) =/(*) -/(a) 7(a, x) = -/(x) + /(a) so that

/(x)

=

/(a)

+

P(a, x)

-

+

'

2P(a, x)

AT(a, x)

(7.7)

From the very definitions of P(a, x), JV(a, x) we note that they are monotonic nondecreasing functions of x [see (7.6)]. /(a) + P(a, x) is monotonic nondecreasing, which proves the theorem.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

280

Problem 1. Let f(x) = sin x for ^ x ^ 27r. Write f(x) as the difference of two monotonic nondecreasmg functions of x for this interval. Problem 2. If f(x) is of bounded variation and continuous for a ^ x ^ b, show that P(a, x) and N(a, x) are continuous for a ^ x ^ b.

An important generalization of the RieThe Stieltjes integral is defined the Stieltjes integral. integral as follows: Let f(x) and g(x) be real-valued functions of the real variable The

7.2.

Stieltjes Integral.

mann

is

=

a

x

^

b.

#o

<

Xi

^

x for a

Subdivide the interval

<

and let 5 be largest form the sum

Xz

<

-

of the

-

-

<

x k -i

(a, b)

<

numbers x k

xk

into

<

<

x k -i,

fc

=

1, 2,

=

n

.

.

b

.

,

n.

Now

n

(7.8)

)]

^

where

any number such that

is

j>-i

^

^

{*

j^.

If

lim 7?

>

8n

exists

OO

independent of the choice of the & and the method of subdivision, provided 5 > 0, we call this limit the Stieltjes integral of /(x) relative to g(x) on (a, b), written b

f f(x)dg(x)

(7.9)

Ja

In the special case g(x) function of

m[g(b)

=

bounded and

If f(x) is

then S n of

x,

-

g(a)]

g

x, (7.9) if

Riemann integral. bounded monotonic nondecreasing

reduces to the

g(x) is a

(7.8) satisfies

m

k [g(x k )

the following inequality,

-

g(a)\

k=l

where 0:^-1,

m

m

and

k is

infemum (greatest lower bound) of f(x) for x k -\ ^ x g supremum (least upper bound) of /(#) for x _i g x g x^, the infemum and supremum of f(x), respectively, for

the

is

the

M

are

AT*

fc

n

a

^

x

^

b.

The supremum

of the

sums

)

nik[g(xk)

g(x k -i)] can be

*-' called the lower Darboux-Stieltjes integral, L,

and the infemum

of the

n

sums ^

M

k \g(xk)

g(xk-i)]

can be called the upper Darboux-Stieltjes

THE STIELTJES INTEGRAL integral, U.

If

these two integrals are equal, and write

281

we say

that the

Riemann-

Stieltjes integral exists

b

= [/=

f(x)dg(x)

f Ja

This definition is easily seen to be equivalent to the one given above. ~ U. If f(x) is continuous in (a, 6), it is a simple matter to prove that L Now if g(x) is a function of bounded variation, it can be written as the It follows immedifference of two monotonic nondecreasing functions. diately that (7.9) exists if f(x) is continuous and g(x) is of bounded variation.

= OforO ^ x < -5-, Let f(x) be continuous for ^ x 1, and let g(x) x ^ 1. For any subdivision not containing x = g we have dg(x] = 0. 1. Thus S n is any The subdivision covering x = ^ yields dg(x) /(), where Example!. 1.

g(x)

=

^

Ifor-g-

number near x =

K

7J

Problem

3.

Let/(z)

=

/() =

hm

Since

\.

we have

i/(x)

=

2

x for

f(x) dg(jr)

Problem

If f(x)

4.

and

b

f(x] dg(x]

f Ja

^

x

f(x) dg(x]

=

/(--)

=

^

Show

1

that

=1

are continuous for a

<7'(^)

/

./O

+ i,

J

/(g-),

^

x

^

h,

show that

b

f f(x)g'(x)dx Ja

The last integral is a Ricmann integral Problem 6. If f(x) and g(x) have a common point Stieltjes integral off(x) relative to g(x)

of discontinuity, show that the does not exist provided the range of integration

covers the point of discontinuity. Problem 6. If h(x) is nondecreasing, f(x) and g(x) continuous with f(x)

^

g(x),

b

show that f

Ja

f(x) dh(x)

^

g(x) dh(x).

[*

Ja

n

Problem

Sn =

Let

7.

Y

g(h)[f(x k )

-

/(arfc-i)],

Xk-i

^ & ^

xk

.

Show

that

i

n-l

with #o

=

a, x,t

continuous at x

Problem a(x)

8.

/3(x) -f

da(x)

=

6.

Assume

a and x

=

g(x) of

6.

Show

bounded

variation, /(#) continuous,

Let /(#) = g(x} H- *^(a;) be continuous for a Show that ^T(^) to be of bounded variation.

P Ja

fir(a;)

d|3(a:)

-

T h(x) dy(x)

Ja

and

that

+ i Ja f* g(x)

^

x

^

dy(x)

+

fc,

i

and assume

^

Ja

h(x)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

282 Problem

Consider the sequence of continuous functions, /(x), n

9.

1, 2,

.

.

.

00

Assume

y

^

fn(x) converges uniformly for a

Show

variation for this interval.

f

b

Jo,

x

^

and

6,

let g(x)

be of bounded

that

00

00

Y

V

f

lm*i

1*4

Ja

/oo

b

under what conditions

f(x] dg(x), arid

i

would

this integral exist?

The Laplace Transform.

7.3.

real variable

t

defined for

^

finite interval is

continuous for

t

all

^

t

Let Let

0.

be a complex function of the

g(t) g(t)

be of bounded variation on the

=

The function e~ zt with z R, R arbitrary. so that the integral

^ t,

x

+

iy

(7.10)

exists for all

complex

may

be that, for a given value of

lim

f*

we

the case,

If this is

exists.

It

z.

e~ zt dg(t)

(7.11)

write

=

6" dg(t)

Jo

z,

Jim f*

<r

(7.12)

dg(t)

(7.12) is called an improper integral, arid the right-hand side of called the Cauchy value of the improper integral.

Equation (7.12)

is

Those values

of z for

which

(7.12) exists define a function of

2,

written

(7.13) f(z) is called the Laplace-Stieltjes transform of g(t). consider now the region of z for which /(z) of (7.13) exists.

We

we

investigate three special cases.

monotonic increasing, and hence

R

Since dg(t)

arbitrary.

fR /

,

. ,

N

,

e -(*+*v)t e *' dt

=

fR

-.

/

Jo

Since

e*

Jo xt increases

e

Let

of bounded we have

dt,

,

,

e e<-xt

cos yt J dt

reader to show that lim

/

for

-

any

I

e

u

.

fR e

i

e

Jo

fixed x,

First

du, so that g(t)

^

variation for

is

et

beyond bound F

g(f)

=

,

xt

t

^

is

72,

sin yt J dt

we leave

it

to the

R e~ zt e et dt fails to exist for

all z.

As a second

THE LAPLACE TRANSFORM

=

let g(t)

example,

exists for all

exists,

-

R

/

Rl

>

z

=

=

with z such that

+

XQ

S

u

cr zt dt

/

=

so that dh(u)

=

/(z) exists for

We

Rl

>

2

0.

f

assume that

e~~

/

Zi)i

dg(i)

A

< A

(7.14)

c-'- dg(t)

(7.15)

dg(()

p- z

"

|o

^(0 =

and

</ST(M)

"

=

Integration by parts (see Prob. K

=

c~ 2t dg(t)

>

z

ctt

z

.

Further, let us assume that a constant

IIJQ.

111 2

e~ (z

then

,

~z

)R

lira ^-^oo

7,

f*

e 1* dh(t).

B

e- dg(t) =

Rl

'

a general case.

h(u)

If

-*&

lim

Define h(u) by the equation

0.

[

=

z

e~ z

e-" tor

-

R

I

er** dg(t)

/ JQ

<

so that

t,

C

lim

Hence

0.

now

Let us consider

exists

#

e~ zt dg(t)

=

Let the reader show that lim

du.

Finally, let g(t)

f

provided x

u

yo

z.

lira

e~~

I

283

"

e-f

|o

Then

dA)

(7.16)

Sec. 7.2) yields

cr*<*

dg(t)

e- (M8) * f

+

-

(z

B

=

e- 20< dg(t)

70

z

f* A(0c-

)

(

"

o)

dt

since

0,

lim R-*

R

and

f-'

f

r

< A

dg(t)

lAWIc-^-'^'rf^

<

for

A/(x

-

oo

x

Moreover

R.

all

for x

),

>

x

I

f

h(t)<r

Allowing

.

/f

(

*- zo)t

dt

become

to

infinite in (7. Hi) yields

(2

for

Rl

z

>

Rl

Rl

exists for

u

1

=

>

#o

+ y

<

Rl

20.

We

=

oo

r-"

(7.17)

rfj/(/)

have shown that,

then/(2) of (7.17)

There

XQ.

<

>

2

iyo,

20)

Thus

20.

f(z)

2-0

-

is

well defined

(7.17) converges for 2

be singularities of /(z) on the and in the half plane Rl 2 < x

may

.

,

if

f or

=

#

+

line

js

iz/

=

provided XQ

+

iy,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

284

The ordinary Laplace transform

=

F(z)

of f(f) is defined as

=

L[/(0]

erf(t) dt

(7.18)

provided the improper integral exists. If the Laplace transform of also exists, one can integrate by parts to obtain

W(*)] = JO L ^"/'(O

=

dt

=

jLF/*'^)]

zZJ/Yrtl

provided

(7.20) will certainly hold

if

z

e-/(0

/

70

/y Xty jgx/ v'*

/(O)

(7.20)

\f(f)\

bounded

is

= =

~ /'(O) - zf(0)

zL[f'(t)]

z*L(f(t)]

We

7.2.

Example

find the Laplace transform of

=

F(z)

by parts

Integration

zi

if

0,

sm

We

at.

/(/) is

the inverse

have

e~ zt sin at dt

I

^.-~ 2

-

sin at dt

a 2 -f

)

z

a cos

(~~z sin at

at}

t=o

we have r

a

oo

e~ zt sin at dt

I

Jo

Table

and

yields

/e~ >

^

(7.21)

the Laplace transform of f(t), we say that of F(z), written f(t) = L" 1 ^^)]. transform Laplace

2

t

-/'(O)

If F(2) is

Rl

for

Further application of (7.19) yields L[/"(0]

If

d*

=

lim e-f(t)

Equation Rl z > 0.

+

e-'/(0

f'(f)

a 2 H;

,

22

Laplace transforms for some simple cases of /().

7.1 lists

Problem Problem

11.

Derive the result* of Table

12.

Show

Example

7.3.

7.1.

+

= aL\f(t)] -f that L[af(t) bg(t)] bL[g(l)]. Let us consider the differential equation

subject to the initial conditions y(0) = i/o, 2/'(0) = t/oAssuming that the solution and its derivatives are such that their Laplace transforms exist, we can apply

of (7.22) (7.19)

and

(7.21)

and the

z*L[y(t)\

,,,

so that

rr /A1 L[y(t)]

=

-

(z .

.

,

(z

result of Prob. 12 to obtain

-

zy*

+ 3zL[y(t)] -

y'

w -1) o

-f4)(z

3/05

2/o

rr

mi

L[<pi(t)]

+ ,

- --+ -

5

4 2/o

3y Q

-

4L[y(t)}

-

Q

z

+ 5

2/o

4

-h

5

2-1

THE LAPLACE TRANSFORM From Table (7.22)

7.1

we

see that

e" 4*,

<f>i(t)

<f> z (t)

285

so that the suggested solution of

e*,

is

y(Q

(7.23)

5

One

It is to be noted that the easily checks that y(t) of (7.23) is the required solution. Laplace-transform method for solving (7.22) introduces the initial conditions in a natural manner.

TABLE

Problem

13.

0, 7/'(0)

Problem

14.

Solve

-j-|

+

2^

7.1

= l~3a; by

the

Laplace-transform

1.

Let /(Q

I

provided the integrals

-

=

for

<

0.

Show that

" *) dt

exist.

/"

6~^/(

-

* 05)

d<

e-

I

e~ rf

method,

ELEMENTS OP PURE AND APPLIED MATHEMATICS

286 Example

Let,

7.4.

an

^

t

We

0.

elastic string

assume

y(,

from x

~

~

=

subject to the boundary conditions y(x, 0) for

wave equation

us attempt to solve the

Inn y(x, a?-* *

t)

=

to x

-

~

t)

-

for

---\

and

for all x,

^

t

0.

t

t

~

f*

8

/ provided we assume T-^ "X J

0*-"

y(x,

y(jc,

t)er

zt

=

dt

of (7.24)

'

,

t)c-

B can be functions of z.

and

^4.

0.

At x

*

=

J

f(t)

we have

oo f

we obtain

t}}

e'^

Thus

dt.

X

Lf?/| satisfies

(7.24)

/,(//]

- Xe

dt

Since

(

/e )

+

Be-<">*

(7,25)

tends to zero as x becomes infinite

t/(ar,

e'" dt

we have

Prob. 14

"

f

--f>

=

we need

f From

-

/

*

where

z*L\ii(x,

t)

is

f

we choose A -

-*

JQ

,L(y]

A solution

=

dt

y(Q,

origin the string is function of time.

move in such a manner that y(Q, t) = /(/), f(t) a given to If we multiply the wave equation by e~ xt and integrate from constrained to

~

-^

Physically,

At the

<, initially at rest.

=

2/(a;,

Oe""

rt

dt

-

c~w / c

" /"

/(O*"*

- a;/c) for t ^ a?/c provided f(t - /c) (7.26) suggests that 2/(z, /) - /(/ x/c (see Prob. 14). It is a simple matter to show that/(* x/c) satisfies the equation and the boundary conditions. Of course one needs the fact that f(t) be

Equation for

t

wave

<

twice differentiable.

Example

7.5.

We wish

to find the function f(t) such that

ftt)e-dt

-

-

2

(2

1)-!

(7.27) 00

Let us assume that f(t) has the Taylor-series expansion.

/

a nt n

Without

.

nt'o tion let us

assume that term-by-term integration

is

Hence

permissible.

n0 Now

for

\z\

>

1

we know

that

1

x/inri

1

^ " 2

Vi -

^ * (i/2)

2

V ^^ U

Wl

(2m)!

1

justifica-

THE LAPLACE TRANSFORM Comparing the two Laurent

we

series,

-

see that a n

-

=*

287 n

if

is

odd and,

if

n

2m,

80 that

jjjoam' oo

=

We

==

note that/(tO

order zero of the

=

One can

kind.

first

^-j~j

(~\

~f~T

f

y

QV

Jo(t),

(

where

is

(0

J"

? 28 ) -

the Bessel function of

start with/(2) of (7.28), justify the interchange (7.27) results.

and summation, and show that

of integration

Problems 1.

Solve

2.

Solve

3.

Solve 2x

^r*

+

-~ -

=

y

(

2

4//

+

^

with

sin x

=

~

=

j/(0)

=

3e zx with y(0)

-

Am.

2z,

0)

2/(z,

y(x,

1

t)

0, j/'(0)

-

=

0.

=

1.

f

0, y (0)

-

1, t/(0,

+

<

for

1,

<

t

By

5.

From

Ans.

r

oo

e~~

/

zi

t

cos a< CM

f(t)

=

22 T~T~I 2

-

1

+

x 2 for

*

>

x2

.

f(t)

-

J

sin a* (see

Table

7.1).

a2

+ o ?T

(2

9 2 2 )

such that

I* e-J(t)dA 7.4.

t)

if

Prob. 4 show that

yo

Find

t).

x* y y(x,

the inversion formula (see Sec. 7.4) solve for/(Z)

4.

6.

for y(x,

The Inversion Theorem.

Z .

2

Let 0(0 satisfy the requirements which

enable one to write

0W (see Sec. 6.16).

choose ^(0

=

t

1 = o-

*

/ zrj -

We

00

^(0 cos

/

y (^ ""

dt

(7.29)

J-oo *

assume that

e-**f(() for

=

T dy

o~

t

^

0,

dv

/

flf(0)

=

dw converges

for

g(t) sin

I

J-

2arJ-*

f(w)

I

we have i

/ .

iw

r-

-i r+-*]' Z*JJo O

t

<

v(w

0.

absolutely,

Since

f)

dt

and

ELEMENTS OF PURE AND APPLIED MATHEMATICS

288

Now

assume " dt

exists for

Rl

x

z

>

0.

IV

F(z)e* dz

Then

=

iy

= on letting

become

z

=

infinite

x+iv ( Jx~iy e

"

e> dz f JO

e-'f(t) dt

w

(7.31)

x + iv, v a new variable of integration. On letting y and comparing (7.31) with (7.30) we see that

/<) = 0=

F(z)e

wz

dz

(7.32)

the Laplace transform of f(w), (7.32) enables one to find /(it?) terms of F(z). This is the Laplace-transform inversion theorem. Equation (7.32) can often be evaluated by the calculus of residues. If F(z) is

in

Example Then from

Let

7.6.

I-

>

with x

=

F(z)

l/(z

+

1)

(7.32)

0.

1

e

wt

We consider U)

the path given

m

Fig. 7.1.

,

dz around

Let the reader

show that, as the radius of the semicircle becomes infinite, the integrals of e wz /(z + 1) tend to zero along BC, CDE, EA. The resiThus due of e m /(z + 1) at z = -1 is e~ w .

f(w)

7.5.

The

= 2~^

I2irt~"1

The Calculus

of

*

*~~

Variations.

calculus of variations owes its

beginning to a problem proposed by Johann Bernoulli near the completion of the seventeenth century. Suppose two points to be fixed in a vertical What curve joining these two points will be such that a particle plane. sliding (without friction) down this curve under the influence of gravity This is will go from the upper to the lower point in a minimum of time? FIG. 7.1

the problem of the brachistochrone (shortest time). A problem of a similar nature is the following: What curve joining two fixed points is such that its rotation about a fixed line will yield a minimum surface of revolu-

THE CALCULUS OF VARIATIONS

289

This is the soap-film problem. A third problem asks the following What curve lying on a sphere and joining two fixed points of the sphere is such that the distance along the curve from one point to the other is a minimum? This is the problem of geodesies. Let us obtain a mathematical formulation of these problems. Let the particle P move along any curve given by 1. Brachistochrone. tion ?

question:

y

=

The speed

y(x) (see Fig. 7.2).

=

or -y

\/2g

of the particle is given

by

v

z

2gy

Hence

y*.

+

ds

dy

1

2

T~(2/T dx

The total time of descent is given by

+

= -!_

7>

(y'Y'

dx FIG. 7.2

(7.33)

To

2.

this

(7.33) a

is

If the curve y(x) lying above the Surface of Revolution. rotated about the x axis, the surface of revolution generated in

manner

is

S =

To

=

y ds

27T

3.

must

find y(x)

(7.34)

such that S of (7.34)

minimum. In a Euclidean space

Geodesies of a Sphere. ds*

For a sphere x ds 2

If

(y'Ydx

27r

solve the soap-film problem, one

a

find the function y(x)

minimum.

Minimum

x axis

is

must

solve the brachistochrone problem, one

which makes

^

-

<^?(0)

=

r sin

r 2 (d#

is

2

+

=

cos

sin 2

dx 2 <p,

d^

z/

2 )

+ = =

dy

2

+

dz z

r sin

r2

1

we have

sin ^,

+

2

sin 2

=

r

(^j

cos

0,

dd 2

so that

.

any curve on the sphere, the distance between two points by ^ = <p(8) is given by

of the sphere joined

L =

dO

(7.35)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

290

To

L

find the geodesies of a sphere, one

of (7.35) is a

Formulas f(x, y, y')

=

mine y

will

(7.33) to (7.35) are special cases of a

be a function of the three variables x, = y'(x), such that y(x), and hence y'

f(x,

x

y')

!/,

=

//,,

y

6,

=

yfa)

= <p,

=

such that

<?(6)

-

2wy[l

For

2/2.

+

1

P,

y

'.

2/0

2/,

to deter-

subject to the restriction that

maximum)

/(*,

Let

case.

We wish

(7.36)

(7.33), (7.34), (7.35)

2

(2/')

more general

?/,

n f(x,y,y'.) dx f j 71

be an extremal (minimum or

=

<p

minimum.

=

y(xi)

attempts to find

=

r

we have

A/T+sin 2

2

z(2/')

dx with

respectively.

Let us see how a problem in the calculus of variations differs from an extremal problem of the ordinary calculus. In the latter case we are given the function y = y(x). For each real x there corresponds a unique Thus y = y(x) maps a set of real numbers into another real number y. The relation y = I/a:, < x g 1, maps the set of real numbers.

< x ^ 1 into the interval y ^ 1. A simple problem in the ordinary calculus is to find a number x which yields the minimum or Now (7.36) may also be looked upon as a maximum value of y y(x). mapping. For any function y(x), (7.36) defines a real number /. Thus

interval

a mapping of a function space [the space of y(x)] into the realOur problem is to select a member of the function space which yields a minimum or maximum value for / of (7.36). To resolve this question, we reduce the problem to one of the ordinary calculus. Let us assume that there exists a 7.6. The Euler-Lagrange Equation. (7.36) is

number

space.

function y(x) which class of functions

makes I

of (7.36)

Y(x, X)

=

an extremal.

y(x)

+

We now consider the

\n(x)

(7.37)

an arbitrary differentiate function such that r\(x\) = 0, and X is a real parameter. We have Y(x\, X) = y(xi) = T/I, r?(x 2 ) = = 2/2. For X = we have Y(x, 0) = y(x). As we vary y(x%) F(# 2 X) X and rj(x), we obtain a family of curves passing through the two = yields the desired given points PI(XI, y\}> Pi(x^ 2/2). Moreover X = The value of / for any member of (7.37) is y(x). curve, y

where

77(2)

is

=

,

/(X)

For any

fixed rj(x]

=

f **f(x, y

Jx\

we know that

7(X)

+ is

9

Xn,

y

+

XT/')

dx

an extremal for X

(7.38)

=

0.

From

the

THE CALCULUS OF VARIATIONS dl

r)/

/)/

=

calculus, of necessity,

291

Assuming continuity

0.

x~o

and

of

dy

.

dy'

we have

d\ fXt

51

_

^-r>dx

upon integration by

it is a

Since

parts.

J / 3f

(| J XI dx \dy'

^ (7.39)

/

simple matter to show that,

rj(xi)

=

77(0:2)

M(x)

if

is

=

we have

0,

continuous on Xi

^

x

2

then the vanishing of

We

M(x)ij(x) dx for art)itrary

/

yri

Hence

leave this as an exercise for the reader.

implies is

if

M(x)

(7.41)

dy

\dy'J

is the important Euler-Lagrange equation. written as a second-order differential equation in the form

Equation (7.41)

a 2/

i/

*-*

./

I

j

^

{

i

we

1

differentiate /

satisfying (7.41),

y

-1 -,

It

=n

_

dx~ y'

dy 'fy'dx If

0.

the required

_da;

=

y(x) satisfy the differential equation

must

solution, of necessity, y(x)

??(x)

^ x^

can be

(7.42)

dy

with respect to x along the curve y(x)

we obtain

~ dx

by

makiiifi;

dij

dx\dy'

dy'

a//'

dx

Thus

use of (7.41).

d (7 43) '

for

an extremal.

of (7.41)

= j-

If/

given by

and

is

-,

explicitly

=

independent

constant.

If

/

is

of y,

one has a

explicitly

first

integral

independent of

x,

(7.43) yields the first integral,

/

-

y'

~

=

constant

(7.44)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

292

One may obtain

the Euler-Lagrange equation from the following point = x* 2x, so that

of view: First let us consider the equation y(x)

y(x

=

dx)

-

(x

Differentiating with respect to X yields

=

(3x*

dy

we have seen that

2)

dx

= c/A

=

dy,

2(x

3(x

+ +

\ dx) X dx) 2 dx

th. differential

can be looked upon as a mapping

(7.36)

2 dx, so

of y.

Now

of a function

space into a real-number space. We call this mapping a functional of y, written 7 = I[y], The first variation, or differential, of 7 may be defined as

=

lim

(7.45)

OA

where 8y(x)

is

any variation

I[y

+ \8y]=

in y(x).

f(x,

\

y

Applying

(7.45) to (7.36) yields

+ \dy,y'+\ dy')

dx

'

(7.46)

dl

= /"'

Jxi r

Integrating

7 8y'

provided 8y(xi) 67

=

=

dx by parts yields

8y(x 2 )

=

0^

Equation

(7.41) holds, of necessity,

if

for arbitrary dy.

In the calculus it is necessary to examine the second derivative of y(x), and, at times, higher derivatives, in order to determine the type of extremal (maximum, minimum, point of inflection) encountered at the fij

point x

for

which

-^-

=

0.

Similarly, in the calculus of variations

necessary to examine the second variation in 7 to determine what type of extremal is obtained from the solution of the Euler-Lagrange Let the reader show that the second variation of 7 can be equation. it is

written as 527

-

'

5y

dx

(7 47) -

THE CALCULUS OF VAKIATIONS Example

Equations (7.33) and (7.34) are special cases of

7.7.

=

/

We have/ =

293

g(y)

-f (y')

[1

~

2

]l,

vi +

g(y)

/

=

so that a first integral

0,

ax

(y')*dx

is

obtained from (7.44).

Let the reader show that (7.44) yields

=

f ^L==^.

=

From tan

we have

-

cos

l/\/l

Moreover

gr'(?/)

dy

=

a sec

(?/)

-j-

=

</(*/)

constant

2

=

a

so that

a sec

(7.48)

=

d0 from (7.48), and dx

tan

dx

=

a sec ----

dy so that

cot

dO

\~

77

g (y)

yielding

-6 + Given

0(t/),

one solves

j as a function of

0.

=

I/a

2 ,

from

Thus

(7.48).

--

g'(y)

Then integration of (7.49) yields (7 48) for y as a function of 0. This parametric representation of x and y as functions of yields

the required curve which extrernahzes In the brachistochrone problem g(y) c

(7.49)' ^

Jo

b

-

/(*/)

=

re 2c

cos 2

\

/.

=

y~~%,

-|//

so that

0d0 =

-

b

Equation y

=

7/(0)

=

6

I

-

C

(1

+

(20

r

=

cos 2

cos 3 0)~ l

,

(r/2)(l -f cos 20),

and

(7.49) yields

re c

d0

-f cos 20)

(1

Jo

Jo x(o)

=

?/

= (-2r*

*

+

sin 20)

(7.50)

cos 20)

(7.50) is the parametric equation of a cycloid

We

Example 7.8. Vanable-end-point Problem. = (p(x), and the functional 1

=

are

fx 2 = b F(x,

/ Jxi

y, y')

given

the

fixed

curve T,

dx

\Ve wish to find the curve y y(x) joining the fixed point A(XI, y\) and B(b, <f>(b)), B is a point on F such that 7 is an extremal. The coordinate x = b is unknown.

where If

we

consider the curve y(x) -f dy(x),

we have fe y

The upper on y

=

+

2/

-

H- fy')

changed since the end point of y(x) Let the reader show that

limit has

^>(#).

8y]

r

4- *y,

/[y]

-

[F(x, y

Jxi

+

by, y'

+

by')

-

F(x,

+

8y(x)

y, y')}

F(x, y

^

-f-

is

constrained to

lie

dx By, y'

+

y') cte

(7.51)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

294 From

(7.51) it

is

logical to define 57

= f

SI

Now

dy

=

at x

=

b

(~ dy + |^ By'} dx + F(x, We

a.

=

y(b)

Thus

<p(b).

y(b -f

mean one

must compute

+

y(b

the law of the

by

+

dx) -f dy(b

-

SJT)

easily

+

y(b)

=

by at x

Sx)

+

dy(b

=

=

dx)

We

b.

+

<p(b

dx

y, y')

(7.52)

have

dx)

+

<p(b

te)

-

*>(&).

Applying

shows that

=

-

[v'(b)

y'(b)} dx

b

Integrating the second part of the integral of (7.52) by parts yields

-JL'K-(S)] *<+['+ 1 "-">]..." If

6/

=0 for arbitrary

8z/,

of necessity

\F L

+

-

(^'

y')

along with the Euler-Lagrange equation. Let us apply (7.53) to the problem of the

We

arbitrary.

y, y')

=

-f (y')*\*

+

have F(x,

+

y\l

=0

IJ1 o^ Ja; = 6

minimum dF

2

fe')

]^,

(7.53)

surface of revolution with

=

<p(x)

2 yy'[l -f fe') ]^, so that (7.53)

^7

becomes !?/[!

or

t/

V

=

1

at x

=

6.

yi/'[i

+ (v'W-W -

=

f

y

)\*-i,

o

Hence, at their point of intersection, y(x) and

<p(x)

intersect

at right angles.

Problems

=

1.

Show

that the solution of the soap-film problem

2.

Show

that the geodesies on a sphere are arcs of great circles.

3.

Derive

4.

Show

7

/

-. dx

-

\dy'J

dy

that there are no extremals or stationary values for the functional

l[y] 6.

a cosh

(7.47).

-

Show

y

that the Euler-Lagrange equation for the functional

dx* 5.

is

T2

=

y dx

/ Jxi

Consider the functioniil

/ Jt\

f(xi, x%,

.

.

.

,

x,

xi t

x2

,

.

.

.

,

xn

,

t)

dt

THE CALCULUS OF VARIATIONS Show

295

that the Euler-Lagrange equations are

why

Explain

it is

f(Xi, Xi,

,

necessary that

Xn

t

XJl, XJ* 2

(7 54) of

7.

Apply

8.

In Chap. 3

,

,

=

X.T rt ,

.

A/(:ri, 0-2,

.

,

XM

,

J'i,

J

i,

.

.

.

,

Xn

,

t)

Proh 6 to

we saw

that Lagrange's equations of motion could be written as

dt

For a conservative system, T

motion.

motion

for a conservative

condition

T

-f

V =

system

is

-f

dqi

V =

such that

to

be an extremal,

=

I[z]

=

z

z(x,

F

II

must

//)

dF

_

I x,

=

2 T dt

is

Show that New toman

h

an extremal subject

to the

with

p r

The Problem

y,

C

z,

as

J its

dy dx, the region of

boundary.

Show

that, foi

satisfy

d ZF

_

dx Op

dz

7.7.

an extremal for Newtonian

dt is

constant /

S integration, S, having the simple closed curve f{z]

L

/tz

constant

Consider the functional

9.

\dqtj

dz = ~>

dx q^

of Constraints.

=

J>*F_ dy dq dz in

Fr

,

dy

In the ordinary calculus one solved

Given the function of two variables, z = f(Xj y), at what point P(x, y) is z an extremal subject to the conWe Enow that if /(x, 2/JT.s continuous on straint <p(x, y) = constant? the closed and bounded curve C, ^(x, y) = constant, there will exist points on C at which z takes on minimum and maximum values. One way to solve this problem is to solve for y from <p(x y) = constant, obtaining = /(x, ^(x)), a one-dimensional proby = \l/(x), and then to extremalize z problems

of the following type:

y

lem.

Another method

so that

dx

Consider as

if

+

=

0-

/(x, y)

+

-^~ -r-

dy dx

w =

is

due to Lagrange.

For an extremal, -jdx

=

0,

This same equation can be obtained as follows: X<p(x,

z/),

X a parameter.

x and y were independent variables.

and

Compute

Setting these

two

partial

ELEMENTS OF PURE AND APPLIED MATHEMATICS

296

derivatives equal to zero yields

dx

dx

dx

^ = ^- + X^ = dy

dy

dij

Eliminating X yields |

dx

Along the curve

y y)

<p(x,,

dy \ d<p/dy

=

~-

~r--j and Laqrange's method

dx

dx

+

TT-

dx

of

X

-JT

=

~T-

0,

so that

multipliers

yields

d<p/dy

-y

-

=

0,

dy ax

The

+

we have -^

constant,

the required equation for an extremal.

simplest constraint problem in the calculus of variations We wish to extremalize the functional

the

is

following:

b

=

/[?/]

f(x,y,y')dx

j Ja

(7,55)

subject to the constraint /

Ja

As

in Sec

and we

7.6

we

<P( X , y, y'}

=

let F(x, X)

y(x)

dx

=

+

X^^x)

+

X 2 ?72, y

constant

(7.56)

+

X 2T7 2 (#),

+

XirjJ

desire to extremalize /(Xi, X 2 )

=

/

Ja

f(x,

y

+

XIT?!

f

+ X^)

dx

subject to the condition

=

/(Xi, X 2 )

<P(XJ

/

y

+

XIT?I

+

X 2 r?2,

+

y'

X^i

+

=

X 2 77 2 ) dx

constant

Ja.

/(Xi,

X 2)

is

=

to be an extremal at Xi

X2

=

0.

Let the reader show that the Euler-Lagrange equation becomes

dx

The

solution y(x, X) of (7.57)

Example

7.9.

'

f

dy

J

\^dy is

substituted into (7.56) to eliminate

Let us find y(x} such that the functional ra

extremal subject to the constraint

/

v + 1

2

(y')

dx

=

I[y]

constant

~ =

fa /

b.

y dx

X.

is

an

Since the

constraint states that the length of arc of the curve v(x] be a constant, this tvne of

THE CALCULUS OF VARIATIONS problem

called

is

We

an isopenmetric problem.

+

(f

A

_ J^L=,

=

X*>)

have /

Vi +

<

+

(f

d ?/

2

+

297

=

\<p

=

A*,)

+

y

X \/l

+

2

(?/')

>

1

-

e</')

so that (7.57) yields

<fo L

or y"/[l

+

=

2 3

(?/')

]

We

1/X.

y(x) has constant curvature,

__ VI +

recall that

Example

7.10.

If

<p a

the (x

l ,

Riemanman x2

+

y" /[I

(y')*\l

and hence must be an arc

.

.

.

,

,

curvature, so that The radius of the

tlie

is

of a circle.

dx

=

b.

metric (see Sec. 3 6)

is

extremahzed subject

which can be obtained from

circle is X,

(?/

xn )

/*i

\/\

/

-f-

(y')i

j

ds,

where

<p a ,

a

==

(18

:

1,2,

3, 4, is

magnetic vector potential, one obtains the motion of a charged particle tiorial arid

electromagnetic; d*x*

d?

to

dx a the electro-

in

a gnivitn-

field,

dx> dx k

+ r ^ "^

W

e

+ m **

dx a

~

X

'

~fc

=

e

m

Problems Find y(x) which exlremalizes the functional

1.

=

/[//)

rb 2 /

(?/)

Ja

dx subject to the

fb

I xy dx = constant. Ja Find the curve of constant length joining two fixed points with the lowest center

constraint 2.

of mass.

The area underneath

3.

the curve y

=

f(x)

from x

=

a to x

/b y dx

=

b

is

rotated about

constant

=

c.

i

Derive

4.

(7.57).

REFERENCES Bliss,

G. A.:

"

Calculus of Variations and Multiple Integrals," University of Chicago

Press, Chicago, 1938.

Carslaw,

II. S.:

"Conduction of Heat

in Solids,"

Oxford University Press,

New

York,

1947. "

Modern Operational Mathematics in Engineering," McGraw-Hill Book Company, Inc., New York, 1944. Doetsche, G.: "Handbuch der Laplace Transformation," Birkhauser, Basel, 1950. Jaeger, J. C.: "An Introduction to the Laplace Transformation," Methuen & Co., Churchill, R. V.:

Ltd.,

London, 1949

Weinstock, R.: "Calculus of Variations," McGraw-Hill Book Company, Inc.,

York, 1952.

New

CHAPTER 8

GROUP THEORY AND ALGEBRAIC EQUATIONS

8.1. Introduction. The study of groups owes its beginning in an attempt to solve algebraic equations of degree higher than 4. The linear The solub/a. equation ax + b = 0, a 7* 0, has for its solution x = tion of the quadratic equation ax 2 + bx + c = 0, a ^ 0, is known to be

~

note that x\ and x* are written in terms of a

4ac).

We

number of operations division, and root extrac-

finite

involving addition, subtraction, multiplication, The operations are performed on the coefficients of the quadtions. We say that the quadratic equation is solvable by radiratic equation. cals. The cubic and quartic equations are solvable also by radicals.

Lagrange attempted to extend this result to algebraic equations of degree higher than 4. He was unsuccessful, but his work laid the foundation which enabled Galois and Abel in the early part of the nineteenth century In general, an algebraic equato grapple successfully with this problem. The equation tion of degree higher than 4 cannot be solved by radicals. 1 = can be solved by radicals, however. It remained for Cauchy x6

The theory of to systematically begin the study of group theory proper. groups plays an outstanding role in the unification of mathematics. Its applications in mathematics are widespread, and, moreover, it has served an important role in the development of the modern quantum theory of physics. 8.2. Definition of

a Group. ^

We

consider

first

some elementary exam-

Let us consider the set of all rationale, excluding zero, ples of groups. note that the product of two subject to the rule of multiplication.

We

rationals is again a rational. If a, b, c are rationale, then the associative The unique rational 1 has the property that law, (a6)c = a(6c), holds. 1

a

=

a

1

=

a, f or all rationals a.

Finally, for

any

rational a, there

a unique rational, I/a = or 1 such that ocr 1 = cr*a = 1. Let the reader show that the four elements (1, i) possess these same 1, i, Let us consider the properties under the operation of multiplication. rotations in a plane about a fixed point. 90, 180, 270, and 360 Let us denote these rotations by AI, A%, A 3, and A 4 = E, respectively. By AiAi we mean a 90 rotation followed by a 180 rotation, etc. We exists

,

298

GROUP THEORY AND ALGEBRAIC EQUATIONS

299

=

note that A*A,A 270 rotation followed by a 180 rotation is A*. equivalent to a 450 = 90 rotation. Moreover A>E = EAi = Ai foi* E is the identity element in the sense that the i 1, 2, 3, 4.

E leaves a body A A = E, EE = 2

invariant.

E, so that

2

rotatldij

A\A* = A *A\ = Ey every element has a unique inverse. From

We

also note that

the fact that

the reader deduce a correspondence between the rotations discussed above and the four elements (1, The i) under multiplication. 1, f, examples above lead us to the formal definition of a group. A discussion of sets can be found in Sec. 10.7. Let G consist of a set of objects A, B, C, An operator, is B. For conassociated with every pair of elements of G, (A, B) = A B = AB. venience we call the operator <8> multiplication and write A The set G is said to be a group relative to the operator if: I. For every A and BofG,AB = C implies C is a member of G. This is the closure property under <S> II. For all A, B, and C of G, let

.

.

.

.

,

.

(AB)C = A(BC) This

is

the associative law.

There

III.

exists a

unique element, #, of G, such that

AE = EA = A A

for all

inG.

IV. For every that

E is called the identity, or unit, element. A of G there exists a unique element, written 1

A"" 1 such ,

AA~ = A~ A = E l

'

We call A~ One can III'.

AX By

l

the inverse of A.

replace III and

For every B,

YA -

choosing

1?

A

and

1

It follows that

A

is

the inverse of A"" 1

.

IV by:

B of G there exist unique X and Y of G such that

B.

= A we

see that every element

A

has a unique right

Thus AEi = A, E 2 A = A. We show = jE 2 A#2 now that EI Now A(S 2 A) = A A = (AJE 2)A, so that A from (III'). Hence # 2 #1 from (III'). We show next that the idenWe have tity element for A is the same as that for B for all A and J5. and

from

left identity,

(III').

.

5#* - #BB - B. NowB(E B A) - (5#*)A - 5A, = A, which implies EB JS^. Let the reader (III'), EaA

AEA - EA A -

A,

so that, from show that (III') implies (IV).

ELEMENTS OF PURE AND APPLIED MATHEMATICS

300

Let x and y be elements of a set such that x* - e, y 3 = e, yxy 8.1. consider the set of elements (x, y, y 2 = y y, xy, xy 2 e), e the unit element.

Example

We

.

,

2

note that

is

t/

the inverse of

=

z

(xy )x (see

Table

y.

(xy*)(yxy)

we

If

=

desire to

xy*xy

=

xexy

We write x own

inverse,

compute

=

xxy

(xy*)x,

=*

x*y

=

x

=

x2 y

-

,

y

and from y 3

we note ey

=

=

x.

Let

,

us construct a multiplication table for these elements. 3 From x 2 s= e we note that x is its y y y = i/ etc. .

=

-

y*,

e

we

that

y

8.1).

We note that

each row and column of Table 8.1 contains the six elements of our set with no repetitions. If x ^ e, y ^ e let the reader show that the six elements are distinct, and hence form a group. t

TABLE

8.1

Problems 1.

2. |a t ,| 3.

of

Verify Table 8.1. Consider the set of square matrices, ||a l; ||, i, j = 1, 2, n, such that Show that this set of matrices is a group under multiplication. 5^ 0. .

.

.

,

Abelian group is one for which AB = BA for all A and B of G. Is the group 8.1 an Abelian group? Show that the group of Prob. 2 is non- Abelian. finite group is one containing a finite number of distinct elements. Show that

An

Example 4.

A

we can

replace (III)

implies

B =

and (IV) by

(III"):

AB - AC

implies

B -

C,

and

BA = CA

C, for finite groups.

5. We define a a = a 2 a a a~*= a 3 etc. A group is said to be cyclic if a single element generates every element of the group, that is, an element a exists such that, n if x is any element of the group, then x = a for some positive integer n. Give an *

,

example of a 6.

Show

7.

If

A

,

cyclic group.

that

and

B

A - (A"

1

)"

1 .

are elements of a group G,

show that (AB)' 1 = B" A" 1

1 .

Generalize

this result. 8.

Show that the

set of rational integers (positive

and negative integers including

form a group relative to the operation of addition. Do the set of rational integers form a group relative to the operation of multiplication? zero)

8.3. Finite

number

Groups.

A

group is a group consisting of a finite deduce now some theorems concerneach theorem with an example.

finite

of distinct elements.

ing finite groups and illustrate

We

GROUP THEORY AND ALGEBRAIC EQUATIONS

THEOREM

301

The order of a subgroup is a divisor of the order of the order of a group is the number of distinct elements H is a subgroup of G if is a group and if, furthermore, every element of // belongs to G. If at least one member of G is not a member of //, we say that H is a proper subgroup of G. The proof of consist of the elements E, A, B, F. Theorem 8.1 is as follows: Let Assume H a proper subgroup of (?, and If // s= G, there is no problem. Construct the set HI of elements let X be any element of G not in H. XF. Let the reader show that these elements are XE, XA, XB, BA~ is a member of H distinct. Moreover, if XA = B, then member is not a of // so that XA ? B. But since H is a group. of is distinct from Hence every element HI every element of H. If the = // of then and exhaust members HI (?, g 2/i, where g is the order of G H. If this is not the the order of h is and case, let F be a member of 8.1.

The

complete group. of the group.

H

H

.

.

,

.

.

.

.

,

X

l

X

not in // or HI. We now construct the set II 2 consisting of YE, YF. One easily shows that the members of 7/2 are distinct YA, and HI. If from each other and are distinct from the elements of If we then 3h. in exhaust continue the same G, g not, //, HI, HZ

G

.

.

.

,

H

G

Eventually we must exhaust

manner.

Thus

elements.

=

g

nh, and h divides

of

G

has a

finite

number

of

g.

The group of Table 8.1 consists of six elements. A subgroup of this 8.2. Another proper subgroup H(x, x* = e). The order of H is 2, and 2 K(y, y y* = e), 6 =3-2. Is it possible for G to have a subgroup of order 4?

Example group

since

6=23.

is

G is

,

THEOREM

Every subgroup of a cyclic group is a cyclic group. The group is given in Prob. 5, Sec. 8.2. Let G consist of = E. Let be a proper subgroup of G with elements

8.2.

definition of a cyclic

A

A,

2 .

.

.

,

,

A\ A Since 61

>

H

A

b

6,

s

.

.

.

,

we have

A 2& b\

.

A

bl

since b in

H.

~ qb

.

,

=

+

qb

yj 61

and

A = E r

.

,

__

<

s

<

1),

<

-

.

<

r

Then

b.

^

= A If s ^ 0, then A is a member of H, a contradiction, was assumed to be the smallest exponent of A such that A b is Hence s = 0, and 61 = qb = 2b, since A b A b = A 2b is in H. The 8

8

.

only elements of // are of the form Example

8.3.

Let

G

are the elements of G. 4 cyclic since a

=

(a

A mb

so that //

,

be a cyclic group of order (a,

IB

^

s,

b

8,

a 2 a3 a4 a5 a6 a7 a8 ,

,

,

,

,

,

,

)

so that

=

e)

Consider the subgroup //(a 2 a 4 a 6 a 8 a 8 = (a 2 ) 3 a 8 = (a 2 ) 4

2

2

is cyclic.

,

,

.

,

,

=

e).

We note that H

ELEMENTS OF PURE AND APPLIED MATHEMATICS

302

A criterion for a subgroup is the following: Let G be 8.3. group, and let S be a subset of G such that the product of any two elements of S is again an element of S. Then S is a subgroup of G. Certainly the closure and associative properties hold for S. Now THEOREM

a

finite

AT be any element of S. Then A, A 2 belong to S. There can exist only a finite number of distinct elements of the type A k Thus A n = A m for n > m and A n ~ m = E belongs to S. Moreover AA n-- 1 = E, so that A n~m~ = A~ l belongs to S. Q.E.D. A

let

.

,

.

.

,

,

.

.

.

.

y

l

Example 8.4. Let us consider the permutations of the integers (1, 2, 3). \\e obtain the six permutations (1, 2, 3), (1,3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). We can consider the particular permutation (2, 1) as being obtained from a subi-J,

(123\/ In this way

we obtain

the six elements

= /123\

123\

fm\ = /123\

Sl

\132/

&4

V213/

A23\ = /123\

/123\

/123

n

\231/>

If we consider a triangle with vertices labeled 1, 2, 3, respectively, then $6 states that we interchange the labels 1 and 3 and leave label 2 invariant. Let us consider any The operation of 82 on f yields function of three variables, f(xi, x%, xa)

tfjfCri,

If

we

follow this

}yy

the operation

Nftjf(.r],

since

and

>S6

it is

permutes

1

into 3, 2 into

natural to define

*S 6 ,

=

J'a, J^.t)

f

*S 6J S 2

=

rs)

3*2,

1,

=

/(.f i, jj, 0*2)

we obtain

=

$bf(f}, ft, xt)

and 3 into

/(o- 3 ,

x-2, j*i)

Thus

2.

~

'

^Se,

written

(

)

(

1097

'

upon 1 ,

)

(

then,

tion of

1

iclds 3

as follows: Starting with the right-hand side,

The left, we see that 1 goes into 3. Again, on the right-hand side, 2 goes into so that the end result is to leave 2 invariant.

moving

to the

into 3.

3 goes into 2, \

)

(

*|' \ol2/ \lo-6/

* 1.

The product

'

yields

(

1

=

Si.

\Wi/"

^

we see that

final result is

;

1

can

goes into

the permuta-

and, moving to the 3 2 followed by 2

3,

'^

left, >

1

It follows that Si plays the role of

and we leave it as an exercise to show that the ele1 ments Si, i 6, do, indeed, form a group relative to multiplication 2, Let the reader obtain a generalizaThe order of the group is 3! = 6 defined above. tion for the permutation group of order n\ often called the symmetric group. the identity element of the group, .

.

.

,

We

consider

now

the function f(Xi,

We

,

2,

Xa)

= Oi -

3- 2

)(j2

-

X.i)(Xt

0*i)

note that Sif = /, Stf - -/, SV = -/, *S</ = /, Xt>f = /, Stf - -/, so that Si, It is / invariant. These elements are called the even permutations.

St, arid S& leave

GROUP THEORY AND ALGEBRAIC, KQUATIONS

303

1

a simple matter to show that the product of two even permutations is again an even permutation. Hence, from Theorem 8.3, the set (S\, St $5) is a subgroup of the symmetric group of order 3!. Do (82, Sz, $e) form a group? These are the odd t

permutations. of a

Any subgroup

symmetric group

is

called a regular permutation group.

Problems 1. Show that the elements of Example 8 4 form a group. Construct the multiplication table for this group Is this group Abelian? Construct all the proper subgroups. 2. Show that the product of two even permutations is an even permutation. Con-

sider the cases of the product of

an even with an odd permutation and the product of

two odd permutations. 3.

We

=

can write S 4

can be written

Y

*S 6

=

= (m)

oo

Do

(1^)(2)

m

the SOIlse tlmt

*

]

2 2 >

*

3 H

the same for $i, $2, Nj, #5

Show that the permutation group of order n contains a subgroup of order n!/2. 6. C is called the commutator of two elements A and B of a group if C = (AB)~ BA. For any element X of G show that A"" ('X is the commutator of X~ 1 AX and X~ BX. 4.

]

1

1

}

8.4.

Let us consider two groups G\ and

Isomorphisms.

(7 2

.

If

a

one-to-one correspondence can be established such that ,1

<-->

B

<->

A'

B

f

implies

AB<-> A'B'

A and B

6 we say that the two groups G and T

(7 2 are isomorphic belong to GV An isomorphism of two groups implies that the two groups are equivalent in the sense that we are using two different languages to describe It is apparent that any theorem obtained the elements of the groups.

for

all

in

}

,

to each other and write Gi

Y

from the fact that 6

i

is

=

(V 2

.

A', B',

,

.

.

a group will also hold for

(? 2 .

8.5. We consider a cyclic group of order 4 with elements A, A z A 3 along with a subgroup of the symmetric group of order 4' Let the reader show that the elements

Example

,

,

A = E 4

1234\ form a

cyclic,

group with 82

=

Y

flj,

<S ],

S.\

/1234

/1234\ S*

= E

S\.

It is

a simple matter to

show that the correspondence 8,*-* is

an isomorphism.

We

A

1

i

=

1, 2, 3,

4

leave this as an exercise for the reader.

An important theorem due to Cayley is stated as follows: THEOREM 8.4. Every finite group G is isomorphic to a regular permutation group. Let the elements of the group be written as E = Ai, A%,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

304 .

.

.

An

,

A k be any member

Let

.

A k Ai A k A<i, 9

Since A* A

A

belongs to

\

and consider the elements

(?,

AkA

.

,

3,

must be equal

(?, it

A k A% =

Similarly

r.

.

.

of

.

.

AkA n

,

A r and by A k we mean

to an

Ak

Let

Akt, etc.

.

\

,

correspond to

''

L) for k

=

2,

1,

.

.

We

n.

. ,

We

morphism.

have from

wish to show that

2

l

l

/I

w\ /

2

VI

'

an

(8.1) represents

iso-

(8.1) '

'

'

'

'

'

?

'\

jn)

J'2

1

'

JM/VA-1

J'2

The isomorphism

which establishes the isomorphism.

was established

of

8.5

Example

in this fashion.

an isomorphism of a group with itself. A simple n be automorphism is as follows: Let A^ i 1, 2, example be any element of G. For conventhe elements of a group (?, and let Let us construct the elements X~ A X^ i ience we choose X ^ E. 1, = A,E = 4 If is n. G Abelian, we have X~ A,X = ^ X~ 2, Let us assume that (7 is non-Abelian. Generally, then, X~ A*X ^ A,. It is a simple matter to prove, in any case, that X~~ A X T X"~ A J i ^ j, for if X~ A,X = X~ AjX, then

An automorphism

is

of an

.

.

.

,

X

1

1

1

1

.

.

.

1

,

X

t.

1

1

a contradiction 2,

.

.

.

belongs

if

i 9* j.

n, contains to G, S == G. ,

A is

X

1

1

'

f

l

1

an automorphism.

t

Since the set

S

elements

of

X~ A l

t

X,

i

n distant elements and since every member We show now that the correspondence <->

X- A,X

We

1

have

which proves the automorphism

i

A

3

<->

=

1, 2,

.

.

.

,

n

= of

1,

S

(8.2)

A^ ^^, 1

of the correspondence (8.2).

Problems 1.

2.

If GI

and

(?2

are isomorphic and

if

G\

show that G* is cyclic. isomorphic to the group of Table

is cyclic,

Find the regular permutation group which

is

8. 1.

GROUP THEORY AND ALGEBRAIC EQUATIONS

305

Show that all cyclic groups of order n are isomorphic (see Prob. 1). 4. An automorphism of the type (8.2) of the last paragraph is called an inner automorphism. Show that the only inner automorphism of a cyclic group is the 3.

i = 1, 2, n. This is called the identity automorphism, A, <-> A automorphism. Hint: If 6. Prove that the product of two automorphisms is an automorphism. A* <-> A( is an automorphism and A t < > A/ is an automorphism, then under the second automorphism we have A( <-> (A|)", so that A <- (A,')" is another corre(A )" spondence, called the product of the two automorphisms. Show that A < represents an automorphism. 6. Show that the inner automorphisms of a group form a group. 7. Show that the automorphisms of a group form a group 8. Find the inner automorphisms of the group given in Table 8.1.

trivial

.

t,

.

.

,

T

)>

t

t

8.5. Cosets. Conjugate Subgroups. Normal Subgroups. Let G be a group and H a proper subgroup of G. Let the elements of 77 be HI, 7/ 2 We consider the ele77 m and let A be an element of G not in H n A. The reader can quickly verify that ments HiA, H 2 A, A ^ H,. If H A = H,, then A = H~ 1 H3 H>A * 3 A for i ^ j, and Thus the set of elements H A i = I, 2, is in 77, a contradiction. This n, are distinct and do not belong to 77 if A does not belong to 77. set is not a subgroup of G since it does not contain the identity element We call this set a coset, written as HA If the elements of 77 and E. HA exhaust G, we can write G = H + 77A. If not, we consider an element B of (7, B not in II and HA. We construct the coset HE and omit the proof that the elements of 777? are distinct from the elements We can continue this process until we exhaust (?, if G of H and 77 A. ,

.

.

.

H

,

,

.

.

.

.

,

H

H

t

t

1

.

J

.

.

,

.

is

Thus

a finite group.

G = Example

We

8.6.

77

+ HA +

IIB

+

-

-

+ HC

consider the symmetric group of order

4!.

(8.3)

Let

H consist of the

2 3 4\ A >3 ) (101 2 1 4 o/

Note that i

SA

1, 2, 3, 4,

member

of

H

5^ >S,A,

are or

i

^

members

HA.

We

j,

and that S

of the coset

consider

%

A ^ S

HA.

}

for all

t,

j.

Now

The elements /SA,

The element 5 =

(

.

J

is

not a

ELEMENTS OF PURE AND APPLIED MATHEMATICS

306

*****

_ /J234\ /1234\ _ ~ /1234\ \234lM2 41 3} \3 12 4/

5sfls "

^

V4231/ /1234\ /1234\ /1234\ _ _ ~ ~ Vl 1 2 3/ \24137 V1342/

&4/ *

=

=

The elements &#,

t

for all

S,.l for all

G =

?,

j,

7*

,

/1234\/1234\ = /1234\

V3412A2413/

/1234W1234\ _/1234\ ~ V1234A2413/ V2413/

2, 3, 4,

1,

members 7* S,# -f- HF.

are

z, .;,

&#

H + HA + HB + HC + ED

THEOREM

The dements

8.5.

of the coset

HB.

for

Continuing,

i 7* j.

Note that S

t

B

we can

5*

5,

obtain

X and Y belong to the same coset if and only

member of //, so that XY~ = HI, H in H. Then .Y //,7, and H,X = HJ],Y = H,Y for all f, with = HJHi. Thus every member of II X is a member of HY. From fly = 7 H^X, H,Y = HJI^X = II k X for all i, we have that every member of # F belongs to HX. Thus flX = 7/F. Conversely, if HX = HY, = H 2 Y so that XY~ = H^H^ then HI and ff 2 exist such that HiX a member of fl. Q.E.D. DEFINITION 8.1. If B = X~ AX, we say that B is conjugate to A,

XY~

if

1

Assume

is in II.

XY~

1

l

a

=

l

l

1

with 4, B, 1.

X in G. We note the following:

Every element

conjugate to conjugate to A, then A

B is = implies A 2.

3.

If

-XjBJST-

and

If ^1

B

is

1

=

1

(Jf- )-

1

is

^-

E~ AE.

A

itself,

1

conjugate to B, since

B = X~

1

AX

1 .

are conjugate to C, then

A and B

We

are conjugate.

have

B = F-VF = Y-*(XAX-i)Y - (A^

A = X~ CX 1

1

F)

H

B,

of G consisting of the elements E, A, Let us consider a subgroup Let .Y be a member of G not in PI. We form the elements .

.

.

.

X~ EX = 1

E,

X~ AX, X- BX, ... 1

1

and designate

this set as

X~ HX. 1

Since

(X- AX)(X-*BX) -

X~ (AB)X

1

l

X~

HX

is a subgroup of G. We have used the fact that // is a group, which implies that AB is in if A and B are in //. We call H and X~ HX conjugate subgroups. By considering Y^HY, etc., one can construct the complete set of conjugate subgroups of H. DEFINITION 8.2. If // is identical with all its conjugate subgroups, then

we have from Theorem

8.3 that the set

1

H

1

H is called

a normal, or invariant, subgroup of G.

