Chow T.L. Mathematical Methods for Physicists: A Concise Introduction

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Example A1.8

n1

1=n

converges since lim

M!1

dx=x

 lim

M!1

1 ÿ 1=M exists.

Problem A1.10

Find the limit of the sequence 0.3, 0.33, 0:333; ...; and justify your conclusion.

Alternating series test

An alternating series is one whose successive terms are alternately positive and

negative u

ÿ u

 u

ÿ u

: It converges if the following two conditions are

satis®ed:

(a) ju

n1

j for n  1; (b) lim

n!1

 0



or lim

n!1

 0



The sum of the series to 2M is

u

ÿ u

u

ÿ u

u

2Mÿ1

ÿ u



 u

ÿu

ÿ u

ÿu

ÿ u

ÿÿu

2Mÿ2

ÿ u

2Mÿ1

ÿu

Since the quantities in parentheses are non-negative, we have

 0; S

 S

S

 u

Therefore fS

g is a bounded monotonic increasing sequence and thus has the

limit S.

Also S

2M1

 S

 u

2M1

. Since lim

M!1

 S and lim

M!1

2M1

 0

(for, by hypothesis, lim

n!1

 0), it follows that lim

M!1

2M1



lim

M!1

 lim

M!1

2M1

 S  0  S. Thus the partial sums of the series

approach the limit S and the series converges.

516

APPENDIX 1 PRELIMINARIES

Figure A1.3.

Problem A1.11

Show that the error made in stopping after 2M terms is less than or equal to

2M1

Example A1.9

For the series

1 ÿ





n1

ÿ1

nÿ1

;

we have u

ÿ1

n1

=n; u

 1=n; u

n1

 1=n  1. Then for n  1;

n1

 u

. Also we have lim

n!1

 0. Hence the series converges.

Absolute and conditional convergence

The series

is called absolutely convergent if

converges. If

con-

verges but

diverges, then

is said to be conditionally convergent.

It is easy to show that if

converges, then

converges (in words , an

absolutely convergent series is convergent). To this purpose, let

 u

 u

u

 u

 u

 u

;

then

 T

u

 u

jj

u

 u

jj

u

 u

jj

 2 u

 2 u

2 u

Since

converges and since u

 u

 0, for n  1; 2; 3; ...; it follows that

 T

is a bounded monotonic increasing sequence, and so

lim

M!1

S

 T

 exists. Also lim

M!1

exists (since the series is absolutely

convergent by hypothesis),

lim

M!1

 lim

M!1

S

 T

ÿ T

 lim

M!1

S

 T

ÿ lim

M!1

must also exist and so the series

converges.

The terms of an absolutely convergent series can be rearranged in any order,

and all such rearranged series will converge to the same sum. We refer the reader

to text-book s on advan ced calculus for proof.

Problem A1.12

Prove that the series

1 ÿ



ÿ

converges.

517

INFINITE SERIES

How do we test for absolute convergence? The simplest test is the ratio test,

which we now review, along with three others ± Raabe's test, the nth root test, and

Gauss' test.

Ratio test

Let lim

n!1

n1

jL. Then the series

(a) converges (absolutely) if L < 1;

(b) diverges if L > 1;

Let us consider ®rst the positive-term

, that is, each term is positive. We

must now prove that if lim

n!1

n1

 L < 1, then necessarily

converges.

By hypothesis, we can choose an integer N so large that for all

n  N; u

n1

 < r, where L < r < 1. Then

N1

< ru

; u

N2

< ru

N1

< r

; u

N3

< ru

N2

< r

; etc:

By addition

N1

 u

N2

< u

r  r

 r



and so the given series converges by the comparison test, since 0 < r < 1.

When the series has terms with mixed signs, we consider u

 u

,

then by the above proof and because an absolutely convergent series is

convergent, it follows that if lim

n!1

n1

 L < 1, then

converges

absolutely.

Similarly we can prove that if lim

n!1

n1

 L > 1, the series

diverges.

