Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations

Подождите немного. Документ загружается.

9.6

TAYLOR

ANO

LAURENT

SERIES

253

(a)

Figure

9.9

(a) Power series withpositive exponents converge for the interiorpoints

circle.

(b)

Power

series

with

negative

exponents

converge

forthe

exterior

points

ofa

circle.

we look at the behavior of the sequence of partial sums

SN(Z) es

Ean(z

zo)".

n=O

Convergence

(9.13) implies that for any 8 > 0, there exists an integer N. such

that

whenever N

> Ns-

uniform

convergence

explained

power

series

are

uniformly

convergent

and

analytic

power

series

can

differentiated

and

integrated

term

In general, the integer

may be dependent on z; that is, for different values

z, we may be forced to pick different N. 's. When N. is independent

z, we say

that

theconvergence is uniform.

9,6.2. Theorem. The power series S(z) =

L~oan(Z

- ao)" is uniformly con-

vergent for all points within its circle

convergence and represents a function

that is analytic there.

By substituting thereciprocal

(z - zo) in the power series, we can show that

L~o

bk/(z

ZO)k

is convergent in the annulus rz < Iz-

zol

rl,

then it is

uniformly convergent for

all Z in that annulus.

9.6.3. Theorem. A convergentpower series can be differentiated and integrated

term

by term; that is, ifS(z) =

L~o

an(z- zo)".then

dS(z)

~=Enan(z-zo)n-I,

S(z)dz=Ea

(z-zo)ndz

n=l

Y n=O Y

for

any path y lying in the circle

convergence

thepower series.

(9.14)

254 9.

COMPLEX

CALCULUS

9.6.2 Taylor and Laurent Series

We now state and prove the two main theorems of this section. A Taylor series

consistsof

terms

withonly positivepowers.A

Laurent

seriesallowsfornegative

powersas well.

Taylor

series

9.6.4.

Theorem.

(Taylor series) Let f be analytic throughout the interior

circle

Co having radius ro and centered at

zoo

Then at each point z inside Co.

f(n)(zo)

f(z)

f(zo)

f'(zo)(z

zo)

+...= L I (z -

zo)".

n=O n.

That is. the powerseries converges to

f(z)

when [z-

zol

roo

Proof From the CIF and the fact that z is inside Co, we have

Maclaurin

series

f(z)

f(~)

d~.

2rrl feo

- z

On the other hand,

1 1 1

- z =

- zo +zo - z =

(~_

zo)

_z -

zo)

-zo

= 1 1 1

ZO)n

~-zo

1-~

~-zo~

~-zo

;

-zo

n_O

The last equality follows from the fact that I(z -

zo)/(~

zo)1

I-because

is inside the circle Co and

is on

it-and

from the sum of a geometric series.

Substituting in the CIF and using Theorem 9.5.2, we obtain the result. D

For

zo = 0 we obtain the

Maclanrin

series:

f(n)(o)

f(z)

= f(O) +J'(O)z +...=L

n=O

The Taylor expansion reqnires analyticity

the function at all points interior

to the circle Co. On many occasions there may be a point inside Co at which the

function is not analytic. The Laurent series accommodates such cases.

Laurent

series

9.6.5.

Theorem.

(Laurent series) Let

and C2 be circles

radii

and

ri.

both

centered at zo in the

Z-plane with

> rz.

Let

f :

--+

be analytic on C1 and

C2 and throughout S, the annular region between the two circles. Then, at each

point z

E S,

f(z)

is given by

f(z)

= L an(z -

zo)"

n=-oo

where

an =

f(~)

2rri f

- zo)n+l

and C is any contourwithin S that encircles

zo.

(9.15)

9.6

TAYLOR

AND

LAURENT

SERIES

255

Figure9.10 The

annular

region

within

andon whose

contour

the

expanded

function

analytic.

Proof

Let Ybe a small closedcontour in S enclosing z, as shown in Figure 9.10.

For

the composite contour

the Cauchy-Goursat theorem gives

f(~)

= J

f(~)

_ J

f(~)

_ J

f(~)

d~,

Yc'

- Z

YCI

- Z

Yc,

- z

-z

where the y and C2 integrations are negativebecause their interiorlies to our right

as we traverse them. The

y integral is simply 2rr:if(z) by the CIF. Thus,we obtain

2rr:if(z) = J

f(~)

_ J

f(~)

d~.