GROUP THEORY AND ALGEBRAIC EQUATIONS

307

Example 8.7. The even permutations of a symmetric group form a subgroup, H. X is any odd permutation, X~ 1 AX is an even permutation if A is an even permutation. Thus X~ 1 HX = H for all X so that H is a normal subgroup of G. If

THEOREM 8.G. The intersection of two normal subgroups is a normal By the intersection of two subgroups HI, 77 2 we mean the subgroup. set of elements belonging to both HI and 77 2 written HI P\ H%. It is ,

obvious that the identity clement E belongs to HI C\ 77 2 If A and B 77 C\ to then AB to and 77 to HI 2 so that AB belongs HI belong 2 belongs to 77 1 Pi From Theorem 8.3, C\ 77 is a 2 of the finite l 2 subgroup .

,

#

H

.

We

must show now that 77 C\ 77 2 is a normal subgroup if HI and //2 are normal subgroups. The set of elements X"~ (Hi C\ H^X belong to HI since any element of HI C\ 77 2 belongs to 77 and, moreover, HI is normal. The same statement applies to 77 2 so that every element of X~ (H l C\ H 2 )X belongs to //i Pi H 2 Thus H L C\ 77 2 is identical with group G.

1

l

1

,

l

.

all its

conjugate groups so that

H

i

Pi 77 2

is

normal.

Problems

The elements

form a subgroup // of the symmetric group of 2 '^J ^j Find the right-hand resets II X of G. Find the left-hand cosets of G. Show that H is not a noi mal subgroup of G 2. Show that the subgroup //, consisting of the elements = <S, 2, 3, 4 of FAamplo 8.5, is a normal subgroup of the symmetric group of order 4' 3. If H is a subgroup of a cyclic group G, show that H is normal. 4. A group G is said to be simple if it contains no proper normal subgroups. Show that all groups of prime order arc simple Show that all simple- Abehari groups are of prime order. 6. Let and N be normal subgroups of G containing only the identity E in common Show that if i and N\ are any two elements of and A respectively, then MiNi = NiMi. Hint Consider C = Mi l N^ l MiNi, and show that C = E. l 6. Consider the set of elements E, Ai, A 2 of G such that A~ XA = for all X of G. Show that this set is an Abehan subgroup of (7, called the central of (7. Show that the central of G is normal. Find the central of the group given by Table 8.1. Find the central of the symmetric group of order 3'. 1.

order

^

^

^

XH

3!.

/

1

,

M

M

M

,

.

r

,

.

l

X

8.6. Factor, or Quotient, Groups. The set of rational integers form a group relative to addition (see Prob. 8, Sec. 8.2). The zero element A proper subgroup of this plays the role of the identity element.

group

the set of even integers including zero. The odd integers yield a coset of the group of even integers, so that 7 = 7 /i, where 7 is the group is

+

of rational integers, 7

odd

integers.

7i is

the group of even integers, and I I is the set of obviously not a group relative to addition since 7 t is

does not contain the identity element.

normal subgroup of

of

If

Let the reader show that 7 is a of 7 to any other element

we add any element

we obtain another element of 7 We may thus write 7 + 7 = 7 we add any element of 7 to any element of 7 b we obtain an element 7i, so that 7 + 7i = 7, + 7 = Ii. Thus if Finally 7 + 7i = 7

of 7 If

7.

,

.

.

X

.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

308

we

we denote

abstract in the sense that

element 7 and denote

all

that the two elements 7

,

all

even integers by the single

odd integers by the single element 7i, we note I\ form a group relative to addition, with 7o

We

serving as the identity element.

N

Let be any normal subgroup with its cosets,

generalize this result as follows: consider group G. along

.

N

We

of the

.

,NA

.

(8.4)

r

N is normal, we have X~*NX ~ N, that every element of the NX belongs to N, and, conversely, every element of N belongs to X-WX. Thus NX XN, and NA A N, = so that 2, the right cosets of N are equivalent to the left cosets of N. Let us conSince

set

X~

is,

1

t

sider the product of

(NA )(NA t

with

N(A A %

3)

two cosets

of

= N(A N)A,

3)

N.

another coset of N. 1

= NN(A A

3 )

We

1,

.

.

.

,

r,

have

N(NA.)Aj = NN(A,A 3 ) =

1

N(A A

i

t

1

Conversely,

= N(NA

J)

V

)A,

(NA

t

)(NA,)

If we look upon a coset as a single element, we note that the set (8.4) forms a group. The element TV is the identity element for this group abstracted from the group G and the normal subgroup N. DEFINITION 8.3. The group whose elements are constructed from the normal subgroup of G and the cosets of is called the factor group, or

N

N

The order, or quotient group, of N, written G/N. of G/N is called the index of the factor group.

THEOREM

N

normal subgroup

of the

H G = with

(N,

=

NA

if

N

is

subgroup

a normal subgroup of G, then A" 77 G. Thus provided

NC

//,

N + NA + NAz + ly

.

.

. ,

NA.).

NA,,

.

.

.

is

a

C

+ NA.

1

Now

+ NA, + NX +

1

(N,

H C G. If C G/N.

C

N + NA + NA, +

G/N =

of elements,

Let 77 be a proper subgroup of G, written G such that TV 7/ C G, then H/N

8.7.

a normal subgroup of Let the reader show that, is

andG/77 =

number

,NA.,NX,

.

.

,NY).

.

-

-

It is

-

+ NY obvious that

G/H C G/N. DEFINITION

8.4.

The product

the set of elements h t k 3 where ,

hi

HK of two subgroups 77 and K of G is ranges over

H

and k3 ranges over K.

THEOREM 8.8. If 77 and K are normal subgroups of G, their product L = HK is a normal subgroup of G. First we show that HK is a subgroup of G and that HK ~ KH. Certainly the associative law holds since H and K are in G. Moreover kihjcr = h s since H is normal, 1

GROUP THEORY AND ALGEBRAIC EQUATIONS

=

so that kihz

and

h^ki

HK

of

is e,

(hih^)(kik^)

hk

309

C

HK.

Finally, for

any element

X

X~ (HK)X = X- (HXX~ K)X = (X~ HX)(X- KX) l

l

The

77JL

the identity element of G. We leave it to ~ KII and that the inverse of every element

again an element of

is

=

(/h/Ci)(/i 2 /c 2 )

HK HK

identity element of the reader to show that

=

1

l

1

of G,

777

and K are normal. Thus 777v is identical with all its conjugate is normal. subgroups so that of G one understands DEFINITION 8.5. By a maximal normal group is not contained properly in any normal subgroup of G other than that since 77

HK

N

N

G

itself.

THEOREM

N

Let

8.9.

and N% be normal maximal subgroups

\

D = NI

their intersection, written

^ A/VZ>

GYAT,

of

(?,

D

AV Then

P\

G/N*

^ AT

(8.5)

2 /7;>

intersection of two subgroups of G is the set of elements common to both NI and AY D is nonvacuous since the identity element obviously belongs to D. The reader should refer to Sec. 8.4 for the definition of an

The

isomorphism.

From Theorem 8.8 the product NiNz is a normal NiNz contains both NI and N z Since NI and A Since any subgroup of necessary that NiNz = G.

Obviously group. are maximal, it is

T

2

.

normal,

makes sense

it

Ni = with A,

7*

A

3

to speak of

for i -^ j,

first

G = NiN

^ ~

L

that 2

+ DA

-

-

and the cosets DA, arc

+

=

NzAi

We

G.

NN 2

1

NzA

i

lf

=

1, 2,

is

(8.0)

r

Let L be the

distinct. .

+NA + Z

a normal group

Now

/D. Z

L = N* show

}

D + DAi + DA +

set of elements belonging to N%,

We

N

.

.

,

+ NiA

Z

so that

r,

r

have

= N,(D

+ DA^ +

-

+ DA

r)

The cosets A^ 2 A z = 1, 2, r, are distinct, for '~ A^ 2 which further implies that A A J 2 ^4 implies N^A^Aj~ a member of A" 2 Moreover A A J is a member of NI since A, arid ~ belong to NI [see (8.6)]. Thus A A J belongs to both NI and A^ 2

since ]V 2

A^A ~ A

N%.

t

.

,

.

.

,

r

is

^

1

1

J

1

,

1

.

1

1

1

~ This implies DA 1 A~ D, 7)^ t diction to (8.6). The correspondence of cosets, DA, <- A^ 2 yields the 2, r, with 7) isomorphism JVi/D and hence to D. .

.

1

.

,

larly JV 2 /D

^ G/Ni.

Q.E.D.

7)^4y, ->

a contra-

A^ 2 ^4 t

= G/N

,

2.

i

=

1,

Simi-

ELEMENTS OF PURE AND APPLIED MATHEMATICS

310 Example

We

8.8.

G of order

consider the cyclic group

6 with elements

a2 a8 a4

a,

,

,

three proper subgroups of G are Ni(a z a 4 a 6 = e), AT 2 (a 3 , a 6 reader the show that Ar and A^ are maximal normal subgroups of G. Let Ns(e). = N* We have also have that JVi A^.

=

a6 a6 ,

The

e.

,

G G = G/tf = G7tf = tf,/D = AT /D = !

2

2

(a (a

a4

2 ,

+

e)

,

[(a [(a [(a [(a

+

e)

,

8

(a

,

a4

,

e), (a

2 3 2

,

,

a3

,

+

(a

3

(a

,

a4

=

e)a*

,

2

e e)

,

(a

3

+ (a, a + (a

,

3

4

e)

,

,

a6)

+

a)

(a*,

a2 )

a')]

,

4

a 2 )]

a), (a',

,

=

a 4 e)a

2

e)a

6), (a,

,

.

(a

8

,

e),

We

i

O

D

=

,

4

),

(a

),

(e)]

),

]

3

an isomorphism, N^/ D == we mean all elements obtained by multiplying elements of (a 2 a 4 e) with elements of (a, a 3 a 6 ). Note that 4 3 2 a 6 ) = (a, a 3 a 6 ) so that the element [(a 2 ), (a 4 ), (e)] is the unit element (a a e) (a, a

The correspondence

G/Ni.

Similarly

<

(e)

(a

2

a4

,

,

(a

e),

N\/D ~G/N>

By

2

3

of

,

,

(a,

,

a3 a6)

is

,

e)

(a,

a3

,

a5)

,

,

,

G/N

>

a4

2

,

,

,

<

)

(a

i.

Problems Find the maximal normal subgroups of the cyclic group of order 12. Consider any two such maximal normal subgroups, and show the isomorphism (8.5) by constructing their cosets and finding their intersection. 2. Find the normal subgroups of the group of Table 8.1. 3. A group is said to be simple if it has no normal subgroup other than itself and the Show that any group of prime order is simple. identity element 4. Show that G/H is simple if and only if // is maximal Let D = H C\ N, 6. Let N be a normal subgroup of G, H any subgroup of G. H /D. is a group, D a normal subgroup of H, L/N L HN. Show that 6. Find the maximal normal subgroups of the symmetric group of order 24. 1.

HN

^

N

8.7. Series of Composition. The Jordan-Holder Theorem. Let be a normal subgroup of G. One then obtains the factor, or quotient, group

G/N = where

NA

lt

i

=

1, 2,

.

[N, .

.

,

NA NA^

r, is

THEOREM

of

8.10.

.

.

,

NA

r]

a coset consisting of the elements

with Nj ranging over the group N.

normal subgroup

.

l9

Let

T =

[N,

NBi,

.

.

.

,

NB

G/N. We prove the following theorem: Every normal group of a factor group G/N

8]

N,Ai be a

yields a

normal group of G; each normal group of G which contains Af corresponds The first part of Theorem 8.10 to a normal group of the factor group. states that the set

H is

=

N + NB +

a normal subgroup of G.

X~ Since mal,

N

is

1

HX = X~

l

= X~

1

normal,

(N

+ NB

1

Now

+

AT^!

+ NB )X

+

8

NX +

we have X~

(NX^-^NB^NX is in

t

T.

1

NX ^ N for all X in G. Thus X~ (BiN)X l

is in

Since

T and

T

is

hence

noris

an

GROUP THEORY AND ALGEBRAIC EQUATIONS

311

HX an element of H for all X of which H is normal. It obvious that G D H D N. Now assume D H D N with both // and AT normal. From H = N + NA + NA + + NA G = N + NA, + NA + + NA. + NA, +1 + + NA

element of H.

Thus X~

1

is

proves that that G

(?,

is

-

-

2

l

8

-

2

r

follows that H/N C G/N. We wish to show that H/N is a normal subgroup of G/N. From the fact that H is normal and that A belongs we have that A~ 1 A^A 3 belongs to H. Thus to

it

%

H

~ (A- AtAj)N since N since N A^A-M^j) = N(NA k AM* l

is is

normal normal

)

Thus the conjugates normal.

N(A~ A

It

l

t

A,)

H/N

of

members

are

A~ A A =

be that

1

may = NE = N.

T

3

of

H/N

so that ///AT

is

E, the identity element, so that

Q.E.D.

DEFINITION 8.6. Let Ni be a maximal normal subgroup of (?, N* a maximal normal subgroup of NI, etc. We obtain a series G, JVi, N 2 .

.

.

,

,

N

=

E, called a series of composition. The factor groups G/N\, Ni/N^

k

N

k -i/N k are all simple groups the Prob. Sec. for of definition a 8.6) % /N l+ i (see 3, simple group. If were not simple, Theorem 8.10 states that a normal group would .

.

.

,

N

exist such that

maximal

N 3 l

M

MD

relative to Af z

JVt+i, a contradiction, since N l+ is assumed These simple groups are called the prime factors The orders of G/Ni, Ni/N z are called the i

.

of composition of G. factors of composition of G.

,

THEOREM 8.11. (Jordan-Holder). For two series finite group G the prime-factor groups are isomorphic. This means that

if

we

.

.

.

of composition of

a

consider two arbitrary series of composition of

G, say

G = G =

//O, //l, //2, /Co,

X

i,

X-2,

-

.

,

.

.

,

H =E K =E r

,

_,

8

with prime-factor groups

Hi/Hi,

then r

=

s

.

.

.

,

Hr-i/H r

and a one-to-one correspondence,

i 4-4

HJH^ ^ K /KJ+l }

with

i

and j ranging from

to

r.

^

can be set up such that (8.9)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

312

The theorem is evidently true if the order of G is prime, for in this case there exists only one normal subgroup of G, the identity element, and G/E G/E. The proof of Theorem 8.11 is by induction. Assume the

=

for any group G whose order, g, can be written as the product of n prime factors or less. We know that the theorem is true for n = 1. Now let G be any group whose order can be written as the prod1 prime factors, arid let us consider two arbitrary series of uct of n

theorem true

+

H

= KI, then G/H l 1 l and composition [see (8.7) and (8.8)]. If the set of elements of HI is a group whose order contains at most n prime By

factors.

H /H

hypothesis

T

therefore, that

HI and KI

From Theorem

8.9

G///I

l+ i

^G/K

^ K /K }

are distinct

^ K./D,

for

J+}

^

i

We

1.

G/Ki

^ H /D l

of necessity,

i

G,

7>n ;; 2|

D

DI,

A'i,

.

2,

.

.

.

.

.

/> w

,

Dm = E

1S

(8.10)

(Hi, H%, tion, as

.

.

do

Theorem

K

.

,

HJ,

(Hi, DI,

the series (Ki,

8.11 holds for the

.

K

.

.

2,

two

.

, .

Pro}). 4, Sec.

= E

,

and the fact that DJDz^D^D*, D 2 /D 3 we see that the two series of (8.11) satisfy Theorem 8.11.

From

of G.

(8.10)

l

where DI = HI C\ KI. Since G/H and G/Ki are simple, Ki/Di and Hi/Di are simple. Hence DI i,s maximal (see Now form the series of composition 8.6). G, //

assume,

maximal normal subgroups

^ D. /D 2

9,

-

IU

etc.,

But the series D m ) satisfy Theorem 8.11 by assumpD TO ). Thus K r), (Ki, DI, .

,

series (Hi,

H

2,

.

.

.

,

.

.

H

r ),

.

,

(Ki,

K

2,

.

.

.

,

so that with (8.10) it is seen that the Jordan-Holder theorem holds 1 prime for any group G whose order can be written as a product of n r ),

+

This concludes the proof by induction. illustration of the Jordan-Holder theorem.

factors.

Problem. are u l v j

(i

-

Example

8.8

is

an

Construct the multiplication table for the octic group whose elements = 0, l)^with u* = 1, v z = 1, vu = u*v Find the different 1, 2, 3; j

0,

series of composition and prime-factor groups for the octic group, and show that the Jordan-Holder theorem applies.

8.8.

ment

Group Characters. a group G we can

s of

Representation of Groups. If to each eleassign a nonvanishing number, real or complex,

written X(s), such that

X(s)X(t)

we say we call

that

X

=

X(st)

(8.12)

we have a one-dimensional representation From (8.12) we have

a group character.

X(s)X(e)

-

X(se)

=

X(s)

of the

group and

GROUP THEORY AND ALGEBRAIC EQUATIONS so that X(e) = 1, where group character is X(s)

Example define

G be

Let

8.9.

the unit element of the group G.

e is 1

313

A

trivial

for all s of G.

a cyclic group of order n with generator

a,

an

== e.

Let us

X(a r ) by the equation

X(a r =

^2*,/

)

rt

)r

=

r

1,

2,

.

We have A"(a )A(a ) = ?<2,n/) V (2T,/). = f ,(2ir,/)cr+.) = A set of characters A\ A\ a character of the group r

8

r

A'^(a

)

=

e<

""

2

ir

M

=

0, 1, 2,

.

,

n

(8.13)

A(a r +'), .

.

.

.

,

.

,

n

so that (8.13) defines

X n -\ -

can be defined by

1

(8.14)

Let us note that

(8.15)

r=l

?=1

The group characters of n-

(8 14) are

orthogonal in the sense that (8.15) holds

AVa-)A->') M=0

now

Let us

elements that

/i,

-

J <'"'"><M=0

o

.

fin

,

We

If'

;

^

(8^)

I }

A n and a group G with X n matrices AI, A 2 assume that a one-to-one correspondence exists such

consider a set of n

</2,

Moreover

n-l

1

,

.

.

.

,

implies

AA t

;

<-> g g, t

easy to show that the set ol matrices is a group, A, isomorphic say that the group of matrices, A, is a representation of the group G. Any matrix A = ||aj can be thought of as representing an affine (linear) transformation y l = a]x (see Chap. 1), so that a representation of a group implies that a group of afhne transformations is isomorphic to the group G. Let the reader show that, if g % <-> A t is a representation of G, then g t <- B~ A 1 B is also a representation If su(;li is

to the

the case,

group G.

it is

We

||

]

1

of G, |B|

^

0.

A

Example 8.10 struct.

The

representation of the symmetric group of order 6

is

easy to con-

'

identity permutation

of coordinates x

=

x, y

=

y, z

=

(

z,

x y z

)

can be looked upon as the transformation

or

= l:r = Ox = Ox

+ + +

O// I// ()(/

+ + +

02

Oz \z

yielding the unit matrix 1

Ai

=

1 1

2 3\ '

(1

J

can be looked upon as the transformation of coordinates

ELEMENTS OF PURE AND APPLIED MATHEMATICS

314

x y z

= = =

= = =

ij

x z

Ox

4-

1X

+ 0/y + O//

Oj-

+

l/y

Oz

-I-

0,2

-f

^

yielding the matrix 1

100 1

the reader obtain the matrices corresponding to the permutations

I.et

Since every finite group is isornorphic to a regular permutation group (see Theorem it follows very simply from Kxample 8 10 that one can always obtain a representa-

8.5),

tion of a finite group.

It

may happen

that a matrix

B

exists such that

t

where

B

r

(A,), r

= is

representation matrices {B r (Ai),

1,2,.

.

.

/c,

,

are matrices

said to be reducible.

B

r

(A 2 ),

.

.

. ,

B

r

(A n )j

One is

=

B^^B

1, 2,

.

.

has the form

.

If this is so,

easily

(8.17)

the original

shows that the

set of

a representation of G. Thus It may be that further reduc-

(8.17) yields k new representations of G. If not, the representation given by (8.17) tions are possible. In this case the correspondence is written irreducible.

+B

n

,

is

2 (A,)

said to be

(8.18)

where the sum

in (8.18) denotes the matrix (8.17). Methods for determining the irreducible representations of a group are laborious. The reader may consult the references cited at the end of this chapter for detailed information on irreducible representations.

Problems Derive (8. 16). 2. Find a representation for the group of Table 8.1 3. If the correspondence g <-> A, is a representation, show that X(g ) = |A| is a character of the group G. 4. Find a character for the group given by Table 8.1 other than the trivial character 1.

l

X(g

l

)

=

1

%

for all g % of G.

8.9. Continuous Transformation Groups. formation of coordinates given by

Let us consider the trans-

(8.19)

GROUP THEORY AND ALGEBRAIC EQUATIONS where

t

is

315

a parameter ranging continuously over a given range of values. (8.19) will be said to be a one-parameter transfor-

The transformation

mation group if the following is true: 1. There exists a to such that

= =

* y

This value of

t,

t

to,

/Or,

//;M

<p(x, ?/; to)

leaves the point

This corresponds

invariant.

(x, y)

to the identity transformation. 2. The result of applying two successive transformations

to a third transformation of the family given that if x 2 = f(xi, 2/1 O, 2/z = ^(^i, yi' ti) then ;

= =

za z/2

where

2

is

4.

/(/Or,

<K/(-^

a function of

0,

/y;

and

t

some

Example

func^tion of

8.11.

The

/y;

;

<p(x, /y; /); Zi)

= /Or, /y; = ?(:r,

J 2)

//; / 2 )

=

(f>(xi, 2/1

we

This implies that

/i)

;

^.

The

=

j-j

x

+

=

//i

/>,

y H-

2/>

satisfy the

requirements

identity transfonnation occurs for b b bj r

=

0,

and the

b

rotations

= =

j*i ?/i

=

form a one-parameter group. j*j

2:1

x2

identical

hold.

one-parameter group inverse transformation is obtained by replacing

t/2

fi)

a unique inverse. x and y such that

translations

for a

The

is

This implies

].

?y

^i

v>(z,

/y; t),

can solve (8.19) uniquely for

with

(8.19).

9

The associative law must Each transformation has

3.

by

=

=

jc

cos 6

.r

sin

t/

+

//

sin

cos

yields the identity transformation Ji cos 0i

sin 0i

//i

0i

+

Vi

(

x cos

(0

+

00

is

additive for two successive transformations

sin

The parameter

'

os

If

then

0i>

sin (0

.v

+

0i

Hi

)

=

x sin

(0,

-f

2)

-f-

/y

cos

(0i

+

2)

A third example is the group x = ax, a y. The identity transformation occurs for a s 1. The parameter is multiplicative for two successive transformations. z

i/i

Infinitesimal transformations can be found as follows: Let

value of the parameter

t

to

which yields tho identity transformation.

be that

Then

^-M//,/) y Tf (8.19) is

continuous

in ,

=

(8>20) <p(x,

/y; /o)

a small change in

/

will yield

a transformation

ELEMENTS OF PURE AND APPLIED MATHEMATICS

316

from

differing

The transformation

by a small amount.

(8.20)

*i-/(*,y;fe yi

=

v(x, y\t*

+ +

)

e)

and

said to be an infinitesimal in the sense that x\

is

by small amounts when

respectively, (8.20)

from

by

/ and

(p

y,

Subtracting

(8.21) yields

5x

If

from x and

y\ differ

sufficiently small.

e is

= =

-

Xi //i

x y

= =

/(;r,

*>0r,

U

y; //;

o

+ +

f(x,

e) e)

are differentiable in a neighborhood of

=

8x

(0"

?/; / )

= ^x

?^

'

=

t

tQ,

one has

^ (8.22)

except for infinitesimals of higher order, Example

(^ oU/ O-o J

=

=

5t.

For the rotation group of Example 8.11 one has

8.12.

x,

e

so that (8.22)

=

-^ ) \o6/ 0=.o

(

y,

becomes 8x

=

y

bt

by

x

bt

Let us consider the change in the value of a function F(x, y) when the point (#, y) undergoes an infinitesimal transformation. One has

8F

=

F( Xl

,

yi )

-

F(x, y)

except for infinitesimals of higher order.

The operator

is t

U

=

dF ~

3x

From

+

dF

^

(8.22)

""Os + 'D"

Sy

one obtains

23 >

defined by

very useful. Equation (8.23) can be written as dF = (UF) dt. If corresponds to the identity transformation, we can replace 6 by 2,

=

remembering that

t

is

It is interesting to

small.

note that

if

one

is

given the system of differential

equations

=

x

=

y

(8.25)

GROUP THEORY AND ALGEBRAIC EQUATIONS

317

then the solution of (8.25), written xi 2/i

= =

/(z, 2/;0

(8.26) <f>(x,y;t)

To show this, a one-parameter group additive in t. curve starting at the fixed point (x, y).

is

(8.26) yields a 2/i

vary

Now

we first note that As t varies, Zi

(see Fig. 8.1). let

us consider the system

2,

2/2)

#2

/2)

2/2

=

Xi at

ti

=

t

at

/i

=

t

x\ at

tz

=

at

tz

=

(8.27)

dy*

If

we

let

tz

=

ti

t,

t

fixed, (8.27)

=

2/i

becomes

,1*^ (2)

Xz

^2)

2/2

=

ctt 2

(8.28) ^2/2

=

2/i

Since (8.28) has exactly the same form and

initial

of necessity, 2

=

f(xi,

2/1; t z )

=

= x\, 2/2 = 2/iAt t ti we have x% same curve as given in Fig. 8.1, and for > t we obtain an extension of T to the point (# 2 2/2). Thus the product of two transformations belongs to the group, and the parameThe establishment ter t is additive.

conditions as (8.25),

-

/Cri,

2/1

It

obvious that (8.29) yields the

s

i>

;

'i

,g 29)

1

,

of the inverse transformation

P

l

(x l9

yj at

t

is left

to the reader.

Let us now begin with a group transformation given by (8.19) and attempt to establish a system of differential

and t. we have 2/i,

equations involving x\, the group property 22 2/2

when t

=

o

From

=/(x, y,p(t,

=

<P(X,

y

ti)

1/3(1, ti

FIG. 8.1

= f(xi, = <P(XI,

2/1

;

h (8.30)

7/1;

We

and 2/1 are replaced by the values as given by (8.19). choose as that value of the parameter which yields the identity element.

Xi

ELEMENTS OF PURE AND APPLIED MATHEMATICS

318

since /(zi, y^ 0) = Xi = f(x, y\ entiate (8.30) with respect to h, keeping x, y, and t fixed.

Of

=

necessity, 0(t, 0)

Evaluating at

t\

=

t,

t).

We

differ-

Thus

yields

dt d<p(x, y,t)

\dtj tl

dt

From

(8.19),

^ = feliJl, at

j, so that

!

dt

(8.31)

where

X(<)

=

(^

If

)

is

t

additive,

we have

&(t, ti)

=

t

+

LVflii/i.-oJ

=

X(0

1

and

(8.31)

becomes

In

(8.25).

0&i, 2/0

any case

i?C&i, 2/i)

so that the change of variable, u = J\(/) dZ, reduces (8.32) to (8.25). Thus one can always find a new parameter such that the group transfor-

mation

is

Example solution

is

additive relative to the parameter.

We consider -~

8.13.

x/(l

Xi

xj,

-~ =

with

xi **

x,yi

**

y &tt

=

Q.

The

Let the reader show that an additive group

=<"+?/.

xt), y\

1,

has been obtained.

Let us now The value of ^(zi, t

=

2/i)

0,

=

consider an arbitrary function of x and i/, say, F(x, y). F(x\, y\) for t small can be obtained as follows: Since

F(f(x, y,

t), <p(x,

assuming that this

is

I

we expand F Thus possible. y,

t)),

dF(xi,

2/1)

F(XI,

Now F(xi,

y t)

*i

,

y).

Also

in a

d^ 7,9

Taylor series about

(8.33)

GROUP THEORY AND ALGEBRAIC EQUATIONS

319

dF(xi, dt

dt

i

dyi dt

dF

dF

[

provided the group transformation

Moreover

additive.

is

dt 2

From

= U 2F(x,

one easily shows that t-o

induction

l9

yi

= UF(x,

dt n

yi)

=

F(x, y)

special case F(XI, yi)

%i(x, y>

Example 8

14.

-2,

y)

can be written in the form

(8.33)

F(x l9

For the

mathematical

can be shown that

it

dF(x

Equation

By

y).

For

+

x

=

f)

x

(x, y)

~

+

(UF)t

=

x\

+

(Ux)t

+

(U*F)

-

(8.34)

we have

+

y, ij(x, y)

Ux = -y

U*x

(U

2

x)

^

x we have

- -x

U*x

**

y

so *U that 4.

x cos

Similarly y\

=

x sin

t

t

-}-

+4! y sin

y cos

/,

3

6

---)-H'~3! + 5!

\

,

21

/^,

i

<

,

t

and the rotation group

is

obtained.

\ )

ELEMENTS OF PURE AND APPLIED MATHEMATICS

320

Problems 1.

Show

that the Einstein-Lorentz transformations form a one-parameter group

-

-

t

_ ~

V/c 2 x

Note that the parameter V is not additive. 2. Find the transformation group obtained by integrating

If Sl(x, y) is

3.

an invariant under a group transformation

that Vil(x, y) = 0. 4. Solve Ull(x, y)

and show that x z

for the rotation group,

=

ll(x, y)

-f y

to(xi,

z

is

y\\ show

an invariant

for this group.

Given

5.

The

6.

differential

formation x

D

=

F

In

y

=

(x, y)

=

I u, v, ~r~ j

In

i/i

=

+ In

=

t

=0, and F

/,

is

J

=

u

Thus F must be independent separation of variables.

of

y,

Integrate ^

8.10.

... ,n

x\, x^,

.

X2

fly --

=

=

dx

=

-

-

t

dxi

^

Let

dxi

The equation -- = /

a

+

v\

-

^2

x

A

=

#3)

and -~ = Xi

f

j

becomes

invariant under the translation

dx

,

,

=

t

_1_

a

( u,

j-

J

=0

if

can be integrated by

^,

--

y

function F(XI,

o; n

x%,

.

.

.

,

xn)

is

a sym-

any permutation on the subscripts

The function

leaves / invariant. f(Xi,

-

x

so that FI

.

.

remains invariant under the group trans-

MI, v

Symmetric Functions.

metric fuiKition in 1,2,

In

~

)

(

since

Vi 4- a,

( u, v,

~

/

-j-

for all

tyi,

the transformation group.

x, find

=

equation

y

txi,

=

y, r)(x, y)

^10:20:3

+

X\

+

X\

+

invariant under the symmetric group of order 6. Xi)(x Xn) we obtain product (x xz) (x

is

X If

we expand the

-

y

(X

-

Xi)(x

'

Xz)

'

Xn)

'

(X

=

xn

-

<TiX

n~ l

+

o- 2

xn

-2

-

<r 3 o:

n- 8

+

-lV

GROUP THEORY AND ALGEBRAIC EQUATIONS with

= = =

(7i (7 2 (7 3

=

i

(7 t ,

x + x + + XiX + 2

'

2

^j

i

ZxtXjXk

XiX 2 X 3

k 7*

;*

*

'

+ X n_lXn

'

3

i

(8.35)

Xn

'

fundamental symmetric apparent that any function of the o-'s is a symmetric A few examples lead us to suspect that the converse is also note that/(x x 2 ) = x? + x\ can be written as

We

.

n, of (8.35) are called the

.

.

,

3 ,

+

(Xi

=

(7i

3

It is

function.

with

+ xn + XiX n + X X +

321

'

'

3

1, 2,

functions.

true.

XiX 2

=

(7 n

The

xi 4-

'

+

x\

3

0*2)

=

^2

2,

XiX 2

=

X 2 X 3)

/(Xi,

The symmetric function

.

xfx

,

=

can be written as/(xi, x 2 x 3 )

+

(xiX 2 x 3 )(#i

,

THEOREM

+

x2

=

x*)

o- 3 <7i.

The Fundamental Theorem of Symmetric Functions. x n ) is a symmetric function (multinomial) in Xi, x<t, If f(xi, x 2 <r n x w ) = g(<ri, o- 2 then /(si, x 2 o: n ), that is, / can be written as a multinomial in the fundamental symmetric functions <7i, 8.12.

.

-

-

,

,

.

.

.

.

-

0*2,

-

.

.

.

,

,

,

,

.

.

,

,

o'n-

-

,

The proof

theorem

of the

is

by mathematical

The theorem

induction.

Let certainly true for a function of one variable since f(xi) = /(0"i). 1 variables. us assume the theorem true for a function of n Thus if is

f(xi,

xz

symmetric, then / = g(<r lt (7 2 Xn] be any symmetric multinomial. is

Zn-i)

.

.

.

,

,

let f(xi, X2,

.

.

.

,

.

x n -i, 0) is a symmetric function in Xi, 0^2, x n -i, 0) = g((ffi) Q tion we can write /(xi, x 2 x2 x2 x n ((n)o = xi where vi = Xi .

.

.

.

.

.

etc., .

that

. ,

/

==

is,

+

(0-^)0

-

+

a(xi

+

x<z

-

x^

+

-

.

(cr 2 )o,

,

.

.

,

.

.

,

obvious that h

h(Xi,

X2

,

.

.

.

Xn)

.

,

It is

xn)

,

Xn-l, 0)

=

X2

3= /(Xi,

/(Xi,

.

.

.

,

,

X2

.

=

is

,

,

a zero of A(XI, x 2

Since h

factors of h.

Thus

h(Xi,

X2

,

.

.

.

is

,

0- 2

g(ffi,

=

2,

.

.

,

.

XiX 2

'

.

,

'

.

,

. ,

k.

con-

<T n

_i).

X n _i, 0) (cTn-l)o)

=

x n ), which implies that x n is a x n _i are also

symmetric, of necessity Xi, x 2

Xn)

0,

1,

Now .

.

Xn )

Now we

,

factor of h.

+

<

of degree

.

so that x n

=

is

also symmetric.

is

for j

0,

assume the theorem true for any symmetric multinomial This is a double mathematical induction type of proof. sider h(Xi, X 2

x n _!

a multinomial of degree 1, that is, then the theorem is certainly true. We ,

x n ),

=

(o- n -i)o),

.

+

evaluated at x n

<r,

.

.

+

+

x^

f(xi,

;

,

,

+

-

the value of

is

Now if f(xi,

n.

,

-

,

Then

x n -i so by assump-

.

.

,

+

.

.

,

Now

a n ~\).

.

.

,

X n 8(Xi, X 2

. ,

.

.

.

,

.

,

.

Xn )

,

=

ff n 8

ELEMENTS OF PURE AND APPLIED MATHEMATICS

322

where

a symmetric function, and

s is

#2,

l,

The degree s(xi,

J*2,

.

=

Xn)

,

(T 2

g((Ti,

.

.

.

+

<T n -l)

,

,

ff n

s(Xi

t

Z2

.

.

.

,

,

XH)

n < k, so that, by the induction hypothesis, x n ) can be expressed in terms of the fundamental symmetric

of s is k .

.

,

functions, yielding 2,

.

=

#n

,

00" 1,

0*2,

+

OVi-l

,

<r n /(Ti,

.

<T2,

.

.

<T n

,

This concludes the proof of the fundamental theorem of symmetric functions.

Example 8 /(a?,,

x2

,

*,)

We

15.

-

r?

+

consider

+

a?i

s? -f

With patience one can show that / ==

f(xi, x 2 , 0)

The function

f(zi, X2, xt)

/ so

that / =

o\

+

x\

a\

=

+

2ar,aj s

+

*l

=

2xix t

^\x\x.

xtf =

+

(ar,

3.r?r>2

-

+

4- J*i^3

We

=

0*2X8)

2

[(r,)

should have the factor ^1X2X3.

3x 1X2X3(2; \xi

-

Let us consider

symmetric.

is

-

2x,r 8 -f 2r 2 o- 3

]

note that

3<r2<rji

3o- 2o- 3 .

<rf

Problems 1.

Show

that (1

2.

Let

,

/3,

-f j-?)(l

-f xj)(l

2

~-~

Referring to Prob.

4.

2,

+

+

1

xz

72

px

-\~

P* 4- T

,

-

<r?

+

z

2

2<r 2

=

+^ H-

<?x

2

2p <y pq

~~

2cr,r 8

+

2 o-

3

Show that

?

V-

7

-f

4pr

r

show that /3

+

<*V

Sl

=

Xi

2

4.

=

arj)

7 be the zeros of f(x)

al+JH

3.

+

2

+

2 )3

72

=

q*

-

2pr

Consider

+

X2

-f

-f

Xn

X\

=

Express a* in terms of

X\

Si, 82, s 3 .

8.11. Polynomials.

We

shall

be interested in polynomials of the type

=

+

n

p(x)

=

Y

ckx

k

Co

Cix

+

dx*

+

'

'

'

+

c nx

n

(8.36)

fc-O

The coefficients c i = The reader is referred t,

0, 1, 2,

.

.

.

,

n, are

assumed to belong to a

to Sec. 4.1 for the properties of a

field.

field F.

If

a set

GROUP THEORY AND ALGEBRAIC EQUATIONS

we can perform the simple operations of The and their inverses on these elements. addition, multiplication, zero element and unit element belong to a field and the distributive laws, The complex numbers a(b + c) = ab + ac, (6 + c)a = ba + ca, hold. form a field. A subfield of the field of complex numbers is the field of The rational numbers form a subfield of the field of real numbers. Let us consider the set of real numbers of the form real numbers. a + b \/2> where a and b are rational numbers. The sum and product The unit element of two such numbers is evidently a number of this set. = is is 1 = 1 + and the zero element We must + -\/2. \/2, show that every nonzero number has a unique inverse. From the fact if and only if a = b = 0. that \/2 is irrational we have a + b \/2 = of elements belong to a field,

+

Assume a [(

2

6) /a

b

2fr

2 26 2 + \/2 7^ 0. Then it is easy to show that [a/a \/2 is the unique inverse of a + b -\/2. Let the reader ]

2

]

solve (a

and

for x

?/,

a

2

+

2b*

b

^

+

V2)(rc 0.

We

y A/2)

=

1

call this field

the field of rationals, written R(\/2). sions of a field in the next section.

We

+

\/2

an algebraic extension of

shall discuss algebraic exten-

If f(x) and g(x) are polynomials with coefficients in a field F, we say that g(x) divides f(x) if a polynomial h(x) with coefficients in F exists such that f(x) = g(x)h(x). An important theorem concerning poly-

nomials is as follows: Given two polynomials with coefficients in F, say, f(x) of degree m, g(x) of degree n, there exist two polynomials with coefficients in F, r(x) of degree < n, s(x) of degree m, such that

<

r(x)f(x)

+

s(x)g(x)

=

d(x)

(8.37)

is the greatest common divisor of f(x) and g(x). All other divisors of both f(x) and g(x) are divisors of d(x). The coefficients of d(x) are in F.

where d(x)

We consider the set of all polynomials of the form a(x)f(x) + with a(x) and b(x) arbitrary. These polynomials have degrees (exponent of highest power of x) greater than or equal to zero. A conProof.

b(x)g(x)

t

a polynomial of degree zero. Let a(x) = those polynomials for which the degree of a(x)f(x) imum, but not identically zero, and let

stant

is

d(x)

Any now

=

r(x)f(x)

divisor of both/(x) and g(x) that d(x) divides both /(x)

is

+

r(x), b(x)

+

b(x)g(x)

f(x)

q(x)d(x)

+

t(x)

is

s(x)

be

a min-

s(x)g(x)

obviously a divisor of d(x). We show If d(x) does not divide g(x).

and

then by division (always possible since the coefficients are in a

=

=

^

degree of

t(x)

<

field)

degree of d(x)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

324 d(x)

^

Hence qd =

0.

qrf

+

=/

qsg

t,

and (8.38)

From

the definition of d(x), (8.38)

d(x) divides /(x), written d(x)/f(x).

show that degree

the greatest

1 is

<

= x 2 + 1, common divisor

=

g(x) of f(x)

x.

and

Then g(x).

1

(x

We

z

+

=

t(x)

0,

so that

Let the reader

Similarly d(x)/g(x).

<

degree of g(x), degree of s(x)

Let f(x)

8.16.

Example that

of r(x)

impossible unless

is

1)

degree off(x).

+

(~x)x

=

1,

so

say that /Or) and g(x) are

relatively prime.

Problems

Show Show

that the integers do not form a field. that the set of numbers x ly -%/3> x an(l y ranging over numbers, forms a field 1 is the greatest common divisor of f(x) = x z -f 1, g(x) 3. Show that x 1.

2.

+

+

all

=

rational

(x

+

I)

2 .

such that r(x)f(x) + s(x)g(x} = x + I. Are r(x] and s(x) unique ? 4. Let f(x) and 0(:r) be polynomials with coefficients in a field Fi, with F\ a sub field of the field F. If f(x) and r;O) are relatively prime with respect to the fipld /<\ show that f(x] and g(x) are relatively prime relative to the field F. How does this apply to Prob. 3? 6. Do the set of numbers a -j- b \/2 + c \/3, a, 6, c rational, form a field? What of a + 6 + c V3 + d >/6 ? 6. Let F be any field. If we label the zero and unit elements and 1, respectively,

Find

r(x)

and

s(x)

V2

and

define

1+1

=2,

etc.,

show that every

The Algebraic Extension

8.12.

field

contains the rationals as a subfield.

of a Field F.

Let p(x)

=

polynomial with coefficients in a field F. We say that p(x) is irreducible F if p(x) cannot be written as a product (nontrivial) of two polynoWe say that p is a zero of p(x) or a root of mials with coefficients in F. Kronecker has shown that one can always extend if p(p) = 0. p(x)

in

=0

the p(p)

field

=

F

number

i

such a fashion that a new number p is introduced, yielding 1 = solution of x 2 involves the invention of the

in

+

The

0.

=

\/~~

!

THEOREM 8.13. Let p(x) be an irreducible polynomial = 0. If q(x) has coefficients in F, then q(p) =

in

p(p)

Proof.

Since p(x)

atively prime.

From

irreducible, p(x)/q(x), or p(x) Sec. 8.11

is

A(x)p(x)

+

B(x)q(x)

and

such that

g(x) are rel-

1

some A(x), B(x). Thus A(p)p(p) + B(p)q(p) =0 = diction. Thus p(x) divides q(x). Two corollaries follow immediately from Theorem 8.13.

for

F

implies that

1,

a contra-

GROUP THEORY AND ALGEBRAIC EQUATIONS

COROLLARY such that p(p)

COROLLARY xn

p(x)

+

1.

0.

we make the

If

2.

c n -ix

DEFINITION

that irreducible polynomial of smallest degree

is

p(x)

=

325

n~ l

+

+

leading coefficient of p(x)

then p(x)

Co,

is

unity,

unique.

The n zeros of p(x), say, pi, p 2 p n are called The zeros of x 2 2, namely, \/2, \/2> arc It follows from Theorem 8.13 that any one of

8.7.

.

.

.

,

,

,

conjugates of each other. conjugates of each other.

the n conjugates of p(x) determines p(x) uniquely, assuming that the leading coefficient of p(x)

THEOREM

Let

8.14.

with coefficients in a

is

unity.

be a zero of the irreducible polynomial p(x) 2 n~

p

The numbers

F.

field

early independent relative to F, since

-1

n

n

1,

F

exist in

nomial in

F

such that

^

p

p,

,

.

.

,

.

,

constants c,

=

i

n-

=

c^p*

k=o n with f(p)

<

if

1

then

0,

=

t(x)

l

p

are lin-

0, 1, 2,

.

.

.

1

y

c*a;* is

a poly*

fc=o

=

0, a contradiction to Corollary 1. the following: Given the irreducible polynomial = 0, one can obtain a field F\ p(x) with coefficients in F such that p(p) We call F\ an algebraic containing p such that F is a subfield of F\.

of degree

An important

extension of

result

F and

is

=

write F\

The elements

F(p).

of Fi(p) are of the

form T

I (8.39)

with

^

6; p 7 7^ 0, a t in F, b3 in F.

j=0 Let the reader show that the elements of the type given by (8.39) form n~ l c n -ip a field. By using the fact that p n one has CQ =

+

P

H

- ~ (Co +

Cip

= - (Cop +

+ 2

Cip

'

'

+

+

'

'

'

11

"

Cn-lp

+

+

powers. P

3

= -(P

+

1)

P

4

if

=

p

3

n

+p+

-P-P

n~ 1 )

C n _i(Co

+

Cip

+

'

'

'

+

Cn-lp"-

1

)

can always be reduced to lower=0, then 1

1

2

P

We

+

'

1

+ For example,

'

)

C n -2p

so that powers of p higher than

'

6

= -p -

show next that the elements a

2

p

3

= -P

2

+p+

of (8.39) can actually

1

etc.

be written as

ELEMENTS OF PURE AND APPLIED MATHEMATICS

326

8

polynomials in

The denominator

p.

h(x)

of (8.39) is

}

b3 p3

Let

.

= y-o

Since h(p) Sec. 8.11

^

0, of

and p(x) are

necessity h(x)

+ +

A(x)h(x) A(p)h(p) so that the inverse of h(p)

is

= =

B(x)p(x)

B(p)p( P )

1

1

n-1 t-0

t=0

Example

=

all

\/2.

cases.

A(p)h(p)

Thus

A(p). r

by using the

=

= A(p)~

a(p)

p

From

relatively prime.

we have

=

fact that p(p)

to eliminate higher powers of

p.

The field R(\/2) is obtained by considering x z - 2 = 0, p 2 - 2 * 0, This is not true for The same field is obtained if we consider p = -\/22 belong to the field of rationale. The elements of The coefficients of x z 8.17.

+

a -h b \/2, a and b rational. bp /2(\/2) are of the type a It is easy to show 3 = 0. Next we consider the polynomial equation p(x) = x 2 that p(x] is irreducible in the field R(-\/2) since \/3 = a -f & \/2, a and 6 rational, is

Let the reader show this. Now the coefficients of x* 3, namely, 1 and belong to R(\/2). Hence we obtain an algebraic extension of R(-\/2) by conb \/2) -f- (c 4- d \/2) \/3> a ^> c d sidering the set of numbers of the type (a This is the field /(\/2, \/3)> and its elements are of the form a -f- b \/2 -h rational. c \/3 d \/6, a, b, c, d rational. In a similar manner we could construct R(\/3 \/2). impossible. 3,

+

)

>

+

}

V3) - J8(V3 V2)

It is evident that fl(\/2,

THEOREM

be any number of the algebraic extension F(p) Then a is the zero of a polynomial f(x) with coefficients in F. We say that f(x) = is the principal equation of x = a. Let the conjugates of p(x) = be p = pi, p 2 p 3 Proof. pn

Let

8.15.

obtained from p(p)

=

o:

0.

,

Since a

is

a

We

,

.

.

.

,

.

in F(p), of necessity,

=

ai

=

a

form the conjugates

= =

Certainly f(x)

s

(re

o

of

+ +

i

ip2 flips

ai)(x

defined

+ +

a 2 pl

x2)

2p3

by

+ +

'

*

*

'

'

'

(x

+ +

n)

an-ipS"

1 1

/o

fln-lpS"

has

i

as a zero.

.QX

We

GROUP THEORY AND ALGEBRAIC EQUATIONS

Now

need only show that the coefficients of f(x) are in F.

=

f(x)

xn

327

-

Moreover 2ct,, S<x,a*, etc., are certainly symmetric in the p, i = i, 2, n, and so can be written as polynomials involving the fundamental .

.

.

,

symmetric functions,

+ P2 4Plp2 + Plp3 +

"

*

4- Pn

Pi

+

'

'

*

'

'

Pn-lpn (8.41)

'

pn

PlP'2

The elements

simply the coefficients of p(x), except for some by expanding (x p2) pi)(x (# p n ).

of (8.41) are

negative factors, as seen

Thus the

coefficients of f(x) are in F.

=

2

+

3

=

-

(x (x

x2

Example

we

8.17,

consider

a.

=

on

=

4 \/2.

3

and

4 \/2,

f(x)

The

Referring to

8.18.

Example

Then

- 32 - 23

ai)(x 3)

2

6.r

=

2)

(x

-

3 4- 4

V2)(x

-

-

3

4

-v/2)

coefficients of /(x) are in the field of rationals.

Example

=

of p(x)

=

f(x)

[x

8.19.

+

x*

-

= x8

(I

Let us find the principal equation of a 1 -f- p 2 where p Let the zeros of p(x) be p = pi, p 2 p 3 Then 4- 2. ,

3x

+ p*)][x - (1 + p )Jlx - (1 -f P + p 4- p -f p ]^ + [(1 4- PI)(! 2 2

2

2

[3

2

From

ai 0"2 <T3

2 2

P1P2P3

=

-f p

)

+

2

(1 4-

p )(l 4- p

2 )

=

pi 4" p2 4" PS

P1P2 4- P2P3

a zero

3 )]

2

= = =

is

.

,

+

=

P3pl

3

2

we have Pl

Also

(1 4- P?)(l

Sec. 8.10).

4- P?

+ pf)(l

2

-

+

1

2(pip 2 4" P2P3 4- P3pl)

<rj

-

2<r 2

4- a\

~

2o"io- 3

+

rj

8 (see Prob.

1,

=

x3

2

2

P 3 ) -f (1 4- pi)(l 4- P 3 )

-f-

2(pf 4- P 2 4- P

+2(-6) 2 8. 4- 3x

3

)

+

4- (pip2 4- PIPS

psp2)

2

2pip 2 pa(pi -f p 2 4- PS)

+9-0 We

(1 4- P

We

2

2-3 - -6

2

==34-

f(x)

(Pl 4- P2 4- P3)

-0 -

+ P!)(! + P?) -

P )(l 4- P 2 ) 4- (1

Hence

=

Moreover

2

(1 -f

+ P2

check that 3

2 )

+

1

4- 3(1 4- P

2

2 )

p

2

satisfies

-8 =

have /(I 4- P

since p 8 4- 3p 4- 2

2 )

0,

- P 6 -f 6P 4 4- 9p 2 - 4 - (-2 - 3p) 2 + 6P (-2 or p 8 = -2 - 3p.

3p)

+ 9p - 4 2

ELEMENTS OF PURE AND APPLIED MATHEMATICS

328

Problems

Show that

1.

= n =

zero of p(x)

m is

1,

Hint: Assume x = m/n is a Show that of necessity n integers m lowest form 1 show that p(x) has no rational zeros Why By testing x =

+

x s -f x

+

xs

x

1.

-f-

=0

1

1,

has no rational roots.

m and

p(x) irreducible in the field of rationals?

The polynomial

2.

ever, p(x)

x4

p(x]

m

reducible

is

the

x3

-f

+

2x 2

-f

z

-f-

How-

has no rational zeros.

1

rationals since

field of

Is this a contradiction ? 2 x 3 -f x -f 1 (see Prob. 1). Let p be a zero of p(x) p Express 1 + p + P /1 as a second-degree polynomial in p n If a 4. Let /(x) = ao -h flnZ + n, integers a, t = 0, 1 2, 2 prime number p exists such that p does not divide a n p divides o for i < r?, /> does not divide c?o, then /(x) is irreducible in the field of rationals. This criterion is due to Eisenstein. Show that x n p is irreducible in the field of rationals, p a prime. Show the field of rationals. that f(x -f 1) = x* + 2x + 2 = (x + I) 2 + 1 is irreducible Why can one immediately make the same statement for f(x) x~ -f- 1 ? 4 2 <r Let the zeros of p(x) be p, Show 6. Consider p(x) = x -f- 3x -f- 9. p,<r,

3.

+p

2

+

i

.

,

.

.

,

,

l

,

m

that the principal equation of a = 6. Consider x 2 - 2 = z3 - 2 7.

A is

Consider f(x)

8.

f(x)

polynomial irreducible

is

is

it is

=

separable. (x

2)

[/i (x)l

8.13.

x

=

ai)(x

=

k ,

+

-

p

=

3

is/(.c)

(j

2) 1

[/(a:)]

.

I)

[(a;

-

),

=

.

r=

.

otherwise f(x)

=

f(a n )

,

2

81

defined

2 ]

.

=

0.

If

is

zero for

by

if

If

(8 40)

then

so that f\(a\(x}}

0,

Hence explain why f(ai(x)) -

the

If /(x) is reduci])lc in F,

which /i(i)

f\(x) is a function for

Why does p(x} /J\(ct\(x})1 =s so that f(a\) = f(a

/W =

-

irreducible, then f(x)

Assume

Op

0, and obtain the field R(\^2, \/2). Show that said to be separable if it has no multiple zeros ;

p(x)

1 -f

=

is

zeio for x

=

pi, P2,

j"

.

.

= .

,

pi.

pn

fi(x).

The Galois Resolvent.

We

a.

say that a

is

Let f(x) be a principal equation of a primitive number of F(p) if f(x) is irreducible

in F.

THEOREM

8.16.

If

a

=

i

is

a primitive

F (a) =

number

then

of F(p),

F(p)

every number of F(#) belongs to F(p), and conversely. a n ) be the principal Let f(x) (x at) (x a\)(x Proof. = Define of a a\. <p(x) by equation that

is,

-

^-^Vx-a.'x-a, The

p,

i

=

Since f(x)

1, 2,

is

.

.

.

assumed

,

,

the a, are not distinct, show that

n, are the

-.,

conjugate zeros of p(x).

irreducible, /'(ai) ?^

(see Prob.

Then

6,

Sec. 8.12).

GROUP THEORY AND ALGEBRAIC EQUATIONS Thus

=

pi

<f>(ai)/f'(ai),

so that pi

is

number

a

329

Any number

in F(cti).

Why? Since i is a polynomial in F(pi) thus belongs to F(ai). follows that any number in F(ai) is a number in F(pi). Q.E.D.

pi,

of it

THEOREM 8.17. Let pi be a zero of pi(x), a\ a zero of pz(x), Pi(x) and can form the fields JP(pi), F(cri). If pz(x) pz(x) irreducible in F. is reducible in F(pi), we consider the irreducible factor of pz(x) having In this way we can form F(pi, a\) (see Example 8.17). (TI as a zero.

We

Let the reader show that F(pi, a\) = F(cri, pi). We show that an irreducible polynomial in F exists yielding a field F(r) such that F(T) = F(pi, <TI). Let Proof.

= =

TI

T2

with a and

in F.

The

=

t

p,,

+ +

api Oipi

2,

1,

.

jSdi /3CT 2

.

. ,

m, are the conjugates of

= 1, 2, = 0. n, are the conjugates of ^(x) 0, and the o-,, t Pi(x) We (;hoose a and so that the r,, j = 1, 2, ran, are all distinct. This can be done since, if ap + fay = ap k + fa then =

.

.

.

,

.

.

l

.

,

t,

,-

P

^

fc

(8.42)

Pk

P^

There are only a finite number of ratios in (8.42), so that a and ft can be chosen such that (8.42) fails to hold for all i p^ A:, j ^ I. For i = k we have faj 7^ fa if <r3 7^ (TI. For j = I we have ap, ^ ap^ for i 7^ k. Next i

we form

= =

g(x)

(x

x wn

n)(x - r - (SrOz- +

r mn )

(x

2)

+

1

(-I)

mn

T mn

rir 2

(8.43)

Any interchange of two p's leaves the coefficients of g(x) invariant, as does any interchange of two o-'s. Thus the coefficients of g(x) are in F. That factor of g(x) irreducible in F having TI as a zero generates F(TI) with

TI

=

We now F(n) =

api

+

fai-

Since

TI

=

show that any number

F(pi,

o-i).

If

r;

is

in F(pi,

+

(60

api

+

fai, F(TI) is

o-i),

T?

is

a subfield ofF(pi,

(TI).

belongs to F(n) so that of the form

of jF(pi,

ry

0*1)

(o

n

1

;-=0

m

+ 1

i=0

+

+ + bm-ipf' )^ + (C + Ctpi + + 1

6ipi '

'

'

1

Cm-ipf- )^?-

1

ELEMENTS OF PURE AND APPLIED MATHEMATICS

330

The mn

with the d l} in F.

by

replacing p\

We

ways.

p2

p3

,

.

.

,

define (p(x)

other conjugates of p\ can be formed by

1

and

pm

.

,

by

<r\

cr 2

,

.

.

0-3,

.

<r n

,

in all possible

by (8.44)

Let the reader show that 771 = <f>(ri)/g'(Ti), g'(ri) 7* 0, since g(x) no multiple roots. Hence 771 belongs to F(r\). Q.E.D. 8.20.