Example A1.10

Consider the series

n1

ÿ1

nÿ1

. Here u

ÿ1

nÿ1

. Then

lim

n!1

n1

 lim

n!1

=n  1

 2. Since L  2 > 1, the series diverges.

When the ratio test fails, the following three tests are often very helpful .

Raabe's test

Let lim

n!1

n1 ÿ u

n1

jj

 `, then the series

(a) converges absolutely if `<1;

(b) diverges if `>1.

The test fails if `  1.

The nth root test

Let lim

n!1



 R, then the series

518

APPENDIX 1 PRELIMINARIES

(a) converges absolutely if R < 1;

(b) diverges if R > 1.

The test fails if R  1.

Gauss' test

n1

 1 ÿ



;

where jc

j < P for all n > N, then the series

(a) converges (absolutely) if G > 1;

(b) diverges or converges conditionally if G  1.

Example A1.11

Consider the series 1  2r  r

 2r

 r

 2r

. The ratio test gives

n1



2 r

; n odd

=2; n even

;



which indicates that the ratio test is not applicable. We now try the nth root test:







2 r





; n odd



 r

; n even

(

and so lim

n!1



 r

. Thus if jrj < 1 the series converges, and if jrj > 1 the

series diverges.

Example A1.12

Consider the series





1  4

3  6





1  4  7

3  6  9





1  4  7 3n ÿ 2

3  6  9 3n



:

The ratio test is not applicable, since

lim

n!1

n1

 lim

n!1

3n  1

3n  3

 1:

But Raabe' s test gives

lim

n!1

n 1 ÿ

n1



 lim

n!1

n 1 ÿ

3n  1

3n  3



()



> 1;

and so the series converges.

519

INFINITE SERIES

Problem A1.13

Test for convergence the series





1  3

2  4





1  3  5

2  4  6





1  3  5 2n ÿ 1

1  3  5 2n



:

Hint: Neither the ratio test nor Raabe's test is applicable (show this). Try Gauss'

test.

Series of functions and uniform convergence

The series consider ed so far had the feature that u

depended just on n. Thus the

series, if convergent, is represented by just a number. We now consider series

whose terms are functions of x; u

 u

x. There are many such series of func-

tions. The reader should be familiar with the power series in which the nth term is

a constant times x

Sx

n0

: A1:4

We can think of all previous cases as power series restricted to x  1. In later

sections we shall see Fourier series whose terms involve sines and cosines, and

other series in which the terms may be polynomials or other functions. In this

section we consider power series in x.

The convergence or divergence of a series of functions depends, in general, on

the values of x. With x in place, the partial sum Eq. (A1.2) now becomes a

function of the variable x:

xu

 u

xu

x: A1:5

as does the series sum. If we de®ne Sx as the limit of the partial sum

Sxlim

n!1

x

n0

x; A1:6

then the series is said to be convergent in the interval [a, b] (that is, a  x  b), if

for each ">0 and each x in [a, b] we can ®nd N > 0 such that

Sxÿs

x

<"; for all n  N: A1:7

If N depends only on " and not on x, the series is called uniformly convergent in

the interval [a, b]. This says that for our series to be uniformly convergent, it must

be possible to ®nd a ®nite N so that the remainder of the series after N terms,

iN1

x, will be less than an arbitrarily small " for all x in the given interval.

The domain of convergence (absolute or uniform) of a series is the set of values

of x for which the series of functions converges (absolutely or uniformly).

520

APPENDIX 1 PRELIMINARIES

We deal with power series in x exactly as before. For example, we can use the

ratio test, which now depends on x, to investigate convergence or divergence of a

series:

rxlim

n!1

n1

 lim

n!1

n1

 x

lim

n!1

n1

 x

r; r  lim

n!1

n1

;

thus the series converges (absolutely) if jxjr < 1or

jxj < R 

 lim

n!1

n1

and the domain of convergence is given by R : ÿR < x < R. Of course, we ne ed to

modify the above discussion somewhat if the power series does not contain every

power of x.

Example A1.13

For what value of x does the series

n1

nÿ1

=n  3

converge?