YCI

- Z

Yc,

- z

Now we use the same trick we used in deriving the Taylor expansion. Since z is

located in the annular region,

ri < [z -

zol

rl.

We have to keep this in mind

When expanding the fractions. In particular, for

E CI we want the

term in

the denominator, and for

E C2 we want it in the numerator. Substituting such

expansions in Equation (9.15) yields

f(~)d~

2rr:if(z) =

I:(z

- zo)"

(~_

)n+l

n=O Cl

I i

( )n+l

f(;)(~

zo)"

d~.

n=Oz-zo

(9.16)

Now we consider an arbitrary contour C in S that encircles

zoo

Figure 9.11

shows a region bounded by a' contour composed

CI and C. In this region

256 9.

COMPLEX

CALCULUS

1(1;)/(1;

- zo)n+l is analytic (because I; can never equal zo). Thus, the integral

overthe compositecontourmust vanish bythe Cauchy-Goursattheorem.

follows

that the integral over C

1 is equal to that over C. A similar argument shows that the

C2 integral can also be replaced by an integral over C. We let

n +1 =

-m

iu the

second sum of Equation (9.16) to transform it into

f: 1 m

1(1;)(1;

zo)-m-l

dl; = f (z - zo)m

1(1;)

~:+1

m~-1

(z - zo)

1'c

m=-oo

1'c

(I; - zo)

Changingthe dummy index backto n and substitutingthe resultin Equation(9.16)

yields

i 1(1;)

n=-1

i 1(1;)

2:n:i/(z) =

L(z

- zo)" (I; _ )n+l dl; + L (z - zo)" (I; _ )n+l dl;.

n=O C

-00

We can now combine the sums and divide both sides by

2:n:

i to get the desired

expansion. D

The Laurentexpansionisconvergeutas loug

as'2

< Iz-

I <

'I.

Iu particular,

if'2

= 0, and if the function is analytic throughout the interior of the larger circle,

then

anwill be zero

forn

-1,-2,

...

because1(1;)/(1;

_zo)n+l

will be analytic

for negative n, and the integral will be zero by the Cauchy-Goursattheorem. Thus,

ouly positive powers of (z - zo) will be present in the series, and we will recover

the Taylor series, as we should.

is clear that we can expand CI and shrink C2 until we encounter a poiut at

which I is no longer analytic. This is obvious from the constructiou of the proof,

in which ouly the analyticity in the annular region is important, not its size. Thus,

we can include all the possible analytic points by expanding CI and shrinking

C2.

9.6.6. Example. Letus

expand

some

functions

in tenusof

series.

Foran

entire

function

there

is nopointin the

entire

complex

plane

atwhichit is not

analytic.

Thus,

only

positive

powers

of (z - zo) willbe

present,

andwewill havea

Taylor

expansion

that

isvalidforall

values

ofz.

(a)Let us

expand

around

20=

Thenth

derivative

of e

is e

•

Thus,

j(n)(O)

= 1,and

Taylor

(Maclaurin)

expansion

gives

tIn)

(0)

e'=

L--z

L-'

n=O

(b)The

Maclaurin

series

forsinz is

obtained

noting

that

ifn

is even,

-r-r-smz =

,~o

(_I)(n-I)/2

ifnisod<!

and

substituting

thisin the

Maclaurin

expansion:

2k+l

sinz = L

(_I)(n-I)/2~

L(-I)k

z .

n odd n!

k~O

(2k +1)1

9.6

TAYLOR

AND

LAURENT

SERIES

257

Figure9.11 The

arbitrary

contour

inthe

annular

regionusedinthe

Laurent

expansion.

Similarly,

we can

obtain

Z2k

cosz =

{;,(_I)k

(Zk)!'

Z2k+l

sinhz =

(;,

""(Z"-k-+--'I'""')!'

Z2k

coshz =

(Zk "

k~O

(c)The

function

1/(1 +z) is not

entire,

sotheregionof its

convergence

limited.