Example

We form n = No two

We

_

+

\/2

The

_= \/2 -

-

(x

-

2

- 3 = = \/2, = with_pi = - V2 + \/3, T = - \/2 -

r2

=_(),

\/3, r 3

-

n)(x

-

n)(x

-

TI)(X

Thus

It

generates F(\/2

g(x)

=

T4)

coefficients of g(x) are in the field of rationale.

irreducible in F.

\/3-

<TI

\/3-

4

Then

of the T'S are equal. 0(x)

is

consider x*

\/3, r 2

has

-f*

-

x4

lOo- 2 4-

1

can be easily shown that g(x} \/3)> which has elements of

the form

+

a

6(\/2

-f

+

\/3)

+

Thus F(\/2

j, y, u, v rational.

+

2 c(A/2 -f A/3)

d(\/2

=

\/3)

/^(

V3) 3 =

-f

V2,

A special case of Theorem 8.17 occurs if pi(#) = The

can be generated.

field F(p\, p^)

x

+ V2

-h

?y

?*

\/3

w\/6

-f

\/3) (see Example 8.16).

Continuing,

p\ ?* a\

=

P2-

adjoin to

F

all

and

p%(x)

we can

the roots of p(x) = 0, obtaining F(pi, p 2 A single number r p n ). The irreducible polynomial exists such that F(r) = F(pi, p 2 p n ). It is a poly0. having r as a zero is called the Galois resolvent of p(x) ,

.

^

n!, since

.

.

,

.

.

,

nomial of degree

.

,

we need only

find constants {a*}

such that

n

and the other

7

the

p^s

obtained from the n\

are different from each other.

Theorem

in

T'S

The

1

permutations of

rest of the proof proceeds as

8.17.

DEFINITION

8.8.

If pi, p2,

.

.

.

,

p n are the zeros of the irreducible

polynomial p(x) such that

F( Pl )

= Ffa) =

= F( Pn

)

=0

is said to be normal and F(p) is called a normal then p(x) or p(x) In this case F(pi) = F(pi, p 2 extension of F. p n ) and p(x) is its own Galois resolvent. ,

Example division

p(x)

-

8.21.

x8

3x

are

pi, 2

-f 1

=

(x

=

52

-f-

3c 2

-f"

2ac.

solution so that pi

and p(x)

is

normal.

=

2

p)(x

2

Vl2 -

- (-1

square in F(p). Assume 12 4 2 that p 8 3p - 1, p = 3p 3

x8

Consider p(x)

It is

p

-f

3x

-f 1

=0.

px -f p 2

3).

We

.

.

.

,

Let p be a zero of p(x).

The

two

other

show that 12

-

2

roots

By of

a perfect 2 2 2 bp -H cp ) a, 6, c rational. 3p = (a Using the fact - p, we have 12 = a 2 - 26c, = -c 2 -|- 2ab 66c, = 2 is a rational easy to check that a = 4, b I, c 2 3p )/2.

+

3/>

is

,

+

p

2 ,

p2

p

2

2,

which implies F(p)

=

F(pi)

F(pt),

GROUP THEORY AND ALGEBRAIC EQUATIONS

331

Problems 2 = 0, p 2 (x) = x 4 Consider p\(x) = x* 2 = 0, with pi(rc), p z (x) irreducible in the field R of rationals. Find 7?(\/2, v^) 2. Find a Galois resolvent for p(x} = x 3 + 2x + 1. 2 3. Show that p(a*) = j 8 3s 2 ) j* + 2r(r* -j- 3s 2 ) is normal for r and s integers 3(r in is irreducible the field of rationals. provided p(x) 3 2 4. The discriminant of p(x} = x 3 Show that the px 9 is A = 4p 27</ discriminant of p(x) of Prob. 3 is a perfect square. 5. Show that 171 *>(ri)/0'(n), ?(z) defined by (8.44). 1.

+

+

+

.

Automorphisms. The Galois Group. Let us consider the field R(\/2) composed of all elements of the form a + b \/2, a and b rational. We consider the one-to-one correspondence between a + b \/2 and its b \/2, written a + b \/2 <-> a 6 \/2. Let us look conjugate, a 8.14.

more

closely at this correspondence.

x

u

+y +v

If

x u

*-+

\/2 \/2

<->

y \/2 v

\/2

then (x

+

y v/2)

+

(u

+

v

\/2)

=

(x

+

u)

+ <-*

(y (x

+ v) \/2 + u) - (y +

= (x

+

y \/2)(u

+

v -x/2)

=

+

(xu

2yv)

->

(xw

(x

+ +

(xv

a =

l

f~ (a

f

),

a'0', for all

such that a <^

a and

attached to

of it,

of the field,

(7/

/3',

-

r

\/2)

yu) \/2 (xv

+

yu) \/2

- yV2)(u -

written a

<->

',

or a!

imply a + ft <-> a' called an automorphism

<is

(x

\/2

v)

+

+

v -s/2)

=

0',

/(a), a/3 <~>

of the field.

b -\/Z is an automapping a + b \/2 <-> a R(\/2). Every field has at least one automorphism the identity mapping, <-, for all a of the field. Not

From above we morphism

a',

y -v/2)

+

2yv)

=

A one-to-one mapping of a field into itself,

-

see that the

one-to-one mappings of a field into itself yield an automorphism. Witness the mapping x <- 2x, with x ranging over the field of real numall

For y <- 2y we have xy <- 2(xy) ^ (2x)(2y). Under an automorphism the zero element maps into itself, as does the unit element, for = a <-> a' + x = a', so that x = 0, and <-> a <-> a', x, imply a + = a <-> a', 1 <-> y, imply a 1 a *-* a'y = a', so that t/ = 1. We say that and 1 are left invariant under any automorphism. Let the reader show that the rationals of a field F remain invariant for all automorphisms of It is also a simple matter to show that the automorphisms of a field F. = fz (y) are automorphisms AI and A^ form a group. If a = /i(/3), we define A^Ai as the automorphism a = fi(fz(y)) = /S(T). Let the reader show that under this definition of multiplication the set of autobers.

-

332

ELEMENTS OF PURE AND APPLIED MATHEMATICS

morphisms

of

F

The

form a group.

unit element of the group

the

is

identity automorphism. Let us return to the irreducible polynomial p(x) with coefficients in F. = F(pi, p 2 shall be interested in the extension field p n ) with = = 0. consider only those autopt i 1, 2, ft, such that p(p ]

N

We

.

.

.

.

.

.

,

,

We

l

,

,

p n ) which leave the elements of F invariant. automorphism is one of this type. Let the reader show that the set of all automorphisms of F(pi, P2, pn) leaving the elements of F invariant form a group. DEFINITION 8.9. The group G of automorphisms of F(pi, p2, p)

morphisms

The

of F(pi, p 2

.

.

.

,

,

identity

.

.

,

.

.

.

,

leaving

F

invariant

group of F(pij p 2

THEOREM

.

.

. ,

,

8.18.

If

=

called the Galois group of p(x)

is

0,

or the Galois

p n ) relative to F. is

p(x)

normal

(see rfec. 8.13), the Galois

group of

p(x) contains exactly n automorphisms. First of all we have Proof. P2

.

,

.

.

,

= F( Pl = F( P z) =

Pn)

= f (p n

'

)

)

^^

since p(j)

assumed normal.

is

Jf

N

p(x] fc

automorphism

of the Galois

TJ

automorphism

of G.

=

Then

N

Thus

q(x).

,

<-

A-

for

any

pi

correspond to a for any

ri

akp\

<->

fc=0

be a zero of

then a*

fc

=

Let

group G.

a^'

(pi <-> pi),

=

a^

y

0,

so that

cr

must

Jfc=0

(pj <~^P2J,

.

.

(PI <->

. ,

p n ) are the

We must remember that (pi *-* p t ) comonly possible elements of 6 pletely determines an automorphism of G since any element of F(pi, p 2 n-l Y

.

,

.

.

.

,

pn)

is

of the

form N

b^pj,

with

fc

?>*,

=

0, 1, 2,

.

.

.

,

n

1

in F.

k*=0

If p(#) were not normal, we could not make this statement. If p(x) is not normal, the order of G is less than or equal to n\ Why? Finally, we shall wish to use the faet that the Galois group G can be made iso-

morphic to a regular permutation group (see Theorem 8.4). THEOREM 8.19. Let p(x) be normal. If a is any element of F(p = pi) which remains invariant under all automorphisms of (?, then a is an ele-

ment

of F.

Proof.

We

We

know

that

all

elements of

wish to prove the converse

ment

of F(pi),

and consider ai

=

bQ

its

if

p(x)

is

F remain

normal.

conjugates.

Then

invariant under G.

Let a

=

a\ be

an

ele-

GROUP THEORY AND ALGEBRAIC EQUATIONS

=

333

and f(x) a n ) is the principal equation of (x ai)(x az) (x a = a\ (see Theorem 8.15), and the coefficients of f(x) are in F. But = a2 = = oi n since a\ is left invariant under G. Thus i '

'

'

*

f(x)

=

n

(x

oi)

xn

nax n

~1

+

+

n

l)

(

an

na is in F, which implies that a is in F. Q.E.D. DEFINITION 8.10. If the Galois group is cyclic, the equation p(x)

so that

is

=

said to be cyclic. 7?

THEOKEM

8.20.

^

If p(x)

dkX

k

is

cyclic

and normal, we can

=0

by a finite number of rational operations and root extractions on elements of F(w), (addition, multiplication, etc.) The degree of p(x) is n, and the where w n = 1, w ^ 1, arg w 2ir/n. It can also be shown that F(w) can be coefficients of p(x] are in F. generated by a finite number of rational operations and root extractions on the elements of F. We omit proof of this latter statement. Since p(x) is normal, G is composed of exactly n elements. Proof. Moreover G is assumed to be cyclic, so that a single element generates G. solve for the roots of p(x)

We represent this element by the permutation P =

(

Let

.

.

^

J.

us consider the set of elements y 1

'2

fc

The

t

P(f 2 ) in

=

i

,

=

P2

=

PI

= -

Pl p,

2,

1,

.

III+

+ + +

.

raising

2

+ W"~ p n + -'>p. l

Wp<>

+

+ wpt + ri

"

n, certainly

.

,

Thus

2.

+ W"pt + ,'p, + -V:< +

=

n~ l

/0 Ar ^ (8 45) '

belong to F(p, w), p

w pi = 2/1^, since w n = 1.

2 [P( 2 )] to the nth power and

Q>n-l

=

_L

-r Pn

P2 4- Pi

=

Now

pi.

w

as a parameter But [P( 2 )] n = P(?) since n then permuting the indices 1,2,

considering

.

.

.

,

equivalent to first permuting the p's and then raising the new entity Thus P( 2 ) = 2 so that ^ is invariant under P of G. to the Tith power. 2 = is invariant From P (?) P[P(2)1 = P(?) = 2, etc., we see that is

for all

elements of G.

The same

result applies to

7,

.

J,

5,

.

.

,

Now 2

=

tto(pl, P2,

+

>

Pn)

Ol(pl, P2,

,

Pn)W

since direct expansion of 2 01, . . . , a w _i are in F so that

by of

iy

,

G we

have a

<->

<-> aj, aj, ai

a2

+

2

n

is

<-^

'

=

'

'

+

a n _i(pi, p 2

We

1.

in /^(ty).

a2

,

.

.

.

,

. ,

.

.

,

Pn)^""

1

wish to show that a Under any permutation

a n _i

,

<->

a^.

Since

^ is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

334

invariant under

we have

(?,

+ = for all

w

wn =

such that

for all Oo, ai,

w satisfying w = n

.

.

.

,

Hence

2 is

+

+

b,x

w

+

-

.

a

1 if

The same

a'^w*-

1

(8.46)

=

kn-i^- 1

Thus

l

.

,

+

type yet has n discan only be true

(8.46) is of this

wn ~ = aj, ai =

2

.

.

,

a n _i are invariant under

in F(w).

aX +

the equation

Equation

w,

1,

~

a[w

Now

1.

roots.

1

tinct zeros, namely,

+

aj

=b +

q(x)

has exactly n

a n ~\w n

a[,

.

(8.46) .

.

,

a n -i

so belong to result applies to ?, (r,

=

Hence

a^-i.

F by Theorem 8.19. 3,

,

&

The

nth roots of elements inF(w), w = e 2?ri/n We now look upon (8.45) as a linear system of equations in the This system has a unique solution provided unknowns pi, p 2 pn the Vandermonde determinant t,

i

=

.

1, 2,

.

n, of (8.45) are

.

,

.

.

.

,

.

.

,

I

w w

D(w) =

1

1

w4

2

(8.47)

wn is

from

different

Let the reader show that

zero.

n-2

D(w) =

Hence the

p

t ,

i

=

1, 2,

.

.

H

(w

n,

. ,

with coefficients in F(w). Example It

is

8.22.

group r

w = 3

1,

w;

5^ 1,

w2

that pipa -f P2P3 -h pspi

= ==

g+

>

i

is

=

x8 1 = 3x is normal. Let the roots of p(x) be

that p(x)

+

=0

of order 3.

Then

PI, PI, pa.

with

w>) T* 0, j

can be expressed as linear functions of the Q.E.D.

we saw

In Example 8.21

also cyclic since its Galois

-

t

^3

P? (pi

+

w

==

P2

P.O

PI

4- pa H- PS

PI -h

^ =

Pl

-f

=

+ W*PJ p2 + ?P3

WPJ

+

f

2

(8.48)

=0. We must find = 1. Now

1

using the additional fact

3, pip2p.<

(W

pi 4-

P2

=

f2

2

+

W' 2 )(P1P2 -f P2p.i

-f (P1P2

+

P'2p.{

+

+

Pa

P3pl)(?

-

?# 2

= 9 - 2(p'] + P 2 4- Pj) - 3(P!P2 + P2P3 -f P'PI -f PIP. -f P = 2(p] + p2 -h p-j) 0(pi -j- p -f ps)(pip2 + P2pt 4= -27 = (& 4- ^) - 4gg - (-27)2 _ 4(9)3 = _ 2)18 7

2)

+

3

2

~ 8 2

^)

2

-^

2

-27 V3*

i)

P?P 8 )

+

4- 27pip2ps

GKOUP THEORY AND ALGEBRAIC EQUATIONS Hence

g = -27

cube roots of

|,

R(w) = J?(\/3

3

must be chosen

now

It is

t).

lL-..

= -27

y-' 6

so that

2

s

=

9.

Note that

gg -

l

and

pi,

p2

Note that

a simple matter to solve (8.48) for

335

,

The

9.

j

belong to

pa,

which we

omit.

Before concluding this chapter let us consider the following: We con= x 4 + 9, irreducible in the field of rationals. Let the roots = be p, cr. From p( p)<r( <r) = 9 we see that p, (7,

sider p(x) 9 of x 4

+

=

=

3 3 4 Hence p(x) is normal. The automor3p /p = Tip p), phisms comprising the Galois group of p(x) are (p<-p), (p<(p <-><r), (p <-> o-), and we can represent (7 by the permutation group

o-

3/p

-

/1234\ _/1234\ P *" P3 _/1234\ ~V3412/ \432iy ~\2143/ V1234/ = = This group is not cyclic. The with pi = p, p 2 = P4 p, PS = certain number of elements of leaves invariant a Gi subgroup (Pi, A) Pl

_/1234\ ~

2

o".

0",

Let us see which of these elements are invariant under G\. PI F(p). 2 8 If a = a leaves all elements invariant. bp dp is left invaricp ant under P%, of necessity

+

+

a

+

bp

+

cp*

which implies that invariant under G\. field

F(p

2

=

d

is

3

=

+ fe(-p) +

c(-p)

2

+

d(-p)

8

Hence the elements a + 6p 2 are left easy to show that these elements form a sub0.

called the Galois

group of F(p) relative

2

)

principal equation of a

= = =

a

D F(p 3 F.

Note that F(p)

The

dp

It is

Gi

of F(p).

)

6

+

+

+

6p

2

is

- (a + 6p )][x - (a + 6 P )][x - (a + - a) - 6(p + * )(x - a) + 6 pV [(x - a) [(x 2

2

[x

2

2

2

2

ba*)][x

-

(a

+

2

6cr )]

2 2 ]

2

+

2

satisfies a quadratic equation [of lower degree than p(x)] with coefficients in F. Thus we can solve a quadratic equation for p 2 2 obtaining p in terms of rational operations and extractions of roots of

so that a

6p

,

Another square root yields p. Of course we know all advance since it is trivial to solve for the roots of x 4 + 9 = 0. Let us note, however, that Gi in the above example is a maximal normal subgroup of G and that E is a maximal normal subgroup of G\. The factor groups G/G\ and G\/E are of order 2 (a prime number). The fact that 2 is a prime number and that a group whose order is prime is cyclic

elements in F. this in

(see

Theorem

8.20 for the importance of this fact) leads to the all-imporwe state without proof.

tant theorem, which

the Galois group of an equation p(x) = relacoefficient field F, a necessary and sufficient condition that

THEOREM tive to its

8.21.

If

G is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

336 p(x)

=

be solvable by radicals relative to G be entirely primes.

F

that the factors of com-

is

position of

It can be shown that in general a polynomial of degree greater than 4 1 = 0, however, can cannot be solved by radicals. The equation x b be solved by radicals. Factoring, we have

-

(x 4 Considering x

so that

upon

and then

+

x3

+

from x

+

x2

x

a:

+

+

8

1

2

+

yx

1

=

+

x*

=0,

1 by x one has

division

for x

+

4

l)(.r

if-

+

x

J)

=

let

+

1

y

=

0.

One

solves for y

0,

Problems

3.

Solvo the quadratic x -f- bx + c by the method of this section. Show that the automorphisms of a field form a group. Show that D(w) of (8 47) is given by D(w) = IT(w - w>), j > i

4.

Solve; for the roots of

5.

Find the polynomial with

1.

2.

2

.r

6

1

.

coefficients in the field of rationals

P

= \/\/2 +

having

1

as a zero.

REFERENCES "Modern Highei Algebra," University of Chicago Press, Chicago, 1937. Introduction to Algebraic Theories," University of Chicago Pi ess, Chicago,

Albert, A. A.: " :

1941.

Artin, E.:

"

Birkhoff, G.,

Galois Theory,"

Edward

and S MacLane:

"A

Arm Arbor, Mich., 1942 Modern Algebra," The Macmillan Com-

Brothers, Inc

Survey

of

,

pany, New York, 1941. Dickson, L. E.: "First Course in the Theory of Equations," John Wiley

New :

&

Sons, Inc.,

York, 1922.

H

Sanborn & Co., Chicago, 1930. Algebraic Theories," Benj. "An Introduction to Abstract Algebra," John Wiley & Sons, Inc., York, 1940

"Modern

MacDuffee, C. C.:

New

Murnaghan, F.

C.:

"The Theory

of

Group Representations," Johns Hopkins

Press,

Baltimore, 1938. Poritrjagin, L.: "Topological

Groups," Princeton University Press, Princeton, N.J.,

1946.

"Modern Algebra," Frederick Ungar Publishing Company, York, 1940. Weisner, L.: "Introduction to the Theory of Equations," The Macmillan Company, New York, 1938. Weyl, H.: "The Classical Groups," Princeton University Press, Princeton, N.J., 1946.

Van

der Waerden, B. L.:

New

CHAPTER

9

PROBABILITY THEORY AND STATISTICS

It appears that the mathematical theory of probformation to the inquisitiveiiess of a professional gambler. In the seventeenth century Antoine Gombaud, Chevalier de Mere, proposed some simple problems involving games of chance to the famous French philosopher, writer, and mathematician, Blaise Pascal. It is to mathematicians Pascal and Fcrmat that probability theory owes its origin, though since their early works a great number of mathematicians have contributed to its development. Its applications range from sta9.1. Introduction.

ability

owes

its

An understanding of probability theory is tistics to quantum theory. necessary to undertake studies of the modern theory of games and information theory. It will be useful, however, to consider some elementary combinatorial analysis before we attempt a definition of probability. Let us assume that we have a 9.2. Permutations and Combinations. collection of white balls numbered 1 to m and a collection of red balls

numbered 1 to and one white

n.

ball?

In

how many ways can we choose exactly one red To each white ball we can associate n red balls.

m

m

white balls, it appears that there are n ways of In this example we note choosing exactly one white and one red ball. that the choice of a white ball does not affect the choice of a red ball, and conversely. The events are said to be independent. We state without formal proof the following theorem: THEOREM 9.1. If there are m ways of performing a first event and n n ways of w#ys of performing an independent second event, there are

Since there are

-

m

performing both events.

The use

of

Given a group

Theorem

9.1 enables us to solve the following problem: numbered 1 to n, in how many ways can we

of objects

order these objects? We have n choices for the object which is to be first in our order. After a choice has been made there are n 1

placed

choices for the object that is to be placed second in the order. Conwe see that there are n(n 2-1 = n' ways of l)(n 2) If we wish to consider the number arranging or permuting the objects. tinuing, of

arrangements or permutations of n objects taken

at the answer n(n

1)

(n

r

+

337

1).

We

r at a time,

write

we

arrive

ELEMENTS OF PURE AND APPLIED MATHEMATICS

338 n

P = nP = P n = r

r

-

n(n

r

n(n

1 )

(n

~

'

2)

(n

1)

-

r)(n

-

-

(n

+

r

-

r

T

+

l)(n

-

1) r) 1

1)

(9.1) r)!

.

.

be9 4

f

,9, using

.

answer.

How many

four-digit numbers can be formed from the integers any integer at most once? We have Pj == 9 /5! = 3,024 as our we were to use the integers as many times as we pleased, our answer would

9.1.

Example 1,2,

If

6,561.

Given 3 red flags, 4 white flags, and 6 black flags, how many signals using the 13 flags? Let us assume for the moment that we can distinguish between the red flags by numbering them 1, 2, 3, with the same statement concerning the white and black flags. It is then obvious that 13! different signals 9.2.

Example

can

we send

Now

the red flags can be permuted in 3' == 6 ways. Each permutaWe must tion, however, yields no new signal if the red flags are indistinguishable. divide 13! by 3! to account for this characteristic of the red flags. By considering the could be sent.

white and black

flags

we

The permutations listed

arrive at the correct answer, 13'/3!4!6!.

of the integers

1,

2,

3,

4 taken two at a time are

below:

12 21

23 32

14 41

13 31

34 43

24 42

A particular arrangement of objects considered independent of their We see that there are 6 order or permutation is called a combination. combinations of the integers 1 2, 3, 4 taken two at a time. Although 12 different permutations, they yield a single combination. If we wish to hire 3 secretaries from a group of 12, we are usually not Now let us see in how interested in the order of hiring the secretaries. we r from order of the choices can choose the objects among n, many ways ,

and 21 are

being immaterial.

combination

Call

will yield r!

this

number

permutations.

n

Cr

"C r

Thus C r H

= r\

C?

=

f

J.

= HP

r

Every

so that

nl ' v (9.2)

r\(n

Note that n

(?o

1

=

=

(

J

n!/n!0! so that 0!

r)!

is

chosen to be

1.

What

does

signify?

among 52 cards in draw poker we are not Thus there are 52 6 combinations of cards which can be dealt. Let the reader show that 62 C 6 = 2,598,960. Let us determine how many full houses (three of a kind and a pair) can be dealt. If we consider the particular full house consisting of a pair of fives and three aces, we note that Example

9.3.

In obtaining 5 cards from

interested in the order in which the cards are dealt.

PROBABILITY THEORY AND STATISTICS 4

there arc

4

Thus

of obtaining three aces.

339

=

4

there are 6

24 different

Ct ways combinations yielding three ares and a pair of fives. But there are 13 choices for the rank of the card yielding three of a kind, and then 12 choices remain for the choice of the rank of the pair. Thus there are 24 13 12 = 3,744 different full houses that can be dealt. Example 9.4. Let us consider a rectangular m X n grid. If we start at the lower left-hand corner and are allowed to move only to the right and up, how many different paths can we traverse to reach the upper right-hand corner? Ail told, we must move Once we choose any m blocks from among the m -f- n blocks a total of m -f- n blocks. for our horizontal motions, of necessity, the remaining n blocks will be vertical motions. Thus there are m4n ( w = (m -f ri)\/m\n^ different paths. Note that the answer is symmetric in m and n. n n a positive integer. In the product Example 9.5. Let us consider (x -f- y) r H will occur. Since a (x -f- y)(x + y} (x + */) we note that terms of the type x term from each factor must be used exactly once, of necessity, r + s = n. The term x r y 8 can occur in exactly n C r ways since we can choose x from any r of the n factors. t

j

'

'

'

i/

Thus

+

(x

y)

n

=

xr y n

2^ ('*)

-r .

If

we choose x =

y

=

1,

we have

2n

=

=0

7

*

^

f

V

r=0

AJ

From

+ x)

(1

Y (") V/ w

=

n

<+->- lO")= fc

Equating

coefficients of

a-*

yields

7-

=

Problems

3.

how many ways can an ordered pair of dice be rolled? Which number will occur most often when a pair of dice is rolled? How many handshakes can 10 people perform two at a time?

4.

How many

1.

2.

In

Ans. 3(> Ans. 7 Ans. 45 five-card poker hands can be dealt consisting of exactly one pair

A ns.

(other three cards different) ? 5. Considering the grid of Example 9.4,

(m

-f-

show that the

l)(n -f 4

1,098,240

grid contains

l)mn

rectangles. 6.

Consider a function of p variables.

formed?

Hint:

by

.

xi,

1.

2,

By

.

.

The ,

xp

grid of

Example

differentiating (1 -f x)

Show

that

How many is

useful

.

n

show that

n 8.

9.4

Y

((~l) r

l

I

=0.

(*)

if

nth distinct derivatives can be

we

designate the horizontal lines Ans. n+ p~ l C p -i

= (* ~

j)

+

(

H

~

1

)'

ELEMENTS OF PURE AND APPLIED MATHEMATICS

340 9.

By

considering

(1 -f x)

n

(l

x)

n

=

x z ) n show that

(1

2k

10. In how many ways can the integers 1, 2, 3, 4, 5 be ordered such that 110 mtegei corresponds to its order in the sequence? For example, 21453 is such an ordering, while 21354 is not, since 3 occupies the third position in the sequence.

An,r'

1

X

'

'

!

Whereas the 9.3. The Meaning and Postulates of Probability Theory. psychologist, economist, political scientist, etc., can, with some measure of success, communicate their ideas to the layman, the mathematician, unfortunately, realizes that he has no hope of describing his chief mathematical fields of interest with a reasonable prospect of being understood.

A modicum

of hope prevails when we consider probability theory from most elementary viewpoint. It is rare indeed to find a man who has not evinced interest at one time or another in breaking the bank at Monte Carlo. However, one need only consider the vast number of persons who believe that someday they will discover an invincible system of gambling, to realize how little the layman actually understands its

the basic ideas of probability theory.

common sense. If this using common sense can

Probability theory has been called the case, it is strange indeed that

the science of

is

two persons

differ so greatly

from each other

in

solving a problem involving probability theory. Let us now investigate the meaning of the word "probability." Webster's New International Dictionary states that probability is "the like-

lihood of the occurrence of any particular form of an event, estimated as the ratio of the number of ways in which that form might occur to the

whole number of ways in which the event might occur in any form." The same dictionary defines "likelihood" as "Probability; as, it will rain

1

Let us consider the question, What To rain on Sept. 9, 1957, in Los Angeles?

1 in all likelihood."

is

that

some mathemati-

it will

the probability

We

There is only one Sept. 9, 1957. cians this question is meaningless. cannot compute the ratio of the number of times it has rained on Sept. 9, 1957, to the total number of days comprising Sept. 9, 1957, rain or shine,

at least not before Sept. 9, 1957. After Sept. 9, 1957, the ratio would be zero or 1 depending on the weather. It is up to the meteorologist to give it will or it will not rain on Sept. 9, 1957, in no front within 1,000 miles of Los Angeles on 1957, the meteorologist would predict no rain with a high degree

us a precise answer.

Los Angeles. Sept. 8, 1

By

If

permission.

Either

there

is

From "Webster's New

International Dictionary," Second Ediby G & C. Merriam Company.

tion, copyright^ 1934, 1939, 1945, 1950, 1953, 1954,

PROBABILITY THEORY AND STATISTICS "

341

"

But the word probability would not be used as a of "probability." mathematician defines probability. Let us now ask, What is the probThis question is, in a sense, ability that a five occur if a die be rolled? very much like the question asked above. Theoretically, if we knew the initial position of the die and if we knew the stresses and strains of the die along with the external forces, we could predict exactly which of the The same statement can six numbers of the die would occur face up. be made about the weather. Unfortunately for the meteorologist there are too

many

variables involved in attempting to predict the exact state The hope of the meteorologist is to reduce the number

of the weather.

of relevant factors to a

minimum

in

an attempt to predict the weather.

The die problem differs from the weather problem in the following way: If we have the patience and time, we can continue to roll the die as often If after n throws we note that the number five has as we please. occurred r n times, we can form the ratio r n /n. One would be naive in calling r n /n the probability of rolling a five, even if n is large. are some who would define the probability of rolling a five as

p =

There

hm n

*

oo

(9.3)

11

limit of a sequence cannot be found unless one knows every term of the sequence (the nth term of the sequence must be given for all n) To

The

.

compute (9.3), one would have to perform an iments, an obvious impossibility.

infinite

number

of exper-

Let us turn, for the moment, to the science of physics. Newton's second law of motion states that the force acting on a particle is proportional to the time rate of change of momentum of the particle. The No particle of Newton's second law of motion is an idealized point mass. such mass occurs in nature. This does not act as a deterrent to the The motion of a gyroscope is computed on the basis that ideal physicist. rigid bodies exist.

The

close correlation

between experiment and theory The math-

gives the physicist confidence in his so-called laws of nature.

ematician working with probability theory encounters the same difficulty. He realizes that a die or a roulette wheel is not perfect. Man, however, has the ability to make abstractions. He visualizes a perfect die, and, furthermore, he postulates that if a die be rolled all the six numbers on " the die are It is, of course, impossible to equally likely" to occur. prove that the occurrence of each number of a die is equally likely.

With these

idealized assumptions

and

definitions the

mathematician can

The success of the gambling predict the probability of winning at dice. houses in Las Vegas is sufficient evidence to the professional gambler that the idealized science of probability theory is on firm ground. The pure mathematician is not interested in games of chance, per se. Probability

ELEMENTS OF PURE AND APPLIED MATHEMATICS

342

theory, to the pure mathematician, is simply a set of axioms and defininow tions from which he derives certain consequences or theorems. The reader is urged to read consider the axioms of probability theory.

We

Sec. 10.7 concerning the union, intersection, complement, etc., of sets. shall find it advantageous to consider a simple example before treat-

We

ing the general case. If a pair of dice are rolled, the following events can happen as listed

below:

E:

(2,6)

(1,6)

E

(4,6)

(3,6)

(5,6)

(6,6)

E

The elements of are called all possible events. is the set of events. subset of events elementary E, say, FI, guaranteeing the occurrence of the number 1 on at least, one of the two dice. is

the collection of

A

/-V

(1,1), (1,2), (1,3), (1,4), (1,5),

(1,6), (2, 1), (3,

1),

(4,1),

(5,

1),

(C>,

1)

Another subset of E is the set F 2 consisting of the events (1, 2), (2, 1), and (6, 6). The union, or sum, of F\ and F is the set of all events or elements belonging to either FI or F 2 or both, written F\ F2 F\ + F 2 In this example FI + F 2 consists of all the elements of FI plus the event The intersection, or product, of FI and F 2 written FI P F 2 = FiF 2 (6, 6). is the set of elements belonging to both FI and F 2 In this example FiF 2 is composed of the elements (1, 2) and (2, 1). If two sets FI and F 2 have no elements in common, we say that their intersection is the null set,

U

.

,

,

.

written

E not

FiF 2 =

0.

What

The complement

of a set

F

is

the set of elements of

the complement of the set FI defined above? The complement of the full set E is the null set 0. Now let G be the set of all subsets of E including E and the null set 0. The elements of G are in F.

is

G is said to be a field of sets in the sense that the called random events. sum, product, and complement are again elements of G. Let us now attach to each element of E (these elementary individual events are also elements of G) the nonnegative number -$. There is no mystery as to the choice of the number -^. First we note that E is composed of 36 elements. The assumption that all 36 events are equally likely implies that any single event has a probability of -$ of occurring. If A is any

E consisting of r events, we define the probability of an eleA occurring as p(A) = r/36. Thus P(Fi) = P(F = ^, = 1. We define P(0) = & = 0. Thus to each set A of G we

subset of

ment P(E)

of

,

2)

have defined a in

PROBABILITY THEORY AND STATISTICS

343

B

have no element

nonnegative number. If A and + B) = P(A) + P(B).

real

common, then P(A

E

be any colFormally, a field of probability is denned as follows Let which are called elementary events, lection of elements x, y, z, and let G be a collection of subsets of E. Assume that the following :

.

.

.

,

postulates or axioms are satisfied I. G is a field of sets.

G contains E and the null set. To each set A in G there corresponds

II.

III.

The number P(A)

written P(A).

ment

of

A

1,

that

B

of

is

a nonnegative real number,

called the probability that an ele-

is,

one of the events of E is sure to happen. element in common, then

G have no P(A

The

is

will occur.

IV. P(E) = V. If A and

E

:

+

= P(A)

B)

+

P(B)

single toss of a coin is a simple example of a field of probability. of the elements G con(for heads) and T (for tails).

H

composed

tains the sets //, T,

p(H, T)

=

1,

p(0)

=

E = 0.

=

p(E)

p(H) = J, p(T) = implies that a head or tail will cer-

0.

(H, T), 1

We

define

,

tainly occur.

Events, such as tossing a coin, rolling dice, spinning a roulette wheel, yield finite probability fields since there are only a finite number of difIf 5 cards are dealt from a pack of 52 ferent events which can occur. 52 It is events which can occur. 6 different cards, there are exactly logical to postulate that the probability of any single event occurring

from among the 52 C & different events be given by p = 1/ 52 CV This is what is meant by an "honest" deal. It is important that the reader understand Postulate V. In rolling a die the event A = (1, 3) means that A occurs if a one or a three is face up on the die. The event B = (4) means that B occurs if a four is face Now A and B have no elements in common, that is, if A occurs, up. B cannot occur, and conversely. We say that A and B are mutually For a perfect die, p(l, 3) = f p(4) = exclusive events. so that Posy

,

,

tulate

V

states that

p(A

+

f -f i = 1- The probability that a the expected answer. A simple working

B)

=

one, three, or four occurs is |, rule for the reader is the following: RULE 1. If A and are mutually exclusive events with probabilities p(A) and p(B), respectively, then the probability that either A or

B

B

occurs

is

p(A

The p(B).

+B) =

p(A)

+

p(B)

reader should give an example for which

p(A

+

B)

<

p(A)

+

ELEMENTS OF PURE AND APPLIED MATHEMATICS

344

Problems

What What

1.

2.

is is

the probability of throwing a seven with two dice? Ans. the probability of throwing a four, eight, or twelve with two dice?

^

Ans. -JFive cards are dealt from a deck of 52 cards. Find the probability of obtaining the following poker hands: three of a kind, a flush, a straight. 88 33 10 4 6 3.

-

A

4.

at

drawing two

of

wrys

numbered from to 20. Two balls are drawn simultaneously What is the probability that their sum is 14? Hint: There are 20 C 2 = 190

set of balls are

random.

1

balls.

Ans. $$ the probability that the sum of the two numbers is 14 if a ball is drawn and replaced and then a second ball is drawn? Ans. -j^ 6. Five coins are tossed. What is the probability that exactly three heads occur? What is the probability that more than two heads occur? Ans. 6-, ^ 6.

In Prob. 4

7.

If

what

is

^

n coins are

tossed,

show that the probability that exactly

r

heads occur

is

Cr /2". 8.

If a coin is tossed

and a pair

occurs and a total of 5 9.

Show that

Postulates

IV and

V imply

-

11.

PA(B)

Show

of dice are rolled,

is

what

is

the probability that a head

Ans.

^

Show

that

rolled?

= by A so

p(0)

Let the complement of A be denoted 1 - p(A). Let A and B be elements of a field G.

10.

p(A)

is

If

p(A)

A + A =

that

9* 0, define

called the conditional probability of the event

B

E.

p A (B) by

under the condition

A

that

p B (A)p(B)

=

Give a geometric interpretation of PA(B], assuming 9.4.

Theorem

of Bayes.

A and B

are point sets in a plane

Let us consider the following simple example Assume that one throws a dart at a target whose area is taken to be 1. :

We

assume further that the dart

is

sure to strike the target and that the dart is equally likely to strike any-

where on the that if

target.

This means

A is any area of the target then

the probability that the dart strikes a point of A is simply the area of A, written p(A). Now let A and B be

FIG. 9.1

overlapping areas (see Fig. 9.1). Let us consider a second person

room who knows The probability that the dart

located in another that a dart will be thrown at the target. falls in

A

will

be given by p(A).

Let us now assume that after the dart

is

PROBABILITY THEORY AND STATISTICS

345

thrown our second person asks the following question Has the dart :

in the area

He

B?

affect this person 's

A?

fallen in

We

fallen

How does this receives the truthful answer, "yes." opinion as to the probability that the dart has also

can answer this question in the following manner It is if it is known that the dart lies in B then the only way :

obvious that

fairly

can also lie in A is for the dart to have struck the region A C\ J3, and the probability of this happening before it is known that the dart has fallen in B is simply the area of A B, written p(A C\ B). Since the dart is known to be somewhere in B, it appears that the ratio of the area it

H

A

of

C\

lies in

B

A

to the area of

that the dart

B

will yield the probability that the dart also

For example,

C\ B.

is in

A

C\

B

if

A

if it is

C\

B is ^ the area of B, then the chance We that it is in B is simply ^.

known

write -

and is

define

known

P(B)

*

pn(A) as the probability that the event

that the event

B has happened.

We may

(9.4)

A

has happened

also say that

if it

ps(A)

is

the conditional probability of the event A under the condition B. It is to be noted that (9.4) is a definition and requires no proof. Experimentally, however, one might attempt to verify (9.4) to some extent. is in JB, he records whether

Every time an observer learns that the dart

A or is not in A. He is not concerned with those experiwhich the dart lies outside of B. The ratio of the number of successes to the total number of tries (the dart must be in B} yields a fraction lying between zero and 1. For a large number of experiments the dart

ments

it is

is in

for

reasonable to hope that this

Since the roles of

A

and

B

number

is

close to the

}

P(A)

Combining

A

and

A

(9.5)

is called the a priori probability called the a posteriori probability that the hypothesis that occurs.

occurs, whereas

occurs under

*

(9.4)

one form of Bayes's theorem,

(9.5) yields

p(B C\ A) = p(A C\

since

that

(9.4)

number p B (A).

can be interchanged, we have from

B).

ps(A)

p(A)

is

B

A group of 100 girls contains 30 blondes and 70 brunettes. Twenty9.6. the blondes are blue-eyed, and the rest are brown-eyed, whereas 55 of the brunettes are brown-eyed, and the rest are blue-eyed. A girl is picked at random. Example

five of

The a girl

at

If we pick a priori probability that she is a blonde and blue-eyed is -j^nr = -J. random and find out that she is blue-eyed, what is the probability that she is

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

346

blonde? Let x denote the quality of being blonde and y denote the quality of being We are interested in determining p y (x). Now p(y) = 3^^ == 0.4, since blue-eyed. 40 of the 100 girls are blue-eyed. Moreover, p(x C\ y) = -J-. Applying (9.4) yields

P(y)

Note that there are 25 blue-eyed blondes and 15 blue-eyed brunettes, so that the appearance of a blue-eyed girl means that the probability of the girl being a blonde is = In information theory one determines the ratio of the conditional probability to the a priori probability and defines the amount of "bits" of information as .

In this example, /

the logarithm (base 2) of this ratio

We

can extend the result of

(9.5) as follows:

elementary probability events Ai, A*, n, mutually exclusive, so that 2, .

.

,

Iog2

E

Let

An

3

=

1.

Iog2 -pf

be a collection of

A

with the

l,

i

=

1,

.

.

,

E = A + A,+ let

B

B

C\

A

%,

i

=

1,

2,

.

.

.

,

n, are

mutually

Thus

p(B)

= p(B

C\ .40

p(B

Applying

Then

be any subset of E.

Let the reader deduce that exclusive.

An = VJ A,

+

-

-

l

Now

.

.

=

H

+

p(B C\ A.)

+

+

p(B C\

A n)

A.)

(9.8)

(9.5) yields

p(B)

Substituting (9.9) into

=

p(A )pAl (B)

(9.9)

t

(9. 6) "yields

Bayes's formula,

P(A,) = ~~

J

p(A )p At (B) t

important to realize that the event B need not be a subset of E For example, the events A x A 2 A n might be different urns containing various assortments of red and white balls. The event B could be the successive drawing of three red balls from any particular urn, each ball being returned to its urn before the next drawFormulas (9.7) to (9.10) would still hold. By B Al = Al B ing. It is

in the ordinary sense.

,

,

.

n

.

.

,

H

PROBABILITY THEORY AND STATISTICS

we mean the composite event

of choosing

347

AI and then drawing three

successive red balls as described above.

Example 9.7. Let us consider two urns. Urn I contains three red and five white and urn II contains two red and five white balls. An urn is chosen at random -1 (p(Ij) = p(I 2 ) = ^), and a ball chosen at random from this urn turns out to be red What is the a posteriori probability that the red ball came (pi(K) - |, pn(R) = y). from urn I? Applying (9.10) yields balls,

?

3

1 _

.__a_j[

^

91

ff

Two balls are placed in an urn as follows: A coin is tossed twice, and placed in the urn if a head occurs, with a red ball placed in the urn if a Let Ao, AI, A 2 represent the events of the urn containing none, one, and tail occurs. two red balls, respectively. Balls are drawn from the urn three times in succession 9.8.

Example

a white ball

is

(always returned before the next drawing), and it is found that on all three occasions a red ball was drawn. What is the probability that both balls in the urn are red? Let B be the event of drawing three successive red balls. We are interested in com-

puting PB(AZ).

Applying

(9.10) yields

p(A

)p Ao (B)

+

p(Ai)p Al (B)

+

p(A<t}pAi(B)

p(A 2 = T, p A9 (B) = 0,p Al (B) = (^) 3 = ?,p At (B) = 1, Fromp(Ao) = ?,p(Ai) = we obtain pB(A 2 =3-. Let the reader show that, if B n represents the successive )

,

)

Is this reasonable to expect? drawing of n red balls, then pB n (A>i) tends to 1 as n * We note that pAl (B) represents the probability of drawing three successive red balls from an urn containing one red and one white ball. We liken this problem to that of The event space for this finding the probability of tossing three successive heads. .

case

is (H, H, //), (H, H, T), (H, T, H), (//, T, T), (T, H, H), (T, H, T), (T, T, H), The a priori probability of the event (T, T, T), involving eight equally likely events. note that, for a single toss of the com, p(H) Is it reasonable (H, H, H} is |-. -%.

We

to expect that section.

p(H, H, H)

=

p(H}p(H}p(PT)t

This question

is

answered

in the next

Problems Three urns contain, respectively, 2 red arid 3 black balls, 1 red and 4 black balls, and 1 black ball. An urn is chosen at random and a ball drawn from it. If the ball is red, what is the probability that it came from the first or second urn? Ans. ^ %f Urn I contains two red and three black balls. Urn II contains three red and two black balls. A ball is chosen at random from urn I and placed in urn II. A ball is then chosen at random from urn II. If this ball is red, what is the probability that a red ball was transferred from urn I to urn II? Ans. y\ 1.

3 red

9.5.

Independent Events. Events Not Mutually Exclusive. we have

If

we

rewrite formula (9.5),

p(A Formula

O B)

= p(A)pA (E)

(9.11) states that the probability that

the probability that

A happens

both

(9.11)

A

and

times the probability that

B

B

happen is happen

will

ELEMENTS OF PURE AND APPLIED MATHEMATICS

348

A

if

What can we surmise if p(A r\ B) = p(A)p(B)t Of PA(B) = p(B), so that the a priori probability of B happening,

happens.

necessity,

p(B), does not depend on the event A. We say that A and pendent events. Two events are independent, by definition,

A

p(Ar\B) = p(A)p(B) A events, AI, A 2

set of probability

independent

,

t

i,

die,

We

j

=

1, 2,

.

.

.

,

(9.12)

are said to be mutually i

*j

(9.13)

n.

the tossing of a coin and the rolling of a of 12 elements,

(H,l)

(H,2)

(#,3)

(ff,4)

(H, 5)

(H, 6)

(T, 1)

(T, 2)

(T, 3)

(7',

4)

(T, 5)

(T, 6)

can assign the a priori probability of head occurs and a six occurs

ability that a

upon the coin,

Tl)

,

= p(AJp(A,)

A,)

we consider, for example, we obtain the event space

If

.

are inde-

if

if

p(A C\ for

.

.

B

rolling of a die as completely

we note

that p(H)

= ,

p(6)

Assigning the a priori probability of the two events are independent.

^ to each event. is

taken to be

The probIf we look

^.

independent of the tossing of a

= ,

^

is

and p(77 Pi

6)

=

=

^

equivalent to assuming that

A die is rolled n times. We compute the probability that a six occur The a priori probability of failing to throw a six on any given toss of the If we assume that each successive roll of the die is independent of the die is -^. Hence the previous throws, the probability of not obtaining a six in n throws is () Example

9.9.

at least once.

Tt

.

probability of rolling at least

=

1 six

m n tosses is

1

(-)"

For n

3,

p

<

-$,

while for

p > Example 9.10. The game of craps is played as follows: A pair of dice is rolled, and He loses if a two, three, or the player wins immediately if a seven or eleven occurs. twelve arises on the first throw. If a four, five, six, eight, nine, or ten is thrown, the player continues to roll the dice until he duplicates his first toss or until a seven occurs. n

4,

.

The numbers two, three, eleven, twelve are disregarded after the first toss. If the We player duplicates his first toss before rolling a seven, he wins; otherwise he loses. compute the probability that the man rolling the dice will win. At gambling establishments an opponent of the roller of the dice (except the establishment) does not win The probability of winning on the first toss is if the player rolls a twelve. p(7)

+p(ll) =

A

ways of rolling a seven, 2 ways of rolling an eleven, and 36 total two dice. The player can also win by rolling a four and then rolling a four before a seven. The probability of rolling a four is ^, and the probability of Thus the probability of rolling a four and then rolling a four before a seven is The same reasoning for the numrolling another four before a seven is -/g % = ^g. since there are 6

possible throws of

.

-

bers

five, six,

eight, nine, ten yields -

49292

PROBABILITY THEORY AND STATISTICS

349

Let p be the probability of success of a certain p the probability of its failure. Assume that the experiment is performed n times, the probability of success remaining constant, while the result of each experiment is independent of all the others. Such a sequence is called a BerThe simplest example is illustrated by the repeated toss of a coin. noulli sequence. We determine now the probability of attaining exactly r successes in n trials. In the case of the coin problem a particular successful sequence would be the sequence H T T\L\ Tn -r if we were interested in obtaining exactly r heads. H\Hi There are obviously n C r different sequences containing exactly r heads and n r tails.

Example

-

Bernoulli Trials.

9.11.

experiment, q

=

1

-

'

n probability of obtaining any particular sequence is (-g-) heads in n tosses of a coin is given by show that r n-r = n

The

.

n

of obtaining exactly r

P

is

r

r

the probability of obtaining exactly It can be shown that the

sequence.

r

Thus the probability Cr /2 n Let the reader .

p q

(9.14)

successes in n trials for the general Bernoulli

number

r

which makes

P

T

a

maximum

for a

given n and p is the greatest integer less than or equal to (n -\- l}p.

We now

consider events which

may not be mutually exclusive. Let E be an event space with subsets AI, 2,

A

2,

.

.

.

.

.

,

.

n,

An

,

.

The A

t,

i

=

l f

need not be mutually

exclusive (see Fig. 9.2).

Certainly

FIG. 9.2

since there is

may

be overlapping regions of

A

A

iy

2,

A

3

Geometrically,

it

easy to verify that

Ai\J A 2

In AI

Ai

s

A* = AI

+A +A 2

3

AiC\ A* - A 2 H A 3 n A + A C\ A 2 r\ A 3

- A

3

I

I

(9.15)

+A +A

we have counted AI C\ A 2 twice; so we substract 2 3 with a similar statement for AI Pi A 3 A% C\ A z In AI A2 we have counted AiC\ A^C\ A 3 three times, and then we have

nA

+A

VJ

2,

,

H

A 2 C\ A$ three times in A 1 C\ A%, etc. subtracted AI A 3 to obtain (9.15). In general AI C\ Az

n

.

+

Thus we add

* <ssl

(9.16)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

350 Example 1 to n,

n are placed at random in n urns numbered the probability that the number of at least one of the urn in which it is placed? Let A t be the event

numbered

Balls

9.12.

1 to

What

one ball to each urn.

is

ball corresponds to the number of having the ith ball in the ith urn regardless of the distribution of the rest of the balls. n. Then p(A t ) = 1/n for i 1, 2, , p(A t C\ Aj} represents the probability that both the ith and jth ball be in their proper urns. From (9.5), .

p(A

t

.

.

n A,)

-

p(A,)pA,(A.)

=

Now pA.(At) l/(n 1), since, if it is known that the jth ball is in the proper 1 Thus that the ?th ball will bo in its proper urn urn, there is one chance in n We note also that there arc "C 2 different A* C\ A 1). l/n(n p(Ai C\ Aj) }

P -

n i

1

W __

_

-

\2/ n(n -

__i 41^

L_j__l

_i-

21+3!

Pn =

Let the reader show that lim n

We

.

easy to see that

it is

Continuing,

+ .

.

W ____- ____ ^ n(n

1)

4. ^

.

(__i)n+i L) (

l)(n

-

4-

2)

_i

n!

l)/e.

(e

*

conclude this section by considering the simple one-dimensional Assume that a person starts at the origin. A

random-walk problem.

coin is tossed, with the result that if a head occurs the person will move one unit to the right and if a tail occurs the person will move one unit to the The process is repeated n times. What is the probability that the left. final position is located at x = r? Let u be the number of moves to the right and v the number of moves to the left that can occur so that v = r, u v = n, so that the final position will be at x = r. Then u n u = (r + ri)/2, v There are C different u r)/2. (n ways in which one can move to the right, the rest of the moves being to the left. The r is thus n C r+n )/2/2 w This problem is probability of ending at x

+

y

.

(

similar to that of

Example

9.4.

Problems 1.

What

is

dice?

the probability that -a seven occurs at least once in n tosses of a pair of n Ans. 1 - (|)

2.

Derive

3.

Show that

(9.14).

P[AI r\ (A 2

What does 4. Show

this

uA

=

3 )i

n AO

P<A!

formula reduce to

if

Ai,

+

nA

A and As 2,

8)

-

P(A,

nA

2

r\

A 3)

are mutually independent?

that the probability of having at least r successes in a sequence of n Bern

noulli trials is

5.

P<AI

P =

Show that

>

t-0

}

f

( i )

t

-

j

p

f

pq

(l

n~i .

- p)~ -

1.

PROBABILITY THEORY AND STATISTICS 6.

A com

351

tossed in a Bernoulli sequence. What is the probability of obtaining tails appear? heads in the first Hint: If there are at least

is

m

heads before n

n

1 tosses of

m

the coin, the

game

is

m +

won.

^

m-\-n

1

I

V

/m+n-l\ (

r

)

r<*m 7.

for

Generalize the random-walk problem to the case of two dimensions with p

-|

each of the possible motions.

8.

9.

+

A and B

taneously,

What

have, respectively, n is the probability that

1

A

what is

and n coins. If they toss their coins simulhave more heads than B? Ans. p = -%

will

the probability of getting exactly 2 sixes in three throws of a die?

10. Coin 1 has a probability of p of getting a head, and coin 2 has a probability of q of getting a tail. start with coin 1 and keep tossing it until a tail occurs, whereupon we switch to coin 2. Whenever a tail occurs on coin 2, we switch to coin 1, etc.

We

What is the probability that the nth toss will be performed on com 1? Hint: Let = P n p -f (1 P n )</, P\ = 1. Show that be the desired probability so that P n

Pn

\

PM-I

- Pn =

(p

-

q)(Pn

- P-i) =

- D(p - q) n+1 Ans P H - gJlJ (P

1

+q -

p

In previous 9.6. Continuous Probability and Distribution Functions. discussions the probability event space consisted of a finite number of elementary events. A simple example illustrates the extension of this idea.

We

consider an idealized spinner whose pointer can assume any ^ x < 2ir, x measured in radians. We say that

one of the directions

zero probability that the direction of the pointer be less than zero or greater than 2ir since we are concerned only with angles between zero

there

is

The direction of the pointer has been put into a correwith the real-number system. The set of all possible direcspondence tions is called a one-dimensional random, or stochastic, variable, usually In this example it makes very little sense to ask the denoted by and 2w radians.

.

following question: What is the probability that a random spin yields ? the direction #o is just one possible event of an infinite number of different events which may occur. It does make sense to ask the follow-

ing question What is the probability that the random event be less than We define the a priori distribution function F(x) as or equal to x? :

F(x)

F(x)

=

for x

<

= P( g x} = = 1 for x ^

-

for

^

x

JjTT

F(x)

2?r

Figure 9.3 shows a graphic representation of F(x). For any x and y we note that (9.17) yields

P(xSy)

=

F(y)

-

F(x]

<

2*

(9.17)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

352

x + Ax) = F(x + Ax) - F(x) = AF. AF is the probability that lies between x and x + Ax. We also note that F(x) is a monotonic nondecreasing function of x with F( <) =0,

For y

=

x

F(+oo) =

+ Ax we have P(x ^

^

1.

FIG 9.3

For the more general case we consider a nonnegative function p(x) lie such that p(x) dx represents the probability that the random event We between x and x dx, except for infinitesimals of higher order.

+

further desire

=

p(x] dx

(9.18)

random

p(x) is called the probability function of the distribution function is simply

= P( ^

F(x)

Note that F(x)

=

x)

nondecreasing since p(x)

is

F(-oo) = dF(x)

A

=

p(x) dx

is

=

an

of a distribution

=

provided the integrals

("^

is

x n p(x) dx

exist.

For p(x)

=

dF(x)

=

p(x) dx

we can n ^n

/ /

p(x)

+

y1

rj~

>

X

dx.

with

by

0, 1, 2,

^A

= 7T

7T

(9.19)

between x and x

is

defined

n =

dx

1

lies

simple example of a probability function

The nth moment

The

.

nonnegative, with

F(oo)

the probability that

is

7>(x)

/"'

J-OO

variable

...

a X ^ only

j~~f~

(9.20)

exists.

From

write

rk n r!F(r} t v**'/ *'-'

n '^

">

1*j

9 ^j

fQ91"!

^J/.^i^l

Equation (9.21) is the Stieltjes integral of Chap. 7. The discrete case can be handled by use of this integral. Let x = 1 represent the occurrence of a head and x = 2 the occurrence of a tail when a coin is tossed

PROBABILITY THEORY AND STATISTICS

353

Then F(x) = for x < 1, otherwise. with p(l) = p(2) = p(x) = = I f or 1 ^ # < 2, /^(x) = 1 for x ^ 2. We can consider that two From dF(x) point masses contribute to the probability function. for 1 < x < 2, dF = J at a; = 2, for x < 1, dF = 1 at x = 1, dF = dF = f or x > 2, we note that ,

,

^0*0

x

<

1

for x

=

1

for

dF = f

for 1

<

x

1

for x

>

2

we note

that F(x)

= Geometrically (see Fig.

9.4),

=

Discontinuities in F(x) occur at #

ing.

1,

is

=

x

<

2

monotonic iiondecreasbecause point masses

2,

F(x)

F(x)=Q 2

1 9 4

Fi(i

To compute

are situated there. a.

Had we

chosen x\

=

=

,

=

x-i

1

,

that

=

;

1

we note

i,

=

we would have obtained

= (-l)i + 0)i =0 mean of x = 1 and x = 2. 01

Note that

|-

is

the

9.13. TTi? Gaussian Distribution. Let us assume that a dart is thrown at plane under the following assumptions: I. If p(x, y) dy dx is the probability that the dart fall in the area bounded by x and x -h dx, y and y -f- rfy, then p(x, y) depends only on the distance of (x, y) from the = tan" (y/x). Thus origin and is independent of

Example

the

#?/

1

We

p(x, y} dy dx

=

p(x, y)

=

q(r*)r dr dB

q(x

2

q(x

2

dx

2

-\-

i/

)

assume further that

II.

III.

We

p(x, y)

1^

*

r

p(x, y) is differeritiable. as x or y becomes infinite.

p(x,y)dydx =

attempt to determine p(x,

/_ oo

p ( x>

^

1.

y).