Solution: Now u

 x

nÿ1

=n  3

,andx 6 0 (if x  0 the seri es converges). We

have

lim

n!1

n1

 lim

n!1

3n  1



Then the series converges if jxj < 3, and diverges if jxj > 3. If jxj3, that is,

x 3, the test fails.

If x  3, the series becomes

n1

1=3n which diverges. If x ÿ3, the series

becomes

n1

ÿ1

nÿ1

=3n which converges. Then the interval of convergence is

ÿ3  x < 3. The series diverges outsi de this interval. Furthermore, the series

converges absolutely for ÿ3 < x < 3 and converges conditionally at x ÿ3.

As for uniform convergence, the most commonly encountered test is the

Weierstrass M test:

Weierstrass M test

If a sequence of positive constants M

; M

; ...; can be found such that: (a)

ju

xj for all x in some interval [a, b], and (b)

converges, then

x is uniformly an d absolutely convergent in [a, b].

The proof of this common test is direct and simple. Since

converges,

some number N exists such that for n  N,

iN1

<":

521

SERIES OF FUNCTIONS AND UNIFORM CONVERGENCE

This follows from the de®nition of convergence. Then, with M

ju

xj for all x

in [a, b],

iN1

x

<":

Hence

Sxÿs

x



iN1

x

<"; for all n  N

and by de®nition

x is uniformly convergent in [a, b]. Furthermore, since we

have speci®ed absolute values in the statement of the Weierstrass M test, the series

x is also seen to be ab solutely convergent.

It should be noted that the Weierstrass M test only provides a sucient con-

dition for uniform convergence. A series may be uniformly convergent even when

the M test is not applicable. The Weierstrass M test might mislead the reader to

believe that a uniformly convergent series must be also absolutely con vergent, and

conversely. In fact, the uniform convergence and absolute convergence are inde-

pendent properties. Neither implies the other.

A somewhat more delicate test for uniform convergence that is especially useful

in analyzing power series is Abel's test. We now state it without proof.

Abel's test

If a u

xa

x, and

 A, convergent, and (b) the functions f

x are

monotonic f

n1

xf

x and bounded, 0  f

xM for all x in [a, b], then

x converges uniformly in [a, b].

Example A1.14

Use the Weierstrass M test to investigate the uniform convergence of

a

n1

cos nx

; b

n1

3=2

; c

n1

sin nx

Solution:

(a) cosnx=n

 1=n

 M

. Then since

converges (p series with

p  4 > 1), the series is uniformly and absolutely convergent for all x by

the M test.

(b) By the ratio test, the series converges in the interval ÿ1  x  1 (or jxj1).

For all x in jxj1; x

3=2

 x

3=2

 1=n

3=2

. Choosing M

 1=n

3=2

we see that

converges. So the given series converges uniformly for

jxj1 by the M test.

522

APPENDIX 1 PRELIMINARIES

. However,

does not converge. The M test

cannot be used in this case and we cannot conclude anything about the

uniform co nvergence by this test.

A uniformly convergent in®nite series of functions has many of the properties

possessed by the sum of ®nite series of functions. The following three are parti-

cularly useful. We state them without proofs.

(1) If the individual terms u

x are continuous in [a, b] and if

x c on-

verges uniformly to the sum Sx in [a, b], then Sx is continuous in [a, b].

Brie¯y, this states that a uniformly convergent series of continuous func-

tions is a continuous function.

(2) If the individual terms u

x are continuous in [a, b] and if

x c on-

verges uniformly to the sum Sx in [a, b], then

Sxdx 

n1

xdx

n1

xdx 

n1

xdx:

Brie¯y, a uniform convergent series of continuous functions can be inte-

grated term by term.

(3) If the individual terms u

x are continuous and have continuous derivatives

in [a, b] and if

x converges unifor mly to the sum Sx while

x=dx is uniformly convergent in [a, b], then the derivative of the

series sum Sx equals the sum of the individual term derivatives,

Sx

n1

x or

n1

x

()



n1

x:

Term-by-term integration of a uniformly convergent series requires only con-

tinuity of the individual terms. This con dition is almost always met in physical

applications. Term-by-term integration may also be valid in the absence of uni-

form convergence. On the other hand term -by-term diÿerentiation of a series is

often not valid because more restrictive conditions must be satis®ed.