Letus

find

the

Maclaurin

expansion

of this

function.

The

function

analytic

within

all circles

radii

r < 1. At r = 1 we

encounter

singularity,

thepointz =

-1.

Thus,the series

converges

forallpointsl! z forwhich[z] < 1.Forsuch

points

we have

/(n)(O)

-[(I

+z)-I]

. =

(_I)nnL

z=o

Thus,

_1_

= f

/(n)

(0)z" =

f(-I)nZn.

l+z

n=O n! n=O

III

Taylor and Laurent series allow us to express an analytic function as a

power

series.

For

a Taylor series

!(Z),

the expansion is routine because the coeffi-

cient

its

nth

term is simply

!(n)(zo)/nl,

where zo is the center

the circle

convergence.

When

a Laurent series is applicable, however,

the

nth

coefficient is

not, in general, easyto evaluate. Usually it

can

be found by inspection and certain

manipulations

other

known

series.

But

use

such an intuitive approach

remarked

before,

theseries

diverges

forall

points

outside

thecircle

lz!

= 1. Thisdoesnotmean

that

the

function

cannot

represented

by a seriesforpoints

outside

thecircle.Onthe

contrary,

we shallsee

shortly

that

Laurent

series,withnegative

powers

ofz - zearedesigned

precisely

forsucha

purpose.

258 9.

COMPLEX

CALCULUS

n+l

= 8kn

Laurent

series

unique

You

can

add,

subtract,

and

multiply

convergent

power

series

to determine the coefficients, can we be sure that we have obtained the correct

Laurent series? The following theorem answers this question.

9.6.7. Theorem.

the series

L~-oo

an(z - zo)n converges to

fez)

at all points

insome

annular

region

about

zo.

then

it isthe

unique

Laurent

series

expansion

fez)

in that region.

Proof

Multiply both sides of f (z) =

L~-oo

an(z - zo)n by

211:i(z

- zo)k+!'

integrate the result along a contour C in the annular region, and use the easily

verifiable fact that

1 J dz

211:i

(z -

zo)k

to obtain

I J

fez)

211:i

fc (z_ zo)k+! dz = ar.

Thus, the coefficient in the power series of

f is precisely the coefficient in the

Laurent series, and the two must be identical. 0

We will look at some examples that illustrate the abstract ideas developed in

thepreceding collectionof

theorems

and

propositions.

However,

we canconsider

a much broader range

examples

we know the arithmetic of power series.

The following theorem giving arithmetical manipulations with power series is not

difficult to prove (see [Chur 74]).

9.6.8. Theorem.

Letthetwopowerseriesf(z)

L~_ooan(z-zo)nandg(z)

L~-oo

(z - zo)n be convergent within some annularregion r2 < [z- zoI< TJ·

Then the sum

L~-oo

(an

+bn)(z

- zo)n converges to f (z)+g(z),and theproduct

00 00 00

L L anbm(z -

zo)m+n

Ck(Z

ZO)k

n=-oo

m=-oo

k=-QO

converges to

f(z)g(z)

for Z interior to the annular region. Furthermore, if

g(z)

0 for some neighborhood

zo.

then the series obtained by long divi-

sion

ofL~_ooan(Z

eo)"

L:'=-oo

bm(z - zo)" converges to

f(z)/g(z)

that neighborhood.

This theorem, inessence, says that converging power series canbe manipulated

as though they were finite sums (polynomials). Such manipulations are extremely

useful when dealing with Taylor.andLaurentexpansions in which the straightfor-

ward calculation of coefficients may be tedious. The following examples illustrate

the power

infinite-series arithmetic.

9.6.9.

Example.

rewriteit as

9.6

TAYLOR

AND

LAURENT

SERIES

259

Toexpandthefunctionf(z) =

22+3~

inaLaurentseriesaboutz =0,

fez) =

(2+

3Z) =

1_)

~(_I)nZn)

21+Z

I+z

L..

n=O

1 2 3 2 1 2

=-(3-I+z-z

-···)=-+--I+z-z

....

Z2 Z2 . Z

Thisseriesconverges for0 <

[z]

< 1.Wenotethatnegativepowersof z arealsopresent.