^ ^

First

we note that

^ /- - P X)

^ dx

'

dy

ELEMENTS OF PUKE AND APPLIED MATIIEM \TICS

354

x

since PI(X

P (2

and

oo

P(x ^

/oo

+

dx,

<

x

<

<

oo,

^ x

+

p(z,

?/) dt/,

^

y

dx, y

<

y

^

<rj

^

+

y r

=

S(//)

/

p(x, y) dy

=

y -f dy)

^

r,

-

oo)

dy)

/

dx,

dx ^ p(x, y)

= P,P Z =

p(z, y)

dy

dy dx

oo

p(x, y) dx, so that

/

?Gr

2

+

2 ?/

(9.22)

)

Differentiating (9.22) yields

.

and dy

yS(j/)

From

(9.23), JB(a;)

= Ae

f

S(y)

Condition II implies that a

1/2*

from condition

(9.25) is the

m

(9.23)

fl

is

III.

,

=

g(x

2

+

2 ?/

If

negative

)

= (V<' i+ * we choose

f

>

a

(9.24)

=

l/2<r

2 ,

of necessity,

Thus

=

o^-, f-d^DCjrt-l-yl)

(

^Trer^

normal distribution of Gauss

0-

where

=

= Be a ^ and

P(X, y)

Equation

constailt

2

,

p(x, y)

C

, dx

o;/2(x)

9 25)

For the one-dimensional case

X/27T

Let the reader show that

represents a displacement from the origin

-

(a;

Thus <r 2 is the second moment of <p(x) relative to the center m. Example 9.14. Tchebyscheff's Theorem. Let be a random variable with a probaLet g() be a nonnegative function of 4, and let S be the set of bility function p(x). 5 will denote the rest of the real axis. The expected points such that g() ^ K > 0. or mean value of g() is denned as E[g(&]

The

- [ J

" g(x)p(x) dx 00

probability that </() be greater than or equal to

simply the probability that

be found

in S, so that

K when

P[g()

is

^ K]

chosen at random ==

/

7*s

- [O g(x)p(x)dx+ Lg(x)p(x)dx JS JS g(x)p(x)

dx^K

p(x) dx

p(x) dx.

is

Now

PROBABILITY THEORY AND STATISTICS

We

and p(x) are nonnegative.

since g(x)

P\a(&

What If

difficulties arise

m is the

first

if

=

g(x) or

moment

1

for

mean

355

obtain Tchebyscheff's theorem,

6 K] 5 j-

irrational, </(x)

and a 2

of

(9.27)

is

=

for x rational,

the variance of

defined

K =

^?

by

m) 2 p(x) dx

(x >

let

the reader deduce the Bienayme-TehebyschefT inequality,

~ w ^

P(\t

if

we choose #() =

2

?/i)

(

>

=

A'

A'V 2

^ ^

ka)

(9.28)

.

Buffon, Needle Problem. A board is ruled by equidistant A needle of length a < d is parallel lines, two consecutive lines being d units apart. What is the probability that the needle will intersect thrown at random on the board one of the lines? Let x and y describe the position of the needle (see Fig. 9.5).

Example

9.15.

7

7

/?f

r.JfLj?.

FKI 9.5 If y ^ a sin x, our situation is favorable, while if d > i/ > a sin x, an unfavorable We assume that x and y are independent random variables Thus case occurs

Pi(x

^

^

x

+

dx)

=

pz(y

^

rj

^

y -f dy)

= '-f (/

7T

and

sm

a

p(x, x),

y)

dy dx

which

is

C TT

Jo

=

fa sm Jo

Let

lies

Example

P(XI) dx\.

x2

+

dx 2

is

dy dx.

We

are interested

in

F(0 ^ x ^

x

}

p(x

>

y)dydx

^

?/

^

Cv

=^d]o

ra

sm

x

d

2a

Jo

between

two experiments (assumed independent)

is

pCrOpOrz) dxi dx 2

Equation

?r,

by

be a random variable with a probability function p(x). The x\ and x\ -\- dx\ as the result of a single experiment is lies between x 2 and If we repeat the experiment, the probability that p(x z ) dxi. The joint probability that both results happen as the result of 9.16.

probability that

"

(l/ird)

defined

(9.29) is a special case of the single

function p(x, y).

We

(9.29)

random variable (,

note that r /

*

-00 7-00 /oo

p(xi)p(x z ) dxi dx 2

=

1

rj)

with probability

ELEMENTS OF PURE AND APPLIED MATHEMATICS

356

We now

ask the following question What is the probability that, as the result of two a constant? In Fig. 9.6 we note that this is just the xz ^ K, :

K

+

experiments, x\

2

=A- X

l

FIG. 9 6

probability that the point (x\ #2) y

P(XI -f x z

From

(9.30)

lie

above the

^ K) =

line x\

/

/ 7-

JK-xi

-\~

K, so that

Xi

p(zi)p(:r 2 ) dx-i dxi

(9.30)

we have

=

Pfa+xt^K+dK)

f

so that

+

X!

x*

K +

f

dK) =

/

(9.31)

from

[

Since

(9.30)

/

J

note that the probability function for

p(w)

=

xi

fK+dK-xi fK+

oo

-

J

by subtracting

(9.31)

[" ^A+d/v

oo

y

i

/

dx^

(9 32)

JK-JTi

oo

K+dK-xi pCr 2 )

K

da: 2

~ p(K

x\)

dK,

xi

= u

xi -f Xz

is

(9.33)

/_

Problems 1.

Show

that

of

satisfies

1.

^(x) (9.26) [ Derive (9.28). 3. Find the four-dimensional joint probability function for the two-needle Buffon problem. i

<

2.

4.

Show that

<r

2

&

/

(z

-

ra)

2

p(z) dx

-

a2

-m

2 ,

a 2 denned by

(9.20).

./-co

= 0). 17 be random variables with Gaussian probability functions (m + 17 has a Gaussian distribution. 6. Find en, a 2 a 3 for ^>(x) of (9.26). See (9.20) for the definition of a t i = 1, 2, 3, 7. If is a random variable with a Gaussian distribution (m =0), find the proba2 Hint: Let u = 2 so that bility function for the random variable and Show that u ~ 5.

Let

,

.

,

.

P(u

-

^

P(\t\

-f

.

.

PROBABILITY THEORY AND STATISTICS Then show that P(t ^ u ^

t

=

+ dt)

1

for

8.

\/2irO

(l/<r

>

j

357

so that

eft,

=

p(Q

for

g

t

be a random variable with probability function p(x).

Let

strictly increasing (or decreasing) function of

for the

e~</

2<r2

random

variable

u

is

=

q (u)

p(x(u))

.

du

Show

Let u

= w() be

a

that the probability function

where x(u}

is

the inverse function

Apply this result to Prob. 7. Consider a distribution of points in the plane with a density function given by each point moves in a direction <p with speed V and with a probaAt t = (9.25). = <p/2w, ^ < 2?r. Show that at time t the probability funcbility function p(<p) of u(x). 9.

<f>

tion in space and direction of motion

is

given by

1 (,-( 1/2(T2)

Show

that p(r, v

9.7.

= V

The

tral-limit

6, <?, t}

cos

satisfies

(p i

+

T

7

~+V

sin

=

(pv)

= V

<p j

cos

Characteristic Function.

[

(<p

witli

0)e r

-f-

^ sm

0)e0

(^>

Theorem.

Bernoulli's

The Cen-

Let J be a random variable with a probability Let us assume that the Fourier integral of p(x) exists.

Theorem.

function p(x).

Then c**p(x) dx is

called the characteristic function of p(x), with

<p(0)

=

/

7

00

p(x) dx

I

I

If

(9.34)

we can

f

y-oo

=

For

1.

real,

x zn e* xp(x) dx

^

n a positive

f y -oo

2n x' p(x)

We

real.

integer,

dx

=

note that

we have

a 2n

differentiate inside the integral, then

= Hence, if we can find from (9.35). Example

t

t

9.17.

Let p(x)

<p(t),

=

n

x n p(x) dx

it will

e~ x for x

^

=

"dt

=

d*<f>

dt*

t-o

(1

-2 -i

i

n

an

(9.35)

be possible to find the nth

0,

p(x) -n

/.'

=

T

=0 otherwise. -^

Then

moment

ELEMENTS OF PURE AND APPLIED MATHEMATICS

358 so that ai

-

a2

1,

Since

2. 00

V

__J_ =

n=0

we have

^>

(n)

=

(0)

i

n

n!

and a n = r(n

Example

9.18.

This result

n'.

+

=

1)

is

also obvious since

=

x n e'*dx

n'

yo

For a Bernoulli sequence of n events we have

Bernoulli's Theorem.

P =

"p

r

r

a ~

P)

n-r

(9.36)

for the probability of obtaining exactly r successes [see (9.14)] The analogue of (9.34) for the discrete case is

(>inp >

2

-oo

Applying

a2

=

>(0)

=

n(n

1,

l)p

2

=

^

-J-

e

n(n w- 2

\-7j-y

-

np|

^ p ^

for

-

-f (1

=

P(l

p)

4

[pc

= ipn, <^"(0) 2 = o er so that 2 -f np, ^>'(0)

Applying (9.28) with k

since p(l

(9.37)

(9 36) yields

= Thus

c*dF(x)

1.

^

^

en)

P)]

n

- I)/) - np Hence = wp(l - p) (see Prob

>

2

0,

4,

= m = p Sec. 9.6).

yields

^

Formula

ai

^

(9.38)

is

^

^n

(9.38)

equivalent to

(9 - 39)

In other words, the probability that Kormula (9.39) is a form of Bernoulli's theorem the frequence of occurrence, /n, differs from its mean value p by a quantity of absolute value at least equal to e tends to zero as n > however small e is chosen. This essentially means that the greater n is the more certain we are that /n differs little

from

p,

where

is

the total

number

of successes in

Let us return to the characteristic function. if <p(t) is

n

trials.

From

(9.34)

we note

that,

known, then i

r t

(9.40)

PROBABILITY THEORY AND STATISTICS

359

Hence p(x) is uniquely determined if the characteristic function is known. and 17 Equation (9.40) follows from the Fourier integral theorem. If are independent stochastic variables with probability functions pi(x), p*(y), respectively, it follows from the method of Example 9.16 that p(u) is

= I

x)

dx

(9.41)

+

the probability function for the stochastic variable

acteristic function of p(u)

<p(t)

-

c ltu p(u)

00

Now

J

f_

J

f

=

#

+

A

?/.

-

dx du

x)

(9.42)

e**pi(x)dx

(

f 7-00

oo

oo

J

=

(x)p 2 (u

c^ r+ ^p

l

(x)p^(y)

00

= ?/

1

<X>

00

?i(<WO =

letting

e^p

f

00

<pi()

so that

char-

du

x

f

J

The

y-

is

= J

by

pi(x)pi(u

co

J

f

00

^

e ttu pi(x)p z (u

-

dy dx

x)

du dx

(9.43)

comparison of (9.42) with (9.43) yields (9.44)

provided the order of integration can be interchanged. Another interesting result is the following: If <pi(t),

...

.

.

<pz(t),

. ,

a sequence of characteristic functions which converges to <p(t), obtained from a sequence of probability functions pi(x), p*(x), it may be possible that p n (x),

<p n (f),

is

.

.

.

,

,

r-*v(t) dt

(9.45)

ao

where

<p(t)

that (9.45)

p(x)

=

and p(r) are limits hold, we must have

lim p n (x) 7}

oo

= Hm

'

n

oo

T

1 7r

of their respective sequences.

00

e~*<p n (i) dt

/

J

<*>

=

1

In order

f"

**R J/

e-*v(0

*

oo

(9.45')

The reader is referred to Sec. 10.22 for a discussion of this type of problem. Let us consider now the following example, which will illustrate one be a aspect of the central-limit theorem involving probabilities. Let the be to we shall consider momentarily specific, variable, and,

random

toss of a coin.

The random

head occurs and the value x

variable

=

1 if

will

a

have the value x

tail occurs.

If

we

=

1

if

a

toss the coin

ELEMENTS OF PURE AND APPLIED MATHEMATICS

360 five times, 1

or

1, i

we can record the numbers (xi, x^ #3, x^, x$) with Xi equal to = 1, 2, 3, 4, 5. The set (#1, #2, #3, #4, x$) represents a point

We

in a five-dimensional space.

we

five coins as often as

=

x\, xi), i

1, 2,

.

.

set of 5-tuples, (x\, x\, x\,

We may look upon the aggregate x\,

.

.

can repeat the experiment of tossing

and obtain a

please

x\, xf,

as a set of results yielding information about the random variable same statement can be made concerning the set x\, x\, x\, x\, .

.

.

.

.

.

.

.

,

The set

etc.

of 5-tuples can also be looked

as defining a

upon

.

The

.

,

new random

For the more general case we let p(x) be the probability funcrandom variable and we consider the n-tuple of independent events (x\, x^ x n ). The probability that a point lie in the volume bounded by x\ and Xi + dxi, 0*2 and Xi + dxt, x n and x n + dx n is given by variable.

tion for the

,

.

.

.

,

p(x n ) dxi dx z

'

'

'

p(xi)p(x 2 )

'

'

'

dx n

n If

8n

is

random

the

N

variable

x

t,

an extension

of

Example

9.16 shows

ii that the probability function for

-

/>

I

(

>Sf w

is

given by

x.)

l-l

n-1

=

'

f~^

f~^

The

'

p(jci)p(x$

characteristic function

'

'

p (u

-

'

^ x^ dxn-i 1-1

'

'

dx l

(9.46)

dx n

(9.47)

is

^(hi

C+** by

letting

=

w

of integration.

#1

+

#2

+

Equation

-**-> f

+

'

(9.47)

P(x n )dx 1

p(x,)

x n du ,

=

-

dx n and reversing the order ,

becomes (9.48)

with p(x)

<p(t)

is

=

w p(x)e

zero so that

^ (0) = hood of t x/

with

/

y

\a(t)\

2 <r

==

o-

If <p(f)

.

0,

<

2

itx

dx.

Let us assume that the

=

moment

=

of

second moment. Thus ^(0) 1, ^'(0) ^7 has a continuous third derivative in a neighbor-

is its

then

A^ 3

first

(see Sec. 10.23).

PROBABILITY THEORY AND STATISTICS

Tn

If

random

the

is

Vn)\ <

with \a(t/a

Bz

z

S n /<r \/X l e t = $ n (<A n (t)

variable

Tn

characteristic function for

is

-

=

z)

/

[2 1

(9 49) it follows

o-

^^

+

z

with

/3(z)

\p(z)\

<

+

(

\~|

f

"-7=

\<7

= -

)

V^/ J

2

o ^

(9.49)

that lim $ n (t)

The

+

In (1

Hence

\/n).

one has that

for z small,

From

the reader show that the

\l/

From

At*/<r*n*.

361

=

e~* 2/2

probability function P(x) associated with the characteristic function

e~ tz/z

is

P

^=i T

e

~'^" X di =

77^z

(

'^ /2

we accept (9.45'), we have shown that if 2, random variables with the same probability function If

(9-50)

1,

,

n,

p(x), with a

are

mean

n

equal to zero and second

moment

equal to

2 o-

,

then

T n = Y f/ t~i

a random variable whose probability function approaches the normal ~ / distribution (l/v ^") e a 2/2 as n becomes infinite. We say that the sequence of random variables obeys the central-limit law. is

"

Problems 2.

Derive Derive

3.

Show

4.

Let p(x)

1.

(9.49). (9.50).

that the

=

1

for

characteristic

J-

^

x

^

function

^, p(x}

<p(t)

for

Cauchy's distribution function,

=0 otherwise.

= -

sin

Show that

g

6. If is a random variable with characteristic function p(t) t show that e~* ta <f>(t) = constant. the characteristic function for the random variable a, a

is

Consider a sequence of 1,000 Bernoulli trials with p = ^. Show that the probaexperimental ratio r/n will differ from % by less than 0.01 is greater than or equal to f6.

bility that the .

ELEMENTS OF PURE AND APPLIED MATHEMATICS

362 9.8.

The x 2

tion of the

From

Application to Statistics. function (see Sec. 4.16) we have

Distribution.

gamma

=

the defini-

e

/

Jo /

a*

I

Jo

/" so that

/

Jo

& r-r

=

1

e-^y*- dy

>

a

1

(9.51)

r(z)

The function /(*/; a, 2)

=

c

~ al/

-^j

z

~1

2/

=

for

?/

>

for

T/

^

/(?/; a, 2)

/oo upon/(?/; a, 2) as a probability function.

(9.52)

=

r/;/y

We

1.

can look

Its characteristic function is

a*

-

(a

Now

i/)T(z) Jo

(a

-

?/)*

-

(1

i//a)

be a random variable with probability function

let

P(*)

= (~\

2 = ( l/V27ro:) e~ x/2 for x > 0, probability function for f is P(x) = If 1, 2, otherwise (see Prob. 7, Sec. 9.6). n are n indeP(x) pendent random variables with the same probability function p(x) given

The

.

above, the characteristic function for the

random

.

.

,

variable

(9.54)

is

the product of the characteristic functions for each random variable n [see (9.44)]. Now the characteristic function for P(x) 1,2,

=

.

.

.

,

-x/2 <S**

dx

=

Too

.-(J-i

2 t

,

is

PROBABILITY THEORY AND STATISTICS

The

2

characteristic function for x

is

-

(1

Comparing

we note

(9.55) with (9.53),

363

probability function associated with the

2^)~

/2

that z

(9.55)

,

random

=

so that the n/2, a = variable x 2 ^/(//; 1, w/2),

=

for ,

zr""<r""

>

?2)

= The

distribution function for

for

/y

(9.,%)

%

(9-57)

is

/'(,'/)

*

^"

The

distribution defined

K n (x)

by

is

2

-

1

1

*

-""

called the x 2 distribution, principally

associated with K. Pearson.

The x 2 test of significance arises in the following manner: Let be a random variable with a known probability function p(x). Let us divide the interval

x^

.

.

.

with

> T

<*>

m _i

,

P = t

^

<

x

x

<

Let

1.

< oo

m

into

oo .

N be

oc

parts, say,

We

have

the

number

of times

<

x

< x^

we sample

,

Xi

^

^

x

the result

=l

The expected of any sample being independent of the previous samples. number of samples which fall in the iih interval is given by NP,. If r is the actual number of samples falling in the ^th interval, then l

(r,

certainly constitutes

and experimental

-

some measure

results.

of discrepancy between the theoretical K. Pearson found that

^yielded a practical

mental

results.

If

means

we

let y,

for

=

measuring the (rt

(9-58)

reliability of the experi-

- NP )/\/WP l

m

%,

then X

2

=

X

V*,

and

mm

ELEMENTS OF PURE AND APPLIED MATHEMATICS

364

m

V

I/,

LI

V VN t=i LI

VP* = -4-

t-i

random

variables

?/ 4

r*

VN V ^ =

~

Vtf -

VN =

LI

t=i are not independent.

The

y%

lie

Hence the

0.

in the

hyperplane

m

/

y%

\^Pi

=

0,

a subspace of the ra-dimensional Euclidean space.

It

t-i

N

can be shown that, as becomes infinite (N is the total number of or then the distribution of x 2 approaches m ~\(x) [see samples trials), The reader referred to Cramer, "Mathematical Methods of is (9.57)]. Thus Statistics," Chap. 30, for proof of this statement.

K

lim P( x 2

^

xS)

---

=

One can determine the value X

2

distribution.

It is

----*

x

y

(m ~ z

of the integral of (9.59)

important to note that

^e-v"dy

(9.59)

from a table

of the

(9.59) is

independent of

the original probability function p(x).

Example 9.19. A coin was tossed 5,000 times and heads appeared 2,512 times. m = 2, Under the assumption that the coin is "true," we have p(H) = -, p(T) -%, N = 5,000, Pi - P 2 - |; n = 2,512, r 2 = 2,488. Hence

= (2,512 *2 "

-

2,500)

2^500

2

+

(2,488 "

-

2,500)

2

2,500

0.115

In a table of the x 2 distribution

found that the probability that x 2 exceed 3.841 that there is no inconsistency in the assumption that the coin is true. The 5 per cent level is taken as a fairly significant level. In the tables one also finds that P(x 2 ^ 0.115) is about 0.73, which means that we have a probability of about 73 per cent of obtaining a deviation from the expected result at We are therefore not too worried about least as great as that actually observed. the fact that experimentally we obtained x 2 = 0.115. is

Since 0.115

0.05.

<

3.841,

it is

we

feel

Problems 1.

A

coin

is

tossed 5,000 times, with heads occurring 3,000 times.

Evaluate x 2 for

this experiment. Would you be suspicious that the coin is "true"? 2. die is rolled 6,000 times with the following occurrences: the number of times

A

6 occurred, respectively, was 1,020; 1,032; 981; 977; 1,011; 979. 2.876. What is your opinion as to the "truthness" of the die?

2, 3, 4, 5,

X2

=

Show

has the probability function p(x) = (l/\/27r)e~x and 466 values of x ^ 1,000 experiments one obtained 534 values of x < you consider that the experiment was biased? 3.

A random

variable

1,

that

*

/2 .

If in

0,

would

9.9. Monte Carlo Methods and the Theory of Games. The method of Monte Carlo is essentially a device for making use of probability theory

PROBABILITY THEORY AND STATISTICS

365

to approximate the solution of a mathematical or physical problem. A few examples will illustrate the method. Let p(x) be the probability function of a random variable, defined ,

on the range a interval (a,

^

x

^

The

fe).

6,

with

/

Ja

p(x) dx

=

1,

p(x)

=

for

x outside the

expected, or mean, value of /(x) has been defined as

/= A sequence

f* f(x)p(x)

dx

of values xi, Xi, x n is chosen at random subject to the lie between x condition that the probability that the random variable dx be given by p(x) dx. For n large one hopes that and x .

.

.

,

,

+

(9.60)

(x>)

rb If

we wish

to find an approximate value of

/

F(x) dx,

we note

that

Ja

I" Ja so that/(x) = F(x)/p(x). The choice p(x) = I/ (6

ment such that

(9.61)

A

simple choice for p(x) is p(x) = 1/(Z> a). a) implies that one can construct an experi-

numbers on the

have equal likelihood of such an experiment can be devised. The toss of a single coin can be used to generate the number An infinite number zero if a head occurs, the number 1 if a tail occurs. of ordered tosses of a single coin, or an infinite number of ordered coins tossed simultaneously, would yield a sequence ao, ai, 2, an or 1 for all n ^ 0. with a n = This, in turn, yields the number occurrence.

all

interval

(a, 6)

It is extremely doubtful that

.

with

^

numbers able

x

^

1.

balls,

.

,

,

.

.

.

,

of random random vari-

Mathematicians have constructed tables

to avoid the

by experiment. N

by use

.

cumbersome process

Let the reader deduce a method for obtaining

of a bowl, containing

and a scoop with

of obtaining a

N+

1

an equal number

ordered holes.

of red arid white

The author chose 25 num-

ELEMENTS OF PURE AND APPLIED MATHEMATICS

366

25

random numbers and obtained

bers from a table of

reasonably close to

We now

\

-%-%

%*

=

0.5030,

x dx.

I

The game to be played conconsider a less trivial example. If a pointer comes to rest in the area

cerns three spinners (see Fig. 9.7).

III

If designated by p, ; we move from the &th spinner to the jth spinner. the pointer comes to rest in the area x i = 1, 2, 3, the game is over. %,

3

The

areas are to be unity so that x l

=

Y

1

p

l3 ,

=

i

1,

2, 3,

and the

;-i

and

p's

z's represent probabilities.

We

ask the following question: If that the

we begin the game at spinner I, what is the probability, /n, game will end at spinner I? We can be successful as follows: 1.

Xi occurs

on the

first spin.

pn occurs?! times in succession, and then Xi occurs, n = 1,2, 3, .... 3. ^12 occurs followed by the probability, /2 i, that if we are at spinner II the game will end at spinner I. 2.

4.

pa

occurs n times in succession, n

Replace pn and/2 i Let the reader show that

of (3)

5.

/ll

=

(4)

^

1,

and then

(3) occurs.

by p i3 and/3 i.

+ pnXi + P U XI +') + (P12/21 + P11P12/21 + PiiPi2/2i +') + (pia/ai + Pnpnfn + Pupnfai + 2

(Xi

X

'

'

')

~

1

By

and

- Pn

the reasoning used above one can readily

show that i

=

i=

1, 2,

3

1, 2,

3

(9.63)

PROBABILITY THEORY AND STATISTICS

From

one obtains

(9.63)

P31/12

we

solve for/i 2

,

+

1)/12

(Pll

P21/12

If

367

+ +

-

(p22

p,{2/22

+ +

P12/22 l)/22

+

-

(P,I3

= = ~^ =

Pi 3/32 P23/32 l)/32

2

we have

Pl3

so that

/i2/#2 is the inverse element of

-

Pll

Pl2

??23

Let the reader show that

(see Sec. 1.3).

for

fzz/xz is the inverse

Thus probability theory can be applied

etc.

p ||A||, mate the elements

-

P33

P32

P31

22

Pl3

P22

P21

of

in the matrix

3021

The quanti-

matrix of a given matrix.

of the inverse

element

to approxi-

=

1, 2, 3, are obtained by experiment. fl} i, j An analogy exists between the diffusion process of heat motion

ties

,

and the two-dimensional random-walk problem. Suppose that a particle located at (x, y) moves to one of the four positions (x + 1, ?/), (x 1, y), (x, = |. What is the probay 1), with equal probability, p 1), (x, y

+

bility P(x, y,

t)

from the origin obvious that y

-

1), (x,

P(x,y;t)

y

that a particle will arrive at (x, y) after t steps if it starts For a particle to arrive at (x, y) in t steps, it is (0, 0) ?

it will

+

have had to arrive at one

-

1), (x

= ^P(x,y - 1;<-

+ Let us subtract P(x, P(x,y\t)

+

%P(x,

1, y),

y\

P(x t

-

1)

(x 1) 1,

+

1,

y)

m t

of the four positions (x, 1

+P(x,y + I;/- 1) + POr + y; *

from both sides

Thus

steps. 1) 1,

t

y;

-

1)J

(9.64)

Then

of (9.64).

-

1) P(x,y;t l,i/;J- 1)

y+l]t-\) -

-

2P(x, y]

In Sec. 10.13 one notes that, f"(x)

=

2P(x,y;t

linr

if

t

f"(x) 2h)

-

is

-

-

1)

ar

1)

-

+

1

continuous f or a 2f(x

+

h)

l,i/;f

t-

;

g

x

-

1)]

1)]

(9.65)

^

then

/>,

+ f(x) (9.66)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

368

In the calculus of finite differences the expression [f(x Thus represents the second difference of /(#).

[see (10.22)].

2f(x

+

1)

+ f(x)]

+

1)

=

/(*

Af(x

+ 2) - f(x +

1)

2)

Thus

(9.65)

may

may

2)

(9.67)

-

be written

AJP)

which

+

(9.68)

be compared with the diffusion equation (9.69)

A very fine subdivision

of space

and time must be used, however,

to freely interchange (9.68) and (9.69).

One

in order

traces the histories of a

large number of particles with initial lattice distributions to obtain the These histories yield an distributions after a given number of steps.

approximate solution to

(9.69).

We conclude this section

with a brief discussion of the theory of games. simple example illustrates the basic factors encountered in the theory We consider a two-person zero-sum game as exemplified by of games.

A

the square array given by (9.70).

(9.70)

X and Y choose a row and column, respectively, and the choice of each is made without any

The number express knowledge of the other's choice. and F, respectively) j'th column (the choices of If a tj < 0, then designates the number of units that Y must pay X.

in the tth

X

row and

X

X

The total sum earned by and Y is zero, pays Y an amount \a%J and it is in this sense that we have a two-person zero-sum game. We assume that will attempt to earn as much as possible and that Y will \.

X

attempt to hold his losses to a minimum.

game the

possible courses of action of

We see

that in this particular F, the consequences of these are fully known. The theory of

X and

and the objectives of X and F games seeks to analyze objectively the conflict of X and F and, moreover, attempts to determine an optimal course of action for each player.

actions,

PROBABILITY THEORY AND STATISTICS

The game given by (9.70) 6. the least he can gain is 4 units. Since 4 > 6, it

369

X

chooses row 1, very easily analyzed. If he chooses row 2, the least he can gain is will always gain at least is obvious that 4 units by choosing row 2, no matter what the choice of F. Since Y will always choose row 2, of necessity, Y will always choose knows that and Y will always column 2 to hold his losses to a minimum. Since choose row 2 and column 2, respectively, with probability 1, we say that and Y use pure strategies. Let us note the following Let f(x, y) denote the quantity in row x and column y. The smallest number in row x is designated by rnin f(x, y). is

If

X

X

X

X

:

y

The

numbers

largest of these

is

written

U = max

min

x

U =

Thus

min f(x Q

,

y

that the choice of x

y)

^ min y

=

f(x, y)

(9.71)

y

f(x, y) for all x.

Let the reader deduce

X

X

will always earn at XQ by guarantees that Let us determine the maximum amount Y can possibly The largest number in lose if he plays wisely whatever the action of X. row y is designated by max f(x, y). The smallest of all such numbers

U

least

units.

X

obtained by varying y

is

F =

min max

Thus

V = max f(x,

yd)

^ max

X

f(x, y)

(9.72)

x

y

f(x, y) for all y.

Let the reader deduce

X

that the choice of y y$ will enable F to hold his loss to an most V. For the square array given by (9.70) we have

U = V = and

=

/(2, 2)

amount

at

4

X

because of this that always chooses the second row and the chooses second column. Let us note that the element always = 4 is the minimum element of second row and the maximum the /(2, 2) element of the second column. It is for this reason that /(2, 2) is called it is

F

Let the reader deduce that the existence of a saddle element implies that U = V. In any case one always has U ^ V. The proof proceeds as follows: Let g(x) = min f(x, y), U = g(xd) = max g(x), a saddle element.

x

y

h(y)

= max/Cc,

U =

g(xd)

y),

= max

V =

h(yd)

g(x)

= max

= min

g

/(x

,

= min/(z y) ^ max/(x, y!) = h(y) = V

[min/(x, y 2/

)

Then

h(y).

y)]

X

The

reader

is

urged to verify every step of

(9.73).

,

(9.73)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

370

We now

U <

consider a case for which

V, given

by

(9.74).

2

1

(9.74)

-1 Let the reader note the noncxistence of a saddle element with

U =

2

< V =

3

X

is sure to win 2 units per game if he always chooses x = 1, while can never lose more than 3 units if he always chooses y = 1. If Y were to notice, however, that 1, then it would be always chose x = for to Y choose 2. it for Can to y change his strategy pay expedient The so that he can realize more than 2 units per game on the average? answer is "yes"! Let us suppose that X and Y use mixed strategies in 2 chooses x = 1 with probability p and chooses x the following sense: = 1 Y chooses with and chooses with probability 1 y probability q p;

Thus

Y

X

X

X

y

=

2 with probability

B = =

+

3p<?

q(6p

-

2p(l 5)

1

+

The

<?

4

g)

-

total expectation of

+ (-1)0 -

p)g

2p

+

4(1

-

X is

p)(l

-

q)

(9.75)

X

now reasons as follows: What mixed strategy should I use in order to guarantee a given expectation no matter what the mixed strategy or p > |, then E is a minimum for 5 > employed by F? If ftp = 0, so that E = 4 On the other hand, if p < f, is a 2p < q 7

.

For Op 5 = or p = f = we have J independent of q. Thus, no matter what the strategy of a unit per game by mixing his can be assured of earning of 7, X chooses row 2 if a six occurs on a die and otherwise chooses strategy. row 1. Let the reader show that Y can choose a mixed strategy such of a unit per game no matter what that he will never lose more than the strategy of X.

minimum

f or

#

=

1

so that

E =

4p

1

<

.

X

-

Let us a

>

0, 6

now

>

consider the following

game

exemplified by (9.76), with

0:

(9.76)

PROBABILITY THEORY AND STATISTICS Let pi

y

+

p 2 ) be the probabilities associated with the choices and let gi, g 2 1 (gi + #2) be the proba-

pz, 1

(pi

1, 2, 3,

respectively,

=

of x

,

=

associated with the choices of y

bilities

371

X is - g + p2(

The

respectively.

3,

2,

1,

expectation of

E =

g\ + + [1 - (Pi + P)][foi + = ?I[PI - P2 + (a + 5)(1 - pi - p + gj-pi + p + (a + 5)(1 - pi - p -

pi(gi

2)

(72)

-

<? 2 )

5(1

-

(0!

+

g 2 ))]

2 )]

of gi

pendent cannot

X

g for 3

=

1

X

and g

gi

pi

For p\ = p 2 = | we have Thus a mixed strategy occurs for

-

p2)

E =

g

X

2.

=

(9.77)

Q indesuch that

has a good strategy (pure) with q\ = g 2 = 0, Why is it obvious that the only good strategy

Y

lose or gain. 2

-

of (9.77).

E

Let us examine

5(1

2 )]

2

1.

the mixed strategy discussed above? However, we can show that good strategies exist for Y other than q\ We can g 2 = 0, ga = 1. write is

- a pj(l + P2[-(1 +

E =

It is

B)qi

obvious that

and pz are

of pi

=

-(!+

+ %i + E

The

=

no

loss for this choice of qi

E =

5)g 2

5]

+

6]

+

solutions of these

that an infinite

+

+

(a

d)( qi

if

+

g 2)

-

5

the coefficients

two equations yield

5),

p 2 )[2(a

+

5)gi

number

of

good mixed strategies

pi

(1

+

)g

~

a

which in turn yields E so that Y can suffer and q%. Thus a good strategy which is mixed Let the reader show that, if gi = g 2 < d/2 (a + 6), then

d/2(a

exists for Y.

+

be independent of p\ and p 2

will

zero.

q\

g2

-

(1

^

6]

for all choices of pi exist for Y.

and p 2 so ,

Problems Generalize the results of (9.63) for the case of n spinners. Let f(x) be a continuous rnonotomc increasing function defined for ^ x ^ 1. a random variable be chosen from (0, 1), show that the probability that/() ^ y Q 1.

2.

If

,

Use this method to find given by p = f~ (yo), so that/(p) = j/ an approximate value of \/2 by use of a table of random numbers. 3. Let us consider the plane of a two-dimensional random-walk problem with

/(O)

^

3/0

^

/(I)

l

is

.

r. To each point (#, f/ t ) we associate a (xi, ?A), i 1, 2, Let V(x, y) be the expected value of a particle whose motion starts and eventually ends at one of the boundary points. Show that

boundary points at

.

.

.

,

value U(xi, y % ). at (x y} t

with

P

t

the probability that the walk terminates at (x vy

i/ t )

if it

begins at

(x, y).

show that 1,

and compare

this result

with

y)

+

^2 F

F(x

-^ +

-

1,

d*V - -

y)

0.

+

F(x, y

+

1)

+

V(x, y

-

1)]

Also

ELEMENTS OF PURE AND APPLIED MATHEMATICS

372 4.

Show that the game of "paper, rock, and scissor" What strategies should be used by X and F?

is

exemplified

by the matrix

below.

REFERENCES Cramer, H.:

"

Mathematical Methods

of Statistics,"

Princeton, N J 1946. Doob, J. L.: "Stochastic Processes," John Wiley

Princeton University Press,

,

Feller,

W.: "Introduction

&

Sons, Inc

to Probability Theory,"

,

New York, 1953. & Sons, Inc

John Wiley

,

New

York, 1950. Mood, A. M.: "Introduction to the Theory of Statistics," McGraw-Hill Book Company, Inc., New York, 1950. Morgenstern, O., and J. von Neumann: "Theory of Games and Economic Behavior," Princeton University Press, Princeton, N.J., 1947.

Munroe, M. E.: "Theory of Probability," McGraw-Hill Book Company,

Inc.,

New

York, 1951.

Uspensky,

J. V.:

Company,

"Introduction to Mathematical Probability," McGraw-Hill Book

Inc.,

New

York, 1937.

CHAPTER 10

REAL-VARIABLE THEORY

We desire to preface the study of the real-number introducing a brief view of our philosophy of mathematics and science. Historians are generally in agreement that in many ways Greek mathematics was the precursor to modern mathematics. One begins the study of geometry with a set of postulates or axioms along 10.1. Introduction.

system by

first

with some definitions of the fundamental entities in which one is interFrom this beginning one deduces through the use of Aristotelian To the novice logic the many theorems concerning triangles, circles, etc. the axioms appear to be self-evident truths. We do not hold to this view a postulate is nothing more than a man-invented rule of the game, the game in this case being mathematics. Actually the mathematician is not interested in the truth, per se, of his postulates. The postulates of a mathematical system can be chosen arbitrarily subject to the condition that they be consistent with each other. The proof of the self-consistency of the postulates is no easy task, if, indeed, such proofs exist. A discussion of such matters lies beyond the scope of this text. Finally, we should like to add that any theorems deducible from the We admit postulates are a consequence of the postulates themselves. this is a trivial and obvious remark. Too often, however, the scientist He is prone to believe that the laws of nature are disforgets this fact. covered. It is our personal belief that this is not the case. We lean toward the more esoteric point of view that the so-called laws of nature are invented by man. Newton described the motion of the planet relative to the sun in a fairly accurate manner by use of f = ma Mercury and f = (GmM/r*)r. Because of this one feels that these two equations are laws of nature discovered by Newton. On the other hand, by postulating that a gravitational point mass yields a four-dimensional Riemannian space and that a particle moves along a geodesic of this space, Einstein also obtained the motion of Mercury relative to the sun. It is well known that Einstein's predicted motion of Mercury is more accurate than Newton's predicted motion. Are we to assume that from the time of Newton to Einstein we had a "true" law of nature and that now we discard this law and accept Einstein's theory as the truth? It is ested.

:

373

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

374

just a matter of time before a physicist will invent a new set of postulates which will explain more fully the physical results obtained by experiment.

Moreover

it is

not our belief that we continually approach the true laws

of nature. 10.2.

The

tem by

first

view.

The

We

Positive Integers. begin a study of the real-number sysdiscussing the positive integers from a postulational point of positive integers will be undefined in the sense that they can

be anything the reader desires subject to the condition that they satisfy We shall attempt to be rigorous, but the postulates set forth below. First we do not feel qualinot be as rigorous as might be possible. attempt this task, and second it would require a treatise in itself to elaborate on the logical and philosophical aspects and implications of

shall

fied to

we

the material upon which

The

shall discourse

positive integers have been characterized

by Peano through the

following postulates: TV- There exists a positive integer, called one, and written P b Every positive integer x has a unique successor x

1.

f

.

:

P

c

There

:

Pd'. If x'

P

e

is

=

no positive integer x such that ?/,

then x

=

x'

=

1.

y.

Let 8 be any collection of positive integers.

:

integer 1, and if, moreover, any integer x of S is the complete collection of integers, /.

The

postulate

Pa

guarantees that there

we may

set of positive integers so that

If

8 implies that

is

at least one

S

contains the

x' is in S,

member

then

of our

agree that there is something involved in the relation x = y

all

we can talk about. The equal sign = ) of Pd means that x and y are identical elements in the sense that we may The replace x by y or y by x in any operation concerning these integers. some from the other P in P differs e is a respects postulates. postulate rule for determining when we have the complete set of integers at hand, and is used chiefly m deducing new laws which the integers obey. As postulated above, the integers could be any sequence (0,1, a*, (

e

.

.

.

,

an

,

.

.

.

)

which the reader has encountered in calculus courses.

It is in

It is obvious that more will have to be said before we can identify the collection of integers defined above with the ordinary integers of our everyday working world. We shall accept as intuitive the notion and idea of a set, or collection, of The Peano postulates will make themselves clearer to the stuobjects.

this sense that the integers are undefined.

dent as he proceeds to read the

text.

The operation of addition (+) by the postulates

relative to the positive integers is defined

Aa Ab

The

:

= +V = a'

:

a

postulate Ab implies that a

+

b

is

a (a

+1 + &)'

an integer (closure property).

REAL-VARIABLE THEORY

We

are

now

375

deduce some new results concerning the

in a position to

positive integers.

THEOREM

Addition

10.1.

is associative

We let S be the

collection of integers x,y, Thus x is in S if (a 10.1 for all a and b.

+

integers a

and

We

b.

(a

so that

1

that (a

+

+

+

+

+

6)

THEOREM THEOREM

S

x

s= /.

+ bY +V + (b +

let

+ 6) + x] [a + (b + #)]' a + (b + x)' a + (b + x')

Law

+

a

from from from

and

x) for all a

(b

1

is

x) for all

Aa

A Aa

4

from

since x

b

This means

fc

in

is

/S

Aa

from from

^1 6

From P

in S.

+

Now

S.

Now

6.

[(a

is

$

is

a.

1

= = =

1

6'

b

c

+

the complete

+

b

=

If

(see Prob. 1).

then

c,

/>

=

c.

if

First,

+

1

c

from Theorem 10.2 from A a from P d

1

c'

c

Now let

+

+b = b+a a + b = a +

commutative a

of Cancellation.

proceed by induction on

Hence 1 belongs to S. whenever x + b = x

=

Q.E.D.

10.3.

+

x

be any element of S.

a*

= = = =

7

+

6)

6)

.

1)

+

Addition

6

a

.

an element of

is

1

(b

10.2.

then

(a

a

S whenever x

so that x' belongs to set of integers,

= = =

1

Now

belongs to R. x = a b} (a

We

+

6)

+

show that

first

.

+ c a + (b + c). which satisfy Theorem

+

(a

z,

This means that

x be any element of S.

c

then b

= = = =

+ #' + x)' c + x x + c

Now

c.

assume x

f

+

=

6

#'

+

c.

Then

+ (b + 3)' b + x x + 6 6

a:'

6

from Theorem 10.2 from ,4 b from P d from Theorem 10.2

c

(c

since x

== c

Thus x' belongs to S whenever x belongs

THEOREM

10.4.

+

we

shall use the

In what follows

Thus Theorem

=

If b

10.4

may

a be

c

+

a,

written &

+

c (See

(=>)

=

by

/S

,

=

symbol a

in

From P e S

to S.

then b

tion in both directions will be represented

b

is

c

+

(<=).

+ a = c + a <=>& =

c

to

s= /.

Prob.

mean

a=>6 = Thus

Q.E.D.

2).

c.

"implies."

Implica-

ELEMENTS OF PURE AND APPLIED MATHEMATICS

376

The reader should realize that b = c=*b + a = c + a does not mean we have introduced the axiom about adding equals to equals. All we have done to the expression b + a is to replace b by c (permissible

that

of equality of two integers). belongs to the class of relations, R, called equivalence note that

from the definition Equality

(

=

)

We

relations.

=

a

=

a

=

a

In general a relation a set of elements (a, b, c,

R

c

an equivalence

called

.

.

= a = c=*a

b,b

is

.

a

b <=$ b

relation relative to

if

)

R\\

aRa

R%

:

aRb^bRa

R-A

:

reflexive

aRb, bRc

symmetric

aRc

=^>

transitive

R enables one to distinguish between various For example, we say that aRb holds if and only if a and b yield the same remainder when divided by 2. Thus any integer belongs either to the class of even integers or to the class of odd integers. The reader should check that the elements of the class of even integers An

equivalence relation

classes of elements.

satisfy

R^ R^

Now we

We

R-*.

write

aRb

<=>

=

a

mod

b

2.

introduce multiplication by the two postulates given below:

M

a

-

Mb'

a

-

1

a

-

b

a f

a

b

+

a

assumes that a b is an integer for all a and b. We shall omit the dot whenever it is convenient to do so. Thus Mb can be written ab = ab + a.

Mb

-

f

THEOREM tion,

a(b

+

10.5. c)

=

Multiplication

ab

+

ac.

Let S be the collection of

S and

for

all

a(b

from

A

,

M

b.

a(b

so that x'

is

in

Thus

+

S

x')

if

x

1

= = =

+

x)

=

ab

+

+

a

.

ax

First

b.

=

1) is

.

.

a and

+

with respect

to

addi-

prove this theorem by induction on ) such that integers (x, y, z, aft

for x in

is left-distributive

We

=

ab'

afr

Now

in S.

let

+ x) a(b + x) + a (ab + ax) + a ab + (ax + a) f

a(b

=

ab 4- ax'

is

in S.

Thus S =

=

a?>

+

x be in S.

from from

a-l

We

Ab Mb

since x

is

in

S

from Theorem from Mb I.

have

10.1

c.

REAL-VARIABLE THEORY

THEOREM THEOREM

=

a

a (see Prob.

377

10.6.

1

10.7.

Multiplication

is associative, (ab)c

10.8.

Multiplication

is

commutative, ab

10.9.

Multiplication

is

right-commutative,

3).

=

a(bc) (see Prob.

4).

THEOREM THEOREM

+

(b

c)a

=

+

ba

=

ba (see Prob. 5).

ca

(see Prob. 6).

DEFINITION that b

is less

a,

a

If

10.1.

than

+

b

written a

>

c,

we say

b or b

<

a,

that a

is

greater than b or We can state

respectively.

four important ordering theorems regarding inequalities:

Oa

Oi\

O Od

c

=>a =

6ora>6ora<6 a>b,b>c=*a>c a>b=>a + c>b + c a, b

'.

:

a

:

>

given

b

=* ac

>

be

We leave these theorems as exercises for the reader A set obeying O a is said to be totally ordered. To

(see Probs. 7

and

12).

place the integers in the realm of our everyday working experiences, integer 1 which occurred in the first of Peaiio's

we now postulate that the

P

,

be the adjective which reasonable, sane, prudent human beings attach to single entities when they speak intelligently of one book, one day, one world, etc. Of course the number 1 as given in Peano's postulates is a noun, not an adjective. B. Russell associates "one" with the class of all single entities. Trouble occurs, however, when one tries to define a class of elements. postulates,

,

will

=

The

+

successor of one, namely, 1' 1 1, will be called the integer written and addition will 2, two, again mean what we wish it to mean, namely, that one book plus one book implies two books. We now feel free to speak of the class of positive integers 7(1, 2, 3,

.

.

.

,

n,

.

.

.

).

Problems .

1.

2. 3.

4. 6.

6. 7.

8. 9.

Prove Prove Prove Prove Prove Prove Prove Prove Prove

Theorem Theorem Theorem Theorem Theorem Theorem

Hint: First prove that a

10.2.

1

-f

a for

all

integers x such that

all a.

10.8. 10.9.

ba

-f-

Ot>,

Oc O d ,

.

a.

Hint: Let S be the class of

> b, show that a > b or a = 6. Using the results of Probs. 10, 11, prove

11. If a' 12.

=

10.6.

ordering theorems that 1 a = a.

that b'a

1

10.7.

Show that 1 ^ all integers a. x ^ 1. Then show that S = /. 10.

-f-

10.4.

Oa

.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

378 13.

define

Prove that every nonempty set of integers contains a smallest

what

meant by a smallest

is

First

integer.

integer of a collection of integers.

Also use the

results of Probs. 10, 11.

10.3.

The Rational

An

Integers.

extension of the positive integers

which includes the zero element and the negative integers is obtained in the following manner: Given any pair of positive integers, we postulate the existence of a difference function d which operates on the positive integers a, b subject to the condition that d(a, b)

=

d(.r,

(10.1)

//)

and only if a + y b + x. Equality as defined by (10.1 certainly is an equivalence relation. We note that the above definition depends only on addition of positive integers (Sec. 10.2). Let us see what properties are possessed by d. Obviously if

)

d(a, a)

+

Moreover d(a

b

c,

+

=

c)

=

d(fe, b)

If

d(a, b).

we

a of ordinary arithmetic, of necessity that addition be defined by

like b

d(a, b)

We now

+

dfcr, y)

=

d(<i

+

desire that d(a, b) behave we are forced to postulate

x, b

+

y)

(10.2)

=

d(a, b)

(10.3)

note that d(a, 6)

+

=

d(c, c)

+

rf(a

c,

b

+

c)

from (10.2) arid (10.1). Hence d(c, c) has the property element for the operation of addition ( + ).

We

define multiplication

=

d(a, b)d(x, y}

What

of the

number

+

+

+

1)

bx, by

1)?

1)-= d(xa

d(x, y)

so that d(a, a

+

d(ay

d(a, a

d(x, y)d(a, a

by the following

=

We

+

of being a zero

rule:

ax)

=

d(x, y)d(a, b)

(10.4)

see that

+ x + ya, ax + + l)d(x, y)

d(a, a

ya

+

y)

(

(L

^

.

}

has the unit property as regards multiplication.

now can easily be shown that the collection of elements d(l, x + 1) when x ranges over the set of positive integers will satisfy all the Peano The set of numbers d(a, b) contains the postulates outlined in Sec. 10.2. It

positive integers as a subclass and in its totality represents the class of rational integers (positive integers, negative integers, and the zero let the reader show that the zero element d(a, a) has the element).

We

properties d(a, a)d(x, y) d(x, y)

+

d(y, x)

= =

d(a, a) d(a, a)

nn uu

. '

b;

REAL-VARIABLE THEORY

379

We

We say that d(y,

x) is the negative of d(x, ?/), and conversely. may, = 0, d(l, a = a, d(a = a, 1) 1, 1) please, write d(a, a) = 6 a. The ordering postulates can easily be extended to d(a, 6) cover the new collection d(a, 6). say that

+

we

if

+

We

d(o, 6)

and only

if

>

d(x, y)

>

a

if

b

+

x

+

y

Let the reader show that d(a, b) d(a, b)

d(a, 6)

> > >

d(x, y) d(x, y}

d(x,

i/)

=> d(a, 6) + d(u, v) > d(x, y) + d(u, v) => d(a, 6)d(z, w + z) > d(x, y)d(z, u + z) => rf(x, y)d(u + z, z) > d(a, b)d(u + z, z)

(10.7)

What

is the ordinary meaning of (10.7)? Let us note some properties of the rational integers as outlined above. Associated First we have a set of elements called the rational integers. with this set of elements are two operations called addition and multipliThe elements satisfy the following properties relative to the cation. operations defined above: 1. Closure properties: a + b, ab arc elements of our set when a, b are elements of the set. 2. Associative laws: a + (b + c) = (a + 6) + c, a(bc) = (ab)c. = a = + a, a-1 = = 1 a. 3. Unit elements: a +

4. 5.

= ac = ab be. Distributive laws: a(b c) ac, (a b)c inverse element exists for each element relative to addition,

+

+

+

+

An

= ( a) a a = for all a. ( a) say that the rational integers form a ring with unit element also note that, if ab = 0, then a = or b = 0. Why? that

is,

+

+

We

We or b

=

1.

We

does not imply a = give an example of a ring such that ab = 0. Such rings are said to have zero divisors. Let us consider the

six following classes: Into class one we place all integers which yield a remainder 1 upon division by 6. Class two contains all integers which Class y has the obvious property yield a remainder 2 upon division by 6. that it consists of those integers which yield a remainder j upon division 6. We represent these various classes by the symbols 1, 2, 3, 4, 5, 0. Addition and multiplication are defined in the ordinary way. Thus 2 5 = 10 = 4 mod 6, since 10 yields a remainder of 4 when divided by 6. 3 + 4 = 7 = 1 mod 6, etc. It is obvious that mod 6, but

by

2-3=0

neither integers

2=0 mod modulo

6 nor

3=0

mod

6.

We have constructed the ring of

6.

The Rational Numbers.

We

can extend the ring of rational integers by postulating the existence of a rational function /?(a, b) where 0. a, 6 are any two rational integers, a postulate that R is to 10.4.

^

have the following properties:

We

ELEMENTS OF PURE AND APPLIED MATHEMATICS

380

R(a, R(a,

b) b)

=

+

We

leave

it

b)

R(a,

b)

=

R(x, y)

=

R(a, b)R(x, y) R(a,

*=>

R(x, y)

> >

=

bx

ay R(ax, bx

+

ay)

R(ax, by)

R(x, y)

<^bx

R(x, y)

*=*

> <

bx

(10.8)

ay

if

ay

if

show that the

to the reader to

set of elements 72(1, b)

has all the properties of the set of rational integers considerations

we may wiite 7?(1, b) == The set of rationals has

?>,

> <

ax ax

and we

b.

For

all

practical

shall obviously write

a further property not possessed R(a, b) = b/a. For of rational the integers. every R(a, b), b ^ 0, a ^ 0, there ring by exists a rational R(b, a) such that 7?(a,

=

b)R(b, a)

=

R(ab, bo)

/?(!, 1)

=

1

Thus every nonzero element has an inverse with

respect to multiplication. form field. the rationals a that say The reader should have no trouble showing that the integers modulo 7

We

form a

The

The

field.

inverse of 2

4 since

is

form a totally ordered

rationals

2-4 =

8

field.

We

=

1

mod

feel,

7.

however, that

the rationals are not complete in the sense that we should like to deal with numbers like \/2, e, w, etc. For the irrationality of \/2> see Sec. 10.5. We shall have to extend the field of rationals. This will be

done in Sec.

10.6.

Some Theorems Concerning the Integers THEOREM 10.10. Let /, g be positive integers. There exist integers r such that / = gh + r, ^ r < g. Our proof is by induction on /. 10.6.

h,

Let/ =

1.

Then 1 1

Now

= 1.1+0 = 0-0 + 1

if

g

=

1

if

g

>

1

assume Theorem 10.10 holds /

= gh + = gh+

for

so that h

/.

=

1,

r

=

Then

^

r

r

<

g

+ 1) Since r < g, we have r + 1 < g or r + 1 = g. If r + I < g, the theorem holds. If r + 1 = g, then/ + 1 = g(h + I) + and Theorem 10.10

/+

and

1

holds. From P e the theorem an exercise to show that the ft, ,

is

(r

true for

r of

THEOREM 10.11. We now derive common divisor of two integers a, b. (>0)

of the

form ax

+

by

>

0.

all

Theorem

+

it

as

a theorem regarding the greatest Consider the collection of integers

Since every collection of positive inte2/0 such that

gers has a least integer, there exist integers xo,

ax

We

leave integers /. 10.10 are unique.

fa/o

=

d

>

REAL-VARIABLE THEORY

381

We now show that d is the greatest assume that d does not divide a. Then from Theorem 10.10. Thus

the least integer of our collection.

is

common

=

a

ds

First

divisor.

+ r,Q<r<d

ax

s

+

by

=

s

=

ds

a

r

and (1 Xos)a + ( y s)b = r, a contradiction since r < d, unless r = 0, which case d divides a, d/a. Similarly d/b. Now any other divisor, Therefore d\, of a and b must divide ax Q + by Q (why?), and hence di/d. d is the greatest common divisor of a and 6, written d = (a, b). As an immediate corollary, if a and b are relatively prime, d = 1 and there in

such that ax

exist integers x, y

THEOREM ax

Proof,

=

6c

10.12.

+

If (a, 6) I

by

+

=

since

+

1.

a/be, then a/c. 1, so that axe

=

b)

(a,

Hence a(xe

ad since a/be.

=

by

and

1

yd)

=

c,

which

in

+

=

fa/c

But

c.

turn implies that

a/c.

THEOREM

The Fundamental Theorem

10.13.

N can be written

An integer

of Arithmetic.

uniquely as a product of primes.

Assume

Proof.

N

=

p^

-

-

=

pr

q8

?i02

obvious that we can consider the p q} as primes since a nonprime be broken into further factors. Continuing this process into primes. Furthermore this can eventually reduces all factors of It is

t

,

factor can

N

done only a

}>e

or equal to 2.

Why?

finite

number

of times since all

From above p\/q\qi

If (pi, qi)

=

1,

'

'

'

then pi/q 2 q*

'

primes are greater than

^ l,thenpi from Theorem 10.12. for an i = 1, 2,

If (pi, #1)

q&. '

qs

continue this process until eventually p\/q^

.

.

=

q^

We .

,

s.

assumed prime, we must have p^ = q We now cancel these In this equal factors and continue our reasoning in the same manner. way we obviously exhaust all the p and #,. Thus, except for the order, Since

</ t

is

t .

%

the

number

N is factored

uniquely as a product of primes.

THEOREM

The \/2 is irrational. 10.14. Assume \/2 rational so that \/2 = p/q. From Theorem have p = pip z pr q = q\q% <?, with the p g, unique. Thus upon squaring we have Proof. 10.13 we

'

'

'

'

'

t

,

N

=

'

2qiqiq 2 q2

'

'

qs q*

=

Pipip^p*

'

'

'

,

prpr

Since the prime factor 2 must occur an odd number of times on the left and can occur only an even number of times on the right, a contradiction occurs.

Q.E.D. Problems

1.

the collection of elements d(a b) satisfies the Peano postulates as the 6 range over the Peano integers.

Show that

elements

a,

t

ELEMENTS OF PURE AND APPLIED MATHEMATICS

382

3.

Trove Prove

4.

Show

2.

6.

7.