Problem A1.14

Show that the series

sin x



sin 2x



sin nx



is uniformly convergent for ÿ  x  .

523

SERIES OF FUNCTIONS AND UNIFORM CONVERGENCE

Theorems on power series

When we are worki ng with power series and the functions they represent, it is very

useful to know the following theorems which we will state without proof. We will

see that, within their interval of convergence, power series can be handled much

like polyno mials.

(1) A power series converges uniformly and absolutely in any interval which lies

entirely within its interval of convergence.

(2) A power series can be diÿerentiated or integrated term by term over any

interval lying entirely within the interval of convergence. Also, the sum of a

convergent power series is continuous in any interval lying entirely within its

interval of convergence.

(3) Two power series can be added or subtracted term by term for each value of

x common to their intervals of convergence.

(4) Two power series, for example,

n0

and

n0

, can be multiplied

to obtain

n0

; where c

 a

 a

nÿ1

 a

nÿ2

a

, the

result being, valid for each x within the common interval of convergence.

(5) If the power series

n0

is divided by the power series

n0

where b

6 0, the quotient can be written as a power series which converges

for suciently small values of x.

Taylor's expansion

It is very useful in most applied work to ®nd power series that represent the given

functions. We now review one method of obtaining such series, the Taylor expan -

sion. We assume that our function f x has a continuous nth derivative in the

interval [a, b] and that there is a Taylor series for f x of the form

f xa

 a

x ÿ a

x ÿ 

 a

x ÿ 

a

x ÿ 

;

A1:8

where  lies in the interval [a, b]. Diÿerentiating, we have

xa

 2a

x ÿ 3a

x ÿ 

na

x ÿ 

nÿ1

;

x2a

 3  2a

x ÿ 4  3a

x ÿ a

nn ÿ 1a

x ÿ 

nÿ2

;

n

xnn ÿ 1n ÿ 21  a

 terms containing powers of x ÿ :

We now put x   in each of the above derivatives and obtain

f a

; f

a

; f

2a

; f F3!a

; ; f

n

n!a

;

524

APPENDIX 1 PRELIMINARIES

where f

 means that f x has been diÿerentiated and then we have put x  ;

and by f

 we mean that we have found f

x and then put x  , and so on.

Substituting these into (A1.8) we obtain

f xf f

x ÿ 

x ÿ 



n

x ÿ 

:

A1:9

This is the Taylor series for f x about x  . The Maclaurin series for f x is the

Taylor series about the origi n. Putting   0 in (A1.9), we obtain the Maclaurin

series for f x:

f xf 0f

0x 

0x



f F0x



n

0x

:

A1:10

Example A1.15

Find the Maclaurin series expansion of the exponential function e

Solution: Here f xe

. Diÿerentiating, we have f

n

01 for all

n; n  1; 2; 3 .... Then, by Eq. (A1.10), we have

 1  x 



 

n0

; ÿ1< x < 1:

The following series are frequently employed in practice:

1 sin x  x ÿ



ÿ1

nÿ1

2nÿ1

2n ÿ 1!

; ÿ1< x < 1:

2 cos x  1 ÿ



ÿ1

nÿ1

2nÿ2

2n ÿ 2!

; ÿ1 < x < 1:

3 e

 1  x 





nÿ1

n ÿ 1!

; ÿ1 < x < 1:

4 lnj1  xjx ÿ



ÿ1

nÿ1

; ÿ1 < x  1:

5

1  x

1 ÿ x

 x 





2nÿ1

2n ÿ 1

; ÿ1 < x < 1:

6 tan

ÿ1

x  x ÿ



ÿ1

nÿ1

2nÿ1

2n ÿ 1

; ÿ1  x  1:

71  x

 1  px 

pp ÿ 1



pp ÿ 1p ÿ n  1

:

525

TAYLOR'S EXPANSION