Using the notation

Theorem 9.6.5, we have

a_2

= 2,

a_I

= 1, an = 0 for n

:::s

-3,

and

an = (_1)n+1

forn

III

9.6.10.

Example.

The function fez) = 1/(4z - z2) is the ratio of two entire functions.

Therefore,by Theorem9.6.8,

itis

analyticeverywhereexceptat the zeros

its denominator,

Z = 0 and z =4. Fortileannularregion (herer2 ofTheorem9.6.5is zero)0 < lel < 4, we

expand

f(z)

in the Laurent series around z =

Instead of actually calculating an, we first

notethat

fez) =

4)·

The second factor can be expanded in a geometric series because Iz/41 < 1:

(~)

4-n

1 _

z/4

L..

4 -

L..

n=O

Dividing this by

cz,

and noting that z = 0 is the only zero of 4z and is exclnded from the

annularregion, we obtain the expansion

Althoughwe derivedthis series usingmanipulations

otherseries, the uniqueness

series

representations assures us that this is the Laurent series for the indicated region.

How can we represent

fez)

in the region for which

Iz[

> 4? This region is exterior to

the circle

lel

= 4, so we expect negative powers

z. To find the Laurent expansion we

write

fez) = -

C_1

and note that ]4/zl < 1 for points exterior to the larger circle. The second factor can be

written as a geometric series:

12This is a reflection

the fact that the function is not analytic inside the entire circle

lz!

= 1; it blows up at z = O.

260 9.

COMPLEX

CALCULUS

Dividingby

_z2,

whichis

nonzero

intheregion

exterior

tothe

larger

circle,yields

f(z)

= -

I>"z-n-2

n=Q

11II

9.6.11. Example. The function

fez)

= z([(z

-1)(z-2)]

has a Taylor expansion around

theoriginIorlz] < 1.To

find

this

expansion,

we write

1 2 1 1

f(z)

=--1

+-2

-1

---1

(2·

-z -z

Expanding both fractions in geometric series (hoth [z]and Iz(21are less than 1), we ohtain

fez)

L~o

z" -

L~O(z(2)n.

Adding the two

series-nsing

Theorem

9.6.8-yie1ds

fez)

L(l-

2-")zn

n=O

for Izl < 1.

This is the unique Taylor expansion

fez)

within the circle Izi = 1.

For1 <

Izi

-c 2we havea Laurent

series.

obtain

this

series,

write

l(z

1 1 (

fez)

l(z

- 1 - 1 -

z(2

-z

1 -

l(z

- 1 -

z(2·

Since hoth fractions on the RHS converge in the annuiarregion (11(zl < 1, Iz(21 < 1), we

get

100

00 00

fez)

= -zL

(z)

- L m= -

Lz-

L2-

n=D n=O n=O n=O

-00

00 00

L z">

Lz-nz

= L

anZ

n=-l

n=O

n=-oo

where

an =

-1

forn < 0 and an =

_Z-n

farn

:::

Thisis the unique Laurentexpansion

of f (z) in thegivenregion.

Finally,

for

Izl

> 2 we

have

only negative

powers

ofz. We

obtain

the

expansion

inthis

region hy rewriting f (z) as follows:

l(z

2(z

fez)

-1

l(z

+ 1 -

2(,

Expanding

the

fractions

yields

00 00 00

fez)

= - L

z-n-I

+ L 2

1z-

L(2n+l

1)z-n-1.

n=O n=O n=O

Thisis

again

the

unique

expansion

f(z)

in theregion[z] > 2.

11II

13We

could,of

course,

evaluate

the

derivatives

of all

orders

of the

function

atz = 0 anduse

Maclaurin's

formula.

However,

the

present

methodgivesthesameresultmuchmore

quickly.

9.6 TAYLOR

ANO

LAURENT

SERIES

261

9.6.12.

Example.

Definef (z) as

{

(I - cosz)/z2 for z

(z) = 1

2 for z =

Wecanshow

that

f(z)

is an

entire

function.

Since 1 - cosZ andz2

are

entire

functions,

their

ratio

analytic

everywhere

exceptat

thezerosof its

denominator.