8.

that the set of elements R(l, b) has

all

the properties of the set of rational

6.

integers 6.

(10.5).

(10.6).

Show Show Show

that the integers modulo 7 form a

field.

that \/p is irrational for p prime. that \/2 4- \/3 is irrational. Consider the set of polynomials with rational coefficients.

Deduce theorems

for

these polynomials analogous to the theorems of Sec. 10.5.

10.6.

The

An

Extension of the Rationals.

The Real-number System.

obey the two following rules: form a field.

rationales

They They

1.

2.

are totally ordered.

On

the other hand, the rationale are not complete. Pythagoras realized that there existed so-called numbers, \/2> e ^c., which are not rational numbers. If we wish to include the \/2 into our number system, we

R. Dedekind gave a fairly satisfacfield of rationals. discuss an extension of the rationals We of irrationals. treatment tory which differs little from Dedekind's approach and which is due to

must enlarge the

B

Russell.

DEFINITION

10.2.

A

RusseM number

the following properties. 1 R contains only rationals and .

2.

R

does not contain

all

is

is

a

set,

ft,

of rationals

having

nonempty.

the rationals.

a is a rational in R and if ft is a rational less than a, then ft is in /?. examples of Russell numbers are given as follows: 1. The set of all rationals ^1 is a Russell number. It will turn out to be the unit of our constructed field of real numbers. 3.

If

Two

2.

less

The negative

rationals

and

all

positive rationals

than 2 form a Russell set or number.

We

whose squares are by the

shall designate it

symbol \/2. DEFINITION 10.3. Two Russell numbers are said to be equal or equivalent if and only if every rational of Ri is contained in 7 2 and conversely, ,

with one possible exception.

The two,

is

set of all rationals less than 2, which defines the Russell number, equivalent to the Russell number consisting of all rationals less

than or equal to 2. The only member of the second Russell number not found in the first Russell number is the rational two.

The Russell number Ri is said to be greater than 10.4. number R^ if, and only if, at least two rationals of Ri are not

DEFINITION the Russell

of R We note that, given Ri and R

members

2.

2y

only one of Ri

= R^ Ri > R 2 R < R 2 ,

l

can occur.

We must now define addition of two R numbers in such a way that the

REAL-VARIABLE THEORY

383

Let R\ and ft 2 be two R numbers. Conresult will yield an R number. struct the set of rationals obtained by adding in all possible ways the leave it to the reader to show that rationals of Ri with those of ft 2

We

.

an R number. What property will the Is this Russell number needed set of all rationals less than zero have? Does every R number have a negative? The student should for a field? readily answer these questions.

new

this

set of rationals defines

We shall define It is a bit more difficult to define multiplication. multiplication for two Russell numbers. ft i, R% which are greater than We first delete the negatives of fti and R 2 the zero Russell number. Then we multiply the positive rationals of R with the positive rationals .

\

of

Ri

To

in all possible ways.

We

this acquired set

we

adjoin the negative

show that the newly acquired An equivalent definition would be the following: set is a Russell number. Let C(Ri), the complement of fti, be the set of rationals not in fti. Show rationals.

leave

to the reader to

it

that

C(Ri) C(R%) we mean the complete set of rationals obtained by the product of rationals of C(R\) with rationals of C(Rz) in all possible ways. We leave it to the reader to extend the definition for the other cases of R} and 7? 2 It is now an easy problem to show that the complete set of Russell numbers form a totally ordered field which contains the rationals as a subfield. Everything hinges essentially on the fact that we reduce our computations to the field of rationals. Let the reader show that the set of Russell numbers forms a totally ordered field. The field of Russell numbers has an additional property not possessed is

a Russell number.

By

-

.

by the

rationals.

To

exhibit this property,

we

define the

first

supremum

upper bound) of a set of numbers (Russell). DEFINITION 10.5. The number s is said to be the supremum numbers S: (x, y, z, .) if and only if: 1. s ^ x for all x in S. 2. If t < s, there exists an element y of S such that y > t. (least

it

to the reader to

less

than

1.

Also,

of

a set of

.

.

show that \/2

is

the

We

leave

the

supremum of the set of all rationals supremum of the set of all rationals whose

1 is

squares are less than 2. Let the reader show that a set of numbers cannot have two distinct suprema. Let the reader also define the infemum (greatest lower

DEFINITION

bound) 10.6.

of a set.

A

set, S, of

Russell

a Russell number, N, exists such that

THEOREM

10.15.

has an upper bound.

N>

numbers x for

is

all

A supremum s exists for every

bounded above

if

x in S.

set of

numbers which

ELEMENTS OF PURE AND APPLIED MATHEMATICS

384

Our proof

is

construct a

sist of all rationals

show that

s is

S be a set of Russell numbers bounded new Russell number as follows Let s conwhich belong to any Russell number of S. We first

as follows: Let

We

above by N.

:

a Russell number.

obviously contains only rationals and is not the empty set. 2. Since contains rationals not in any of the Russell numbers of S, there are rationals not in s. 1.

s

N

3.

Let a be a rational of

s

and

Thus

s is

strate that s 1.

s

Si, of

a rational <a.

ft

Russell number, say, Si of S, then to s from the definition of s.

ft

number (see Definition supremum of the set S.

a Russell is

the

Since a belongs to a

also is in Si (why?), so that

10.5).

We

ft

belongs

next demon-

^ all numbers of S, for otherwise there is at least one element, S such that Si > s. This implies that Si contains a rational not

which is impossible from the construction of s. Let r be a Russell number less than s, and let ft, y be elements of This is possible since s > r. Now ft, 7 belong to at least 8 not in r. one set, Si, of S from the construction of 8. It is easy to see that Si > r.

in

s,

2.

From Definition 10.5, s is the supremum The set of Russell numbers thus 1. Forms a field. (A)

2.

Is totally ordered.

3.

Satisfies the

property that a

of the set S.

supremum

exists for every set of

numbers bounded above.

Any

set of elements satisfying the three properties of (A) is

unique they differ from the Russell numbers in name only. Thus the real-number system as set forth above is unique as regards its

in the sense that

construction.

To show

this uniqueness,

we

with the supremum property. the unit elements.

let

S and S be two

We

first

0<->0

Then we match the

rational integers,

n

Then we match

the rationals,

totally ordered fields

match the zero elements and

REAL-VARIABLE THEORY

385

Now consider any element, s, of S which has not as yet been put into We consider the set one-to-one correspondence with an element of 8. It is obvious that s is the supremum of all rationals of S less than s. The

of this set.

S which correspond to the above-mentioned be bounded above. Why? A supremum S

rationals of

S

set of rationals of

will

We match s <- s. Thus every member of 8 can will exist for this set. be mapped into a number of S. We leave it to the reader to show that every number of S is exhausted (mapped completely) by this method. Under the above mapping we realize that

implies

a

+

b

->

ab

<~>

=

(JT+~B)

+

a

b

db

(ab)

Such a correspondence, or mapping, is called an isomorphism, and this sense that we say the sets 8 and S are equivalent. We can now prove the Archimidean ordering postulate by use supremum. Let r > 0, and consider the sequence of numbers 2r, 3r,

r,

We

maintain that there

not

so,

.

.

. ,

nr,

.

it is in

of the

.

.

m

is an integer such that mr > 1. If this were the sequence constructed above would be bounded above and so a supremum s would exist for the sequence. From the properties of s we

have

^

s

nr for

such that pr

exists

Now

n.

all

>

t

=

consider

s

=

t

s

+

Thus pr

r.

r

r

<

>

s

s.

An

or (p

+

element pr l)r

>

s,

a

contradiction.

We

list

the Archimidean ordering postulates.

lates implies the existence of the others. => 3 integer n =5 nr 1. AOi'. r

>

>

Any one

of the postu-

(3 == there exists,

=3 == such

that.)

AO*: a A0 3 a :

A0 then

4

ft

> =

=

^

3 rational

r

3

a

3

a

>

r

3 rational

(Eudoxus)

:

ft

> >

If

r

> >

>

est of

>

ft.

for every rational r

>

ft

it

follows that r

>

a,

a.

We now prove A0 2 from AOi. If a > AOi an integer n exists such that n(a /3

r 0.

> 0, then a > 0. From > I or na > nft + 1. Now Let m be the smallso that an integer m exists such that m > ftn. 1, Hence Thus m > ftn ^ m all such integers. na

>

nft

+

1

^

ft

ft)

m>

so that

.>=>,

nft

ELEMENTS OF PURE AND APPLIED MATHEMATICS

386

Problems If b is

1.

Mri =

b- l b

a Russell number

=

5^ 0,

how does one

find the

number b' such that 1

1?

2. Extend the definition of multiplication of two Russell numbers for the cases for which both numbers are not positive. 3. Show that the Russell numbers form a field. 4. Show from the definition of the supremum that it is unique for a set of elements. 6.

Define the infomum of a

6.

Prove AO* from

10.7. Point-set

set.

AO^ A0 Theory.

4

from AOs, AOi from

A0

4.

For convenience, we

shall

consider

only

Any set points or elements of the real-number system in what follows. The field of real numbers can of real numbers will be called a linear set. be put into one-to-one correspondence with the points of a straight line in the usual fashion encountered in analytic geometry and the calculus. All the definitions and theorems here proved for linear sets can easily be extended to finite-dimensional spaces. DEFINITION 10.7. The set of points \x\ satisfying a ^ x g &, a, b If we omit the end points, that is, finite, will be called a closed interval. consider those x which satisfy a < x < b, we say that the interval is ^ x ^ 1 is a closed interval, open (open at both ends). For example,

< x < 1 is an open interval. DEFINITION 10.8. A linear set of points there exists an open interval containing the

while

if

will set.

be said to be bounded It must be emphasized

that the ends of the interval are to be finite numbers, thus excluding

and

oo

+ oo

.

An

A

alternative definition would be the following set such that bers is bounded if there exists a finite number :

N

S

of real

num-

N<x<N

for all x in S.

The

numbers whose squares are

less than 3 is certainly bounded, then 2 x 2. < < 3, obviously However, the set of numbers whose cubes are less than 3 i unbounded, for # 3 < 3 is at least satisfied by all the negative numbers. This set, however, is bounded above. By this we mean that there exists a finite number such that x < if 3 = 2 does the trick. Generally speaking, a set S .r < 3. Certainly of elements is bounded above if a finite number exists such that x <

for

if

set of

x2

<

N

N

N

N

for

all

x in S.

N

Let the reader frame a definition for sets bounded

below.

We shall, number

in the main,

be concerned with sets which contain an

of distinct points.

The

rational

<

form such a collection. DEFINITION 10.9. Limit Point.

of

a set S

numbers

infinite

in the interval

<

x

I

A point

every open interval containing of distinct elements of S. if

p

will

be called a limit point

p contains an infinite

number

REAL-VARIABLE THEORY For example,

let

S be the sequence '

r

'

numbers

of

/iii '

'

3' 4'

V 2'

387

i

\

n

)

any open interval containing the

It is easy to verify that

con-

origin

is a limit point of elements of S. Thus does not belong to S. the set S. Note that in this case the limit point It is also at once apparent that a set S containing only a finite number of points cannot have a limit point. Let the reader show that a point q is not a limit point of a set S if

tains an infinite

number

of

at least one open interval containing q exists such that (except possibly q itself) are in this open interval.

DEFINITION

10.10.

A A

Neighborhood.

no points

neighborhood of a point

is

of

S

any

N

deleted neighborhood p of p is open interval containing that point. a set of points belonging to a neighborhood of p with the point p removed.

The

^

set of points x such that of x = -nnr.

<

x

<

T&, x

^

-nnr, is

a deleted neighbor-

hood

The

definition of a limit point can be reframed to read: p is a limit S if every deleted neighborhood of p contains at least one

point of a set point of S.

DEFINITION of

^

10.1

S

point of a set

if

1.

A point p is said to be an interior

Interior Point.

a neighborhood

N p of p exists such that every element

Np belongs to S. or

Does p belong to S? If S is the set of points ^ x are not interior points since every neighborhood of contains points that are not in S. All other points of this set, how-

then

1,

1

and

1

ever, are interior points. DEFINITION 10.12. Boundary Point.

A point p is a boundary point of every neighborhood of p contains points in S and points not in S. If S is the set ^ x ^ 1 then and 1 are the only boundary points. A boundary point need not belong to the set. 1 is the only boundary point of the set S which consists of points x such that x > 1. a set

S

if

,

10.13. Exterior Point. A point p is an exterior point of not an interior or boundary point of S. DEFINITION 10.14. Complement of a Set. The complement of a set S the set of points not in S. The complement C(S) has a relative mean-

DEFINITION

a set is

S

ing, for is

if it is

it

depends on the

the set of real numbers

set

T in which 8 is embedded. If S for example, ^ x ^ 1, then the complement of S relative y

1

to the real-number system is the set of points \x\ 1 ^ x ^ 1 1 x ^ 1 relative to the set

>

1.

The complement

the null set (no elecomplement of the set of rationals relative to the reals is

of

is

ments). The the set of irrationals, and conversely.

DEFINITION

10.15.

Open

confused with open interval)

Set. if

A

set

S

is

said to be open (not to be is an interior point of S.

every point of S

ELEMENTS OF PURE AND APPLIED MATHEMATICS

388

For example, the set of points {x\ which satisfies either 6 < x < 8 is an open set. DEFINITION 10.16. Closed Set. A set which contains For example, the set points is called a closed set. /

V is

2'

r

closed since its only limit point The set of 10.17. is

all

x

<

its

1

or

limit

4'

DEFINITION to Si or 82

<

is 0,

all

called the union (Si VJ

$

which

contains.

it

points (or elements) which belong 2 ) or sum (Si 2 ) of the two sets

+

The shaded area below is S Si + 2 (see Fig. 10.1). The union, S = VJ S a of any number of sets, Si, 82.

,

()

the set of

is

{S a \,

in at least

points {x}, x

all

one of the

Sa

.

DEFINITION 10.18. The set of all points which belong to both Si and /S>2 simultaneously is called the

S (JS2 l

intersection,

nS

5i Siy

area

S%. is

2

or

or Si

written

product, 82, of the

Graphically, Si C\ Sz.

The theory cation

-

two

sets

the shaded

and its appliand mathematics

of sets

to logic

was given great impetus by the Engmathematician, George Boole

lish

(1815-1864). FIG

Two sets Si, 82 are said to be equal = 82, if every element of >S\ contains every element of $ we say

10.1

,

81

2 belongs to Sz, and conversely. If Si If 82 is a subset of Si and if, that 82 is a subset of Si, written Si S%. furthermore, at least one element of Si is not in 8%, we say that 82 is a proper subset of Si, written ,

D

Si

Some obvious (1) (2) (3) (4) (5) (6)

D

or

8,

facts of set theory are:

A +B = B

+A

(A+B)+C

= A

+

(B

+

C)

AB = BA

(AB)C = A(BC)

A + A = A C A +

A, B,

AA = A AB C A, AB C B

S

2

C

Si

REAL-VARIABLE THEORY

A C

(7)

CC=*A +B CC

B

C,

389

A^C,B2C=*ABD<r

(8)

AB

(9)

(I

A(B

(10)

A(B

+

+

C)

= AB + AC

C)

We proceed to prove (10). Let x be any element of A(B C). Then x belongs to A and to either B or C. Thus x belongs to either AB or AC AC. Thus A(J5 so that x is an element of AB AC. C) (I ,4

+

+

+

Conversely,

AB

(I

+

A(

AB + AC C A (5 +

AC C A(# +

0),

C), so that

+

A(B

= AB

C)

+

C) from

+

From

(9")7

(7),

AC.

Problems 1.

What

are the limit points of the set

^

x

^

Is the set closed,

1?

What

open?

are the 2.

3.

boundary points? The same as Prob. 1 with the point x = ^ removed. Show that the set of all boundary points (the boundary

of a set) of a set

S

is

a

closed set 4.

of

Prove that the

S and

its

set of all limit points of a set S is a closed set. The set Is S a closed set? is called the closure of S.

S

consisting

limit points

Prove that the complement of a closed set is an open set, and conversely. WT hy is a set S which contains only a finite number of elements a closed set? 7. Prove that the intersection of any number of closed sets is a closed set. 8. Prove that the union of any number of open sets is an open set. 9. A union of an infinite number of closed sets is not necessarily closed. Give an example which verifies this 10. An intersection of an infinite numbei of open sets is not necessarily open. Give an example which verifies this 11. Show that (A + ) (A + C) = A + BC. 12. The set of elements of A which are not in B is repiesented by A B. Show 6.

6.

that

-

(A

A(B - C) = AB - AC + (B - A) = (A + B) - AB

B)

(B-A}=A+B

A +

A(B - A) =

13.

Show

elements) 14.

If

that,

is

A

A + X =

if

represented by

C S,

~~

we

S

call

AX =

R,

0,

the null set

then

X

=

The

f(A).

null

set

(no

0.

A

C(A) the complement

of

A

relative to S.

Show

that

A

C

B

& C(B) C C(A)

C(T+B) =

C(A}- C(B)

C(AB) = C(A)

+

C(B)

10.8. The Weierstrass -Bolzano Theorem. We are now in a position to determine a sufficient condition for the existence of a limit point.

THEOREM

10.16.

A

limit point

p

exists for

every

infinite

bounded

linear set of points, S.

The proof proceeds place

all

as follows:

We

construct a

points which are less than an infinite

new set number

T.

Into

T we

of points of S.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

390

T is not empty since S is bounded below. Moreover T is bounded above, S is bounded above. From Theorem 10.15 a supremum p exists for the set T. We now show that p is a limit point of the set S. Consider for

any neighborhood

of p, say,

>0

p-e<x<p+5 Since p

an

is a member t of T such that t > p e. Since t is e is less than number of elements of S (why?), p number of S. Thus an infinite number of points of S lie to

<

e

than an

less

infinite

6>0

there

p,

infinite

+

the right of the point p 5 has On the other hand, the point p e. 6 only a finite number of points of S greater than it, for otherwise p would be a member of T, a contradiction, since p is the supremum of the

+

set T.

Thus the open

interval

<

p

number

contains an infinite

of

x

+

< p

8

elements of S.

Since

e,

d

are arbitrary,

every open interval containing p contains an infinite number of elements This proves the existence of a limit point p. of S. Problems

Show

1.

that the limit point p of

Theorem

JO. Hi is the

supremum

of

all

limit

points of S.

Prove the existence of a least or smallest limit point

2.

for

an

infinite

bounded

linear

set S. 3.

Show

Prove that a bounded monotonic increasing that the sequence

n =

0+0'

set of points lias a

unique

limit.

1, 2, 3,

has a unique limit. We call it e. Consider the totality of 4. Let Si and $2 be two closed and bounded linear sets. distances obtained by finding the distance from any point si in S\ to any point 2 in S<2 s Show that there exist two points 5 J; s 2 of Si and <S 2 respectively, such that |s 2 is the maximum distance between the two sets. Define the diameter of the set, and show 5. Let S be a closed and bounded set. is the maximum distance between that two points *i, s^ of S exist such that \sz i| .

}

,

any two points 6.

of S.

Consider the infinite dimension linear vector space whose elements are of the 00

form a

(ai,

a2

,

.

.

,

a,,,

.

.

.),

the a t real, such that

y

af converges, written

00

V

a t2

i-i

If

b

\

*=

(61,

6 2,

,

&n,

.

.

)

with

>

t-i

6?

<B

j*

oo,

then by definition

REAL-VARIABLE THEORY a

=

b

-f

+

(01

a 2 H-

61,

62,

.

.

391

a n -H b n

,

,

.

.

.)

xa n and xa. We can define the distance .). (xai, xa z two elements of this space (Hilbert] by the formula ,

.

.

.

,

.

,

.

=

p(a, b)

)']*

n

Show

l

sum

Hint: Consider the

that p(a, b) converges.

V

+

(\a n

between

~

(&

^=

[

p(a, b)

W

2

wl oo

or

(

^=

n

X

+

a*)

2

d

a,,/>,,)

inequality,

-")'

show that

Also (1)

P (a, b)

(2)

P (a, b)

= =

P (b, a)

(3) P (a, b)

+

P (b, c)

Show

t=* a

=

b

^

P (a, c)

that the infinite hounded set of points

P2

-

gn

-

e,

cannot have a limit point.

we mean the which

(i, o, o,

(0, 0, 0,

P

x

=

.

,

o,

.

0,

.

,

.

.

,

1, 0,

.)

.

.)

.

.)

spherical neighborhood of a point

By a

-

.

(0, 1, 0,

(pi, P2,

,

p M|

)

set of points (xi,

,*,...)

.

1-2,

satisfy

Let

L

ments from

be a limit point of a set S, say,

s\, s z , Sa,

.

.

.

Show

S. ,

sn ,

.

Hm n

>

.

.

sn

,

that one can pick out a sequence of elesuch that

L

o

in the sense that every neighborhood of L contains We call such thermore s n +i is closer to L than s r,

toL.

+

X

(^ n*=l

1

and prove the Schwarz-Cauchy

7.

oo

ao

2

.

an infinite number of the s t Furan approach a sequential approach .

ELEMENTS OF PURE AND APPLIED MATHEMATICS

392

M

... be a Nested Sets. Let Mi, Af 2 n such that bounded sets and 2 M$ MI D D 3 D sequence We show that there is at least one point p which belongs n D First we choose an element p\ to all the sets MI, M%, n .... The set of points from MI, then p 2 from 71/2, n etc. p n from This set is bounded since MI is ) belongs to Mi. pn (pi, p 2 bounded. From the Weierstrass-Bolzano theorem the set has at least one limit point p which belongs to MI, since MI is closed. Similarly p is These points belong to a limit point of the set (p 2 PS, ) pn Theorem

10.9.

of

.

M

'

.

.

M

.

,

.

,

M

.

.

,

.

.

,

.

so that peAT 2

so that If

,

peM n

,

,

.

.

,

,

2

,

.

.

,

M

.

M

'

'

.

.

,

of closed

Finally p

.

is

,

a limit point of the set (p n p n +i,

the diameters of the sets

M

n

)

,

to each 3f, i = tend to zero, then p

Thus p belongs

.

1,

is

.

2,

.

.

,

n,

.

.

.

.

(see Prob. 5,

unique

Sec. 10.8) for the definition of the diameter of a set. Some mathematicians believe it necessary to postulate the existence of

More generally they desire the followthe set (pi, p 2 ). pn Then there exists ing postulate: Let S = {8 a be any collection of sets. a set P of elements \p a such that a point p otP exists for each set /S of S, p This is essentially the axiom of choice (Zermelo). belonging to the set S % .

.

.

.

,

,

,

}

l

}

t

t

.

10.10.

and

The Heine-Borel Theorem.

Let $ be any closed and bounded

T

be any collection of open intervals having the property that, if x is an element of S, then there exists an open interval T x of the collection T such that x is an element of T x The Hcine-Borel theorem states that there exists a sub collection T of T which contains a finite set,

let

.

f

of open intervals and such that every element x of S is contained one of the finite collection of open intervals that comprise T' Before proceeding to the proof we point out the following: 1. Both the set S and the collection of open sets T are given beforehand, since it is a simple matter to choose a single open interval which completely covers a bounded set 8. 2. 8 is to be closed, for consider the set 8(\, 1, 7, 1/X .), and let T consist of the following open intervals:

number

in

.

.

.

.

.

,

Tn JL

It is

-

.

easy to see that we cannot reduce the covering of

any of the given T n for there is no overlapping Each Tj is required to cover the point l/j of S. ,

.

S by

of these

eliminating

open intervals.

REAL-VARIABLE THEORY

393

N^x^N.

Proof of the Theorem. Let S be contained in the interval This is possible since S is assumed bounded. Now divide this interval into the two intervals (1) ^ x ^ 0, (2) ^ x ^ N. Any element is in S, then lies in both x of S lies in either (1) or (2). If the origin The points of S in (1) form a closed set, as do the points of (1) and (2). S in (2). Why? Now if the theorem is false, it will not be possible to

N

S in both (1) and (2) by a finite number of open which form a subset of T. Thus the points of S in either (1) or (2), or possibly both, require an infinite number of covering sets of T. Assume the elements of S in (1) still require an infinite covering; call cover the points of

intervals

Do the same for Si by subdividing the interval these points the set Si. into two parts. Continue this process, repeating the ^ x ^ argument used above. In this way we construct a sequence of sets

N

S

D

S,

D

8*

D

D tin D

such that each S ^ is a closed set (proof left to reader) and such that From the theorem of nested sets as n > oo diameters of the S n * (Sec. 10.9) there exists a unique point p which is contained in every /S\, and ptS. Since p is in S, an open interval Tp exists i = 1, 2, which covers p. This T p has a finite nonzero diameter, so that eventuWhy? But by ally one of the /S,, say, S m will be contained in T p assumption all the elements of S m require an infinite number of the {T} This is a direct contradiction to the fact that a single to cover them. .

.

.

,

.

,

T p covers them. Hence our original assumption was wrong, and the theorem is proved. Note that our proof was by contradiction. Certain mathematicians from the intuitional school of thought object to this type of proof. They 7 r exhibited, wish to have the finite subset of T\ say, TI, 7 2, such that every point p of S belongs to at least one of the T3 j = 1, 7

T

.

.

.

,

,

,

2,

.

.

.

,r.

Problem. Assuming the Heine-Borel theorem, show that the Weierstrass-Bolzano theorem holds. Hint: If S has no limit points, it is a closed set. If p is a point of S and is not a limit point, a neighborhood N p of p exists such that N p contains no points Continue the proof. of S other than p itself.

We

now show that, if we assume the Heine-Borel theorem true for any linear point set, then the existence of the suprernum can be established. be an infinite bounded set of real numbers. can we dispense Let

Why

A

with the case for which

S

contains only a finite

number

of

elements?

bounded above by an integer N, we can speak of the consisting of all numbers x such that x ^ every s of S, x ^ N.

Since set

T

S

is

is

bounded.

Is

T

closed?

Yes, for

if

tis a limit point of T,

set

T

The and

if

ELEMENTS OF PURE AND APPLIED MATHEMATICS

394

<

then we can find a neighborhood of t which excludes so, so that T in this neighborhood are less than SQ, a contradiction. Thus 7 and 7 is closed. Every member of T satisfies the t ^ SoeS, so that ttT first criterion for the existence of a supremum. Hence, if no member of T is the actual supremum of the set S, the second criterion for the existt

So,

SoeS,

the points of

r

,

ence of a supremum must be violated for every member of T. Thus such that no member t every point teT is contained in a neighborhood These neighborhoods form a covering of T. of S is in this neighborhood. Assuming the Heine-Borel theorem, we can remove all but a finite number of these neighborhoods and still cover the set T. Let the finite

N

collection of neighborhoods be designated

N:

at

<

x

<

bl

=

i

,

by 1, 2, 3,

.

.

.

,

r

r. We assert that a Let a u be the smallest of the a, i = 1, 2, this. Then the to T. Let the reader neighborhood to verify belongs which a u originally belonged has the property that no point of S is in this neighborhood. This is a contradiction since those points to the left of a u (<a u ) are not covered by the neighborhoods N. Hence there must be a supremum of the set S. Q.E.D. 10.11. Cardinal Numbers. We conclude the study of point sets with a brief discussion of the Cantor theory of countable and uncountable sets. DEFINITION 10.19. If two sets A and B are such that a correspondence exists between their elements in such a way that for each x in A there .

.

.

,

corresponds a unique y in B, and conversely, we say that the two sets are in one-to-one correspondence and that they have the same cardinal

number. Thus, without counting, a savage having 7 goats can make a fair trade with one having 7 wives. They need only pair each goat with each wife. DEFINITION 10.20. A set in one-to-one correspondence with the set of integers (1, 2, 3,

DEFINITION

.

10.21.

.

.

,

n}

is

said to have cardinal

number

n.

A set which can be put into one-to-one correspond-

ence with the Peano set of positive integers

is

said to be countable, or

denumerable, or countably infinite, or denumerably infinite. We note that the set of integers I 2 2 2 3 2 , . . n 2 is, countably infinite and at the same time is a subset of all the positive integers. ,

The

cardinal

number

.

,

,

,

.

.

.

of the positive integers is called aleph zero (Xo).

DEFINITION 10.22. An infinite collection which said to be uncountable, or nondenumerable.

is

not denumerable

is

A countable collection of countable sets is a count10.17. Let the sets be Si, $2, Sn This can be arranged since we have a countable collection of sets {$}. Since the elements of THEOREM

able set.

Si are assumed countable,

.

.

.

,

,

.

.

.

we may arrange them

.

as follows:

REAL-VARIABLE THEORY

Similarly

Now

consider the collection of elements

011, 012, 021, 013, 031, 022,

2(-l),

01r,

,

,

0r,l,

we have a first element, a second sequence exhausts every element of {S}. This Thus the proof of the theorem is is done by a diagonalization process. demonstrated. We can now easily prove that the set of rationals on the interval ^ x ^ 1 is countable. Into the set S we place all fractions m/i, This collection

countable since

is

But

element, etc.

this

l

m^

^

i

There are a countable number fractions. of (0

g

x

The complete

^

=

i

.

S each containing a

of

Now

being countable.

1), this set

.

.

<r<

oo

numbers

O.an 012 013

S2

= =

Sn

=

0.0 n i

O.a2l 022 023

'

'

'

'

'

*

0n3

n2

'

*

02n

"

=

Now bl

the

Sj, j 1, 2, 3, consider the number s

=

.

2

011

-2 occurring

=

if

62

=

M

=

-

.

. ,

=

n,

.

.

.

of

8 or

9.

:

'

*

*

'

'

'

1

of

'

9.

bn

...

2

x

Then the elements

*

*

0nn

set.

^

are written in decimal notation.

O.&i 62

022

a countable

is

in the interval

01n

where the a mn are the integers zero through

The

number

the reader show that

let <*>

not a countable set. Assume the set countable. ^ x ^ 1 can be arranged in a sequence S}

finite

set of all such fractions yields the set of rationals

the set of rationals, {r}, for which Cantor showed that the set of real is

1, 2, 3,

,

bn

The number

b

= is

where nn

2

...

certainly not

among

from every one of these Sj. But the collection {$,} was supposed to exhaust the real numbers on the interval ^ x ^ 1. Thus a contradiction occurs, and

any

of the s3 j ,

1, 2, 3,

.

.

.

,

n,

.

the real numbers are not countable.

.

.

since

it differs

ELEMENTS OF PURE AND APPLIED MATHEMATICS

396

Problems 1.

Show

3.

Consider the Cantor set obtained as follows:

that the set of algebraic numbers, those numbers which satisfy polynomial equations with integral coefficients, is a countable set. 2. Why are the irrationals uncountable?

From

the interval

^

x

^

1

delete

the middle third, leaving; the end points ^, ^. Repeat this process for the remaining The set that remains after a countable intervals, always keeping the end points.

number of such operations is the Cantor set. Since each removed set is open, show If the numbers that the Cantor set is a closed set. ^ x ^ 1 are written as decimal expansions to the base

spondence with the

3,

show that the Cantor set can be put into one-to-one correon ^ x 5s 1 when these numbers aie written as

set of reals

decimal expansions to the base 2. Show that every point of the Cantor set is a limit point of the set. 4. Show that a limit point exists for a nondenumerable collection 8 of linear points.

The

set

need not be bounded. Hint' Consider the intervals N ^ x ^ N + 1 for the At least one of these intervals must contain a nondenumorable

rational integers N. number of 8. Why? 6.

Show

that a limit point exists and belongs to a nondenumerable collection

S

of

linear points.

Limits and Continuity. Let G 10.12. Functions of a Real Variable. be any linear set of real numbers. G may be an open or closed set, a finite We write G = {x}, set of points, the continuum, a closed interval, etc. x of Assume that to each of G there correis element G. where x any {y} be the totality of sponds a unique real number y, x > y, and let H numbers obtained through the correspondence. This mapping of the set G into the set // defines a real-valued function, usually written

real

y

=

f(%)

x

mG

We

say that y is a function of x in the sense that for each x there exists a rule for determining the corresponding y. The notation f(x) simply means that / operates on x according to some predetermined rule. For 2 3 ^ x < 5, the rule for determining y given example, if y = f(x) = x x the set 3 5 simply is to square the given number x. of < x ^ any Consider Table 10.1. The set G consists of the numbers x = 1, 2, 5, 14, ,

TABLE

and the

set

respectively.

10.1

consists of the corresponding values y = 3, 8, 2, 0, The table defines y = /(x), x in G. The reader is familiar

H

with the use of cartesian coordinates for obtaining a visual picture of 3/

=/(*)

REAL-VARIABLE THEORY

An important

397

concept in the study of functions

is

the limit process.

One reason why

the student of elementary calculus encounters difficulty in understanding the theory of limits is simply that the concept of a

The statement lim x 2 =

limit rarely occurs in physical reality.

x\

is

X-+XQ

To

the novice this statement appears to It must be understood XQ? xl an abstract entity invented by man. We

quite confusing to beginners.

be

when x =

2 trivially true, for is not x

that the statement x

-

Xo is

visualize a variable x which

is

orced to approach the constant XQ in such

a manner that the difference between x and XQ tends to zero. us to the following definition

This leads

:

DEFINITION less If

An infinitesimal is a variable which approaches said to be an infinitesimal if \tj\ becomes and remains

10.23.

zero as a limit.

77

is

than any preassigned

and

77 i

for arbitrary

772

e

e

>

0.

are infinitesimals, then eventually \TH\ > 0, so that \rii < e, ^ \rji\ 772] 772!

+

+

<

c/2,

and

771

|r/2

+

772

<

e/2,

is

also

an infinitesimal. Let the reader show that the difference and product of two infinitesimals are again infinitesimals and that the product of a bounded function with an infinitesimal is again an infinitesimal. Let x be a variable which m succession takes on the values 1, 3, ,..., as n becomes Certainly x is an infinitesimal since l/n > 1/n, For any value of x the sum S x -\- x -\- x -{'is infinite. infinite. On the other hand each term of $ tends to zero. Hence an infinite sum For another example, let x of infinitesimals need not be an infinitesimal. be an infinitesimal, so that # 3 /(l + x 2 ) n is also an infinitesimal for n = 0, .

.

.

.

'

1, 2,

....

'

Now

an infinitesimal. An infinite sum of infinitesimals may Let the reader show by specific examples that the quotient of two infinitesimals may or may not be an infinitesimal. Let us return to the statement lim x 2 = #. To show that this so that or

may

S

is

also

not be an infinitesimal.

X

>0

statement holds, we must prove that x 2 x$ is an infinitesimal whenever 2 = \x an infinitesimal. Now is x #o \x \x x\\ XQ\ #o|, and x is an infinitesimal (by Since \x a: + XQ\ is bounded for x near XQ. 1 our choice), of necessity x 2 x\ is an infinitesimal, so that x x\ > 0, or x 2 * .TO, whenever x > XQ.

+

\

|

Let the reader prove the following results:

lim g(x)

=

N

If

lim /(x) x*a

= M,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

398 then (1)

lim [/(x)

(2)

Ito [/(x)0(x)]

g(x)}

M N

=

= MAT

M

N

N

5*0

A knowledge of limits enables us to introduce the concept of continuity We consider the functional relationship y = /(x), and

of a function.

we let x = x be a point of G with y Q = /(x we say that f(x) is continuous at x = x ? any point if

of

G

which

is

close to x

,

the corresponding value y will

y

What

).

shall

we

we mean when

if x is then /(x) will be continuous at x = x be close to y$. Actually we desire yo to be an infinitesimal if y

x

y =/(*)/

Intuitively

feel that,

XQ is an infinitesimal. DEFINITION 10.24. /(x)

to be continuous at x

any

>

e

=

there exists a

said

is

x

if

5(e)

for

>

such that |/(X)

-/(So)

|

<6

x < 6, x in G. |x Let us look at the graph of Pig. For any c > we construct 10.2.

whenever

horizontal

the FIG. 10.2

y

tinct vertical lines x in the

between /(x

The

=

)

c,

/(x

+

XQ

open interval XQ

6 )

maximum

size of 6 will

If

lines

y

=

y

+

c,

we can draw two dis-

x = XQ ~ 6 such that, for every point x of G x < XQ + 5, /(x) has a numerical value lying we say that /(x) is continuous at x = XQ.

,

must

distinct vertical lines

yQ

.

5,

<

+

=

|

exist for every

depend on

c.

The

e

smaller

>

0.

e is,

Note that the

the smaller

6 will

The p"oint x will influence the value of 6. the the smaller 5 becomes. steeper curve, Two equivalent definitions of continuity are as follows: /(x) is conhave to

be.

=

x

lim /(x)

=

tinuous at x (a)

Moreover the

if:

/(x

),

x in G.

X-+XO

there exists a For any c > < e whenever Xi and X2 are

(b)

f(xz)

|

6

in

neighborhood of x such that |/(xi) G and belong to the 6 neighborhood

of XQ.

Functions are discontinuous for two reasons: 1. lim /(x) does not exist. x *xo

2.

lim /(x) exists but X

is

not equal to /(x

).

REAL-VARIABLE THEORY Consider /(x)

10.1.

Example

-

+

x

hm

1,

+

(x

*

x

-

2,/(2)

-

1)

399

We

5.

have

3

x-+2

However /(2) = 5 ^ 3, so that /(a;) is discontinuous at x = 2. If we were to redefine the value of /(x) at x = 2 to be 3 instead of 5, we would remove the discontinuity. In this example f(x) has a removable type of discontinuity. Ol/x 5* 0, /(O)

hrn

7-37^. -T 2 1/T

>

2 1 /* 9 2 1/x

lim x->0

1

2 1/z

=

1

-

We

1.

have

=

_>() 1

"T

z>0 Hence lim x-0

x

Consider /(x)

10.2.

Example

j<0

/(x) does not exist for all

manner

approach of x to zero

of

It is

obvious

that f(x) does not have a removable type of discontinuity at x ~ 0. Example 10.3. f(x) = sin (l/x), x ^ 0, /(O) = 2. Since /(x) oscillates between = 0. > 1 and -h 1 as x 0, lim /(x) does not exist and f(x) is discontinuous at x x-*0 /(x) discontinuous at x

is

is

f(x)

2

x 5^ 0, /(O) since division

2

l)/(x

41

Some authors

consider that \/x

undefined.

In this example

zero

by

is

=

since hrn l/x does not exist. On the other hand z->0 2 is continuous at x = 1 even though x 7^ 0, /(O)

1),

1) is

l)/(x

=

l/x,

=

discontinuous at x (x

/(x) (x

=

10.4.

Example

=

undefined at x

1.

^ x ^ 1, we note that /(x) is continuous everyExample 10.5. If /(x) = x 2 for and x = l? If a; =* c is a where on this interval. Why is /(x) continuous at x = 2 2 = \x r' so that |x -f c\ < 3|x c\ point of this interval, we have x c| 2 = 6. We have found a value of 5 independent c 2 < * whenever |x r| < e/3 |x We say, therefore, that /(x) = x 2 is uniformly continuous on the of the point x = c. 1. x interval ^ ^ Example 10.6. Let/(x) = l/x for a ^ x ^ 1, a > 0. It is easily seen that/(x) is c we have continuous for every x of this range. For x |

|

1

__

x J

ent of x =

c,

if

|x

-j

<

we have uniform

6

<

x

^

1 is

as

>

whenever

\x

a c|

e

x

^

For any

d for all c

not a closed

or

|x

c|

1

2

=

6

5.

Since 5

= a 2 e is independ-

the other hand /(x) is also continuous is not uniformly continuous on this

But /(x)

.

>

e

<

On

continuity.

<

0.

>

<

<

c|

at every point of the interval

range since

c

c

1

1

Thus

x

1

we cannot

<

on the range

find

a

^

1.

x

5

>

such that

x

c

Note that the interval

set.

DEFINITION 10.25. Let f(x) be defined over the range G. f(x) is said to be uniformly continuous on G if for every > there exists a 5

=

5(c)

such that l/(*i)

whenever

\xi

THEOREM

#2!

10.18.

<

5,

If

x\

G

is

-/(a*)!

and #2

<

c

in G.

a closed and bounded set and

tinuous at every point of G, then /(#)

is

if

f(x) is con-

uniformly continuous on G.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

400

We

Proof.

base the proof of this theorem on the Weierstrass-Bolzano assume that f(x) is not uniformly continuous on

First let us

theorem.

This means that there exists at least one there will exist Xi, Xi such that |/(xO such that for any 5 > We take 5 = 1 and obtain the pair of Xi\ < 6. /(xi)| > eo with |xi and obtain the pair (x 2 x 2 ). For numbers x\, x\. Then we take 6 = 5 = 1/ri we have the pair (x n x n ) such that \f(x n ) /(x n )| > e with the closed and bounded set G.

=

,

-J-

,

xn

\x n

X2

(Xi,

<

|

.

.

.

,

l/n = Xn ,

,

and bounded

By

5. .

.

we generate the two sequences

this process

(

.),

5 1, X 2

.

,

.

Xn

,

ShlCC

)

,

G

IB

& dosed

xn the sequence (xi, x 2 .) has a limit point x = c belonging to G. We pick out a subsequence which converges = c. The corresponding points from the sequence [x n sequentially to x set,

.

,

.

.

.

,

.

,

\

have the same limit, x be designated by

{x'n

=

\x'n

\,

xw

c,since|o;n

Since /(x)

}.

- /Ml < o/2, |/($ n - /(c)| < a contradiction, -~/(Zn)l < c > Q.E.D.

I/GO I/OO

)

o,

6

/(*n )|

<

|

is

Let these two sequences continuous at x = c, we have l/n.

eo/2, for

n

sufficiently large.

Thus

since for all pairs x^, x'M |/(xn )

.

THEOREM

10.19.

and

If /(x)

are continuous at x

0(x)

/(x) g(x), /(x)0(x), f(x)/g(x) with 0(a) The proof is left to the reader.

THEOREM G, then /(x)

10.20. is

If /(x) is

^

=

a,

0, are continuous at x

then

=

a.

continuous over a closed and bounded set

bounded.

Assume

There exists an x\ such that /(x) is not bounded. an x such that an x n such that /(x n ) > n, > 2 /(x 2 ) 1, /(xi) 2, The set (xi, x 2 has a limit point x belonging .) x, to G since (? is a closed and bounded set. There exists a subsequence Thus from .) which converges sequentially to x. (xj, Xg, x^, Proof.

>

.

.

.

.

.

.

.

,

.

.

.

.

.

,

.

.

,

.

.

,

continuity lim /(x;) Xn'

But /(xn )

>

value at x

=

THEOREM

co

as x^

>

=

/(x)

Z

a contradiction, since /(x) has a specific real

x,

x.

continuous on a closed and bounded set such that /(x ) S /(x) for all x of G. From Theorem 10.20/(x) is bounded above as x ranges over G. Proof. Thus the set of values {/(x) has a supremum, say, S. Thus S ^ /(x) for all x of G. Choose Xi such that /(xi) > S 1, x 2 such that /(x 2 ) > x S such that S .... n I, /(x n ) > l/n, Why is this possible? The set (xi, x 2 xn We .) has a limit x belonging to G. pick out a subsequence {xn which converges sequentially to x, with x^ = x m Now/(x m ) > /S 1/m, so that 10.21.

If /(x) is

G

G, then a point x of

exists

}

.

.

.

,

,

.

.

.

.

,

,

.

}

.

lim f(xm)

^

lim

-

- = S (*-!)V /

(

}

S

REAL-VARIABLE THEORY

From

n

S ^

=

continuity, lim f(x^)

=

/(x), so that /(a)

^

But

S.

oo

=

/(z), so that/(s)

THEOREM

lim f(x m )

m * Q.E.D.

> oo

401

5.

Let f(x) be continuous on the range a ^ x ^ 6. If there exists a point c such that /(c) = with a < c < b. < > 0,/(6) 0, /(a) T the Let be set of a of x 6 such that ^ ^ Proof. points /Or) < 0, and We speak only of x on the range if f(xi) < 0, x z < Xi, then /(z 2 ) < 0. a ^ x g Z>. T is not empty since x = a belongs to T. T is bounded above since x = b does not belong to T. Hence T has a supremum; call

itx

=

Since f(x)

10.22.

Now/(c) =

c.

0,

continuous at x

is

-

or/(c)

=

c,

>

0,

we can

or/(c)

<

>

Assume /(c)

0.

0.

find a 5 such that

> 0. Thus/(z) > /(c)/2 > whenever |x - c| Let the < 5. But /(a;) < for x < c, a contradiction, so that/(c) g cannot occur, so that /(c) = 0. reader show that /(c) < whenever

|jc

c\

<

5, 5

Problems For the following functions find a x z < 5:

1. \x\

6

=

5(e)

such that

f(x 2 )}

\f(xi)

<

c

whenever

\

(a) /(JT) (b) /( X ) (r)

f(x)

Show Show

2.

= =

=

2x* sin

for

x

for

-v^l"^

a:

2

for

-1 g JT ^ 1 ^ j- ^ 27r - 1 g a; ^ 1

that g(x) = x sin (I/a*), x 5^ 0, /(O) = 0, is continuous at x = that y = sin (1/jr) is not uniformly continuous on the range

0.

< x ^ 1. Find an example of a function /(x) which is continuous on a bounded set, but /(a) not bounded. 6. Find an example of a function f(x) which is continuous on a closed set, but f(x) not bounded. 6. Prove Theorem 10.19. 3.

4.

is

is

Discuss the continuity of f(x)

7.

tional, /(O)

=

=

when x

is

rational, f(x)

=

x

when

x

is

irra-

0.

when x is irrational, f(x) Show l/q when x = p/q, (p, (/)=!. discontinuous at the rational points and is continuous at the irrational Hint: Show that there are only a finite number of rational numbers p/q with

Let f(x)

8.

that f(x) points.

is

=* 1 such that > 0. l/q > e for any Prove Theorem 10.18 by making use of the Heine-Borcl theorem. 10. Prove Theorem 10.20 by making use of Theorem 10.18. 11. Let f(x) If f(x) and g(x) are continuous for g(x] for x rational.

(PJ #)

9.

that /(x) 12.

f(x)

^

=

all x,

show

g(x) for all x.

^ x ^

Let f(x) and g(x) be continuous on the range a g(x), h(x)

=

g(x)

if

g(x)

^

f(x).

feW = iCfW

+

Show flf(x))

+ i|/(or) -

Also show that h(x) is continuous for a ^ x ^ b. 13. If /(x) is defined only for a finite set of values of at these points.

6.

Let h(x)

=

f(x)

if

that

x,

0(x)|

show that f(x)

is

continuous

ELEMENTS OF PURE AND APPLIED MATHEMATICS

402

=

>

>

exists such that, 14. If f(x) is continuous at x c with/(c) 0, show that a 5 for \x - c\ < 5, f(x) > 0, x in G. 15. If f(x) is continuous on a closed and bounded set G, show that a point X Q of G exists such that f(xo) ^ f(x) for all x in G.

10.13.

The

say, {x n }.

Let/(x) he defined on a range G, and let c be G which converges to c,

Derivative.

We

a limit point of G.

consider any sequence of

Next we compute /(c)

(10.9)

c

and is independent of the particular sequence which to we represent c, say that /(x) is differentiable at x ~ c. converges this limit or first derivative by /'(c). If this limit exists

We

The

definition

above

equivalent to the statement that for any e > there exists a 6 > and a number,

is

y

such that

f'(c),

-

f(c)

x

whenever

<

c

0< \x

c\

<

6,

x in G.

follows almost immediately that /'(c) exists then/(x) is continuous at It

it

=

.1

for

c,

o 10.3

as

The reader

is

well

aware

tangent line

T

=

L joining P

defined as

is

>

/(c) as

a:

>

c.

of the geometric interpretation of the first

A*

the slope of the secant line of the

c implies /(a;)

c)

In Fig. 10.3,

derivative (see Fig. 10.3).

PR =

>

JT

-

~/

\f(x)

FKJ

to

/' (x),

-

+

f(x

Q

f(x)

-/(a?)

is"

and

RS =

f(x) Ax

is

the slope called the

differential of /(x).

Example

10.7.

Let /(x)

=

x 2 sin (1/x), x

0, /(O)

.

lim

lim

= hm

_

X

Hence /'(O)

=* 0.

0.

To

see whether f(x)

is

=0, we compute

differentiable at x

Let the reader show that/' (x)

that lim f'(x) does not exist.

x->Q is

=

x sin X

not continuous at x

=

by showing

REAL-VARIABLE THEORY The reader can

if

verify that,

jx

f(x)

-

0(*)]

[/(*)

-

d (constant) ^ dx dx n dx d- sin x y dx -

10.8.

Example

known

=

x

exists at

We

xo.

=

Jo,

u

.-

dx d In x j dx If y

no.

=

ex

=

-

1

x

and assume that

<P(XO),

=

and u

/(M)

we say

<p(x),

exists at

-^

that y

u =

MO,

-?

show that

_

dy dx at x

(10.10)

0(f)/'(c)

.

cos x

=

Let MO

x.

c,

to hold:

>^

=

then

d cos x dx

Implicit Differentiation.

depends implicitly on

+

=0n

,

=

+ g'(c)

f(c)g'(c)

following; differentiation rules are

403

g(x) are differentiable at x

f(c)

=

lr-e

The

and

df

dy du dy dx

_

d<p

du dx

From AM) -/(MO)

-f

/(w

A?/

we have A//

_

Ax as

If,

Ax

0,

_

/(MO -f A?^) _

since the limit of a product

the other hand,

this case,

however,

we have *H .

is

)

A// p^

Ax

A?/ ~

,

=

(lu.ll

)

J

AM

infinitely often, then eventually

lim

)t

^

7^ 0,

and

<

-

the product of the respective limits, provided they exist. 0. For (10.11) if AM == infinitely often as Ax

we cannot apply -j-

dx

= hm A

fM-

Ax

=

^(x

Ax

AM does not vanish dti -7-

On

-f Ax) __

/(MQ) y?(x

AM

_

.

Ax

= Ax o,

But

0.

for those values of

so that lim AZ-+O

Ax

-

0.

Hence

Ax

for

which AM

dx

dy dx

still

=

holds

since both sides of this equation vanish.

Example

10.9.

The Rule of

d dx

,

.

-j- (uv)

If

we

differentiate again,

The

Leibnitz.

we obtain

=

u

derivative of a product u(x)v(x)

dv -j

dx

-r-r (MV) 38

ox

.

is

du dx

h v -j-

= u -r-9z ax

4-

~

2 3^ ax ax

+ v -~ ax z

Let the

ELEMENTS OF PURE AND APPLIED MATHEMATICS

404 reader prove

by mathematical induction dn

(UV)

5?

with -77 dx

=

=

w, -TT

dx

l}

d "v

_ W "

\

(

_L

d^ +

or otherwise that

~

( n \ du dn

lv

+ Vl/ Sd?=

As an example

f.

(}

d

n

d* u d "~ 2v

fn\

W

_

(x 2

d*- 2

dx*

l)

.

we

n

>P

(x)

+

-i.

'

'

*

consider the Legendie

=

1,

n =

0, 1, 2,

.

.

.

.

(IX

i

We

_L

of Leibnitz's rule

1 - ^-. Til

polynomials P n (x), denned as P n (x)

,

have

n -W + 2*-in \

/

1

i

.

(n

dx

dx

-

x

^r*M

-

nP n -i(Jc)

4.

(1013)

Moreover r^

,

dH

1

x

._

(

f/

(

r 2 __

i \

dx n

2"n! L

(^

l

^

IV*" 1 4- 2r>r ;

T2

(r 2

dx"- 1

l

w + .g 1

(

1 IV'" j

+fJ

..

,

w

,,,

oinbining (10.13) with (10.14) yields

^ [(1 Equation (10.15)

is

- x)

^^]

+

W(TI

(10.15)

Legendre's differential equation (see Sec. 5.13).

From Theorem 10.21 we readily prove Theorem 10.23 due to Rolle. THEOREM 10.23. Rolle's Theorem. Let f(x) be continuous in the interval a

g

x

^

6,

and

this interval.

If /(a)

such that

=

/'(c)

0.

=

let f(x) possess

f(b)

=

0,

a derivative at every point of = c, a < c < 6,

there exists a point x

REAL-VARIABLE THEORY

From Theorem

Proof.

^

that f(c)

f(x) for a

^

10.21 there exists at least one point x

x

/O

=

/(c)

c

such

# + *>--&-> 5=0

fc-o

+

=

Now

b.

Ac) -lim since /(c

405

ft

Moreover

.

Of necessity, /'(c) = 0. Q.E.D. The condition be weakened by assuming that /'(c) exist at x

of the

theorem can

with

/(c) a local Let the reader give a geometric; interpretation of Rollers theorem, and let him construct an example of a continuous function, /(#), /(a) = /(&) = 0, such that nowhere is f(x) = 0, a ^ x ^ b. Law of the Mean. Let f(x) and <p(x) be differentiable functions on the interval a ^ x ^ b. We show that a point x c exists such that

supremum

or inf emum of f(x)

(

The

-

v (b)

c,

.

*(a)]f (c)

=

-

[/(&)

f(a)] v '(c)

(10.16)

function

-

-

/(a)]

/(*)[?(&)

-

*(a)]

- v (a)f(b) +

<t(b)f(a)

=

a and x b. $(x) was obtained by finding A, B, C = a, x = 6. Applysuch that $(x) = A<p(x) J5/(x) + C vanishes for x As a special ing Rollers theorem to \f/(x) yields the theorem of the mean. case, choose <p(x) = x, so that (10.16) becomes vanishes for x

+

/(&)

- /() =

x

/

(c)(6

~

a

a)

<c <b

(10.17)

Let the reader give a geometric interpretation of (10.17). L' Hospital' s Rule. From (10.16) we have

- /(a) _ - <p(a)

/(b)

provided <p(a)

=

<p(a) -^ 0,

<p(b)

^>'(c)

?^ 0.

a

<

c

<

b

(10.18)

Let us assume that /(a)

=

0,

0, so that

and

lim

=

<

c

<

6.

We

=

lim ;>_

since a

/'(c)

lim

(p'(c)

f

_ ^ '(C)

can rewrite (10.19) as lim

^f( =

lim

f& '

(10.20)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

406

= 0, L'HospitaFs rule. We cannot apply (10.20) unless /(a) = = 0? If /' (x) and <p'(x) What can be done if /' (a) 0, /(a) *(a) = 0. are continuous at x = a, and if <p'(a) 9* 0, then (10.20) can be written which

is

(10.21)

=

provided /(a)

0,

Example 10.10

able.

1

Reapplymg ..

x

-

x

+

sin

z 3 /6

r"

-h

=

-

cos ---a*

,

lim -0

+

1

5.r

= 5x both

# 2 /2 and h(x)

--

cos r

hm

0.

--

sm

lim

Now g(x) =

=

<p(a)

4

vanish at x

4

=0 and are difforonti-

(10.20) yields

x

x 7 x

+x

3

/Q

*

.

Inn *->o

- -sin x

-f

x

x --r- = hm hm sm ,

,

.

.

J^H-O

=

,.

lim

cos x ^r120

-

+

cos -~ x ---1

=

1

J^T 120

We complete this section with a discussion of the second derivative. Let f(x) be differentiate in a neighborhood of a: = c. The second derivative of f(x) at x = c is defined as

provided this limit

f'(c)

=

lim^

f"(c)

=

/'(c '

+

=

lim*^-----^

*)

= Um

-

^-- so that

+

+ /(c)

A->0

lim

nm The

f'(.r)

ft

/.-.O

and

Now

exists.

i

im

order in which

/(c

+

we take

+

k

h)

- f(c +

these limits

however, that under certain conditions

one limit process

is

/" (c)

lim

f(c

-

f(c

important.

ft)

It

can be shown,

we can

replace k by h so that only T e would thus have

W

necessary.

-

is

k)

+ 2k)-2f(c + h) + f(c) (10>22)

We now investigate under what conditions (10.22) yields /"(c). - /(c), U(c + K) = f(c + 2h) - f(c + h),so that U(c) = f(c + A) U(c

+ A) -

C7(c)

=

/(c

+

2A)

-

2/(c

+ A) + /(c)

Let

REAL-VARIABLE THEORY Applying the law

=

hU'(c)

mean

of the

+

h[f'(C

-

h)

as given

/'(c)]

=

by

/(c

407

(10.17) yields

+

-

2/0

+

2/(c

h)

+ /(c)

Applying (10.17) once more yields

<

with c

<

5

c

-

hT(c) =

f(c

+

obtain (10.23),

To

h.

+

neighborhood of x = c. we note that (10.22) results.

in a

2ft)

+

2/(c

+ /(c)

ft)

we assumed is

moreover, f"(x)

If,

(10.23)

that f"(x) exists

continuous at

a:

=

c,

Problems 1.

Derive the results of (10.10)

2.

Show

that

if

x m/n ,

y

= m and n integers, then (IX -^ _____

for

n

any cii

positive integer n. COS x i-

^ that hm

3.

