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
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
$j(z) is analytic at z
=
*,(*)
where
of a
Com-
In this case
-
a)"fc(*)
j
=
1,
2
(5.109)
of (5.108)
(5.110)
Now
a; #/(a) 5^ 0.
= so that
Thus j
and
-
<f> 2 <f>{'
-T-
(^2^1 /
x
p(z)
^1^2'
+
^1^2)
p(^)(^2^{
=
~
--
= - ^7
=
<Pi^ 2 )
p(z)(<f>*<P\
^w/
i
1
r
2
dW(<pi, j dz
'
N
<pz)
(5.H2) ^ '
ELEMENTS OF PURE AND APPLIED MATHEMATICS
210
Making use
show that
of (5.110) to (5.112), let the reader
=
P
+ P*M
r
(5.113)
with p*(z), g*(z) analytic at
=
=
The same
a.
result occurs for
<p(z)
of
We
(5.109). z
z
notice that p(z) of (5.113) has at most a simple pole at = a. This suggests the q(z) has at most a double pole at z
simplify matters, we assume that z = 0. ZQ transfers the singularity to the origin.
To
regular singular point.
The transformation Let w(z)
=
analytic at 2 = 20find a solution of (5.114) in the neighborhood of a to attempt zo)*q(z) is
-
1)
+
= 2a + zp(z) = 2a = a(pi + P& +
ap Q + q + p + p& + p& 2 ) + 0! + q& +
u(z) satisfies (5.116), then w(z)
=
zu(z)
(5.117)
+
'
'
'
will satisfy (5.114)
provided a satisfies (5.117). The existence of a solution of (5.114) hinges on the existence of a solution of (5.116). If u(z) satisfies (5.116), u(0)
= w
DIFFERENTIAL EQUATIONS
211
then a double integration of (5.116) yields
zu(z)
=
-
[P(0)
and
=
u(z)
With
-
[P(0)
=
B(Z, r)
We leave
+
l]u*
[* Jo
-
+
P(r)
+\j B
I]u
2
-
[2
(z >
+(r -
P(r)
(T
r
-
z)(Q(r)
MO dr
P'(T))]I*(T) dr
*
*
-
*)[$(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
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.
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,
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. ,
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.
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.
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,
> 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
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
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
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.
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
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
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
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
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
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,
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
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