Theonlysuchzerois z =

Thus,

Theorem

9.6.8

implies

that

f(z)

is analyticeverywhere exceptpossiblyatz =

Toseethebehaviorof

f(z)

atz = 0,

we lookatits

Maclaurin

series:

z2n

cosz =

L(-I)n_-,

n~O

(2n)!

whichimplies

that

The expansion on the RHS shows

that

the value of the seriesat z = 0 is !'

which,

definition, is f(O). Thus,the seriesconverges for all z, andTheorem9.6.2saysthat

f(z)

entire.

III

A Laurent series

can

give information about the integral

a function around

a closed contour in whose interior the function may

not

be analytic. In fact, the

coefficient

the first negative

power

in a Laurent series is given by

a_I

= .2., J

f(~)

d~.

21f1

(9.17)

Thus, to find

the integral

a (nonanalytic) function around a closed contour sur-

rounding

zo,

we writethe Laurentseries for the functionand read

off

the coefficient

the

I/(z

zo)

term.

9.6.13.

Example.

As an illustrationof thisidea.Iet us evaluatethe integralI =

dz/

[z2(z - 2)], whereC is the circleof radius I centeredatthe origin.Thefunctionis analytic

in the

annular

region

0 < [z] < 2. Wecan

therefore

expand

it as a

Laurent

series

about

z =0 in thatregion:

I I (

(z)n

z2(z - 2) = - 2z

I -

z/2

= - 2z

2];

=-~

C2)

....

Thus,

-1,

and

dz/[z2(z

- 2)] = 21tia_l =

-i1t/2.

A directevaluationof the

integral

nontrivial.

fact,

wewill see

later

that

find

certain

integrals,

itis

advantageous

to cast them in the form of a contourintegraland use either Equation (9.17) or a related

equation. II

262

COMPLEX

CALCULUS

Let f : C

C be analytic

zoo

Then by definition, there exists a neighbor-

hood

ofzo

in which f is analytic.

particular, we can find a circle Iz-zol = r > a

in whose interior f has a Taylor expansion.

zero

order

k 9.6.14. Definition.

Let

f(n) (eo)

f(z)

= L I (z -

zo)"

Lan(Z

- zn)".

n=O

Then f is saidto have azero

oforderk

atzo

iff(n) (zn) =

Oforn

= 0, 1,

...

k-l

but

f(k)

(zo)

that case

f(z)

= (z -

ZO)k.

I:~o

ak+n(Z- zo)n, where ak

aand [z

-zol

< r.

We define

g(z)

= L ak+n(z - zo)"

n=O

where [z -

zol

< r

the

zeros

ofan

analytic

function

are

isolated

simple

zero

and note that g(zo) = ak

Convergence of the series on the RHS implies that

g(z)

is continuous at

ZOo

Consequently, for each E > 0, there exists 8 such that

Ig(z) - akl < E whenever [z -

zol

8.lfwe

choose E = lakl/2, then, for some

80>

0, Ig(z) - akl < lakl/2 whenever [z - zo] < 80.Thus, as long as z is inside

the circle [z - zn]

< 80,

g(z)

cannot vanish (because if it did the first inequality

would imply that

lakl < lakl/2). We therefore have the following result.

9.6.15.

Theorem.

Let

f : C

be analytic at zo

and

f (zo) =

Then there

exists a neighborhood

zo throughout which f has no other zeros unless f is

identically zero there. Thus, the zeros

an analyticfunction are isolated.

Whenk = 1, we say that zo is a simple zero

To find the order of the zero

of a function at a point, we differentiate the function, evaluate the derivative at that

point,

and

continue

the

process

untilwe

obtain

nonzero

valueforthe

derivative.

9.6.16.

Example.

(a) The zerosof cosz, which are z = (2k + 1),,/2, are all simple,

because

coszl = - sin[(2k +

I)::]

z~(2k+I)n/2

(b)Tofindtheorderof thezeroof f (z) = e

- 1- z -

z2/2

atz = 0, wedifferentiate

f(z)

andevaluate f'(O):

f'(O) = (e

-I-z)z=o

Differentiating

again

gives f"(O) = (e

l)z=O =

Differentiating

once

yields

(0) =

(eZ)z~o

Thus,thezero is oforder3.

III