Show

4.

Let w(z)

=

2, 3, 4,

5.

Show that

=

From

x.

by the use

,

2

1

/2

=

~l

by assuming

.n

1

r^-

tan^ 1

.

+X

1

wa

x m/n fl

e

f(x))

(e

z 2)

-7-^ -f

dx

2x

~ dx

find

-^ dx n

^->

/) 2

at z

for

of Leibnitz's rule

ax -j

+

(1

aT

(D

-f a)

with

(n)

f(x)

D ~

=

-r-v

t

fc-o

Find an expression for D [f(x} cos \x]. Prove Leibnitz's rule by mathematical induction. = constant If f(x) Qfor a x b, show that/(x) Show that (10.17) can be written as /(6) -/(a) = n

6. 7.

8. 9.

<

e

<

for

a

(6

-

^

x

^

a)/'(a

b.

+

0(6

-

a)),

1.

10.14. Functions of Two or More Variables. We say that /(x, y) is continuous at x = #o, y = 2/o if for any > there exists a 5 > such

that

-/(zo,

l/(z, y)

whenever

|

+

(#?,

x^,

z

|x

continuous at such that

2

<

2

|y .

2/o| .

.

,

j

\

J/nJ

I

<

c

In general

x^) if for

/*.

t..

If J/2,

5.

2/0)

any

/f/yO 'rO J \<i) Jsty

e

>

f(xi, #2,

.

.

,

#w )

there exists a 6 /.(r\ | Jsft)

I

|

s* V.

ig

>

,

n

whenever

Y

2

\Xi

x$\

Theorems 10.20 and

<

8.

10.21 can be

function of n variables, n

finite.

shown to be true

for

a continuous

ELEMENTS OF PURE AND APPLIED MATHEMATICS

408

An example y but

of a function which is continuous in x and is continuous in not continuous in both x and y is as follows Define

is

:

xy

=

0)

r->0

at

0= /(O,

=

lim

2

7*

0), so

that f(x, y)

is

continuous in x

r->0

The same statement

(0, 0).

y

=

/(O, 0)

Then lim f(x,

+

x2

applies to the variable

since lim f(x, y) does not exist.

However,

y.

This can be seen by letting y

= mx,

x->0

V-0

m

7

and noting that lim

0,

/(x, y)

fTLX^

=

lim -^2 *-+o *

*-() i/-o

m is

+ ,

ra

, 2

=

9

x2

-m

Vfli

t

Since

^

:

arbitrary, lim f(x, y) does not exist.

One follows

defines the partial derivatives of /(x, y) at the point (x

,

2/0)

as

:

/iOo,

Z/o)

=

lim

==:

/AX

Ax

o

A?/)

(10.24)

-

A//

provided these limits

we

Further derivatives can be computed, and

exist.

write

/u = Example

d 2/

,

d 2/

_

UJU

:

d 2/

,

= 2

Consider

10.11.

x 3y /(O, 0)

Then

f /22

dy

/i(0, 0)

-

/i(o, y)

-

/2l(0, 0)

=

^

V

*+l

=

1

*

/>-><)

^

/

0,0

L MLIL^^T /_*n

"

o,

A(o,o)-|

0,'

0)y

-

0,0

=

o^

o

,. lim T A-^O'1

=

A

lim r

r-~V-

dxdy

o)

lim

Ji_kft "-

0,0

fc_*n

n

=

REAL-VARIABLE THEORY a 2/

In this example,

U =

f(x /(*

*(y

Ay)

-f

=

dx dy

0,0 ==

+ +

/(a, -f

+

Ax, y

Ax, y -f

and

if fi%

/2i,

+

f(x, y

-

Ay)

+

/(x

Ax, y)

+ /(z,

y)

/(x, y)

/(x, y 4-

A?/)

=

+

*(y

mean

the law of the

-

A?/)

-

Ax, y) (7

we apply

can show, however, that,

0,0

/2 i.

so that If

We

.

dy dx are continuous, then /i-2 Define Proof.

409

to U,

-

Ay)

Ay) *(y)

we have

p.

Again applying the law of the mean to the variable x constant as far as x is concerned) yields

U = AJ By

Arr

+ ^ Ax,

d 2/(*

y

+

Ay)

B,

Ay can be considered as a

(y -f 0i

dx dy

< ^ < < 62 <

I

<

1

]

interchanging the role of x and y one easily shows that

+

A. Ay A - Ax

6*

AX

>

^^

+

04

A '^)

Equating these two values of U, dividing by Ay

Ay

0,

Ax,

and

3

<

finally allowing

dy dx

z 4-

Az Az

we apply

= = =

+ Ay) + Ax, y + Ay) - /(x, y) - /(x, y + [/(x + Ax, y + Ay)

/(x

the law of the

=

mean

M* +

y

partial

+

A ?/)

Ax

[/(x,

we

-

f(x, y)}

obtain

+

*/(* y

.

y -f Ay)

*

*y)

Av

Oy

Under the assumption

of continuity of the first partial

we have

+

0i

Ax, y _

+

Ay)

y) df(x, __ _ _____

-r-

^2

Ay)

_

2/

>

+

^x

df(x

ei

Ay)j

to both terms above,

* A *>

0<^i<l,0<02<l.

derivatives

first

/(x -f Ax, y

Az

where

0,

Now

derivatives.

with

->

dx dy

provided these partial derivatives are continuous. Example 10.12. Total Differentials. Let z = /(x, y) have continuous

If

Ax

we obtain

0, c 2

as

>

A,

=

Ax

0,

Ay

>

^M

Ax

+

>

0.

a/(x, y) 2

Hence Az becomes

3/M AJ

,

+

1

Ax

+

2

Av

(10 26 ) .

ELEMENTS OF PURE AND APPLIED MATHEMATICS

410 The

If z

principal part of

f(x, y)

where dz

is

3

x, dz

A

=

is

defined to be

do;.

Af ~

dz

=

i/

=

1

ox

TT ay

,

-

0,

and Ax =

+ fdy fdx dx dy

=

called tho total differential of z

jZdx dx

If

dz.

+

Similarly

Ay =

dy, so that

dy %dy

(10.26)

x and y arc functions of

/,

then from

(10.25)

^

=

If

~ and

we

exist,

A

<VC*V'/) " "

A/

since

>

i

0,

e2

*

The

^

,

as A/

if

-

0.

->

,

df dx

_ '

.

.

.

,

,

that

if

A

^= -

{

T~~{~~'

dy dx

z n ), the total differential of u

\= dx-~\

}+dy-jt[

rf((fe)

=

+

rfa:

[dz

rfx

we can

Symbolically, Jt

2

d*

,

^-LaS^ + ai In general

of necessity

>

is

Ax

* 0,

defined as

,

dy

(10.29)

}

where the bracket can represent any differentiate function

ox dy

A/

is

d[

provided

*2

A<

df dy

Remember

u = /(x 1 x 2

operator form of (10 20)

and

^/

,

l

A*

a//

obtain cte

At/-+0. In the general case,

<V(,j[) AV "^

,

A^

d.r

-f

of x

and

d?/

write (10 30) as

y.

In particular,

(10,30)

REAL-VARIABLE THEORY 10.13.

Example

of coordinates given

From

v.

411

Change of Coordinates. Let z = f(x, y), and consider the change x(u, v), y by x y(u, v), so that z depends implicitly on u and

(10.25)

^ Aw

d/(*,

//)

dx

A Aw

Ay Aw

"" a/(j, y) ,

dy

"^ Cl

Ax Aw

A Aw

//

Allowing Aw to approach zero yields

dx du

du

dy dy

Similarly, dz

=

dv

dx df_ dx dv

ay

_^ df_

Symbolically,

where the brackets can represent any function

any variable

of x

r

*\~.

dz\ ~~ _ dx d_ (dz\ ~ '\~ I a-. fk~.l M i dx \dx) du ^i/

\

\

_ ~

d*z_

'dx 2

/dz\ = ~ dx d /dz\ u \dy) du dx \dy) dx

10.14.

dz

dz dx

dr

dx dr

Consider

.

dz

=

dt/

_j_^ dy dr

=

z

dz .

(

dx d

d

dr

osu ln ..,

=

_

z(x, y},

os

fdZ\ (dx)

-. ,

+ .

In particular

/^A

,d]/d_ (dz\

du dy \dy)

d 2 z dy

.

-f

=

x

depending on

du dy\dx) d*z dy dy dx du

dx du

dx dy du

^

f

#//

,

//

"i

d

Example

and y with x and

Of course ~ and ~~ must, in general, be continuous.

dy* dn r

cos

y

0,

=

Then

sin 6

r

dz

Sm

d

Q

dr

~"

,

dydxdr ^

(dZ\ (dy) '

\_

d

dxdydr

ya r j

COS

Let the reader show that dz

^. r

dO

_= d*z

dO*

=s

-r

sm

n

dz

dx dz ,,~

cos

dx

.

f_

r cog

dz n __

dy

-

dz *, r

cos

dy

-

/

sm

r

0\

-

d*z

.

r

sm

+ ,

a 2c

72

[a -

r

2

ax ay

r

~i

cos

sin

^

f)

2

Z

-f 7

ay

; 2

?*

cos

1 I

J

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

412

Problems 1.

A

2.

Let T

=

z

=

i

Let u

x

2

y

2

o

o 2

xz

?/

=

x

,

,

=

v

sin

y

e',

x

-j-

r,

=

2x

oZJ

3.

Let x

^

-

^ a#

T/.

cos

r sin

<f>,

=

i

=

y

dx dy dx --i and dy dv du du dv *

2x

is it

Tr

,

>

n

=

,

Why

etc.

2y ay

findj

.

-f

du 7 du

=

by two methods

Find

t.

dx du

J

Hint

>

O?/ - -

rt

2?/

>

du

=

true that ?-

0?

dt>

sin p, 2

r sin

r

cos

If

0.

V =

V(x,

y, 2),

show

that

,*!r

*!i 2

*!r

+ ^

d2 2

a?/

dV

'

4.

Let

/(xi,

X2

.

,

.

.

d

d

x n ) be homogeneous of degree

,

Differentiate with respect to

set

/,

=

t

1,

and prove a

k,

that

is,

due to Euler,

result

*/

6. 6.

Prove (10.31) by mathematical induction. x n ) is homogeneous of degree k x2

If f(xi,

.

,

.

.

,

(see

Prob.

4),

show by mathe-

matical induction that k(k

-

-

1)

(k

-

+

p

10.15. Implicit -function

THEOREM

'

y

=

?/o,

?

dx

say, for \x

=

x

,

y

=

yo satisfy F(x, y)

=

0,

and

us

let

*\ T7I

r\ 77T

that F. suppose *^

Theorems

Let x

10.24.

I)

and dy XQ\

<

are continuous in a neighborhood of x &

I,

\y

y<>\

<

1.

^

If

at x

=

XQ,

y

= =

x

',

y<>,

there exists one and only one continuous function of the independent with ?/o = /(ZQ). In variable x, say, y = f(x), such that F(x, /(#)) = other words, if Xi is any number near XQ, there exists a unique ?/i such that /(#!,

2/i)

Proof.

the case

hood

=

It is in this sense that

0.

Assume

<

0.

dF

>

From

at

P(x Q y Q ).

2/0)

,

such that -^

> Q >

'

0,

of y as a function of x.

The same reasoning

continuity of

SF of P(x*>

we speak

Q

applies to

there exists a neighbor-

fixed.

We

can decrease the

REAL-VARIABLE THEORY size of this ~~ \y

l>

2/o|

neighborhood, if necessary, so that it is bounded by \x XQ\ < < 1. In this new neighborhood of x = XQ, y = 2/0, given by

Q >

XQ\

<

0.

Let

\y

IQ,

R >

For any x and

=

F(x, y}

=

Let y

+

-

F(x

<

t,

- ITX )-(

Since eQ

>

0,

we

r

^

>

ox

dF

=

-

?/)

<

2/o

F(x*, y)

IQ

neighborhood.

we have

+ F(x,

y)

-

F(x,

*/)

Then

.

dF( 3

F(z,

in this

dx

\y

IQ,

>

continuous and ^ oy dy

+

'

>

e ^'

xo)

shall

-

1

<

XQ\

\x

2/0)

,

<

e

if \x - x < eQ/R. Simiyo + e) > < eQ//?. From Theorem 10.22 a We have shown that, for any x satisfying

have F(x,

<

e) larly F(x, 2/0 exists such that F(x, y)

\x

F,

be any upper limit of

F<V a. i (X, y 0+t

/-

we have

/o,

satisfying

2/

F(x, y)

y

<

yo\

A//

AJ?

f$J? a; |

413

for

=

\

a-

|o:

0.

?y

We say 0. IQ, a number y exists such that F(x, y) a function of x and write y = /(x), so that F(x /(#)) = 0. Next we show that y is a single-valued function of x. Assume that, XQ\

that y for

< eQ/R ^

is

t

< eQ/R ^

xo any number x\ satisfying x = ?/i and 2/2 such that /'X^i, 2/i) |

and

bers

law of the mean to F(x

>

Since

show that y F(x z

2

2/2)

,

_

=

0, of is

0,

flW

is

=

a continuous function of

=

To

/(x) is single-valued.

we note that

x,

num-

Applying the

0.

yields

that y

s

2/i>

=

7/ 2 )

if

F(x\, yi)

0,

then

_

F(x 2> 2/2) -xi), 2/2)

-

F(X!, .

,

.

+ F(x

2/2)

.dF(x lyyi

-

-

2/2)

l}

i

F^!,

__

+e

2 (y 2

-

=

2/1)

y,))

_

rlJ?

^

and

oy

This

1/2

=

F(XI, y\)

y-z)

necessity

l

Since

} ,

there exist two

IQ,

F(XI,

is

ox

it is

bounded,

obvious that

2/2

*

y\ as # 2

>

x\.

exactly the property of continuity. 10.24 is a special case of Theorem 10.25.

Theorem

THEOREM conditions 1.

F

2.

The

is

10.25.

Let F(XI, z 2

,

.

.

.

,

xn

z)

,

satisfy

the following

:

continuous at the point first

P(a,i,

partial derivatives of

F

a2

. ,

.

exist at

.

P

,

an

,

ZQ).

and are continuous.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

414 3. F(di,

dF 2L

4.

a2

*

dz

an 2

.

.

.

,

=

)

,

,

0.

at P.

used in Theorem 10.24 it can be shown that a neighbora n ) exists such that z = J(x\, x z x n ) and On) With ZQ = /(i, 2 ,X n )) = ,X n ,f(Xi,Xz xn a continuous function in the variables xi, #2,

By the method hood

of Po(ai, a 2

F(Xi,Xt,

.

.

.

.

.

.

.

,

,

,

.

.

Moreover 2 is For the case F(x,

f (a?,

.

,

.

,

.

.

.

.

9

.

,

>

.

.

,

/y,

?y,

=

2)

-

2)

=

0, z

/<>,

/Or,

//),

F(a,

=

6, c)

z

if

=

c)

fe,

=

we have

/(.r,

/y)

so that

F(x,

-

y, z}

F(a,

+ F(a, y, z) -

y, z)

F(a,

b, z)

+ F(a, 6, z) ~

F(a,

=

6, c)

Applying the law of the mean to each difference yields

^ ^ ^ - a)

dF (

+

e ^x

~

a )>

^(^ b + ^_M oj

y> z ^ 4. -h

.

y

=

6,

'

v

1

If

^(?y

-

ft),

g)

/

dz

we have lim

A^

=

=

^

lim

7~Z~^

=

~~

air /a*

(10.35)

Similarly

Example have F(l, z as

1,

fy Consider F(x,

10.15. 0)

-

0,

and ~-

a function of x and

=

xe*

+

y

+

xe*

y, z]

at

7*

** f(x, y), for a y, z

yz

At the point

y.

(I, 1, 0).

Thus

neighborhood of

(1, 1)

F(x,

y, z)

such that

= xe' -f y, so that = = e* we use (10.35). Thus To compute oz ox ox ox Notice that we do not claim to solve for z explicitly in terms of x and y. t

THEOREM

Let F\{x,

10.26.

involving the variables are satisfied: 1.

FI(XQ, UQ, VQ)

=

0,

u, v)

(x, u, v).

(1,

-

0)

1,

we

defines

=/(!,!). xe

7~\~~

y

= be two equations that the following conditions Suppose =

^2(^0, Mo, VQ)

0,

=

F%(x, u, v)

0.

FI and F% have continuous first partial derivatives for a neighborhood of (XQ; MO, ^o). Of necessity FI and F% are continuous. 3. The Jacobian of FI and F 2 relative to u and v, written 2.

% 77T

or

does not vanish at (x^ u$,

VQ).

*\ JTf

i

or

du

dv

du

dv

i

KEAL-VARIABLE THEORY

Under these conditions there

415

one and only one system of con-

exists

tinuous functions

=

u such that F(x, VQ

=

<pi(x),

=

<? 2 (x))

0,

F 2 (x,

with UQ

s=

<f>i(x), <? 2 (x))

=

<PI(X O ),

^> 2 (Xo).

If

Proof.

we take

2

^

-^

some neighborhood hood

of (XG; UQ,

^

on

of (x

WG,

;

2

or

-

dv

we note

Also

0.

2

then

7* 0,

./

that J

^

0.

is

continuous, so that J

and, moreover, -^-

*>o),

ed

v

?^,

dFi

From

u.

2

constant

for a neighbor-

dFi

F\(x, u,

x = constant = constant

"i

= constant

= constant

as a unique

/>

=

Hence

0.

v), v

=

/(x, u)

i

constant

|x

r_

x = constant v constant

for

5^

so that

,

y

^

10.25 we can solve for such that F%(x, u, /(x, u)) /(x, u),

=

by considering F as a function of .r and we note that FI is a function of x and

i

loss of generality

From Theorem

VQ).

function of x and u,

di/

Without

oifVdw

r- constant -constant L i

^rr /n OP %/ OV

so that from

Theorem

dF 2/dv Since

0,

we can

=

v

/(x, u)

=

/(x,

Example (x,

i/,

10.16.

w, v)

.

If

=

^(x)) .

,

x

ss ^>(w, v)

in the

==

,

u, v)

^/

=

Q.E.D.

=

0,

<PI(X), <p\(x)

Theorem

F 2 (xi,

x2

.

=*

0.

We

l)u

IhT

.

,

same manner, making use

f(u, v}, y

=

u

for

<p 2 (x).

xn

. ,

The proof proceeds

2

=

solve FI(X, u, /(x, u))

consider FI(XI, x 2

F

ifa **>/(*>"))

we have

Hence unique. 10.26 holds if we

. ,

of

10.24

xn

,

u, v)

Theorem

=

0.

10.25.

then Fi(x, y, u, v) s /(w, v) x = 0, can solve for u and v as functions of x and

<f>(u, v),

i/

provided

du

dv

dx du

d<p

d<p

dy

dy_

du

dv

du

dv

dx dv

(10.36)

du along with the condition that the

first

dv

partials of x

and y be continuous.

If

x

=

r

cos

0,

ELEMENTS OF PURE AND APPLIED MATHEMATICS

416 y

=

then

r sin 0,

J

For

we can

r 7^

any point

=

o

(

^

)

solve for r

cos

ro

r

=

0o, 2/0

dx

dx

dr

d0

cos

cty

dy

sin

dr

d6

=

and

=

(x* -f

uniquely

fo sin 9o.

cos

m terms of x and y in the neighborhood of

Indeed,

=

2 2/

r sin r

)*

tan- 1 ^ x

-TT

<

^

TT

mathematical induction one can extend the results of Theorem We state Theorem 10.27 but give no proof. x m ?<i, y>2, THEOREM 10.27. Let f = f (xi, x 2 ^n), = 1,2, n satisfy the following requirements: n has continuous first partial derivatives 1. Each /t i = 1, 2,

By

10.26.

%

i

.

.

.

.

.

t

.

,

,

,

,

.

,

.

.

.

.

.

,

,

at the point P(a\, a 2 1 2. /t = at 7 f =

.

aw

,

,

1, 2,

,

.

61, 6 2

,

.

.

b n ).

.

,

,

n.

.

.

,

at P.

3.

There

one and only one system of continuous functions,

exists

which

a satisfy (2) such that <p t (ai, a 2 Example 10.17. For the coordinate transformation .

we note tion of

.

.

,

F =

l,hat

7/1,

y2

fi(xi, x<2 ,

t

.

,

.

.

,

xn )

,

?y

=

771

,

0, so

=

)

that

6t

,

=

i

we can

1, 2,

.

.

.

,

n.

solve for x as a funct

y n provided dF\

assuming that the first partials are continuous. For the system of linear equations y = a ax a l

1

\a]\

*

we can solve

for x*,

i

=

1, 2,

.

.

.

1, 2,

.

.

.

we have

n,

,

Kl

dx>

so that

i

,

,

n, as

a function of y 1

n

2 ,

t/

,

.

.

.

,

t/

,

provided

(see Sec. 1.2).

THEOREM

10.28.

Let u

t

=

/,(xi,

x2

,

.

.

.

,

x w ),

=

i

1,

2,

.

.

.

,

n,

have continuous first partial derivatives at P(xJ, #, ,x). A necessary and sufficient condition that a functional relationship of the form wn) s exist is that F(UI, u<t, .

.

.

.

.

.

,

(10.37)

REAL-VARIABLE THEORY Proof.

First

assume that F(UI, u z

.

,

.

.

417

=

un)

,

Then

0.

n

dF ~

5*

V

dF du

2,

a

=

Since

tions in the

n

"

10 ...,n J-1,2,

n

*

(10.38)

can be looked upon as a linear system of n equa-

quantities,

-dF a

=

>

du a

1.2.

.

.

dF du a

.n, with

.

^

for at least

one a of necessity, z*.

=

(10.39) has been shown to be a necessary condition functional for the existence of a relationship involving the u^ i 1, 2,

Hence

(see Sec. 1.4).

.

.

. ,

n.

Why

Conversely,

dF is

let

^

-

for at least

one a?

us assume that (10.39) holds, so that

dx l

dx<2

a/2

a/_2

dx n

=

(10.39)

dfn

dx n

For the sake

of simplicity

we assume that the minor

of

^

uX n

does not

vanish, so that 3/1

d/2

<tf*

a/2

dXi

8X2

dXn_i

(10.40)

dx n -i

Now

let 1/1 1/2

= =

Ui U%

= =

/i(Xi, X2,

.

.

,

Xn)

,

Xn)

.

/2(Xi, X2,

(10.4

n_l /i i

yn

= W n -l = ff ^n

/n-l(Xi,

X2

.

,

.

.

,

Xn )

ELEMENTS OF PURE AND APPLIED MATHEMATICS

418

From

and

(10.40)

we have

(10.41)

a/i

a/i

6x2 dfr

a/2

(y\, y*,

j

y*\ Xn)

>

\Xi, Xt,

.

.

.

,

dx n

=

dfn 1

From Theorem tions of the

?/,

10.27 z

,

Xi

=

=

1

,

we can 2,

?t (2/i,

along with Ui

so that

dF T

=

=

w2 =

1/1,

-

.

2/2,

.

,

-

-

.

3n)

,

.

and w n =

,

=

i

2/)

,

i

,

1, 2,

.

.

,

=

^(2/1,

=

^n-i

1, 2,

.

.

-

2/2,

.

,

,

n

From

2/n-i [see (10.41)].

Thus

0.

10.18.

Consider

u - x u, *>

Tt is

(10.42)

2/n)

1/2,

functional relationship predicted

Example

n, as func-

.

that

n, so

,

2/2,

Un = /n (Xi, X 2

and

solve for the X T .

.

.

obvious that G(u,

v,

w)

+

v,

w\

2/>

z/

by the theorem. y

+

z,

v

1

1

x

y

Q.E.D. z,

w =

2x.

1

200 -1

1

**u+v

-1

wzzQ. Problems

m

1.

Let x

2.

In Prob.

r sin

1

cos

solve

$>,

=

y

r, 0, ^>

in

r sin 6 sin

terms of

<p t

x, y,

z

z.

r cos

0.

(? is

Show

that

Then

the

REAL-VARIABLE THEORY

in

3.

Let u

=

4.

For J

j&

terms of

u,

=* xyz, v

+

xy

+

yz

in Prob. 3 solve for

w since /

v,

du /i/l.

_ ~

,

6.

Let u

=

x -f

?y

-f

2,

^

that

we can

solve for

x, y, z

differentiation with respect to w,

dz

dy ).

,

.

(

)-

dz <)?/ __ -!___

__

Solve this linear system for

dx )- + .

g

(v

Show

~

xy

-. O . dw _ dx _ _ __ dv

z.

/)/9

.A/JJ

yz

Thi

+

y

Hint: Assuming that

.

du

,1ldx

_ ~

+

x

***

we have, upon implicit

7* 0,

l

w

zx,

419

du

=

x?/

+

Find a functional relationship between 3 2 rr a* 6. Consider a function of (x the Hessian of F by

2/

,

Show that

2 -

u, v, w.

1

,

+2

2

-f-

,//2

.

,

,

.r

M

1

),

say,

/(a;

,

x2

,

.

.

,

x n ).

We define

ay Under the coordinate transformation .

.

.

,

y

n

Show

).

dy

The Riemann

10.16.

1

dx

Integral.

f(x) defined over the range a

but we desire

r

a

=

the function / becomes F(y l

t

y

z ,

that

\f(x)\

<

A.

We

1

Let us consider any bounded function b. f(x) need not be continuous,

^ x ^

subdivide the range

(a,

into

6)

n

arbi-

trary subdivisions,

a

=

#o

<

x\

<

<

Xn

=

and form the sum (10.43)

where

M

%

is

the

supremum

of f(x) for the interval x% -i

^

x

^

xt

.

The

sum Sn

obviously will depend on the manner in which we subdivide the interval (a, 6). The infemum of all such sums obtained in this manner is called the upper Darboux integral of /(#), written

&

(10.44)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

420

Sn

Since

A (b

<=

a)

we can form

Similarly,

,

the inf emum

the

=

sn

m

with

^

the

infemum

Darboux integral

m^x,

bounded functions.

exists for all

-

(10.45)

ov_i)

for the interval x % -\

of f(x)

sums obtained

of all such

supremum

S

sum

in this

manner

^ is

x

^

of /(#), written

x)dx

(10.46)

The reader can easily verify that S ^ S. If S S, we say that /(x) is Riemann-integrable we write l Ja

1.

=

n

=

c(x

ov_i)

t

2.

c(b

2/n

-

<

-

-^

i

Sn ~

The infemum be

=

Y

c

fa

Xi-i)

=

c(fe

a)

a)

^

i/n

<

i

i

(

of all

=

^

x

cdx.

I Ja

We

1.

<

-

l

\-"

1.

1 2

<

consider the subdivision

l/n

Then

+ l)_l/i Vf-*(n2^ 2

V

sums obtained by equally spaced

|

divisions

^ is

seen to

Similarly

^.

The supremum of all such sums is for all

manners

1

x dx 3.

f(x) is jft-integrable.

a)

S = 8 = c(b Let/(z) = x for

<

which

T-l

so that

<

for

(10.47)

n

Y t=l

6'

and

constant n

Sn =

f(x)dx

some conditions = c. Then

investigate

Let/(;r)

(7^-integrable)

b

=

We now

The

av_i.

called the lower

A

=

.

Let the reader prove that S

= S =

-g-

of subdivisions (not necessarily equally spaced), so that

^.

continuous function

is

always R-integrable for the finite range

(a, b).

REAL-VARIABLE THEORY Since f(x)

Proof.

is

^

a

Hence

for

>

any

it is

continuous,

x

421

uniformly continuous for

^

b

^

there exists a finite subdivision of a

that the difference between the interval of the subdivision

\S n

~

supremum and infernum

is less

sn

\

<

than e/(6

-

(b

a).

r-^~d =

a)

x

^

of f(x)

6 such

on any

This yields

6

$ n < e, |& necessary we subdivide further so that |*S Sn| < e. can this be done? One must recall the definitions and properties This is left for the reader. Hence of the supremum and infemum. If

Why

<

S\

\S

S =

S

Since

3e.

=

0,

can be chosen arbitrarily small, of necessity It must be remembered that *S and are

e

Q.E.D.

.

numbers.

fixed

Any bounded monotonic

increasing or decreasing function is R-iutegrable (see Prob. 10). 5. An example of a function which is not /Mntegrable is as follows: 4.

f(x)

=

for x irrational, f(x)

1

=

We

^

for x rational,

show that 0. 1, S list some properties of the Riemann

reader

integral.

x

1.

:g

It is

Let the

assumed that

the integrals under discussion exist.

fcf(x)dx Ja

(1)

b

=

c

f(x)dx

[ Ja

b

fa f(x)dx= ff(x)dx b = f f(x) dx r f(x) dx +

(2) (3)

Ja

Ja

b

f

f(x)

Jc

dx

b

(4)

lif(x)

^

"

(5)

(f(x)

[ Ja

0,

+

<p(x)}

[ Ja

dx

=

\f(x)\

dx

^

f f(x)

Ja

b

ff(x) Ja

for b

dx

+

^

a.

b

( Ja

*(x) dx

dx,b^a b

2

^ f' [fa v (x)f(x) dx]

(7)

^

b

b

(6)

then f f(x) dx Ja

<p\x) dx

fa

f*(x)

dx

Proof. b

l Ja

\fr,(x)

+ /(x)]

2

dx

=

b

X 2 f <p\x) dx Ja

+ for 6

>

+

a.

25X

If

+

b

2X f Ja

<p(x)f(x)

dx

+

b

( Ja

p(x) dx

= AX 2 + 2J5X + C ^ for all real values = has either two equal real roots or no C y

^

of X, then real roots.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

422

B - 4 AC ^ 2

Of necessity

which yields

0,

2

<p(x)f(x)

[f* (8)

r/ie Firs/

(9)

^ M\b

f(x) dx

/

Ja

on the range a

^

x

dx

<p*(x)

f' /

2

(x)

dx

(10.48)

M

= suprenium of f(x) for a fg x g 6 Mean for Integrals. If /(x) is continuous

a|,

Theorem of

^

g I*

dx]

the

then

ft,

b

-

[ f(x)dx

-

(6

a

a)/(f)

g

^

fr

y

The proof (10)

We (a, 6),

of (9) is left as

['f(x)dx Ja

an exercise

for the reader (see Prob. 8).

=

should like to emphasize that, we can write /

if

b

=

f

f(x)dx

Ja

is

f(x)

on the range

/t'-integrable

b

=

f f(t)dt

Ja

depend on the variable which is Of course 7 depends on the upper arid lower limits, Moreover / does depend on the form of f(x). b and a, respectively. Once f(x) is chosen, however, we need not use x as the variable of intesince the value of the integral does not

used to describe /(x)

If /(x) is 72-integrable for

gration. for a

^

.

^

x

Each value

6.

the range

(a, 6),

then

\

Ja

of x determines a value of

I

Ja

f(f) dt exists

f(f) dt,

so that

a single- valued function, F(x) can be defined by t

=

F(x)

We

could write F(x)

(10.49).

From

+

F(.r

=

A.r,)

=

(10.49)

but confusion

f(x) dx,

/

Ja

(10.49)

f'f(t)dt

Ja

is

avoided

if

we use

we Have

r^

X

f(t)dt

Ja

and F(x

+

Ax)

-

F(x)

=

+

[' Ja

"f()

-

dt

dt

r/(t)

Ja

dt

Since

/(.r) is

bounded

for a

|F(x

Hence F(x)

is

g

+

x

^

Ax)

6,

-

continuous since F(x

(10 50) *

we have F(x)|

+

Ax)

^ >

^.

Ax

F(x) as Ax ~>

(10.51) 0.

REAL-VARIABLE THEORY

From

(10.50)

we note

+

F(x [see (9) above],

-

Ax)

that

F(x)

provided /(x)

dx

= /() Ax is

x

^

continuous on

^

(x,

Ax

AS-+O

x

+

x

+

Ax Hence

Ax).

AT-+O

=

.. "X 7ff(f)dt a

We

423

(10.52)

f(.r)

We

can obtain (10.52) without recourse to the law of the mean.

have W(v \**

*

I

x

1^

F(^\ \Js)

AT*"| LA^/y

l

Az

/*z-f

1 A

j>/ ,_\

//j\

/

x + Az

[/(O

If f(f) is

Choose

continuous at x|

<

F(x

+

|

6

t

=

x,

we have

c

-

/(x)]

/(x)|

<

r/

/

\

^j

A for

\t

x\

<

8.

so that

Aa;)

-

F(x)

-

Ax Since

\f(t}

A

jj

1

/(a-)

r

can be chosen arbitrarily small, we note that

=

F'(x)

lim

We obtain the fundamental theorem of the integral calculus as follows Let G(x) be any function whose derivative is/(x), /(x) continuous. Then = F'(x) for a g x ^ 6. From Prob. 8, Sec, 10.13, we note that (?'(x) :

=

F(x)

For x

For x

G(x)

=

=

+

C,

C =

is

Hence

+C =

so that

a we have G(d)

6

we

f(x)

;

G(x)

~G(a) =

G(b)

-G(a) =

obtain

Hence, to evaluate tive

constant.

/

Ja

/(x) dx,

we

b

( f(x}dx

(10.53)

ya

find

any function G(x) whose deriva-

the difference between G(b) and G(a) yields

/

Ja

/(x) dx.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

424

Problems n 1.

From

+

(k

-

2

I)

fc

2

2k

+

-

n]

n

Y

we have

1

K*

+

*)

2

-

Afl

-

w

+

2

Y

** so

n

V

that

k

=

+

^[(n

and show that

I)

-

-

2

fr

2

1

=

+

w(w

-

+

I)

3

(A;

-

n(n

+

l)(2n -f

1)

3

/c

Consider

0/2.

+

3A: 2

3A:

+

1

*-i 2.

Without recourse to the fundamental theorem

x 2 dx

/

3.

-g

show

^

that/(.c)

=

(a, 6), b

show that

0,

^

x

>

b

6.

a,

show that

1

and

if

f(t) dt

/

Ja

=

for all x

on

0.

continuous on

^

/(j-)^(o-)

/

^

continuous for a

If f(x) is

fb

of the integral calculus

by making use of the result of Prob.

If /(x) is

(a, b),

4.

=

>

=

/(a-)

and

a,

on

if,

for

any continuous function

^(rr),

(a, b).

Ja 6.

If /(x) is

^

continuous for a

^

x

6,

/>

>

a,

and

^

if /(.r)

on

(a, b),

b

= f f(x)dx

Ja

show that/Cc) 6.

^

Let

a?

^

/(a;)

= = 1

on

(a, b).

^

for

x

^

5, /(.T)

contradict (10.52)? 7. Consider the interval tional, /(x)

=

I/?

x

if

<p(x)

^ x f L = p/(?, p and g ^ 1.

b

( /(x)*>(x) dx

Ja 9.

^ < x g

and consider

1,

not exist at x

= ^?

Does

If /'(x)

and

-

b

f Ja

f(x)<f>'(x)

dx

=

^

/(

^(z") are 72-mtegrable,

this

Define /(x) as follows: /(x) = if x is irraShow that /(x) is integers in lowest form.

on ^ x Prove the first theorem of the mean for integrals. If /(x) is continuous on 72-mtegrable on (a, 6), and <p(x) of the same sign on (a, 6), then

72-integrable 8.

2 for

Why does F'(x)

Obtain a graph of FOr).

1.

=

^

a

*>(x) rfx

Ja

(a, b),

^

show that 6

/(x)^(x)

a

-

-

r vW(x) dx

Ja

/(a)^(a)

-

b

f v(x)f\x) dx

Jo

Equation (10.54) represents an integration by parts. 10. Prove that any bounded monotonic function is -R-integrable.

(10.54)

REAL-VARIABLE THEORY

b

=

F(t)

Let us consider the integral

Parameter.

10.17. Integrals with a

425

U ^

f f( x t)dx ,

Ja

^

t

(10.55)

ti

integrated with respect to x, what remains depends on the ^ i ^ t\- Of course F(t) depends also on a and b, but for parameter /, the present we assume that these numbers are fixed and finite. For

After f(x,

i) is

tQ

example,

is

a well-defined function of

We now We desire

t

for

t

>

0.

determine some conditions for which F(f)

\irnF (t)

=

F(t)

=

If

-

and

exists

is

at

x for

U ^

i

=

t\,

be continuous

b

l Ja

t-St

will

'

f(x,

t}

dx

(10.56)

a bounded /f-mtegrable function with respect to

then b

[f(x,t)

-f(x,l)]dx

^ M\t -

= \t-t\

t

(10.57)

From (10.57) we see /(#, t). by applying the law of the mean to/(x, t) that lim F(t) = F(t). A less stringent condition which does not imply t->~t

the existence of -^-

and which enables us to prove the continuity

be uniformly continuous in t with respect to x 6 > 0, 5 independent of x, such that whenever \t 6. One notes immediately that t\ <

F(t) is as follows: Let/(x,

so that for \f(Xj t)

any

f(x,

e

T)\

> <

t)

there exists a

e

-

\F(t)

b

F(t)\

^

(

\dx\

=

e\b

-

a\

can be chosen arbitrarily small, we see that F(i) > F(t) as Of more particular interest is the possibility of showing that

Since

c

dF(t)

= ~

t

> t.

(

dt

Let us assume that

of

e

-

'

is

continuous for a

^

a:

g

6

/ ;

^

^

^

<i-

We

ELEMENTS OF PURE AND APPLIED MATHEMATICS

426 define 0(t)

by

G(t) dt

/

b

= f

f

Then

Jto

f df(x J /

Jto

Ja

P

=

/

The interchange is

|<|

<

obvious that 2

T

,

/(*,

*o)]

=

f

1

/ ./Jo

dx (10.59)

)

From

can be justified since the

we obtain

(10.52)

.

~(

We

consider

=

F(t)

It is

F(i

b

I

of the order of integration

c?^

10.19.

f

=

dt

7a

-

[M

continuous. dF(t)

Example

dx

O*

= F(0 integrand

t} J

\!

/

~-

t(>nn:r

(e

)

so that

=

*">**

j

cos x

^

f'

/v

-j7

/

dt

tosj

(is

\t\

< T

continuous

\Q

cos x e1

COB x

(10.60)

in x

and

t

^

for

x

2S TT,

dj

Jo

It is easy to justify a further differentiation so that

d*!F_

_/""., C S a;e (oos*^,. 8a!

""

rf?

Jo /

F(t)

F(t)

sin 2 x

+ /sin Jo

xd

e*

(

08 *

T

dx

eto BX -

1

7

-o

*

Bessel's differential equation (see Sec. 5.13) for z

=

i<,

we have

-p

-f

-

w?

-^-

0,

which

]

/

\<

is

/ I

cos x

,

e*

CU8 x

dx

Jo

n =

satisfied

is

-f -j^ -j^ a z 2 oz

by F(t)

-f

w

0.

If

of (10.60).

Problems 1.

Show

that f J

/6 1

tive integer.

cos /x dx

dx

1

(

/"6

(1/0 (sin

6<

sin a<)

compute

/

JO

x2

''

cos to

(/x,

p a

posi-

REAL-VARIABLE THEORY Show

3.

427

that

x dx

sin

~

a cos

_

+a

a;

< a

2

1

2

a

^

= f y*(fl

Given F(t)

4.

for certain rost notions

/(*,

on

dt,

-l^-'--

<

show that

^(/) ? ^(/).

By two methods show that

5.

2

f'

J 10.18. a:

>

Improper and

=

0, /(O)

1, is

As

>

e

>

0, F(e)

1,

for

we

<

4< 3 4- 3/ 2

-

4/

The function

Integrals.

^

are not certain that

^

e

g

x

x

^

1.

it is

=

f(x)

or*,

Since /(x)

is

/Mntegrable on

we have

1,

2 and we define

r^. Jo V% We

-

dx

Infinite

^

x

However,

this range.

+

well defined on the interval

^

not bounded on

(2x

=

lim r:2

r ^

*

=

v*

notice that /(O) could have been defined in

2

any way we please withri

out affecting the value of the improper integral

({

/

x -

Let the reader

1

show that the improper

A

c

>

If for

1.

/

e-0 ^ a + >0 2.

f dx Jo % /

fails

to exist.

Let/(x) be #-integrable for the range a Define \l/(x) by

Practical Rule.

for all

lim

integral

0.

11

<

1

we have

<

\\l/(x)\

M

=

constant for a

+ ^ e

x

f(x) dx exists.

If f or

jit

^

1

we have

\$(x)\

> m >

lim f

fails to exist.

/(x)

for a

dx

g

x

^

f),

then

^

x

b,

g

6,

then

ELEMENTS OF PURE AND APPLIED MATHEMATICS

428

These results are fa p

that lim

-o

If f(x) is

-

-,

/

(x

J a+t

exists

-r-

if

<

\i

and

1

fails to exist if

b

^

=

f f(x) dx

^

x C

~(

provided these limits independently.

However,

ra

We consider

f(x)

= (a ^ x ^

xY

important that

*(a -f

so

.r)*,

= i

/

Ja

f(jr)

dx

dx

_ ~

r ^ i

yo

1

+ ,

o

=

g

we say that f(x)

From

I

^

Ja

f(x) dx

x^

One

define

(10.62)

approach zero

5

=

limit

(a

may

give us trouble.

=

x)*f(x)

(a

+

x)*

is

TT

g X,

and

is

defined as

X arbitrary. We define (10.63)

[*f(x)dx

Ja

vim tan" 1

-

/

j

l

2

1

7o

v A

cos x dx does not exist.

/

=

If

/

70

TT

-

Ja

\f(x)\

dx

absolutely integrable.

(10.03) \ve note that for

I

or

is

x

X-*

P r dx

r lim x-+

Let the reader verify that exists,

we

For example,

exists.

x-

6,

2

lim

[~f(x)dx= Jo

/

<//O)

Va^^P

follows: Let/(x) be 72-intograble for a

provided this limit

and

called an infinite integral

is

<

c

the integral exists and, actually,

/"a

integral

e

The upper

x2 that

-y/a 2

Jo

The

<

a

Anf

-r=^=-

/

Since M

a.

c,

./

It is

exist.

Jo

for

1.

lim f f(x) dx c+5 d (5-0 &>Q

e>0

bounded

>

/u

b

+

dx

f(x)

J

_>

=

except at x

b

lim I a

Ja

20.

easily sees

a)*

continuous for a

Example 10

One

as an exercise for the reader.

left

any

- P/Cr) Ja f(x)

sees immediately that

<

dx

if

/

Ja

>

e

an XQ

<

dx

for

e

\f(x)\

exists such that

for

X

^

X

X

(10.64)

dx exists then

\

Ja

f(x) dx exists,

since

\[~f(x)dx J

I

A

-A-

f"\f(x)\dx

J -A.

*

simple test for the convergence or existence of

follows: If a function <p(x) exists such that

/

Ja

\<p(x)\

/

Ja

f(x) dx is as

dx converges, and

REAL-VARIABLE THEORY if

>

\<p(x)\

^

for x

|/Cr)|

then

a,

429

We

f(x) dx converges.

\

Ja

leave the

proof of this statement as a simple exercise for the reader. Another simple test for the convergence of an infinite integral

= u(X)KX) -

u(x) dv(x)

we note

if

P

-

u(a)v(d)

lim u(X.)v(K) exists and

if

that,

is

based

From

upon the integration-by-parts formula.

Ja

du(x)

then

r(x) du(x) exists,

/

x-

t>(x)

Ja

dv(x) exists.

/u(x) i

x

sin

,

t,

=

lim

We

10.21

Example

We

1.

r

X

sill

Jir/2

X

cos

X =

f

TWT

i-

lim

^7

X

X-+co

The

dx.

/

origin

need not concern us since

,

have

r

=

dx

I

Now

consider

/*

X

,

,.

arid

cos x)

(

/"X cos---x

,

lim

/

X ->

X

COS

I

- a

I

X2

Jir/2

/W2

S

/"

^r X

JTT/2X ,

X

,

dx exists since x

,

dx

X*

dx

/" /

~.

omce

r exists,

JTT/2X

m

COS x

/

J r /2

2

.

**

Jo exists,

we

see that

/

We now

dx exists

x

Jo

Its value

is

2

consider an infinite integral involving a parameter.

Let

<p(t)

be defined by

^(0

We

f( x

/

t)

i

dx

to

^

^ ^

t

(10.65)

assume that the integral

range for

~

U ^

any

t

^

>

e

of (10.65) exists for each value of our attention on a specific value of t, such that

If

t\.

an XQ

I

The value

of

we

fix

/v yX

/(x,

/g(x)

dx

t.

exists,

then

for all

t

and

We

10.22.

since

/

Ja

i

Example

(10.66)

t

range with respect to t. Let the reader show that == t

X

^

on the

we have

Xv (for a fixed e) may well depend on If, for any e > 0, XQ independent of such that (10.66) holds for all on the ^ <i, we say that the infinite integral converges uniformly

there exists an o

X

<

ds

t

if

f(x,

consider F(l)

t

g(x) t)

=

^

e~x dx

t)\

for

tQ

^

t

^

fa,

and

if

dx converges uniformly, 2

/

e~* cos xt dx.

*

/

\f(x,

\/ir/2 exists,

F(0

Since e~ x *

exists for all

<.

^

x*

\e~

cos

xt\

ELEMENTS OF PURE AND APPLIED MATHEMATICS

430

We now show that if f(x,

X

finite

/

f(x,

t) is

but arbitrary, then t)

X, uniformly continuous in t for a rg x of (10.65) is continuous in t provided

<p(t)

We

dx converges uniformly.

have

t

,

[f(x,t

+

U) -f(x,t)]dx "

+ /x From uniform convergence we have f(x,

From

Sec. 10.17

t

+

we note

X

<

finite.

This

5.

is

any

f(x,

>

that for any e/3

+

*

>

e/3

M) dx

[/Or,

for |A|

for

+ A<) dx -

i

/(*,

A0 -

a

an

t)

such that

dx

>

d

X

dx

/(x,

exists

such that

f(x, 0] cte

the property that

f(x,

/

Ja

t)

dx

is

continuous in

/,

Hence \<p(t

+

A)

<

<p(0l

6

<

for |A| 5, d > 0, so that <p(t) is continuous. It is interesting to find a sufficient condition

Q.E.D. which enables one to write dx

Let the reader

first

show

that,

if

=

G(t)

*

/

converges uni-

ot

7a

(10.67)

formly, then /*<

/v

I

/oo

G(0 provided

^

is

>

^^

^

./fc

^

^=

continuous for

a;

(10.68)

^

a,

^

<

<

^

<i

[see (10.59)].

Dif-

ferentiation of (10.68) leads to (10.67).

Example

10.23.

We

consider F(t) given in /(x,

we have

~ vt

2x6"** sin

2a;<,

t)

To show

e-*

2

Example

10.22.

From

cos 2x<

that

-

/

JO

-

dt

- dx converges uniformly for

REAL-VARIABLE THEORY all

t,

we note

that

dj(xt) dx

I

L

i

d

yo

431 Hence

jo 2

2:ce-*

sin

o r

=

oo

= ~2/

sin 2xJ rfe- x2

/

g" xl cos

yo

=

-2lF(t)

Ae~ iZ =

Integration yields F(t]

e~*

/

9

At

cos 2xt dx

t

=

we have

so that x * cos 2xt dx e~ e ~" '

/o"

Problems

/i

^^

O SJKnw *' bh W that

aH*-

I

^

-

1

7=r_:

I

V(l ~

JO

^

dt exists for

<

1

exists

~

T 2 )(l

a

frX)

/7T/2 3.

Show

that

dx exists

In sin x

/

Jo 4.

Show

that 7r/2

-

tan"

1

.C COS ___

t

6.

Show

/

Let

that

x sin

/

ex

=

/(JT,

xt.

Consider F(y}

10.19.

t

^

all/.

Evaluate this integral

dx

=

xe~ x d(

/

cos P*) exists.

Jo

Show

/o" /-i .

//>-'* sin x dx,

integration.

Jo 7.

(1

converges uniformlv for

^j.

/oo

by contour

r

=

Methods

I

that

f(x

-

*i

*^

>

~

dx,

and show that

We first

of Integration.

consider the special indefinite

integral

In the elementary calculus the reader was told to expand 1 (x 2)" in partial fractions, obtaining

Ax

1

(x

2

+

l)(:r

-

2)

x2

+B +1 (A

2

(.r

+

I)"

1

C '

x

+

-

2

C)x

2

+

(B

- 2A)x + C -2B X

, 9 L)

(10.70)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

432 Since

holds for

(10.70)

C - 2B =

so that

1,

all

A =

one has

x,

-|,

B =

-|,

A + C = 0, B - 2A = 0, C = |. The integral (10.69)

can be written

/dx (

X2

_i_

_

I

~~

~~

_

f)TjT

5

2)

= -

^r lu

x dx

f

"""

+1 (x +

x2

/

5 1)

1 ,

x2

/

-

2

In

dx

2 f

+

tan" x \ o 1

+

~

dx dx__

f

2

/

x

In ^ o

(x

5

1

-

+ The x

2)

constant

Since x 2 results of Sec. 8.11 justify the expansion (10.70). 2 are relatively prime (there is no polynomial of degree ^

divides both x 2

+

1

and x

there exist

2),

+ 1

1 and which

two polynomials P(x\ Q(x)

such that

Thus P(x) = 2

(x

+

C, Q(x)

l)(z

To

2).

= Ax B, and (10.70) results upon division by evaluate the integral

+

x

+

2

(.T

cto

-

1)(X

2)

one notes that x 2

(x

+

~ _

x2

1

5 x2

l)(x"- 2)

+

x

2

__

2

5

1

a:

+

"

1

,

5

1

___ - 2 a:

x

-

2

so that

T

-Z

If

C!T

=

-

i tan-' x

$ In (x*

+

1)

+ f In

(z

-

2)

the zeros of a polynomial P(:r) are known, one can evaluate

(10.71)

$$** provided Q(x)

" P(x) with a k

x

+C

rk

=

is

One

a polynomial.

w^

r^ + ^=7; +

lim

A

i

i

*

p(

and performs a

writes

fc

=

1,

2,

.

.

+ j_

.

*

T=T +

~i~

.

.

.

k

.

.

.

,

n.

One then

series of simple integrations for k

+ ^~=Tn j_

divides Q(z)

=

1, 2,

.

.

by

.

,

n.

REAL-VARIABLE THEORY If

a polynomial such that x

is

P(x)

=

433 a

r is

A;th-fold zero of

P(x)

we have

and

P

=

P'(x)

- r)*Q(oO (x - r^-WOr) +

=

P( X )

Thus

(x

* r)Q'(z)]

-

= provided k > 1. For fc = 1, P'(r) ^ 0, so that P(x) and have no zeros in common if P(x) has no multiple zeros. In this case P(x) and P'(x) are relatively prime so that polynomials A(x)

P'(r)

f

(x)

latter

and B(x)

exist such that

+

A(x)P(x)

To

Q(r) (x

B(x)P

r

(x)

=

1

(10.72)

evaluate an integral of the type

J provided P(x) has no multiple zeros, we make use of (10.72). C(x)

A(x)C(x)

[P(x)]

(P(x)]<-i

""

Thus

B(x)P'(x)C(x)

[P(xf]-~

and f C( x} J fP(x)]" -

i

m has

Since of the

r

=

been reduced by

10.24.

/ " (X (x>)

1

+

rfr

"*"

1

^KI(?)

1

[P(x)]"-'

-"m

we can

1,

(V

+

We

repeat this process until integrals

consider z

~

f Jo

dt 2

(<

(</2)(2<)

1)

z

~

+

l)"

=

1.

f Jo

+

/i(x)

dt

+

2

(<

Integration (t*

From

__J

type (10.71) occur.

Example

since

A^^)

f J [P(x)]"-

tan" x one has / 2 (z) 1

1)'-"

2(n

=

1)(

^ tan" x 1

/2 )

f Jo

1

by parts

I*

-

1

-

l)"-

2 (<

"

l)"

yields

^

f*

+

I)--

+ ^x/(x + z

dt

Repeated applica-

1).

tion of (10.75) yields

^W T

where

/2(x) is

i

\

-

1

3

5

(2n

2-4.6..

a rational function of

j.

.

-

3)

(2n-2)

^^ * + R(X) .

2t di

+

.

,

,

v

ELEMENTS OF PURE AND APPLIED MATHEMATICS

434 If

F(x)

is

a polynomial,

can be written in the canonical form

it

F(x)

with X\j X%, Since

X\

.

.

. ,

Xm

= X,X\

polynomials relatively prime to each other. X%, we have prime to X\ ,

-

is

relatively

X

-

-

X

A(x)Xi+B(x)Xl ..

~Y +

B(x)

.

so that

*\

A(x) ^

,

i

Y ^Y3

^\

2 2

.

.

.

3

^Y mm

-

Continuing this process by noting that X\ one can write

X%

=

I

1

Y 22 ^Y 1^ is

AY mw

relatively

prime to X\

y

'

if f(x) is

also

'

X,

F(x)

'

XZ

X\ The

a polynomial.

evaluation of

dx F(x)

reduced to an evaluation of integrals of the type given by (10.73), which in turn can be reduced to the simpler form given by (10.71). Let R(x, y) be a rational function of x and T/,

is

,

y)

-

'~^r

2 2

(10.76)

MV

y_0 t-0

To

evaluate

TB)

= \/Ax

do:

(10.77)

+

2

= Ax

+

with y #, one notes that the change of variable t B, = (2tdt)/A, x = (t 2 - B)/A, reduces (10.77) to an integral of the type discussed above.

dx

Example

10.25.

To

evaluate

dx / ~^f^

we

let

aad

P -

x

4- 1, 2< dt

/

-

dx, so that (10.78)

(10J8)

becomes

V*i ^.2x^+1+ ^1^-1+ C Vx + + *

to

1

1

REAL-VARIABLE THEORY

We

435

can evaluate

V^nr^+~c) dx

fR(x,

provided

R

so that # 2 == the point (a,

A a2

+ Ba +

ft) is

y

ft

t(x

=

satisfy

of the straight line through the slope. The intersection Ax'2 Bx C is easily seen to

with

a.)

curve

of this straight line with the

y

a,

The equation

C.

=

=

Let x

a rational function.

is

(10.79)

=

2 /y

/

+

+

yield the coordinates

=

y

(B

+

ft

+ 2Aa -

x, T/, and da: are rational functions by the methods discussed above.

10.26.

We

evaluate

/

B

**

0,

C = p

and we choose

,

2

2pt

so that

f

y

1

The

integral (10.79)

p*

is

We

+

/

=

dt

]

[ P J

!

-

From

2 ?/

*

x2

-P ^

^

+

p

2 ,

A -

1,

(10 80)

2

!

In

i

*

p

t

have

p.

-P!--/i

-^L^ +

x \/x 2

80)

one can integrate (10.79)

t,

2

so that &

'-i-7.

of

x + p v-^=^.

x

/

2

'

2-_L~4

Since

Example

(1

20t)t

i

>"P + r

a special case of the following type of integral.

Suppose R(x, y) is a rational function of x and ?/, with y depending If the curve f(x, y) = implicitly on x through the relation f(x, y) = 0. is unicursal in the sense that we can describe the curve parametrically by x = <p(t), y = \l/(t) with v(t) and \(/(t) rational functions of t, then JR(x, y) dx

and

this integral

=

fR(v(t),

WW() dt

(10.81)

can be evaluated by the methods discussed above.

A by

rational function of the trigonometric functions can be integrated the use of the following change of variable :

.

2

=

X tan -

"

rrr^ 2<

=

/I

COS X

(10-82)

ELEMENTS OF PURE AND APPLIED MATHEMATICS

436

For example,

The

elliptic integral of the second kind appears in a natural manner one attempts to find the arc length of an ellipse. Let x a sin <p, b cos <p, ^ v? < 2ir be the parametric representation of an ellipse. y Then ds 2 = dx 2 + dy* = (a 2 cos 2 <p + b 2 sin 2 <p) d<p 2 so that if

,

L = withe 2 = (a 2 6 2 )/a 2 kind is denned as

E The

2 (k,

<

62

a2

.

2

sin 2

The

(p

d<p

elliptic integral of

~ W smT7 d<p

\/l

j

e

\/\

/

if

1

=

<?)

period of a simple

kind, given

<

a

pendulum

\k\

<

1

the second

(10.83)

leads to the elliptic integral of the

first

by |fc|

sn

<

1

(10.84)

Extensive tables can be found in the literature for the evaluation of these

important

elliptic integrals.

Problems Integrate the following:

-f

/"

,

.

10.

^

]

a

y

a sic sin ex

-\-

b cos cz

Let a and

+

for a

bx -f f or

>

0,

a

<

c

ai

>

52

a2

<

52

b cos cz

be real roots of Ax*

+

Bx

-f

Bx

+C-

(a?

+C -

ft)

0.

Show

\!A

:

3:

that

REAL-VARIABLE THEORY and that

=

t

-~~

^A

+

evaluate / \/x*~~- 3x 11.

this result to

Apply

t.

Show that

x dx.

sin

/

as a rational function of

a:

2 dx.

fir/2

=

Let I m

defines

437

yo L2P

=246

with (2p)t

(2p), (2p

+

A

Sequences and Series.

10.20.

Sl, S 2 , S3,

=

1)!!

Jlptl

2

(2p)M 1

+

(2p 5

3

-

+

(2p

l)!l

1).

sequence of constant terms -

-

,

Sn,

.

.

(10.85)

.

simply a set of numbers which can be put into one-to-one correspondence with the positive integers. To completely determine the sequence, we must be given the specific rule for determining the nth is

term, n

=

2, 3,

1,

.

.

.

We may

.

also look

upon the sequence,

of (10.85) as a function defined only over the integers, so that/(n)

n = I, 2, 3, .... DEFINITION 10.26.

number S

The sequence

terms

of

{s n

}

is

{s n },

=

said to converge

sn ,

if

a

exists such that

lim n

for each

Equation (10.86) states that Y

<

sn

= S

(10.80)

* oo

> there exists an integer AT(e) In general, the smaller e is chosen, The existence of an integer jV for e

^ N(e). |iS the greater is the corresponding N. the given e implies that the sequence becomes and remains within an distance of S. such that

sn

\

t

for n

+

e

1 Example 10. 27. It is easy to show that the sequence 2, l-J, 1-J-, 1/n, We have w - 1| = |l/n| < eif n > 1/c, so that N(c) = [l/], converges to the limit 1. where [l/e] is the first integer greater than l/. .

.

.

,

.

.

.

|.s

In most cases we cannot readily determine the limit of the sequence even though the sequence converges. Cauchy obtained a criterion for the convergence of a sequence which does not depend on knowing the limit of the sequence. Cauchy' s Criterion. sn

\s n+p

\

<

e

for

If,

^

n

for

any

N and

all

e

>

p ^

0, 0,

an integer

a unique limit S. This criterion is certainly necessary, for

if

lim s n n

< <

/2 for n e f

fies

p

^

^ N,

\S

orn ^ N, p ^

Cauchy's 0.

0.

criterion.

Hence

s n+p

\

<

e/2

forn

jV exists

the sequence

S

[s n

=

S,

\

such that

converges to

then \S

sn

\

sn

\

*<

N, p

^

0,

so that

\s n+p

-

Conversely, assume that a given sequence satiss n < 1 for n ^ NI, Choose e = 1, so that \s n + p \

\SN I+P

sjyj

<

1

for

p

^

0,

and

all

terms in the sequence

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

438

There are only a finite following SN, are within a unit distance of SN^ number of terms preceding SN V so that the sequence must be bounded. From the Weierstrass-Bolzano theorem at least one limit point exists for the infinite set of numbers exist, say, |.$ n

$n

+p

S <

S and \

T, for n

<

*z

{s n

T.

N, p

Let us assume that two limit points choose e = (T - R)/2 and note that

}.

We ^

This means that after SN the terms

0.

sequence are clustered within an e distance of each other. Let the reader conclude that, if S is a limit point of the set {s n j, then T cannot be a limit point, and conversely. Thus Cauchy's criterion guarantees of the

Km

that the sequence has a unique limit, S, with

n

sn

=

S.

It

should be

oo

>

we showed that the sequence understood that by first considering e = we Then that bounded. deduced was only one limit point could exist. 1

10.28.

Example

we

If

return to the sequence of

Example

.P n(n + p)

we note

24,

that

.

if

n

>

l/, p

^

0, since

p/(p

-f n)

<

Again

1.

n

N

[1/e].

By applying the Weierstrass-Bolzano theorem it is very easy to prove that a bounded monotonic nondecreasing sequence of real numbers always As an example of the use of this result, we consider the converges. sequence

n = 1,2,3, ...

We

show

that the sequence

first

bounded.

is

We

have from Newton's

binomial expansion

Sn

~

l

JL

-

1 l

-h

+ I

~

1

2) 42!~ ^ + 3! ^+ n) 1 1 H- -l-l/*.i:(l-l/n)(l-2/n) + + ....-h 2! 3|

+

U

-

-

"

}

'

'

1

+^ -L

,

so that 5n

<l +

Every term

<l + l+| + ^ + ^+

l+^ + ^+ of the

sequence

is less

than

3.

Next we show that

so that the sequence is monotonic increasing. if a. and ft are any two numbers such that a >

we obtain

=3

'-

sn

>

s-i,

First let us notice that, ft

^

0,

then from

REAL-VARIABLE THEORY

Now

a =

let

+

1

n

l/(n

>

fi

-

- na

an

>

1) n~

=

ft

-

l

(a

+

1

=

ft)

n

1/n,

a^a

439

> 1, so that - n(a - ft)]

yields

=

Sn

(

+

1

\

>

-) n/

_i (

Thus the sequence converges lim

( 1

\

n-+

The

+

1

n

\

~~

=

T )

n

1

to a unique limit, called

+

-

U2

+

=

J

e

>

n

_!

1

I/

=

2.71828

+

Un

e.

n/

infinite series

+

Ui

with u n well defined f or

n =

'

1, 2,

'

.

sn

{

converges to

}

'

(10.87)

can be given a clear meaning

.

n =

i.

the sequence

.

+

We define Si =

reduce the series to a sequence.

If

-

-U:

1, 2, 3,

HI, s 2

HI

+

HI,

if .

.

...

converges to a unique limit $,

S and

=

we .

,

(10.88)

we say

that the series

write 00

S = If

the sequence

fails to

converge,

y

w,

(10.89)

we say that the series diverges. In when we consider the convergence

general, one of three things will occur of a series:

The The The

1.

2. 3.

series converges.

beyond bound, and so diverges properly. and does not converge.

series increases series oscillates

00

Example

_|_

I

-

^) m' j-0

8n

.

.

.

.

.

_|_

The

10.29.

i

(1

.

_|>

.

series

ra +1 )/(l

.

f

-

increases

m m < j

,

wi)

->

7j (1

converges to the limit 1/(1

1,

x

as

w->

.

.

beyond bound

since s n

=

n.

.

The

1

series 1

1

and

+ 1

n+1

since

1 -f

+

1

1

fails

to

= ^ uiy n =

1,

^[1 -f

_j_

converge.

series

m)

does not converge since (_i)n+i _|_ The terms of the sequence oscillate between

.

_j_

The

oo.

m)

l)

(

]

zero.

n

we apply

If

2,

.

.

.

,

the

Cauchy

criterion to the sequence s n

we note that a necessary and

sufficient condition for the con-

ELEMENTS OF PURE AND APPLIED MATHEMATICS

440

vergence of the series y u3

is

that for any

>

c

an integer

N exists such

that

" <

u,

e

(10.90)

n-\-p

n ^

for

>

p

AT, all

=

since s n + P

0,

= Y jI any e >

u jy

>

;=1 equivalent to the statement that for

is

If

we

Rn

we can

an integer

(10.90)

N

exists

n

for

e

^ A

7

write

.7

with

Formula

?v

"

<

such that

sn

y find

=

1

J

we note that

Uj,

an integer

Af

=

l

the series converges

such that

all

R nj

remainders,

for

if

any

>

c

are less than

e

n ^ N.

for

If

for

we apply

any

e

>

(10.90) for a convergent series with

an integer

N exists such

that

\u n+ i\

=

p

<

e

1,

for

we note

n ^ N.

that

Thus

a necessary condition that a series converge is that the nth term of the series tends to zero as n becomes infinite. However, the harmonic series 00

V / i

diverges even though the nth term tends to zero as n becomes

L^mJ

n-i oo

-

Zl

diverges

is

seen

by writing

Yt

n-1

---++--a+ We now positive 1.

consider

1

+

some

1

^

0,

*)

+

i

(i

+ +

i

+

(i

+

i)

+ A) +

practical tests to determine whether a series of

terms converges or

Since u n

+

=

not..

we note that the sequence

u}

sn

,

n =

1.

2,

j-i 3,

...

,

is

monotonic riondecreasing.

If

a constant,

M,

exists

such that

REAL-VARIABLE THEORY

=

sn

Thus the

M for

<

u3

V

series

the sequence

all n,

441

bounded and hence converges.

is

i

)

converges since

-^

=1 n

<! + >+;!-,-... =2

=

,.

;p

for all n. 00

w

the comparison

00

and N

^ un ^

If v n

2.

vn

=

converges, then N n

1

exercise for the reader.

The

u n diverges.

series

n=l 1 ,

2,

.

vn

a

<

1,

is

and

is trivial

^

0,

and

if

\

as an

is left

vn

diverges,

<

=

a diverges, since \/n

>

l/n

1

.

The Integral

Test.

function defined for x

2>

1

Let &(x) ^ be a monotonic nonincreasing Let the reader show that .

^ From

^

/ n

=

3.

un

This

w n converges. 1

00

/

for n

if

Conversely,

OO

then

The proof

of Weierstrass.

test

=

(10.91) the reader can

+

^(2)

+

^(3)

deduce that

<p(n)

}

-

-

+

(10.91)

^>(w)

converges or diverges

l_j

W

according to whether

/

(p(x)

=

l

dx exists or not.

00

The

V

i

)

series

,

a

>

.

seen to converge since

is

1,

(p(x)

=

i

is

mono-

H=l tonic decreasing

and

-

lirn

^oo

/

Jj

xa

-

-

a

for

-

a

>

1

.

\

Suppose that a constant d

D'Alembert's Ratio Test.

4.

=

exists

such that

00

f or

n

^ A we have r

a n+ i/a n

<

If

d.

d

<

1,

From

^

a n converges.

<

d p aAr, ... we note that

a AT +1

nl a^,

aAT +2

<

da N +\

<

d 2aN

.

an

<

ajv

y r

=

.

.

,

,

00

a^ +p 00

rf

r

=

a^/(l

d).

Thus y a n

is

bounded and so con-

ELEMENTS OF PURE AND APPLIED MATHEMATICS

442

00

verges.

>

a n +i/a n

If

^

d

n ^ N, the reader can show that

for

1

a

^ n-l

At times

diverges.

be useful to consider

may

it

?=! = d

lim

<

d

If

(10.92)

^n

oo

TJ

then a n +i/a n < d' < 1 for n ^ JV. Why? Thus the series If d > 1, the series can be shown to diverge. The case indeterminate since examples can be found for which both con-

1,

converges.

=

d

is

1

vergence and divergence exist. 5. The Cauchy nth-root Test.

Assume that lim \/a n = n

f or

^ N we

n

<

have \/a n

<

fc'

<

so that a n

1,

<

1.

Then

n ^ N.

Since

fc

*

n for (fc')

00

00

w

)

L<

(fc')

The reader should

) a n converges. L*

converges, of necessity

n-V

n-l

00

show

that,

=

lim \fa n

if

t

rl

fc

>

1,

then

/ an T?

__

~

^

9

The

diverges.

^^J

QQ

converges since ^/o^

00

series

/

L.^J

^ l

,:TI

^

00

6.

We

Raabe's Test.

consider the series^

an a n

^

,

>

0.

If

n-l

(10.93)

with lim n/(n)

=

0,

then

or

or

^

n

From

1.

a n converges or diverges according as

y +4

n-

n n ~ \n~ a n+

>

1

so that

~

<r

=

1

n fjL or

^

Let n

We

-

AT,

obtain

]\T

+

1,

[na n

AT

(n

+

+

-

2,

fc

1

>

_ (

.

er

-

^

N

J 0.

n

+

(n

.

=

1)

i

[

Assume a

>

we have

(10.93)

lim

o-

1

.

,

+

For n

>

1)

I)a n4-i]

N+

p

we have

I

>

-

n+i

1,

(10.94)

in

(10.94),

and add.

REAL-VARIABLE THEORY

443

N+p-l

<

a n+1

^

? (Na N

- (N +

<

p)a N + p ]

\ k

Na N

N + p-l Since (2/k)NdN

V

a fixed number, the series

is

a n+

is

i

bounded

for

n?N 00

Hence

all p.

N

n ^

^

an

^

bounded and so converges.

is

we have a n +i >

N aN+2

aAT '

o

<

a

If

1,

then for

Thus

+ n +1 ^ N "~ ^ aAr4 n

,

-,-

.

1

"

x

N

^ >

-

so that

00

and

y

a n diverges.

It

can also be shown that N

diverges

ar,

a

if

=

1

.

n-l

V

As an example we consider

lim n

^"

We

^

n2

a W 4-i Since

i

>

=

have

n2

and

o-

=

2

>

1

,

n2

n

the series converges.

The

series

oo

V y

-i di diverges since

(7

=

1.

Of particular interest are those

We

negative terms.

series

with alternating positive and

consider

(10.95) 00

In order that

N

(

l)

n+1 a

n

converge, of necessity, a n

Let us assume further that a n+ i

g

a n for w

=

1, 2,

.

.

.

>

.

as

n

oo.

We show that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

444

First let us note that

the series given by (10.95) converges.

= (i = ai

S*n

+

02)

as)

(a 2

Hence the sequence

+

a4 )

(a 3

+

82, 8*,

.

$ 2n ...

.

.

^ a 2n -i)

(&2n-2

as)

(cu

a 2n )

(a 2w ~i

,

^

a 2n

#1

monotonic nondecreas-

is

,

~"

Moreover ing and bounded, and thus converges to a unique limit S. lim $ 2n +i = S since lim a 2n -fi = ct2n+i, so that by $2n+i = Sz n

+

n

Thus the

assumption.

8

n

oo

oo

alternating series converges to a unique limit

0. 00

00

If

(~ l) M+1 a n converges but y

/ n

-1

n

we say that the

a n diverges,

=

series is

1

oo

^

If

conditionally convergent.

a n converges,

we say that the

alternating

n~l series is absolutely convergent.

Problems

Apply the

1.

integral test to the sei les

>

verges or diverges according as a

or a

1

TT->

y

and show that the

series con-

j-r.

^

1

.

7+4

n+1 - -f

2.

Show

that the series

3.

Show

that a necessary and sufficient condition for the convergence of

^-f^ Z 3

1

--f

l)

(

converges.

n

00

n

that for any

m ^ 4.

n

e

>

N exists such that

an integer

Show

that, for

|r|

<

lim

1,

Show

that lim \l -7

=

r

so that

>

(

,

if

7

+

>

a

-f-

+

1)0(0

Q

Show

1)

.

that the series

,

,

>

w

__ n \/n .

,,

V>

w nl

(2n)!!

(2n

1)!!

- 1-3-5

a(

+

!)(

+

<

n |r

+ DQ8 + + 2)3!

2)/8tf

T"(T -f 1)(7

n Consider the convergence of

-

|r|

and diverges otherwise.

n**l

8.

n |r|

l

1)2'

00

7.

=

"^

"7(7 +

T converges

=

that the series

^

,

"*"

n+ l \r\

converges.

.

L4 n\ n

Show

//m:

0. 00

=

*n\

n-4oo

i

<

oo

__

6.

flm+i|

-

(2n

-

diverges.

L

-

l)!!4n +31 2 = with ^ &n-\- L\ (^fiW;!!

|"(2^

- 2-4-6 1).

,

...

-

(2n)

2) -r

'

'

is

l

^ ^ W

6.

+

a n +i -f a n + 2 H~

an

y =

'

c

for

REAL-VARIABLE THEORY

445 00

Assume a n ^ a n +i >

9.

=

n

for

2, 3,

1,

.

.

.

and further assume that

,

y an n-l

oo

Show

converges. oo

Let

10.

Why

0.

does

-

n diverge?

^/

n-l n

oo

converge to S, y n=

sn

y n

=

that lim na n n-+>

=

1

hm

and show that

=

Define <r n

converge to T.

tn

(

n

*)

/

/

(

1

1

^)> 1

.;

ST.

cr n

00

11.

Show

V

that the series

l)

(

/

n+1 is

^

Lj

Is

conditionally convergent.

it

possible

n=l

would converge

to rearrange the terms of this sequence so that the resulting sequence to TT or any other number? 12.

<

Consider the sequence

ki

<

1,

?

=

1, 2,

.

Then prove that hm l-^

.

.

=

fc

k\,

k^

.

.

.

ku

,

,

.

Show that k l+1 >

.

.

.

k>

,

defined

<

and

by

fc,

1

for

+

|

\

|

(a

+

-

bi)

Ml

(on -h

<

.

.

.

.

S]

+

S)) -h

(S 2

(3

S Z)

+

'

*

'

2, 3,

....

If

show that the > an integer

6,

^ N

forn

c

14. Show that the convergence of a sequence can be vergence of a series. Hint:

10.21.

1, 2,

oo

Consider the sequence of complex numbers, a n b n i, n = 1, the sequence ja n converges to a and the sequence {b n converges to sequence \a n -h ^n*J converges to a + 6i in the sense that for any N(e) exists such that

=

-f k lt

\///l

=

%

1.

13.

Sn

2

k l+ i

<

made

to

+

-

(

Sequences and Series with Variable Terms.

depend on the con-

S n _i)

Let us consider a

sequence of functions /i(x),/,(x),

with /n (x), n = 1, 2, number x = c of the of the

.

.

,fn (x),

.

.

.

(10.96)

defined on the range a ^ x ^ 6. For any interval (a, b) we can investigate the convergence .

.

.

,

sequence of constant terms /i(c),/ s (c),

If

.

.

.

the sequence of (10.97) converges,

,/(c),

.

we can

lim fn (c)

= A

c

.

.

.

(10.97)

write

(10.98)

n

A c is a constant which obviously depends on the number x = c. the sequence of (10.96) converges for all x on the interval (a, 6), we obtain a set of numbers [A x which defines a function of x on (a, ft), where If

\

ELEMENTS OF PURE AND APPLIED MATHEMATICS

446 for to

=

lim fn (x) n

= Ax .

.

hm

.

We

^

x

b

sequence

obvious that

it is

|/n (/)l

we have /,,(()) =

At x -

1.

^

For x

the

consider

(10.99)

limiting function for this example is/(jr) = 0, wJ ), w = 1, 2, 3, /()! with/nCr) - 1/(1

+

1

hm H

>

/(())

= hm

oo

The

0.

1

=

lim

JT

oo

x/(\

.

.

.

,

=

r-^ * nx

+

nx), so that

0.

The

'

we consider the sequence I, we note that

^

1

oo

>

71

=

Inn fn (x) *

j& 0,

If

1.

,

1, 2, 3,

n->

1

.

=

==

>

^ x ^

=

with fn (x)

0, n

lim /n (.r) *

n

oo

whereas, for x

write

*

^

x

,

0.

/n (0)

a

f(x)

We

.

.

10.30.

Example n - 1, 2, n->

Ax

each x there corresponds a unique

limiting function in this latter case

is

n

This function is discontinuous at x by /(x) = OforO < x ^ 1 /(0) = 1 Let the reader graph a few terms of this sequence. Example 10.31. Let us see how rapidly the sequence \x/(\ -f nx) converges to as 0, the convergence occurs immediately since f (0) = x ^ 1. At x f(x) For x ^ we have for M 1, 2, 3, .... defined

;

}

:

fl

-

!/(*)

l

+

> and arbitrary, provided (I (10.100) will hold since l/e > 1/e

c

shall

have

\fn (x)

is

<

its

/x

If

The sequence

10.27.

do.ioo)

*

Thus for n. > 1/c 1/cor n > A I/JT A is the first integer greater than 1/e, we 1

if

for

\fn (x)

important to note that an integer JV jWe say that the sequence ^ 1

is

^

limiting value f(x)

to converge uniformly tof(x)

<

r

1

< * for n ^ N It independent of x for

converges uniformly to

DEFINITION

nx)/x

f(x)\

can be found which

~

=

/(x)|

s

{fn (x)\ defined for a

any >

an integer

e

-f(x)\

<

x

^

t

b is said

(10.101)

all x on (a, b) provided n ^ N Uniform convergence is essentially the following:

holds for

^

N exists such that

.

If

one imagines that

surrounds f(x) throughout the a circular tube of arbitrary radius e > of definition of then uniform the f(x), convergence guarantees that length

N

N

an integer can be found such that, for n ^ fn (x) will also lie inside In other words, all terms of the sequence the tube for all x on (a, b). from the Nth term onward lie inside the e tube throughout the length of the tube. The value of N, in general, depends on c. Since the curves = n are two-dimensional, the tube need only be /n (x), 1, 2, 3, two-dimensional. It is not necessary that /(x), and hence the tube, be .

.

t

.

,

continuous. Example for

-

10.32.

x

-J.

Let us consider the sequence

From

Prob.

4, Sec. 10.20,

jl

x n \, n

we have

that

=

1, 2, 3,

lim x n

.

-

.

.

,

defined

if \x\

|.

REAL-VARIABLE THEORY Thus

J(x)

-

-

lim fn (x) n

lim n

>

(1

|z|

for

1

-

447

^

We

.

wish to deter-

= |x| n < (|-) n have |/n (o?) f(x)\ Hence we can make \fn (x) if we f(x)\ < < e. This can be done if we choose n > Thus the sequence 1, we can choose n ^ 1

We

uniform.

is

our range of definition. choose n sufficiently large so that ( ) n In 3) for In /(ln 2 < < 1. If e ^ since

-

xn )

oo

mine whether the convergence

^ ^

-

is

,

[fn(x)\ converges uniformly to f(x) on the range J- ^ consider the sequence \nxe~ nT Example 10.33.

We

^

x

^-.

denned

\

^ x ^

for

Let

I,

the reader show that f(x)

=

nxe~ n *

lim n

=0

^

x

^

1

*

We show, however, that the sequence does not converge uniformly to f(x). note that

we

^x ^l

=nxe~*

-/(*)!

\fn (x)

First

=

the convergence were uniform, then for e 0.01 we would be able to find an integer arid for all x on such that nxe~ nx < 0.01 for n ^ ^ a: ^ 1. In particular Nxe~ Nx would be less than 0.01 for all x on If we choose x = we have If

N

N

l/N,

(0, 1).

N

N e~ > 0.01, a contradiction. (l/N)e~ (l/N) ^ x ^ 1 verge uniformly to f(x) on the range '

Hence the sequence

l

fails to

con-

.

The

following theorems will emphasize the importance of uniform

convergence

:

THEOREM

Let (fn (x)} be a sequence of continuous functions If the sequence converges uniformly to f(x) on (a, b), then/(#) is continuous on (a, 6). The proof is simple. Let x = c be any point of (a, 6), and choose any 10.29.

^

defined for a

6

>

f(c

^

=

[/(c

6.

From

0.

+

x

-

ft)

/(c)

+

-

h)

fn (c

+ h)] + [U(C +

-

h)

+

fn ( C )]

(fn (c)

^

/(C)]

we have

+

|/.(c

+ A) - /.(c)| +

From uniform convergence we note that an f(x) < e/3 forn ^ N and for all x on \fn(x) \

|/m ( c )

integer (a, b).

N

-

(10.102)

/(c)|

exists

such that

Applying

this result

to (10.102) yields |/(c

Since /N(X) |A|

<

6,

<

5

is

> for

+

h)

~

continuous at x

0.

=

Hence for any

<

<

/(c)|

e

|

c,

>

+

|/^(c

we have a

6

>

+

h)

\/N(C

+

exists

- /*(c)| h)

/AT(C)|

such that

<

|/(c

/3 for

+

h)

Q.E.D. Example 10.33 shows that f(x) can be continuous even though the convergence is not uniform. Uniform convergence is a sufficient condition for continuity to occur, but is not necessary.

/(c)|

6

|&|

8.

ELEMENTS OF PURE AND APPLIED MATHEMATICS

448

THEOREM

Let \fn (x)} be a sequence of continuous functions If the sequence converges uniformly to/(x), then

10.30.

^

defined for a

^

x

6.

b

The proof

dx

f(x)

f Ja

b

=

ous, for

lim

_>

f Ja

__>,

+

/(*)

fn (x) dx

R*(x)

(10.104)

continuous from the previous theorem, R n (x) is the difference of two continuous functions.

also continu-

is

is

it

(10.103)

R n (x) by

of (10.103) proceeds as follows: Define

= Since /(x)

b

=

lim fn (x) dx

f Ja

From

(10.104)

we have b

b

=

f f(x) dx

l Ja

Ja b

or

\[ Ja

f(x)

-

dx

I

b

From uniform convergence we a)

t/(b

an integer all x on (a,

0,

N and for

^

n

>

b

n

^

that

Theorem

Example

10.34.

i

Inn

_

We

b

1

shown

x - 7 dx

-.--.

1+nx

/I

f(x)

\

r \R

f fn (x) dx

Ja

Q.E.D.

= rhm

dx

-1

1

f

/\1

/

(

=

0,

V

<

e

/

fn (x) dx (Converges to

It

should be emphasized

Ja

Example

n^^njo

dx

n (x)\

to be true provided a

consider the sequence of

f / JO

dx

Thus

Ja

10.30 was

R n (x)

b

note that for any e > 0, and hence for such that \R n (x}\ < t/(b a) for

f(x) dx, so that (10.103) results.

f

dx

l Ja

This means that the sequence

N.

R n (x)

exists

b).

-

f f(x) dx

Ja

for

N

f Ja

=

fn (x) dx

f Ja

b

+

fn (x) dx

and

We

10.31

l

.--

b are finite.

have

\ dx i I

1+nx/

which checks the

results of

Theorem 10.30

since the convergence was seen to be uniform. nx Example 10.35. Let the reader show that the sequence \nxe~ *\ does not converge s x ^ 1 but does converge to f(x) 2= 0. uniformly on the range

Now

l

so that

lim n

On

>

*

nxe~ nx dx

f

oo

-^" n ' *

f '^

nxe~ nx dx

f

*

=^ ^

= f

|(1

- -)

f(x) dx

=

'^

the other hand, the sequence \nx/(\ -\- n 2 x 2 ) does not converge uniformly on the ^ x ^ 1 but does converge to f(x) s (), and

range

j

REAL-VARIABLE THEORY Uniform convergence

is

449

a sufficient condition to apply (10.103), but

it is

not a neces-

sary condition.

THEOREM which

Assume

Let {/(#)} be a sequence defined over a

10.31.

known

is

to converge to the constant /(c) at x

=

c,

a

^ x^ b ^ c ^ b.

further that the sequence {/(#)} converges uniformly to g(x)

on (a, b) with/^(x) continuous on (a, 6) for n = 1, 2, .... We show that the sequence \fn (x) converges uniformly to a function f(x] such that /'Or) = g(x) for a ^ x ^ 6, that is, }

=

lim f'n (x) n

=

f'(x)

(

(*X

> oo

Since the sequence 10.30 that

\f'n

lim fn (x)] n

(10.105)

oo

converges uniformly to g(x)

(x)}

we have from

Theorem

g(x)dx

From

=

lim

=

f'n (x)dx

Jc

n -+ x

lim [fn (x)

-

fn (c)}

(10.106)

n __>*

(10.106) the reader can readily deduce that

= f'g(x)dx+f(e) Jc

lim fn (x) n->oo

Thus lim /n (^) =

/(x)

and since

exists,

f(x) = g(x). From /'() = /;() let the reader show that the sequence

g(x)

is

continuous,

we have

+ R n (x)

If

/n (x), n

1,

2, 3,

.

.

.

is

with |B n (a?)| < e for n ^ ^V converges tof(x) uniformly. defined for a ^ x ^ J>, we say that the \fn (x)

}

oo

series

oo

N /n (x) converges at

a:

=

c if

the series of constant terms

n-l

Y

/n (c)

n-l 00

converges, a

^

c

^

b.

We

=

write /(c)

Y

/n(c),

where

n-l

/(c)

=

lim

^-

/4 (c) ffi

00

If

/n(^) converges for all x

^

on

we

(a, 6),

write

n=l n

oo

f(x)

= Y ^.1

/(*) =

n

It

is

convenient to express f(x) as a

lim n

>

V MX)

(10.107)

,^, fc=l

finite series

plus a remainder which

ELEMENTS OF PURE AND APPLIED MATHEMATICS

450

number

obviously contains an infinite

of terms,

R *W with

R n (x) =

fk (x).

DEFINITION on

f(x)

(10.108)

The series

10.28.

(a, b) if

for

any

\R n (x)\

e

>

<

e

^

an integer

n

for

is

fn (x)

N

said to converge uniformly to

such that

exists

a

N,

b

.r

In terms of the Cauchy criterion we note that uniform convergence exists exists such that if for any c > an integer

N

-

\S n+p (x)

<

8 n (x)\

(10.109) n

for

n ^ N,

all

p

>

0,

and

for all x

on

with

(a, 6),

s n (x)

Y

fk(x).

The

results concerning continuity, integration, and differentiation of uniformly convergent sequences apply equally well to any uniformly con-

To prove

series.

vergent

these results, one need only change the series

into a sequence [see (10.88)].

Before constructing a few tests for determining the uniform converun gence of a series we prove a result due to Abel. Let Ui, Uz, be a monotonic nonincreasing sequence of positive terms, u3 +\ ^ u3 a n be any set of numbers, with Let ai, a 2 and m the Uj > 0. .

.

.

,

.

.

.

maximum and minimum,

,

M

,

,

respectively, of the set n

&1,

We

Oi\

&2, #1 H-

-f-

C&2

4" ^3)

tt <

/

)

show that n

mu,

^ Y OM g Mui

(10.110) n

Let

Proof.

a2

=

2

~~ Si,

n

si

a3

= =

ai,

6*3

2

=

$2,

i

+

02,

,

aw

.

.

.

,

=

sn

sn

= N

a t so that ai

=

-,

We write

$ n -i.

n

Z^r\ djUj

=

/

/w

(Sj

Sji)U3

So ==

'

-f

+

(S n

~

n-l)Wn

si,

REAL-VARIABLE THEORY

^

Since u3+ i

u}

=

for j

1,

2,

.

.

.

,

w

w2)

+ M(^

u2)

+

1,

>

and since w n

0,

we note that

+

+

MS)

2

451

71

V 3

3

L,

> =

u s)

m(u z

+

which proves (10.110). A. AbeUs

Test for

Uniform Convergence.

a n w n (x) converges uni-

y r/-l

formly on

(a, 6) if

00

]

N a n converges.

.

n=l 2.

w n (x)

3.

|wi(x)|

> <

and u n (x) ^ u n +i(x), a for all x on (a, &).

Since N

Proof.

g

x

^

^

n

fe,

1.

A;

a converges, an integer AT exists such that rt

n-AT+1

<

c//c

for all

>

p

From

0.

(10.110), Abel's

lemma, we have

N+p a n u n (x)

for all

>

p

0.

Q.E.D.

B. The Weierstrass

M

x)

Assume

Test.

^

~

Ui(x)

|w n (^)|

^

M

n

=

constant, for

00

n =

1,

2,

3,

.

.

. ,

a

^

^

x

6.

(a,

n

\

Proof.

6).

Y

N Af n converges, then

If n

converges uniformly on such that

00

For any

e

>

an integer

exists

M

n

<

for all

p

>

But

C.

Q.E.D.

^ u n (^)' n-l

;

n(^) is

uniformly convergent on

(a, b) if:

w n (x) 1

N

ELEMENTS OF PURE AND APPLIED MATHEMATICS

452 n 1.

u

/

t

M

<

(x)

=

constant, for

'A

all

n and

for all x

on

(a, 6).

ao

^

2.

y n + i(#)

|

uniformly convergent on

v n (x)\ is

(a, 6).

n-l v n (x)

3.

uniformly as n

>

*

QO

.

n

Let

Proof.

= }

s n (z)

u

l

\s n (x)\

(x)

<

M from

(1)

t-1

Now = =

Sn)V n +l

(Sn+i

SnVn+1

+

'

Sn+l

'

*

+

S n+pV n+p

n+p-1

- Sn ^ \

w

|

|

|i'

+

n +l|

n + p-1

Since

uniformly as n

y w (a;)

~

Vr

kn

>

we have

oo ?

Vr+i\

|v w+ i|

<

e/3Af,

|^ n +p|

<

n + p-l

e/3Mforn ^

From

A^I.

V

(2),

~ ^il < e/3Mforn ^ ^

1^

r = n-f 1

A^ equal to the larger of NI, \S n + P

Example

for

<

8

^

^

have

(1)

(2)

K -,!+I

00

Y

we have N,

all

p

>

Q.E.D.

consider

We

R.

2

n^

e

\

We

10.36.

x

- Sn <

N

|

(

a,

+

<p(j

}

a n v n (x) with wi-

=

<

2

/nH

x

1)-

Jn

Zv 00

n

(

+1)

~ R

Z

n-1

>

2.

For

REAL-VARIABLE THEORY

453 00

00

Since

\ n-l

<

verges uniformly for

^

5

^

z

type of series

we apply the

ratio test,

<

=

we note

\x\

We

call

R

vn

\v n+ i

con-

\

<

5

^

x

5*

uniformly

/2.

n > a nx L-t n-O

(10.111)

that the series of (10.111) converges

lim or

^ n-l

the power series

is

P(x)

If

n~* ->

so that

(C), <?(x) converges uniformly for

An important

test states that

#.

n~ x ^ n~ 5 ->

(3)

From

M

n ~^ 8+l) converges, the Weierstrass

if

a nx n lim

(10.112)

the radius of convergence of the

The word "radius"

series.

00

comes

into play

if

we

= Y

consider the complex series P(z)

anz n

,

n-O z

=

x

+ iy, which converges for

<

\z

R.

If

P(x) converges for

n

also converges for

\x\

<

/^,

00

follows that the series Q(x)

it

= }

\a n \x

< R

\x\

n=

The

[see (10.112)].

ratio test fails to tell us anything concerning the oo

points x

=

R, x

=

The convergence

R.

must be examined by the methods

of

Y a n R n and Y a n n0 n=0

of Sec. 10.20 or

for

1

If

<

x

P(x)

10.37.

<

1

.

=

(1)

(3)

n

00

y n\x n=0

V}

R)

(

by other means.

00

Example

end

oo

n

converges only for x

=

0.

(2)

x n converges

y

n=0

xn

converges for

a n x n converges for

\x\

all finite x.

<

R,

it is

very easy to show that

n=

a n x n converges uniformly for

n-O

x\

^ R\ <

fi.

We

have

for

\x\

^

/?i

ELEMENTS OF PURE AND APPLIED MATHEMATICS

454

00

that \a nx n

^

\

=

n

kn|/Zj,

1, 2, 3,

.

.

.

and since }

,

\

a *\R* converges,

n-O

M

Since each term of test yields uniform convergence. continuous, P(x) is continuous inside its region of converLet the reader show that P'(x) has the same radius of conver-

the Weierstrass the series gence.

is

00

Why

gence as P(x).

= Y na nx n- lt!

that P'(x)

is it

n-l

A

result

due to Abel concerning power

Y

If

series

can be stated as follows

:

00

00

a n converges to

Y

then

s,

a nx n

uniformly convergent for

is

n-O

n-O 00

^

x

g

1,

Y

and lim

a nx n

W n

*-!

=

8.

The

proof depends on the result

00

given by

(10.110).

N a n converges, we know that for any n-O

Since

an integer N exists such that |a n + a n +i + ^ a; ^ 1 we know that x w n p ^ 0. For monotonic nonincr easing sequence, so that

all

+ =

^ Hence the

[see (10.110)].

series is

ex"

^

a n +p|

<

.

>

n ^

e f or

0, 1, 2, 3,

,

c

.

.

,

AT, is

a

^

1

c

g

uniformly convergent for

x

oo

sothatPOr) =

Y

a nx n

is

continuous on the interval

^

x

^

Hence

1.

n-O oo

lim P(x)

=

P(l)

= Y

X >1

an

=

For example,

s.

~*

it is

known

that

n

to

(i+*).V(-i)^ n-l

V^ for

\x\

<

1.

Since

-

(_]_)n-fl -

-

>

V"V

converges,

we have

In 2

n-l

=

)

-

/_^]n-fl ----

-

n-l 00

The converse

of Abel's

theorem

is

not true.

If

P(x)

= Y

a nx n con-

n^O 00

verges to s as x -*

1,

we cannot say

that

Y n-l

a n converges to

s.

An

REAL-VARIABLE THEORY

455

00

example due to Tauber

=

as follows: P(x)

is

^

l)

(

=

n n

x

+ x)

1/(1

9

00

and lim P(x) =

However,

.

*->!

^ w-0 }

(

l)

n

is

not convergent.

Problems oo

Show that

1.

-M

1/(1

=

2 )

(-l) n < 2w converges uniformly

/^

for

s

t

<

1.

n-O Integrate,

and show that T 2n-fl

n0 ,.

-^5rn

y (-1)-^ 2n 4^

4

1

00

lim \/ch, w n

If

2.

T

show that P(z] =

exists,

a n (z

y

zo)

n

converges for

^rt

-2c|

|z

The series ^ n=0

3.

?V N

(x) is

said to be absolutely convergent for a

^z ^

b

if

y

nO ao

Z#

n-1 for

a: |

|

^

1

and that S(x)

n (1

2

.

~t~

not uniformly convergent for

is

is

x $)

absolutely convergent

^

\x\

1.

00

4.

Show

Sm

that

)

.

zn

Z^

By considering ^ 3ir/4.

=

x

n

ir/2

is

,

H- 1

show that the

uniformly convergent for series

is

7r/4

<

x

^

3^/4.

not absolutely convergent for T/4

<j

x

5.

Show

that

In

i

/"

/

Jo

=

I

x x

-

,

rfx

=

V) T Lt Jo n0

1

X n In i

/

a?

j dx

V)

Li

n-O

1 r

7

(n

+

rr^2 I)

=

TZ 6

00

6.

7.

Prove that the

series

Prove that

Let S(x)

^

iT^

rrI

~r

z-i is

w x

uniformly convergent for

all x.

.

o

8.

V^ >

v^ >

n-l

I

x x

4

w n (a;) converge uniformly for a

x

&,

and assume further

ELEMENTS OP PUEE AND APPLIED MATHEMATICS

456

eo

that lim

Ln

un (x)

u n (x) not necessarily continuous.

for all n,

If

Ln

/

con-

^-*

a->e

n-1 OB

verges,

show that lim S(x) = X

^

Let a mn

9.

>fi

m

for

=

y^ L n n-1

1, 2, 3,

.

.

.

K exists such that

Show

M, N.

that

00

00

^

}

n

,

1, 2, 3,

.

.

.

,

and suppose a constant

M

N

for all

.

a mn exists and that

*mn n=

Write the elements {a mn

l

lm

1

T/I

as a square array,

=1 n

(10.113)

1

and interpret

(10.113).

00

-

10. If P(z)

y

a n z n and

n-O P(z

+

w) -

r-0

a=0

converges absolutely, that

^ ^ n-0 r =

is,

|a n

T |

Apply (10.114) to E(z)

+

w)

-

|?i;|

-

converges, show that

a s+rz'^

and show that L^

|0|

00

00

P(z

n~r

r

rj

J57(

+

w;)

-

(10.114)

E(z)E(w).

fi

n=0 00

A

11.

series

^ w n (^)

is

said to be

"boundedly convergent"

for the interval

nl 00

a

^

^

x

6

if

y u n (x)

<

M for

converges for

all

x on

(a, 6).

all

x on

The

(a, b)

and

if

a constant

series is said to

M exists such that

be uniformly continuous on

n^l (a, 6)

c

x except for the point c if the series converges uniformly on the intervals a c 8 ^ x ^6, however small 5 may be, 8 > 0. Show that if a series

8,

+

is

REAL-VARIABLE THEORY

457

uniformly convergent on a ^ x ^ b except for a finite number of points and if the series is also boundedly convergent then the series may be integrated term by teflm. 12. For what ranges of x do the following series converge uniformly?

V n-l 13.

Second

Law

Mean for

of the

Y J_

*

+n

L, x*

2

Lf x*n*

n-1

Consider

Integrals (Bonnet).

n-l /

f(x)<f>(x)

dx

} y-o

Let

<p(x)

be positive and mono tonic nonincr easing on

and consider the

(a, 6),

set of

numbers

S Q =f(a)(xi -

-

Si

-

f(a)(xi

a)

a)

,=0 Let

A and B be the minimum and maximum values of the set S

=

i

t,

0, 1, 2,

.

.

.

,

n,

show that

apply Abel's result of (10.110), and

dx Ja If f(x) is

continuous, show that b

f jo

Consider

^>(x)

f(x)<f>(x)

-

P/W ^

^(a)

a

Ja

^(x) monotonic nondecreasing

3* 0,

<p(b)

dx

dx

=

^

^

^

and

6

positive,

and show that

6

v(a)

+

P/(*) dx

^(6)

/W dx

[

(10.115)

A simple example shows that (10.103) does 10.22. Dini's Conditions. not necessarily hold if one of the limits of integration is infinite. Let us /n n = 1, 2, 3, consider the sequence j/n (x) with fn (x) (2x/n*)e~** = f(x) s Oforx ^ 0. ,withx ^0. It is easy to see that lim fn (x) *,

}

.

.

.

n

By

setting f'n (x)

for

x

f or

n ^

^

=

occurs at x

N=1+

limit f(x)

=

for

= n/\/2

[1/e].

x

^

0.

so that |/n (a?)|

/

n-*w JO

/n (x)dx

maximum

^

value of fn (x)

V^A (l/^)

<

1/w

<

Hence the sequence converges uniformly to

On

the other hand,

r

lim

>

one easily shows that the

=

lim n

/

or

/

^e-* ^

JO

!/

"'<fe

=

1

=

dx

=

lim n

>

"

*

I yo

lim /(*) -

do;

=

/

yo

1

c

its

ELEMENTS OF PURE AND APPLIED MATHEMATICS

458

We now

and prove a

state

result

We

due to Dini.

can write

(10.116)

n^ I

n^\ the following conditions are 1. u n (x) is continuous for #

if

=

lim v n (x)

2.

yn

()

fulfilled 2> a,

exists for

:

n = 1, 2, 3, n = 1, 2, 3, u n (t)

Y

3.

u n (x) converges uniformly for a

^

.

.

.

.

.

with

.

.

,

dt

^

x

-Sf ,

X arbitrary but finite.

n-l 00

4.

2,

v n(x)

converges uniformly for x

*

a.

n-l Proof.

From

and

(3)

(2)

we can

oo

/X I

write oo

eo

V\/

f /y

/7 /V* t*nv"k/ ^**k

/)/

V* \

"""

|

/

Lf

L-4

nl

Hence

/V

/

y

\

Wnv*^)

^ dx

=

r lim

T

X

I

V y

/\^_rlim

V

^n\x) dx

y

00

To

prove- (10.116),

y yn() = Lj y

Lj

exists

Y

that lim

X~~>8 n-l u n (x)dx.

\

n-l

n-l

we must show

00

= Y

y n (X)

It is first necessary to

show that

of (2)

and

(4).

From

(4)

(),

since

y

()

Li

Jo.

by making use

yn

n-l

an integer

N

yn

n-l can be found

such that

N

oo

C

n-l

^ 4

n-l

for all

^

X AT

From

(2),

tfn()] the limits.

The

lim

<

v n (X)

c/4 for

=

Z

yn (oo)forw

^

1,2,3,

... ,tf,sothat

Zo, since the limit of a finite

sum

identity

N n-l

=

n-l

n-l

N

N

n-l

n-l

is

^

the

K(X) sum

of

REAL-VARIABLE THEORY

shows that

for

>

any

we can

an integer

find

459

N and then

an XQ such

that

N ^n(oo)

(10.117)

<|

n-l for

X

^

XQ.

By

we have

the same reasoning

N+P

V

i;

n (X)

V

-

n-l

.

n-l

X *z Xi and p a positive both XQ and Xi yields

for

fixed integer.

X larger

Choosing any

<

6

than

(10.118)

n-l

Now (10.118) holds independent of an integer so that for any c >

any

N

all

>

0.

This

^n()

exists.

p

exactly the

is

X since no X appears in (10.118),

exists

Cauchy

such that (10.118) holds for

criterion for convergence, so that

00

/

Hence

for 6/2

N of

an integer NI

~Z" n-l

n-l

Choosing the larger with (10.119) yields

>

(10.117)

and

00

00

n-l

n-l

exists

such that

B

(10.119)

(10.119)

and combining (10.117)

(10.120)

for

X

^

XQ.

Formula

(10.120)

is

just the statement that

lim

Example

We

10.38.

Q.E.D.

consider the Bessel function

n-0 and show that

/

yo

e"*Jo(x) dx

I/ A/2.

The nth term

of the series c""/o(a;)

is

ELEMENTS OF PURE AND APPLIED MATHEMATICS

460 u n (x)

e~*

/

,

>

nlnl

,(-)-

and

and we see that u n (x)

lim *(*) 3_no

Let the reader show that lim r

that e~*J

o(a;)

n converges for

"""

~

vn (

continuous for x

is

0.

X

arbitrary.

By

Also

0.

2 2n n!n!

j\j

-

)

^

the ratio test

it is

easy to determine

K*> |z|

^

X,

Finally

it

remains to show that

00

converges uniformly for x

v n (x)

y

^

If

0.

we

write

n-0

n0

n

we cannot show uniform convergence

since the series of constant terms in (10.121) However, through integration by parts the reader can readily verify that

diverges.

M /o* so that

y n (0)

0,

a series of alternating terms with

v n (x) is

y n

BS

0, 1, 2,

.

.

.

.

For such a

series the

|t>

n +i(:c)|

<

|t>

n (#)|

for x

>

0,

remainder after n terms of the

00

series

v n (x)

y

has the property that

\Rn(x)\

02n + 2 /;

\Vn+l(x)\

? nT/J

.

1 S"l

Since the right-hand side of (10.122) tends to zero as n becomes infinite,

converges uniformly for x

v n (x)

^

0.

Applying (10.116) yields

n-0 /"* e

J

Jo

W

tf:c

-

V

(^D n (2n)

2,

2 2 n!n!

n-0

In the next section (Example 10.40) we shall show that

n-0 For x

-

* 1

we have

/

yo

e~ x Jo(x) dx

-

(10.122)

we note that

REAL-VARIABLE THEORY

461

Problems go

1.

=

Let S(x)

00

y Li

*

n-l 2.

We

>

,

^

a;

3

(x 4- n)

Show

0.

that f JO

S(x) dx

- i 2

y Li

1n*

n-l

wish to determine f(x) such that 1

s.

oo

Assume

=

f(x) 3.

f(x)

=

JQ(X).

a n :c n integrate formally,

y

make

,

Then

Consider fn (x)

your work.

justify

= n

2

use of (10.123), and show that

xe~ nx

n =

,

3,

2,

1,

.

.

.

,

show that

lim n >

/n (x) dx

I

^

^

oo

/

/

jo 4.

lim /n (#) rfa;, and determine which one of Dini's conditions is not fulfilled, M n*x 2 ), n == 1, 2, 3, Consider fn (x) = nx/(l and show that

+

lim

Do

.

.

=

/n (z) dx

/

.

,

f li lim /n (x)

\

JO

da;

n

Dini's conditions hold for this case?

10.23. Taylor Series. such that f'(x),f"(x), We note that, for

Let f(x) be a function denned on c ^ x ^ d n (x) exist on (c, d) with/< >(:r) fl-integrable.

JM c^a^x^dorc^x^a^-d, .

/

.

.

/ (n) (0

* =

/->(0

dt

f

(n

~ l)

(x)

-

f

-

/

(n

- l}

(a)

Ja t* /

dx

yo I

Ja

dx

f

x

/

=

2

/<"- >(z)

(

"- 2 >(a)

-

Ja

/

(

"-"(a)(a;

-

a)

dx

\

Ja

- /<-(a)(* -

a)

-

Continuing this integration process yields X

dx t dx Ja

r

"V-HO

Ja

dt

=

f(x)

- /(a) -

f'(a)(x

- a)

a

"' V

so that

/(re)

=

V /W(

'

3! a)r

(x

^

)

+

'

(10.124) x

.(*)

=

r dx Ja[

Ja

dx

r Ja

x -

f f^(t)dt Ja

ELEMENTS OP PURE AND APPLIED MATHEMATICS

462 If

M

is

any bound

on the interval

of |/ (w) (#)l

* -

r\dx\ r\dx\ i Ja Ja

\R n (x)\

R n (x)

of

r M\dt\

-

=

we have

M

can be obtained

*

\*

7

q|

(10.125)

">

Ja

Ja

Another form

(a, x),

if

we note that F(x) defined by

1

(n

(-*

-

1)!

Ja

yields

=

p>(x)

/*

==:

v*^/

/

7

Q\"f

oj

(^71

by applying the

!

fM(t)(x

^

^

/

'

-

^*^

1)"^ dt

FW(x) =

==

^

Integrating

/""(a;)

x) yields F(x)

(a,

\^/

'

=

Ja

results of Prob. 4, Sec. 10.17.

times over the range

F'(a)

= R n (x),

so that

(10.126)

The

inequality of (10.125) results immediately from (10.126). known that f(x) has derivatives of all orders and if

If it is

shown that lim series

R n (x) =

it

can be

then (10.124) yields the important Taylor-

0,

=

expansion of /(#) about x

a,

00

/(*)

must be emphasized that > as n f(x) unless R n (x) It

(10.127) occurs

if

a

=

0,

=

origin.

is

(r)

()

^r^

(10.127) cannot be used as *

oo

(see

Example

10.41).

(10.127)

an expression

A

for

special case of

with

f(x)

Equation (10.128)

V/

=

/

(r)

(0)

(10.128)

the Maclaurin-series expansion of j(x) about the

REAL-VARIABLE THEORY Example \x\ ^ X,

for

X arbitrary.

The reader can

=

Consider f(x)

10.39.

For

=

r

|/

463

All the derivatives of f(x) exist

0.

X

\x\

lim n_>flo

verify that

with a we have

e*

(n

-

>(z)|

so that

A simple proof of this statement

0.

ni

V Xn

) p which is seen to converge by w! n*0 Thus |/? n (x)| necessity, must tend to zero.

one considers the

ex,

\e*\

series,

the ratio test

occurs

if

Hence

LJ

the nth term, of

as

n

for all z,

>

00

"

*

and

^ ^ since /

r-0 Example 10.40.

w

0, 1, 2, 3,

.

{r)

() -

Consider

.

.

1

for r

/(a;)

(1

for

|o;|

^

<r

<

1

0, 1, 2,

.

.

.

.

The reader can

-h a?)~i.

verify that

Moreover

.

1,00(3)1-

and

-

M.',

we have

R n (x) =

Let the reader verify that lim

Thus

0.

n-0 The

series of (10.130) converges for

verify directly that lim n

Example

1

and diverges

for x

1.

The reader can

-

g- 1 /*^ ^

^

0, /(O)

-

We

0.

have

we note that

;

'(0),

x_>0

It

x =

0.

>

Consider /(z)

10.41.

To compute /

R n (l)

-

converges to zero for

all

for

a;

3.^0

n

can be shown that / (n >(0)

-

a:

0,

1,

2, 3,

.

.

.

.

Hence the Maclaurin

00

series

V/ 4-4

/ (a) (0)

xn ;

fl!

values of x since every term of the series

n-0 zero.

Rn (x)

Returning to (10.124), we note that f(x) does not tend to zero as n becomes infinite.

-

e"* 1/*

8

- R n (x)

The Maclaurin

for all

is

n so that

series of f(x)

does

ELEMENTS OP PUEE AND APPLIED MATHEMATICS

464

not converge to f(x) for this example. We note that the convergence of a Taylorseries expansion of a function /(z) does not guarantee that the series converges to /(a;).

One must always

Another form

+

a

R n (x).

investigate the remainder,

of the

Taylor

can be obtained

series

we

replace x

by

so that

hj

<r)

we allow a

If

if

by replacing a by

to vary

/(*

(a)*

we

#,

(10.131)

obtain

+ A) -/<"(*)

(10.132)

*J

r-0 (r) Equation (10.131) is very useful if we know / (a) for all r and if we wish to find an approximate value of f(a + h). If only a finite number of terms of the Taylor series are used in approximating /(a + A), an estimate of the size of the error can be obtained from (10.125).

In Example 10.39

10.42.

Example

laurin-series expansion of

we

let

x

-

*,

\x\

^ X.

and note that \R n (l)\

1

we saw that \R n (x)\ ^ e*\x\ n /n\ for the MacIf we wish to find an approximate value of e, For n = 8 we have |# 8 (1)| < e/n\ < 3/n!.

7

<

3/8!

0.0001, so that

}

Actually

The

e

-

e

accurate to three places.

....

2.71828

more than one variable is x n ) as a function of the n h n be any set of constants,

Taylor-series extension to a function of

not very

Consider /(x 1 x 2 xn Let h 1 /i 2

difficult.

variables x 1 x 2 ,

and define

= =

2.718 yields a value of

-^

=

r

f(x

/(u

+

1 ,

u

.

.

.

,

,

,

,

.

.

,

.

.

,

by

p(f) l

,

.

. ,

.

h l t, x*

+

2 .

,

.

.

,

h*t,

u

r

.

n

.

,

u*

)

=

xn

x

+ hn +M

t)

i

i

=

1, 2,

.

.

.

,

n (10.133)

x as constants temporarily so that p(t) is We consider x x*, looked upon as a function of the single variable L Let us assume that = 1. We tp(t) has a Maclaurin-series expansion which converges for t n

1

,

.

.

.

,

have

<"

n!

n-O

Now *

du" _ <b_dj_ ~ dt

df

du* ^T~~toT"

REAL-VARIABLE THEORY and, at

=

t

0,

-* and

(

For

t

= -^2 )

=

1

465

tt

we

h a W.

i

-

Continuing in this manner yields

obtain

\

.

.

.

+

x"

,

A) =

f(x\ x\

.

.

.

,

-

-^ For a function

f(x

of

'

+

dO.134)

two variables (10.134) becomes

+ h,y + k)= f(x, y)

+

where

* r-O Problems

00

1.

Show

that sin x

=

V

-775

(*n

n-O 00

2.

Show

that cos

=

I)

(

>

Li

f

Y* ) n=

^

n x 2n+1

+TTTT 1)1

_ l)a;2n /0 N (/n;

.

holds for

holds for

all

all

values of

values of

3.

Show

that sinh x

00

v^

) xlv

n0

x.

i

00

=

x.

x 2n+1 75 (^Jw

7-=^-. -t~

and cosh x

-

1;!

V

a;

2n

hold for w) 755-^ v* w n=0 /

Of X. 00

4.

Show

that In

(1 -f x)

-

Y

C^ 1 )*"" 1**

holds for

-1 <

a:

g

1.

all

values

ELEMENTS OF PURE AND APPLIED MATHEMATICS

466

5.

Show that In ~-f~ - 2 > n-O

holds for

-qrj

N<

If tf

1.

-

j-^~

>

0,show

< x < 1. How many

Determine In 5 accurate to four places. terms in the Taylor-series expansion of sin x about x ir/6 are needed to determine sin 31 accurate to six places? Evaluate sin 31 accurate to six that 6.

places. 7.

8.

By integrating Show that

1/(1

+

x 2 ) find the Maclaurin expansion of tan"

1

x.

ao

10

Find the

np-

2

I

n-0

(2n -f 1)(2*

+

*

>0

!)**+*

four terms of the Maclaurin-series expansion of In cos the series valid? 10. Find the Taylor series of cos 2 x about x ir/3. 1L lff(n) (x) is continuous on (a, x), show that

9.

what range

first

of x

/&.()

by making use 10.24.

x.

For

is

-

f (n) (a

-h 0(s

-

(x

~

a)

"

^

a))

1

of (10.126).

Extrema

The Lagrange Method

of Functions.

of Multipliers. exists

that f(x) has an extremum at a point x = c if an 17 - f(c) has the same sign for \h\ < 17. If /(c such that/(c h)

One says

>

+ h) ^ /(c)

+

has a local maximum at x => c. If A) ^ /(c) for |h| < 17, we say that /(c) has a local minimum at x = c. /(c If an extremum of /(x) at x = c exists and if /(#) is differentiable at x = c, we have for

|A|

<

17,

we say that

f(x)

+

(10.136)

or

= assume f(c) = (c) =/<-D( c ) = 0, (n) / (c) 5^ 0, and further assume that / (a;) is continuous in a neighborhood of x = c. Applying (10.124) and the result of Prob. 11, Sec. 10.23, we have so that /'(c)

=

f

Now

0.

-

-

(n)

Since / (n) (c)

5^

small so that /

(n)

and since (c)

+

is

c

>

as h

one-signed.

0,

we can choose h sufficiently

Hence

/(c

+

h)

/(c) will

be

REAL-VARIABLE THEORY

n

one-signed provided

necessary and

is

even

(h

467

Thus a

can be positive and negative).

have an extremum at x = c nonvanishing derivative of f(x) at x = c (w) (c) < 0, 0, a minimum occurs, while if /

sufficient condition that f(x)

and the first that /'(c) = be of even order. If / (w) (c) > a maximum occurs. Why? For a function of two variables we have is

/(*

+ *, + *)- f(x,

y)

2fxvhk

If

y

we

2 8 neglect the higher-order terms (A A fc,

.

,

.

flf

+

2fxV hk

be one-signed provided

f(x, y) will

k)

+/

2

yi,fc )

^

.),

=

we note that/(#

~

+

r)f

0,

0,

and

(fxxh

z

ft,

+

does not change sign for arbitrarily small positive and k. Let the reader deduce that f(x, y) has an

negative values of h and

ftf

extremal at

fxx

>

provided ux

(x, y)

>

or/yy

a

0,

r)/

=

minimum

-

0,

=

oy

occurs,

0,

and

and

if

fxx

<

or/^

<

0,

a maxi-

mum occurs. Why? =

be extremalized under the a and bounded curve, closed <p(x y) and if /(x, y) is continuous at every point of the curve <p(x, y) = c, we know that there will be a point Po on the curve p(x, y) = c such that Remember that we restrict f(x, y) will take on its maximum value at Po. = on the curve lie c. The to same statement applies if ^(x, y) P(x, y)

Let us consider the function

=

condition

constant.

y

z

f(x, y) to

If p(x, y) is

we consider the minimum value of /(x, y). Now in the elementary calcuwe would solve <p(x, y) = c for y as a function of x, say, y = ^(#),

lus

substitute y

We would

since

-^

=

$(x) into z

set -=-

=

dz

_ -

ax

+ -~

~-

due to Lagrange.

We

0.

df

=

0.

=

/(x,

also

f(x, ^(#)),

and then

~ v/dxl df\-d

df

/i

1

can obtain (10.138) by another procedure

Consider the new function

U - f(x,

We

=

have

dfdy

We

to obtain z

?/),

y)

+ X*(x, y)

differentiate J7 with respect to x

and

2/,

(10.139)

assuming that x and ^ are

ELEMENTS OF PURE AND APPLIED MATHEMATICS

468

independent variables, while X

_

dU =

A dU = ,-

considered as a parameter.

is

Setting

A 0yields .

.

,

+x

==

(iai4o)

5+ 5 = *dy

dy

This

Eliminating X in (10.140) yields (10.138).

is

Lagrange's method of

multipliers.

More generally, if y = /(xi, # 2 to the conditions ^(#i, x 2

.

,

#n )

.

.

,

.

.

.

,

,

#) =

ct

is

i

,

=

to be extremalized subject 1, 2,

.

.

,

m, we form

,

#n)

.

m

U = and we

/(#!, #2,

te

.

.

df

.

#2,

.

,

.

,

xn )

/.-,-

=

ct

i

,

i

,

,

Example

10.43.

We wish to

U

t

(#i,

2,

ira; !/

.

;

.

.

,

.

.

. ,

=

1,2,

.

.

-f X(27ro;i/

Eliminating X yields y/x volume.

irx

2

(10.141)

?/.

S

y

1.

,rc

find the ratio of altitude to radius of a cylindrical glass have for a fixed surface area.

+ trx

=

.

m, along with the equations ra, yields the values of xi, #2,

We

maximum volume

V 2

-

= 1, 2, = 1, 2,

x n which extremalize

(open top) having a

From

X t ^>

,

a~

.

elimination of the X t

^(#i,

+ /

set

dU

The

#n)

,

2 )

2irxy

+ irx = 2

constant

we have

It is

easy to show that y/x

=

1

yields a

maximum

Problems 1. Consider a cylindrical buoy with two conical ends. Let x be the radius of the Show that cylinder, y the altitude of the cylinder, and z the altitude of the cones.

x:y:z

-;r-:

1

:

1

yields a

maximum volume

for

a

fixed surface area.

Find the maximum distance from the origin to the curve x 3 -{- y* 3xy = 0. 8. Derive (10.137) if fxx h* + 2fxv hk + fyy k 2 does not change sign for arbitrary values of h and &. 4. A rectangular box has dimensions x, y t z. Show that x y:z = 1:1:1 yields a minimum surface area for a fixed volume. 2.

:

REAL-VARIABLE THEORY Consider the set of values y Q y i} y z

6.

,

-

y(x)

m <

Show

n.

+

a

that the set of

}

satisfy

fo

with

S*

=

a 2x

0, 1, 2,

.

n

n

= ^

xj, fo

t-0

+ a m xm

+ .

469

y n and the polynomial

,

2

= ^

w s l+k a l

+

aix

a*, i

,

.

.

.

.

,

w, which extremalize

yyx*.

y-o

Numerical Methods. Let us assume that we wish to find a root 0, and let us further assume that /'(x) exists. We can plot a rough graph of y = /(x) and attempt to read the value of x at which the curve crosses the x axis, y = (see 10.25.

=

of /(x)

Fig. 10.4). If we choose a point Xi not far removed from x, f(x) == 0, we note that

'V

the equation of the tangent line at (xi, /to)) is y -/(scO =f'(x 1 )(x-z l ). This line, L, intersects the x axis at the point x 2 = Zi f(xi)/f'(xi), /'(zi)

I

FIG. 10.4

obtained by setting

5^ 0,

2/

=

0.

This process can be continued, with

=

Xn+1

If

the sequence

limit

c,

.

Xi, X2,

c

xn

.

.

(10.142)

f'M ,

.

.

,

can be shown to converge to a

.

then lim x n +i

and

- ~

Xn

=

=

=

f(c)/f'(c) so that /(c)

c

tinuous at x

=

Thus

/'(c) 5^ 0.

c,

-

lim x n

c

f(x n/)

lim y,

provided /(x) and /'(x) are conlim x n It is apparent that

0,

= x =

.

n

difficulties will

Example

10.44.

occur

if

/'(x) is small

For/(x)

=

x

Xn+I

From

(10.143)

we note

deduce that xj +1

>

2

x\

>

=

x.

-

.

,

_

^

xn +i

<

xn provided x%

<

that if

near x

>

2

2.

we

2

Thus

if

.

have/'(x)

.

=

2x,

and (10.142) becomes

iL2

we choose

(10 143) .

>

2,

xn

>

xi

-

2,

(10.143) will yield a

0.

Let the reader

ELEMENTS OP PUKE AND APPLIED MATHEMATICS

470

mono tonic

decreasing sequence bounded below by zero. The sequence must converge 2 0. The computation may be arranged as follows:

to a solution of # 2

The

x

desired root to five decimal places

An iteration process may = <p(x). We shall assume

is

=

x

1.41412.

be used to find a root of x that

\<p'(x)\

n =

fn lying

NOW

Xn

We

1.

...

1, 2, 3,

between x n -i and #, by applying the law

Kfi -

xn

=

(Xi

+

XQ

XQ

+ Y

=

\x n

\

X

)

-

+

<

|^({)|

a? n -i|

(2

+

l)

'

'

\x n

'

+

mean.

of the

~

x n - \A

<

1,

from

Xk~~i

|x*

~

(#

(10.145).

-

converges by the ratio test since l-^l"" Xn

Thus n

=

x

+

n

From

(10.144)

is

=

the desired root of x

Example

We

10.45.

The computation

is

Y

w

*W

<

(x

and the continuity

lim x n

and f

lim

x

=

find a root of

=

fc

x k -i)

of <p(x)

=

^

we have

lim

<p(x).

x

arranged as follows:

-J-

sin

s

-f 1, ^(a?)

^-

sin

Thus

n-i)

\

lim x n

exists.

by

(10.145)

t

k*i .4

or

(10.144)

00

and

=

<p(x)

define x n +i

Thus

with Xi arbitrary.

with

< A <

x

-f 1,

<

!L

Xn - 1

'

REAL-VARIABLE THEORY

471

We started with xi = and obtained the desired root, x decimal places. Had we graphed y = x and y = -J- sin x -h that Xi = 1.5 would be a good starting point.

1.4987, accurate to four 1,

we would have

noticed

It is often useful to replace a given function by a polynomial which approximates the given function to a high degree of accuracy. Such a polynomial is called an interpolating function. Let f(x) be a function defined for xo ^ x ^ x n and let us assume that we have evaluated f(x) x n obtaining the n + 1 values y% = /(x), i = 0, 1, at x Xi X2, A simple way of constructing a polynomial of degree n, 7i. 2, L n (x), which satisfies L n (x t ) = y^ i = 0, 1, 2, n, is due to Lagrange. Let us note that the polynomial ,

,

.

,

,

.

.

.

.

9

,

.

(x (x

=

has the value zero at x

A sum

x2) x 2)

Xi, #2,

(x

-

-

xi)(x

,

x 2)

.

(X

,

-

Xn)

x n and has the value

such polynomials yields Lagrange

of

L n (x) =

- xi)(x - Xi)(x

.

?/o

at x

=

XQ.

9

s interpolating

polynomial,

Xn)

(X

2/0

Xg)

(x

x )(x

(X n

Xo)(x n

(X

^n

One i

=

'

'

'

-

X n -i)

Xi)

(10.146)

X n _i)

(x n

how closely L n (x) approximates /(x) for x ^ x To answer this question, we construct F(x) defined

naturally inquires

0, 1, 2,

.

.

.

,

n.

t

,

By

= /(x) - L n (x) -

F(x)

R(x

-

x )(x

-

(x

xi)

-

xn)

(10.147)

=

R

a constant. Let us note that F(x) vanishes for x Xo, Xi, x 2 n. Now let us choose x n since /(x) = L(x ), i = 0, 1, 2, xn any point x of the interval xo ^ x ^ x n other than Xo, Xi, X2, and determine R so that F(x) = 0, that is,

with .

.

.

,

,

.

t

,

.

.

,

.

/(*)

Up

- L n (55) -

R(Z

-

x )(f

-

*i)

(

- *) -

to the present no restriction was placed on /(x).

/(x) is differentiable at every point of

x

^

x

^

If

.

,

(10.148)

we assume that

x n then ,

.

F (x)

is differ-

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

472

If we apply the entiable on this range. theorem) to the intervals (#0, #i), (zi, 2 ),

(x n -i,

x), we note

that

F

/'"(x),

.

.

.

,

/

(w+1)

(s) exist

=

vanish at a point x

(x) will

we have

a polynomial of degree n,

+

n

s

,

n

XQ

y

mean

of the

1

(or Rollers

#), (#, x*+i),

(#t,

,

that F'(z) will vanish at

assume that f"(x), (n+1)

law -

If

points.

we

-

-

,

further

on x g x g n we note ^ s ^ x n Since L(x) is = 0. From (10.147) we a?

,

.

^

obtain

F<+() = ^n+l since -r-^+i [(#

R =

=

f (n +(s)

#o)(#

/c-+i)()/(n

+

- #(n +

=

x n )]

(x

Xi)

1)!

z

^

+

1)1.

(n

^

*

*

Substituting

1)! into (10.148) yields

/(n+D/o

(- x.)

xO

(10.149)

Equation (10.149) holds for any value of x in the interval XQ ^ x although (10.149) was obtained by choosing x ^ x, i = 0, 1, 2, for one notes that/(x ) = L n (x ) is consistent with (10.149). is any upper bound of |/ (n+1) (^)l on XQ ^ x g xi, we have If .

t

.

^ . ,

xn n,

l

M

L n (x)| ^ The

\x

Example -3),

The

10.46.

x(x "*"

-

a;

;

^

8

-

-

2-

3z 2

-f

l)(x

-

For x

*f

=

*-/

0\

t*/

40 or

/|

ir.

Since

^

sin

,

-

bound

xn

\

(10.150)

to the differ-

(0,

5),

x(x-2)(x-2)

g(ar

-

l)(x

-2)

3-2-1

a;

Af

dx 7

=

TTTT

,

=* 0>

we have

wi

we have

(?)

< Thus the use

|x

Let LI(X) be the polynomial which coincides with sin x at x

10.47.

r/6, T/3, T/2, |TT, |r,

"4

-3)

(-1) 4x"- 5 1

7

I

-

third-degree polynomial passing through the points

2)(*-3)

Example

-

sxl

|

(3, 7) is

-1),

(2,

|

L n (x).

ence between f(x) and

(1,

x

useful in finding an upper

is

inequality (10.150)

-

of L(fir)

() (i) (i) (j)

0.00005

would yield a value of

sin

QQ

.

i (i)'

40 accurate to four decimal places.

REAL-VARIABLE THEORY

473

h^x^h. We

Let us consider a function f(x) defined on struct the polynomial Li(x\ which coincides with f(x) at x

From

we have

(10.146)

4r

and

L,(x) dx

^

t

well inquire

This difference

how

i(tt\

=

y-/

We may

A'

-

[/(-A)

=

con-

h, 0, h.

-ft)

,

r

2

+

4/(0)

+ /(fc)]

the value of (10.151) differs from

(10.151)

/

J

h

f(x) dx.

is h

= ! J-

We f

(lv)

=

f(x)

dx-

o

+

[f(-h)

4/(0)

Let us assume that f(x\ f"(x), /'"(x), and Differentiating <?(/&) with respect to A yields

note that

<p(0)

(x) exist.

0.

+ 4/(0) |

and

^'(0)

=

0.

-/'(-A)]

"

=

f(-ft)]

Differentiating again yields

-f(-A)] - I

so that <^"(0)

-

[/'(A)

0.

-

[/'(A)

[/"(A)

-/'(-A)]

|

+/"(-*)]

Differentiating once

more

yields

-

"

/'"(-A)]

Now /"(A) dA = If

M

is

any upper bound

of |/ (lv) (^)l

on

dh ft

g

dh x

g

(-I)A^'(A) dh ft,

we have (10.152)

ELEMENTS OP PURE AND APPLIED MATHEMATICS

474

The

inequality (10.152) yields an estimate of the error

Simpson's rule as given by (10.151) to approximate f

if

we use

We

f(x) dx.

note that Simpson's rule is exact if f(x) is a polynomial of degree less than or equal to 3 since f (x) = 0. Since (10.151) depends only on the values of /(x) at the equally spaced points h, 0, /&, we can extend (10.151) to the interval a ^ x ^ b by subdividing (a, b) into an even

M

number

of intervals

a,

a

+

a

hj

+

+

2h, a

3A,

.

.

.

,

a

+

(n

l)h,

a

+

nh = b

applying Simpson 's rule to the intervals (a,

a

+

and adding.

'/(x)

d*

+

2A, a

4/0,

.

.

.

,

(a

+

(n

-

2)A, a

+

nA)

This yields

=

+ -

+

[/(a)

|

+

|

|

4/(a

+

[/(

+

2/0

[/(a

+

(n

+ -

+ /(a +

h)

4/(a

2)/0

+

+

3*)

4/(a

2/01

+ /(a + +

-

(n

+

4A)J

1)/0

+ /(ft)] +

+ ^(a +.A) + 2/(a + 2/i) + 4/(a + 3fc) + 4/(6 - A) + /(&)] + ^ + 2/(a + 4A) + [/(a)

|

i^i^-^r^^l = Mhi nh = Mh\b || g

...

with

M

+

2/0, (a

w m -

3

[

(10.154)

jgQ

J

(10.153)

/1A1K ^

a)'

t.

any upper bound of |/ (lv) (x) on the interval a ^ x ^ b. The greater number of subdivisions chosen and hence the smaller the value of A, the more accurate becomes Simpson's rule for approximating an integral. is

\

the

Example \E\

g

^^

10.48.

If

we apply Simpson's

-

6

Yon

C

-

1) 4

<

^( 1+ O

+

=

r2

fi

/

Ji

0.0002, so that

accurate to at least three decimal places.

ln2w

rule to In 2

we can expect

>

x

n ~

10,

we note

that

to obtain a value of In 2

Applying (10.153) yields

o + o + O + T^^I^ + Tj^O"f r9"f 2)

0.69314 In the tables one notes that in 2

We conclude this section formula.

0.69315 accurate to five decimal places.

with a discussion of the Euler-Maclaurin sum Let us suppose that f(x) has continuous derivatives at least to

REAL-VARIABLE THEORY

g

the order r on the interval

f*f(x)

=

dx

x/(x) I

=

-

/(I)

Bz(x) ing

z (x)

^(x

X

dx

dx

x/'(x) /(I)]

+

/(I)]

+i

i[/(l)

=

(x

x

-

4",

^by

-

/(O)]

dx

/'(*)

-

Replacing x

x).

we integrate by parts, we obtain

If

1.

/o of (x)

such that B'2 (x)

2

by parts

g

+/(!)]

i[/(0)

We define B

-

x

475

-

-

dx f* xf(x) dx

*)/'(*)

=

#2(1)

x/'(x) ete

0.

(10.155)

Integration yields

and integrat-

B((x) in (10.155)

yields l

fQ

f(x)

=

dx

i[/(0)

+ /(I)] + fQ

l

B*(x)f"(x) dx

(10.156)

B s (x) such that B (x) = B 2 (x) + 6 2 with the stipulation = JB (1) = 0. Thus B (x) = z 3 /6 - x 2 /4 + b x + c, and that 5 (0) = = c ^2 we have upon integration A- From -82^) = BS(^) 0, 62

We now

f

define

9

8

3

z

2

by parts l

fo

f(x)

dx

=

i[/(0)

+/(!)]

-

6 2 [f(l)

-

/'(O)]

-

dx J* Bt(x)f'"(x)

This process is continued so that a sequence of polynomials (Bernoulli) and a sequence of constants (Bernoulli) are generated. We have

= **_,(*) + B 4 (0) = B 4 (l) =

B'k (x)

6 4 _x

A;

S3

and /(I)]

-

6 2 [/'(l)

- /'(O)] +

bj/"(l)

B^,(x)/

(r)

-

(x)

dx

(10.158)

In the generation of the Bernoulli polynomials and constants no mention of J5o, -Bi, 61, 60. For convenience we choose B Q (x) = 0,

was made Bi(x)

=

x, 6

=

1,

61

Let us see whether such that

=

0.

it is

possible to find

r

two functions, <(x,

t),

<p(t),

~ (10.159)

r-O

ELEMENTS OF PURE AND APPLIED MATHEMATICS

476

If $(x, t) and <p(t) exist satisfying (10.159), we call them the generating functions of the Bernoulli polynomials and Bernoulli numbers, respec-

The

tively.

(see Sec.

series

expansions are simply the Taylor series expansions

=

10.23), so that b r

(r)

*>

(0)/r!,

B

= 4l r!

r (x)

y

\

ff dr

^)

For-

.

/t-o

mally we have

r-O oo

=

B((x)t

=

+

(x

t&(x,t)

+

+

B't(x)t*

-

+

2

l)^

+

t<p(t)

B'r (x)t

I-

t

+

(B.(x)

6.

^

-

(10.160)

so that

Integrating with respect to x yields

=

$(*,*)

with

A(0 an

arbitrary function of

=

so that A(<)

Finally $(1,

Hence

[*>(0

A(t)e*

-

*(*)

Now

+^

<i>(0,

t)

=

since

=

Br(l)^

-

1)

=

r

=

1,

*

since JS^l)

=

1,

S

r (l)

=

r

0,

0,

consider

<p(f)

and

^

1.

and

(

One can now

=

fi r (0)

^, and

<p(t)

-}/](e

t.

-

$(a;, t) of

(10.161)

and

10 161 > -

justify the series

expansions (10.159) along with the term-by-term differentiation given in

REAL-VARIABLE THEORY

One then easily shows that (10.157)

(10.160).

BI(X)

=

=

feo

x,

1, bi

=

We

0.

property of the Bernoulli constants

-

- -

_

-

*

and that B

An

this fact.

discernible

is

if

te

-

*-'-!""

2

valid

is

omit proof of

477 =>

0,

that

*

-

2

we note

(x)

important

1

____ + __ __- + +__ t

.

.

*

.

*

*

,

,

Hence If

an even function so that &2r+i = for all r [see (10.159)]. 2s + 1 and apply (10.157) to (10.158), we obtain

<p(t) is

we take l

f f(x) dx

=

r

=

-

+/(!)]

i[/(0)

~

W(D

/'(O)]

-----

-/'"(O)]

B If

we apply g(x)

(10.162) to g(x)

=

dx

f(x

+

=

+

f(x

l)dx =

1),

we

2.

^[/"

+ iW/

<2 ' +1)

W dx

(10.162)

obtain

*f(x) dx

Continuing this process for the intervals (2, 3), yields the Euler-Maclaurin formula,

(3, 4),

.

.

,(

.

!,

n)

and adding (x)

dx

=

+ i/(n) -

6 2 [/'(n)

-

-

- 64[/'"(n) - /'"(O)] -(n) -/<- (0)]

/'(O)]

6i.[/

(1

dx

The

(10.163)

Bernoulli polynomials and constants of (10.163) can be calculated It can be shown that if / (28+1) (x) is monotonic (10.161).

from (10.157) or

^

decreasing for sign

on

^

x

^

x

1)

+

derivatives of f(x) have the

n,

then the true value of f f(x) dx always

+

1

lies

between

+/(!)+*

terms of the series following ^/(O) For this case we need not be concerned

/(w) in (10.163).

with n-l

m+1

2 /m m-0 Example

10.49.

same

n

the sums of s and s

+ f(n

^ n and if the odd

Stirling's

Bw(x - nW-(x) dx

Formula.

For f(x)

(x

In (x -f 1)

we note

that

ELEMENTS OF PURE AND APPLIED MATHEMATICS

478 x

for

2a

n and / (lrhl >(aB)

(n

+

1) In (n

+ ln

-

(n

1)

*

-

n

-

f

+*ln

n

-

6,

(n

+

+

1)

Now (n

+

1) In

n

(

+

1)

-

^

monotonic decreasing on

is

parts yields

+

In (x

-

(^ -

1) In

[n (l

-

dx

1)

In 1 64

l)

^

a;

+

(n

1) In

Urn n

(n

1)

(n

+

+

1) In

1) In

+

(l ^

* oe

Neglecting those terms which tend to zero as n (n

+ T)

n

ln

~

n

ft>

2)

-

n

we note from

> ,

-v n!

In

nn n!

Formula meaning

;

(n

+

I) In

(l

+

~

1

+

(10.164) that

C

where C is the sum of the constant terms in (10.164). In 2*. Thus with (10.165) to show that C ,

(10.164)

since

w) /

ln n!

by

4-

----

-

(^fyy,

+ ^)] -

Integration

+ In 2 + In 3

+ and, for large w, (n H- 1) In (n -H

n.

(10.165)

Let the reader apply (4.47),

n

2am

V2irnn n e-

=

ft

V^m (-)

(10.166)

n

(10.166) is Stirling's formula for approximating w' for of (10.166) is that

The

large.

true

'

lim

Problems

and x = 1, accurate to six 3 -f 1 1. Find a root of x* lying between x decimal places. In (x 2. Find a root of x 1) 4- 1, accurate to five decimal places. - 2, p(3) 3. Find a polynomial p(x) such that p(0) - 0, p(l) 1, p(2) 3,

=0

+

p(4) 4.

same

-

11.

Apply Simpson's for

/

V% -

rule with 10 subdivisions to

How accurate can you

,

Jo 6.

Show that

6.

Show that

approximate

sin

/

-

--

Do

the

expect your answers to be?

x

- -yiir, B^(x) (x* 2*In C [see (10.165)].

64

2x*

+ x*).

10.26. The Lebesgue Integral. It is rather obvious that not all functions are Riemann-integrable (J?-integrable). consider the interval 1 and define /(x) = 1 if x is irrational, f(x) = 2 if x is rational. g x

We

g

The upper Darboux

integral

(see Sec. 10.16).

integral is

1

However,

let

is

seen to be

Thus f(x)

is

us consider the following: It

2, whereas the lower Darboux not fr-integrable on ^ x ^ 1. is

known th&t the

rationals are

REAL-VARIABLE THEORY

we can enumerate the rationals by

countable (see Sec. 10.11), so that

g

interval

x

g

479

1,

fi,

r3,

7*2,

.

.

.

,

rn ,

.

=

/i(*),/(*),

with /n (a;)

fn (x)

=

=

1

=

except at x

It is

2.

*

g

define /i(x) on 2

on the

.

.

x g 1 by /i(z) = 1 for x = /(r 2 ) = by / (z) 1, x 5^ ri, r 2 ,/(ri) a of functions we construct sequence manner,

We

define /2 (x)

rationals

and we designate the

ri,

-

-

r2 ,

.

,/(*),

n, /i(n)

=

2.

(10.167)

-

rn ,

and at these rational

1

for

2

f or

x irrational x rational

.

.

,

We

Proceeding in this

2.

points,

apparent that

=

lim fn (x)

=

/(x)

Moreover the sequence \fn (x) is monotonic nondecr easing as n increases, and we write fn (x) /*/(z). The reader can easily prove that every func}

tion of the sequence \fn (x)

We

is

}

fi-integrable,

with

fn (x) dx

I

1 for all n.

define the Lebesgue integral of f(x) as L(jf)

Even though

l

=

f JQ

l

SE

f(x)dx

lim n

f w JQ

fn (x)dx

=

lim

1

=

1

(10.168)

n-

not /Wntegrable, it is Lebesgue integrable (Z/-inteof its Lebesgue integral is defined by (10.168). grable), More generally, we have the following situation Let C be the class of all J?-integrable functions on the range a ^ x ^ 6, and let f(x) be the of functions belonglimit of a monotonic nondecreasing sequence, {/n (#) is

f(x)

and the value

:

}

ing to C.

We

define the Lebesgue integral of f(x) b

=

L(/)

=

f f(x)dx

jo,

provided the limit

Let

exists.

which are L-integrable.

C denote

Any = fn (x)

Thus C

f f(x)dx

J"

fn (x)dx

such functions J?-integrable is automati-

the class of is

(10.169)

all

/" f(x) and

b

lim n-K

b

lim

f n-t* Ja

function which

cally L-integrable since /(x)

,

by

=

b

f J*

f(x)dx

a proper subset of C. Any function which is JB-integrable is and the values of the two integrals are the same. It can be shown that if /(x) is a limit of a monotonic nondecreasing sequence of is

L-integrable,

functions belonging to C then/(x) also belongs to C. An analogous situWe define the irrational numation occurs in the real-number system.

bounded sequences of rational numbers. The limit bounded sequence of irrational and/or rational numbers is again a number, rational or irrational.

bers as the limits of of a real

ELEMENTS OF PUKE AND APPLIED MATHEMATICS

480

To we Ci

C

extend the class

C

in order to

consider the following:

A

>

find

if

for

any

e

we can

embrace a larger

function f(x)

is

two functions,

class of functions, said to belong to the class

g(x)

and

h(x), belonging to

such that g(x)

S

for a

h(x)

f(x)

x

g

6

and If /(x)

belongs to Ci (/)

we

m

define its Lebesgue integral b

f f(x) dx

J*

geC

and h(x) satisfying

for all g(x)

= supLfo) =

(10.170).

by

inf L(h)

The reader can

f(x) belongs to C then f(x) belongs to Ci and L(f) set Ci contains all Lebesgue-integrable functions.

that

if

(10.171)

h*C

easily verify

=

<(/).

The

The

ideas contained above lead to the concepts of measurable funcand measurable sets. In most texts one begins with the definition of the measure of a set, and afterward the Lebesgue integral is defined. We omit a discussion of measure, but we do emphasize that the theory The of measure is of prime importance in modern probability theory. Lebesgue integral plays an important role in modern mathematical analytions

sis

with

theory,

its

applications to the Fourier integral, probability theory, ergodic

and other

fields.

REFERENCES Burington, R.

S.,

and C. C. Torrance: "Higher Mathematics," McGraw-Hill Book

New York, 1939. Courant, R.: "Differential and Integral Calculus," Interscience Publishers, Inc., New York, 1947. Franklin, P.: "A Treatise on Advanced Calculus," John Wiley & Sons, Inc., New Company,

Inc.,

York, 1940. Goursat, E.:

"A

Hildebrand,

F.

Course in Mathematical Analysis," Ginn & Company, Boston, 1904. B.: "Introduction to Numerical Analysis," McGraw-Hill Book

Company, Inc., New York, 1956. Householder, A. S.: "Principles of Numerical Analysis," McGraw-Hill Book pany, Inc., New York, 1953.

Com-

Kamke, E.: "Theory of Sets," Dover Publications, New York, 1950. Milne, W. E.: "Numerical Calculus," Princeton University Press, Princeton, N.

J.,

1949.

Munroe, M. E.: "Introduction to Measure and Integration," Addison- Wesley Publishing Company, Cambridge, Mass., 1953. Newman, M. H. A.: "Topology of Plane Sets," Cambridge University Press, New York, 1939. Titchmarsh, M. A.: "The Theory of Functions," Oxford University Press,

New

1939.

Widder, D. V.: Wilson, E. B,:

"

Advanced Calculus," Prentice-Hall, Inc., New York, 1947. "Advanced Calculus," Ginn & Company, Boston, 1912.

York,

INDEX A A

posteriori probability, 345 priori probability, 345

Bessel's equation, 231 Bessel's inequality, 247, Beta function, 170

Abelian group, 300 Abel's lemma, 450 Abel's test for uniform convergence, 451 Absolute convergence, 444, 455 Absolute tensors, 89 Absolute value of complex number, 123 Acceleration, 50, 52 Addition, parallelogram law of, 38, 123 of tensors, 89 of vectors, 38 Affine transformation, 313

Aleph

zero,

Amplitude

Bianchi's identity, 106

Bienaym^-Tchebysheff inequality, 355 Bilinear transformation, 125

Binomial coefficient, 339 Binomial theorem, 339 Binormal, 53 "Bits" of information, 346 Bonnet, second law of the mean for integrals, 457 Boole, G., 388

394

Algebraic extension of a Algebraic numbers, 396 Alternating series, 443 of

field,

Boolean operator, 6

324-326

complex number, 123

dz

,221

Calculus of variations, 288 problem of constraint in, 295 variable end-point problem in, 293

Argand plane, 123 Argument of complex number, 123 Arithmetic n-space, 84 Associated Legendre differential equa-

Cancellation, law

of, 375 396 Cantor's theory of sets, 394r-396 Cardinal numbers, 394 Cartesian coordinate system, 93 Cauchy, convergence criterion for quence, 437

Cantor

230

Associated vector, 94

Automorphism, 304 of a field, 331 of a group, 304 inner, 305 of choice,

z

Brachistochrone, 289, 293 Branch point, 149 Buffon needle problem, 355

Angular velocity, 46 Arc length, 52 Archimedean ordering postulates, 385 Arcs, rectifiable, 277

Axiom

=

Boundary of a set, 389 Boundary point, 387 Bounded set, 386 Bounded variation, 277

Analytic continuation, 150-151 Analytic function, 130 Angular momentum, 115

tion,

258

set,

inequality

392

of,

se-

391

integral formula of, 143 integral theorem of, 136

nth-root test of, 442 Cauchy's distribution function, 361 Cauchy-Riemann equations, 148 Cayley's theorem, 303 Central of a group, 307 Central-limit theorem, 359-361 Character of a group, 312 Characteristic function, 357

Bayes's formula, 346 Bayes's theorem, 344-346 Bernoulli numbers, 475-477 Bernoulli polynomials, 475-477 Bernoulli trials, 349 Bernoulli's theorem, 358 Bessel function, 232

481

ELEMENTS OF PURE AND APPLIED MATHEMATICS

482

Characteristic roots, 24 Characteristic vector, 24 Chi-equared distribution, 362-364 Christoffel symbols, 96 law of transformation of, 98 Christoffel-Darboux identity, 244

Closed interval, 386 Closed set, 386, 388 of orthogonal polynomials, 249 Closure of a set, 389 Coefficients, Fourier, 245, 252 Cofactor, 9

Combinations, 338

Commutative

law, of integers, 375

of vector addition,

38

Commutator, 303 Comparison test for series, 441 Complement of a set, 387 Complete set of orthogonal polynomials, 249-250 functions, 125 continuous, 126 differentiability of, 127

Complex

(See also Complex variable) Complex-number field, 122 Complex numbers, 122

absolute value of, 123 argument of, 124 conjugate of, 124 modulus of, 123 vector representation of, 123 Complex variable, functions of, 125 analytic or regular, 130 derivative of, 127 integration of, 131-136 Taylor's expansion of, 145

Components, of a tensor, 89 of a vector, 41 Composition, series of, 311 prime factors of, 311 Conditional probability 345 Conditionally convergent series, 444 Confluence of singularities, 223 Conjugate of a complex number, 124 Conjugate elements of a group, 306 Conjugate numbers, 325 Conjugate subgroups, 306 Conservative vector field, 67 ,

Continuity, 126, 398 uniform, 399

Continuous function, 398 Continuous transformation groups, 314315 Contour integration, 160-162 Contraction of a tensor, 90 Contravariant tensor, 89

Convergence, absolute, 444, 455 conditional, 444 of Fourier series, 251, 256 interval of, 386, 453 of power series, 453-454 radius of, 146, 453 of series, 439 tests for, 440-444 Cauchy's nth-root, 442 D'Alembert's ratio, 441 integral, 441 uniform, 446, 450 tests for, 451-453 Abel's, 451 Coordinate curves, 95 Coordinate systems, 40, 84 Coordinates, cartesian, 93 curvilinear, 62, 65 cylindrical, 65 Euclidean, 93 geodesic, 104 normal, 32 Riemannian, 105 spherical, 63 Laplacian in, 103 transformation of, 84, 411 Cosets, 305 Cosh 2, 150, 158 Cosine 2, 131 Countable collection, 394 Co variant curvature tensor, 106 Covariant derivative, 100 Covariant differentiation, 100 Covariant tensor, 89 Covariant vector, 87 Cramer's rule, 10 Craps, game of, 348 Criterion, of Eisenstein, 328 of Lipschitz, 183 Cross or vector product of vectors, 45 Curl, 59 of a gradient, 60 of a vector, 59, 83, 102 Curvature, radius of, 53 scalar, 106 Curvature tensor, 106, 108 Curve, unicursal, 435 (See also Space curve) Curves, family of, 174 Curvilinear coordinates, 62 curl, divergence, gradient, in,

65

Cyclic group, 300

D'Alembert's ratio test, 441 integral, 419

Darboux

Laplacian

INDEX Definite integral, 131, 419-424 mean-value theorems for, 422, 457

Del

57 De Moivre, formula of, 125 Density or weight function, 237 Dependence, linear, 20, 187 Derivative, 402 co variant, 100 of a determinant, 8 of an integral, 142 (V),

of matrix, 13 partial, 408 of a vector,

of,

of,

10

multiplication of, 9 properties of, 7-10 solution of equations by,

,

192

covariant, 100 of Fourier series, 262

403 408 rules, 61, 403 of series, 450 implicit, partial,

of,

352

Distribution function, 351 Cauchy's, 361 Distributive law, 376 Divergence, 58, 102 of a curl, 61 of a gradient, of series, 439

1

Diameter of a set, 390 Dice (craps), 348 Differential, 402 operator, 410 total, 409

59

Divergence theorem of Gauss, 75 Dot, or scalar, product, 42, 88

Differential equations, 172

Eigenfunction, 24 Eigenvalue, 24 Eigenvector, 24 Einstein, Albert, summation convention

Bessel's, 231 in complex domain, 201

definition of, 172

175

of, 1

exact, 178 first-order,

ax

Differentiation, 402

moment

Wronskian, 187 Determinants, 5

of,

=

Gaussian, 353

6

Vandermonde, 334

degree

wave, 172, 286 Differential operator, s

Discontinuity of a function, 398 Distribution, chi-squared, 362-364

of matrix, 5, 13

order

Differential equations, system of, 190

of vectors, 50, 61 Diffusion equation, 367-368 Dini's conditions, 457-459 Direction cosines, 43 Directional derivative, 56

50 Desargues' theorem, 40 Determinant, cof actors of, 9 derivative of, 8 expansion of, 9 minors

483

Einstein-Lorentz transformations, 320

175

Eisenstein, criterion of, 328 Electric field, 165

homogeneous, 177 hypergeometric, 221 integrating factors of, Laplace's, 226 Legendre's, 201, 222

Electrostatic potential, 76, 165 1

79

Elliptic integrals,

linear (see Linear differential equa-

Equally likely events, 342 Equation, Bessel's, 231 differential (see Differential equations)

tions)

nonlinear, 271 partial (see Partial differential equations)

Euler-Lagrange, 290 exact, 178-179 indicial,

second-order, 199-202 separation of variables

integral, in,

175

simultaneous, 190 singular point of, 207 solutions of, existence of, 183 general, 182, 191 particular, 182

uniqueness

436

Entire functions, 145

of,

185

Sturm-Liouville, 200

210 203

Laplace's, 76, 78, 226 Legendre's, 230, 404 linear homogeneous, 17 principal, 326 wave, 172, 286 Equations, Cauehy-Riemann, 128 of motion, Euler's, 115

Hamilton's, 113

ELEMENTS OF PURE AND APPLIED MATHEMATICS

484

Equations, of motion, Lagrange's, 110 solution by determinants, 10

systems

of, 2,

4

homogeneous

linear, 17

Equivalence relation, 376 Error, in Simpson's rule, 473-474 in Taylor's series, 462 Essential singularity, 157 Euclidean coordinates, 93

Euclidean space, 93, 107 Eudoxus, 385 Euler's equations of motion, 115 Euler-Lagrange equation, 290 Euler-Maclaurin sum formula, 474-478 Even function, 256 Exact equation, 178-179 Expansions, in Fourier series, 252, 260 in Maclaurin's series, 462 in orthogonal polynomials, 246, 251 in Taylor's series (see Taylor's series or

expansion) Expectation, 354 Exterior point, 387 Extrerna of functions, 466-468

expansion

in,

251, 256

of,

252, 260

264 minimal property trigonometric, 252

integration

of,

partial,

of,

248

Fourier transform, 268 Frenet-Serret formulas, 52-53 Frobenius, method of, 215 Function, analytic, 130 Bessel, 232 beta, 170 of bounded variation, 277 characteristic, 357 of a complex variable (see

Complex

variable)

conjugate, 124 continuous, 126, 398 piecewise, 258 sectionally, 258

uniformly, 399 density, or weight, 237

145

even, 256

expansion

Stirling's formula for, Family of curves, 174

477-478

380

of, 331 165 extension of, algebraic, 324-326 normal, 330 of probability, 343 of sets, 342 steady-state, 41 vector, conservative, 67 irrotational, 69, 73 solenoidal, 77 steady-state, 41 uniform, 42

automorphisms

electric,

First-order differential equation, 175 First theorem of the mean, 422

Flat space, 107 integral, 143

interpolation, of Lagrange, 471 Stirling's,

248

differentiation of, 262

entire,

Factorials, 168-171

Force moment, 116 Formula, Cauchy's

of,

Fourier remainder, 247 Fourier series, convergence

differentiate, 127, 402 discontinuous, 398 distribution, 351

Factor, integrating, 179 Factor group, 307-310 index of, 308

Field, 122, 323,

Fourier partial sum, minimal property

477-478

Fourier coefficients, 245, 252 Fourier integral, 265 Fourier partial sum, 245, 258

extrema

of,

of,

factorial, or

145

466-468

gamma, 168

generating, 234, 476 homogeneous, 177

hypergeometric, 222, 223 integrable, Lebesgue,

478-480

Riemann, 419-424 square, 246, 257 limit of,

397-398

memory, 270 of more than one

variable, 407

odd, 256 potential, 111 probability, 352 rational, 434 of a real variable,

396

regular, 130

symmetric, 320, 321 transfer, 270 Functional, 292 Functions, orthogonal, 237 (See also Complex functions) Fundamental theorem, of algebra

(Gauss), 145 of arithmetic, 381 of the integral calculus, 142,

423

INDEX Galois group, 332 Galois resolvent, 330

Game

theory, 368-372 Gamma function, 168-171

Gauss divergence theorem, 75 Gauss fundamental theorem of algebra, 145 Gaussian distribution, 353 General solution of differential equations, 182, 191 Generating function, 234, 476 Geodesic coordinates, 104 Geodesies, in a Riemannian space, 95 of a sphere, 289 Gradient of a scalar, 55 Greatest common divisor, 323 Green's theorem, 77

Group, Abelian, 300

automorphism inner,

of,

305

character of, 312 conjugate elements of, 306 conjugate subgroups of, 306 continuous transformation, 314-315 cyclic, 300

298 307 index of, 308 finite, 300 Galois, 332 octic, 312 order of, 301 quotient, 307 index of, 308 definition of,

factor,

regular permutation, 303 representation of, 313 irreducible, 314 reducible, 314

simple, 307, 310 of,

Hermite polynomials, 233, 239, 241 generating function for, 234 Hermitian matrix, 34 Hessian, 419 Hilbert space, 391 Homogeneous function, 177 Homogeneous linear equation, 17 Hyperbolic cosine, 150, 158

Hyperbolic sine, 150 Hypergeometric equation, 221 Hypergeometric function, 222 of Kummer, 223 Hypersurface, 95

Identity theorem, 151

Impedance, 270 304

central of, 307

subgroup

485

Helix, 53

301

invariant, 306

normal, 306

maximal, 309 symmetric, 302 Groups, intersection of, 307 isomorphism of, 303 prime-factor, 311 product of, 308

Implicit differentiation, 403 Implicit function theorems, 412-419 Improper integral, 282, 427

Cauchy value of, 282 Independence, linear, 187 Independent events, 347-348 Index of factor group, 308 Indicial equation, 210 Inertia, moment of, 117 tensor, 116 Infemum, 383 Infinite integral, 428 uniform convergence of, 429 Infinite series, of constants, 439 of, absolute, 444, 455 conditional, 444

convergence test for,

440-444

uniform, 450-454 expansion in, 246, 252, 461, 465 of functions, 449 theorems concerning, 447-450 of trigonometric functions, 252 Infinitesimal, 397 Infinitesimal transformation, 315 Infinity, point at, 158, 218 Information theory, "bits" of information, 346 Integers, commutative law of, 375 positive, 374r-377 of, 380-381 378-379

properties rational,

Integrable-square functions, 246, 257

Hamilton-Cayley theorem, 35 Hamiltonian, 113 Hamilton's equations of motion, 113 Harmonic series, 440 Heat flow, 227 Heine-Borel theorem, 392-393

Integral, Darboux, 419 definite (see Definite integral) of, 142 436 equation, 203 formula of Cauchy, 143 Fourier, 265

derivative elliptic,

ELEMENTS OP PUKE AND APPLIED MATHEMATICS

486

Integral, improper, 427

429 Lebesgue, 478-480 line, 66 mean-value theorem, 422, 457 Riemann, 131, 419 Stieltje, 280 test for convergence, 441 theorem of Cauchy, 136 Integral equation of Volterra, 203 Integrals with a parameter, 142, 425-426 Integrating factors, 179 Integration, of a complex variable, 131136 contour, 160-162 of Fourier series, 264 of Laplace's equation, 78 methods of, 431-436 numerical, 474-477 by parts, 143, 424 of a real variable, 419-420 of series, 450, 458 term-by-term, 448, 458 Interior point, 387 Interlacing of zeros, 199, 244 Interpolation formula of Lagrange, 471472 Intersection of sets, 388 Interval, closed, 386 of convergence, 386, 453 open, 386 Invariant, 320 subgroup, 306 Irreducible polynomial, 324 Irrotational vector field, 69, 73 infinite, 428,

Isomorphism, 1585 of groups, 303 Isoperimetric problem, 297

Lagrange's method of multipliers, 296, 467 Lagrangian, 111 Laguerre polynomials, 224, 239, 241 associated, 224 ordinary, 226 Laplace transformation, 282 inversion theorem of, 287 table of, 285

Laplace's equation, 76, 226 integration of, 78 solution of, 226 Laplacian, 59, 103 in cylindrical coordinates, 65 in orthogonal coordinate system, 65 in spherical coordinates, 103 in tensor form, 103

Laurent expansion, 154 Law of the mean, for differential calculus, 405 for integral calculus, 422, 424, 457 Lebesgue integral, 478-480 Legendre polynomial, 230, 239, 404 associated, 230 Legendre's duplication formula, 169 Legendre's equation, 230, 404 Leibniz, rule of, 403 Length of arc, 52, 95

L'HospitaPs rule, 405 Limit, 397-398 Limit cycle, 273 Limit point, 386 Line element, 93 Line integral, 66 Linear dependence, 20, 187 Linear differential equations, with constant coefficients, 191 definition of, 175

exact, 178 existence theorem first-order, 175,

Jacobian, 62 Jordan curve, 131

Jordan-Holder theorem, 310-312

of,

183

177

general solution of, 182, 191 second-order, 199, 202 indicial equation, 210

ordinary point, 202 properties, 199-201 Kepler's first law of planetary motion, 52 Kinetic energy, 111

Kronecker, 324

Kronecker

delta, 3, 9

Kryloff-Bogoluiboff, method of, 271 Kuinmer's confluent hypergeometric function, 223

Lagrange's equations of motion, 110 Lagrange's interpolation polynomial,

471-472

regular singular point, 210

systems of, 190 Linear equations, 2

homogeneous, 17 Linear independence, 187 Liouville's theorem, 145 Lipschitz criterion, 183 Ln 2, 148

M test of Weier&trass, 451 Maclaurin

series or expansion,

462

INDEX Matrices, 11

comparable, 12 equality of, 11 multiplication of, 12-13 sum of, 12

Matrix, characteristic equation

of,

24

column, 20 derivative

of,

13

determinant of, Hermitian, 34

5,

13

identity, 14

inverse of, 16, 29-30 multiplication by a scalar, 12 negative of, 12

orthogonal, 22 skew-symmetric, 14 spur of, 15 square, 11 nonsingular, 16 subdivision of, 32 symmetric, 14 trace, 15 transpose of, 14 triangular, 28 unitary, 22 zero, 12

487

Nonlinear differential equations, 271 Normal to a surface, 56, 71 Normal acceleration, 52 Normal coordinates, 32 Normal derivative, 56 Normal distribution of Gauss, 353 Normal extension of a field, 330 Normal subgroup, 306 maximal, 309 Normalizing factors, 240 Null set, 389 Numbers, algebraic, 396 Bernoulli, 475-477 cardinal, 394 complex, 122-124 conjugate, 325 primitive, 328 real, 382 triples, 81 Numerical integration, 474-477 Numerical methods, 469-478

Octic group, 312 function, 256 One-to-one correspondence, 394

Odd

Maximal normal subgroup, 309 Maximum-modulus theorem, 151 Mean, law of, 405, 422, 424, 457

Mean or Memory

expected value, 405 function, 270

Methods

of integration,

431-436

Open Open

interval, set,

386

387

Operator, of Boole, 221 of continuation, 154, 207 differential, 192 vector, 57, 60 Ordered set, 377 Ordering theorems, 377 Ordinary differential equation, 172 Ordinary point, 202 Orthogonal curvilinear coordinates, 62 Orthogonal functions, 237 Orthogonal polynomials (see Poly-

Metric tensor, 93 Minimal property of partial Fourier series, 248 Mixed strategy, 370 Moment, of a distribution, 352 force, 116 of inertia, 117 Momentum, angular, 115 generalized, 113 Monodromy theorem, 153 Monte Carlo methods, 364-368 Xlorera's theorem, 145

Orthogonal trajectories, 176 Orthogonal transformations, 22 Orthogonal vectors, 82 Orthonormal set, 241

Motion, equations of (see Equations) in a plane, 52 of rigid body, 46-47 Multipliers of Lagrange, 296, 467 Mutually exclusive events, 343

Paraboloidal coordinates, 95 Parallel displacement, 108

Navier-Stokes equation of motion, 117 Neighborhood, deleted, 387 spherical, 391 Nested sets, theorem of, 392 Newton's binomial expansion, 339 Nonessential singular point, 157

nomials)

Paperitz, 221

Parallel vectors, 38, 108-109 Parallelogram law of addition, 38, 123 Parameters, integrals containing, 425

variation of, 197 Parseval's identity, 265 Partial derivative, 408 Partial differential equations, Fourier's method of solving, 226-228

ELEMENTS OF PUKE AND APPLIED 'MATHEMATICS Partial differential equations, heat tion,

equa-

Power

368

series,

convergence

of,

Laplace's equation, 226 wave equation, 172, 286

uniform, 453-454 differentiation of,

Partial differentiation, 408 Partial fractions, 431-432 Partial sums, 245, 258 Particular solutions, 182, 194-198

Peano's postulates, 374 Pearson, K., x* distribution of, 363 Permutations, 337-338 Picard's method of successive approxi-

mations, 184, 203 Picard's theorem, 157 Piecewise continuous function, 258

454

expansions in, 462 functions defined by, 151 integration of, 455 solutions of differential equations by, 206, 212-217 Pressure, 119 Prime factors of composition, 311

Primitive number, 328 Principal directions, 118 Principal equation, 326 Probability, a posteriori, 345

a

Point, boundary, 387

345

priori,

branch, 149

axiomatic definition

exterior, 387 at infinity, 158,

central-limit

interior,

continuous, 351

singular, 156, 157 of differential equation,

of,

theorem conditional, 345

218 387 x limit, 386 neighborhood of, 387 deleted, 387 ordinary, 202 set theory, 386-389

field of,

343

function

of,

joint,

343 359-361

of,

352

distribution, 351 Gauss, distribution of,

353

355

Pursuit problem, 54

207 J*

157

essential,

nonessential, 210 regular, 210

Poker hands, 338

Quadratic forms, 21 positive-definite, 28 Quotient group, 307 index of, 308

Poles, simple, 157

Quotient law of tensors, 90

Polynomials, 322 BernouUi, 475-477 Hermite, 233, 234, 239, 241

Raabe's

removable, 157

irreducible,

radius of,

453

324

Lagrange's interpolation, 471-472 Laguerre, 224, 226, 239, 241 Legendre, 230, 239, 404 normal, 330 orthogonal, 237

test,

442

Radius, of convergence, 146, 453 of curvature, 53 Random variable, 351

Random-walk problem,

350,

367

Potential, electrostatic, 76, 166 Potential function, 111

Ratio test of D'Alembert, 441 Rational function, 434 Rational integers, 378-379 Rational numbers, 379-380 Real-number system, 382-385 Recapitulation of vector differential for mulas, 61 Reciprocal tensors, 92, 94 Rectifiabte path, 277 Recursion formula, 202 Regular function, 130 Regular permutation group, 303 Regular singular point, 210 Relative tensors, 89 Relative velocity, 54

Power

Remainder

closed sets of, 249 complete sets of, 249 completeness of, 250 difference equation for, 243 expansions in, 246, 251 zeros of, 242 relatively prime, 324 separable, 328 Tchebysheff, 241 Positive-definite quadratic form, Positive integers, 374-377

series,

453

convergence

of,

453

28

462

in Taylor series expansion,

INDEX Residue, 158, 160 Ricci tensor, 106

Riemann

Series, Taylor's (see Taylor's series or

integral, complex-variable

theory, 131 real-variable theory, 419-424

Riemann

surface, 149 Riemann-Christoffel tensor, 106 Riemannian coordinates, 105 Riemannian curvature tensor, 106 Riemannian metric, 93 Riemannian space, 93

geodesies in, 95 Rigid body, motion of, 46-47 Ring, 379 Rolle's theorem, 404 Rule, Cramer's, 10 Simpson's, 473-474 Russell number, 382-385

Saddle element, 369 Scalar, 37

gradient of, 55 Laplacian of, 59, 103 Scalar curvature, 106 Scalar product of vectors, 42, 88

47

triple,

Schwarz-Christoffel transformation, 163 Second law of the mean, 457

Second-order differential equations, 199, 202 properties of, 199-201 Sectionally continuous function, 258 Separation of variables, method of, 227 Sequence, 437

convergence 437

of,

Cauchy

criterion for,

Sequential approach, 391

439

alternating, 443

comparison test

for,

441

of composition, 311

convergent, 439 differentiation of,

450

divergent, 439 Fourier (see Fourier series)

harmonic, 440 infinite (see Infinite series)

integration

of,

450, 458

Maclaurin, 462 oscillating, 439

sum

(see

of,

Power

439

Simple closed curve, 131 Simple group, 307, 310 Simple mappings, 124 Simpson's rule, 473-474 Simultaneous differential equations, 190 Sine

z,

131

Singular points (see Point) Singularities, confluence of, 223

Sinh

2,

150

Slope, 402

Space curve, 52 arc length of, 52

convergent, 437 uniformly, 446

power

expansion) uniformly convergent, 450 Set, boundary of, 389 bounded, 383, 386 Cantor, 396 closed, 386, 388 closure of, 389 complement of, 387 countable, 394 denumerable, 394 diameter of, 390 infemum of, 383 limit point of, 386, 389 nondenumerable, 394 null, 389 open, 387 ordered, 377 orthonormal, 241 supremum of, 383 totally ordered, 377 uncountable, 394 Sets, Cantor's theory of, 394-396 field of, 342 intersection of, 388 nested, theorem of, 392 of, 388

un^|

Schwarz-Cauchy inequality, 391

Series,

489

curvature of, 53 Jordan, 131 tangent to, 52 torsion of, 53 unit binormal of, 53 unit principal normal of, 53 Spherical coordinates, 63 Steady-state vector field, 41 Stieltje's integral,

280

Stirling's formula, 170,

477-478

Stochastic variable, 351 Stokes's theorem, 70 Strain tensor, 118 Stress tensor, 118

Sturm-Liouville equation, 200 Subgroups, conjugate, 306 series)

invariant, 306

normal, 306

ELE\ Rub

-act*'''

Su>

of a

i

t^

.tors,

38

439

^pnvention,

t

OP PUKE AND APPLIED MATHEMATICS

ITS

1

v

fe,

l

.premum, 383 70 normal to, 56, 71 Symmetric functions, 320 fundamental theorem of, 321 Symmetric group, 302 System of equations, 4 orface,

Tan- 1

Transform, Fourier, 268 Laplace, 282 Transformations, affine, 313 bilinear, 125 continuous group, 314-315 infinitesimal, 315 Laplace (see Laplace transformatw orthogonal, 22 Schwarz-Christoffel, 163 Triangular matrix, 28 Trigonometric series, 252 Trihedral, 41, 53, 63 Triple scalar product, 47 Triple vector product, 48

z y 150

Tangent to a space curve, Tauber, example of, 455

52,

84

Taylor's series or expansion, 145, 461-

465 for functions of

n

variables,

464-465

Weierstrass

Union

of,

characteristic,

89

differentiation

451

of parameters, 197 Vector, 37 associated, 94 binormal, 53

24

column, 20

Tensors, absolute, 89 addition of, 89 product of, 89 quotient law of, 90 reciprocal, 92, 94 relative, 89

Termwise

for,

388

der Pol's equation, 273 Variation, bounded, 386 calculus of (see Calculus of variatioi

Riemann-Christoffel, 106

weight

M test

sets,

Vandermonde determinant, 334

mixed, 89 Ricci, 106

118 118

two

Van

metric, 93

strain,

of

test for,

Unitary matrix, 22

91

stress,

of

194 Unicursal curve, 435

Uniform continuity, 399 Uniform convergence, Abel's of a sequence, 446 of a series, 450

remainder in, 462 uniqueness of, 147 Tehebysheff's theorem, 354 Tensor, components, 89 contraction, 90 contravariant, 89 covariant, 89 curvature, 89 density, 89 inertia, 116 iso tropic,

Uncountable set, 394 Undetermined coefficients, method

components

and

integration,

448-450, 458 Theorem of the mean, differential calculus, 405 integral calculus, 422, 423, 457 Theory of games, 368-372

Torque, 116 Torsion of a space curve, 53 Total differential, 409 Totally ordered set, 377 Trajectories, orthogonal, 176 Transfer function, 270

of, 41,

81

physical, 87 conservative, 67 contravariant, 84 curl of, 59, 83, 102 derivative of, 50 of, 58, 102 irrotational, 69, 73

divergence

length of, 37, 94 normal, 53 operator del (V), 57 parallel displacement potential, 77 solenoidal, 77 space, 41

tangent, 52 unit, 37, 81 zero,

37

of,

108

INDEX otor field (see Field) -

tors,

addition

of,

Velocity, relative,

38

Volterra integral e&

j^

etJ

-ngle between, 42

Hfferentiation of, 50, 61 tquality of, 37 linear combination of,

108-109

scalar or dot product of, 42, 88

38 product of, 47 triple vector product of, 48 unit, fundamental, 41 vector or CTORS product of, 45 elocity, 50 angular, 46 subtraction

triple scalar

It!

equation, 172, 4. Weierstrass approximat thuW ^" 250 "** Weierstrass test, 451 Weierstrass-Bolzano theorem, 389 L Work, 112 Wronskian, 187 1

39

orthogonal, 82 parallel, 38,

Wave

M

4

*'

'<-

of,

Zermelo postulate, 392 Zero, Aleph, 394 Zero matrix, 12 Zero sum game, 368 Zeros, interlacing of, 199, 244 of orthogonal polynomials, 242

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