7-advanced calculus book - chapter five

8/12/2019 7-Advanced Calculus Book - Chapter Five

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402 5 Series of Funct ions

O f c o u r s e as remarked above, an absolutely convergent series is conver

gent. In th e case o f absolu te convergence we ca n p o se a comparison test,just as fo r series o f number s .

Theorem 5.1. (Comparison Test) L et {fk} be a sequence of continuous

functions defined on a set X. Suppose there is a sequence {pk} of positivenumber s and an integer N > 0 such that

(0 \\fk\\ N

00

(ii)Z

Pk k l

Then /fc converges absolutely.

Proof. T h e verification is th e same as tha t fo r n u m b e r series (Theorem 2.3).

Examples

1. zk VU z) uniformly and absolutely in {|z| r} fo r an y 1. F o r zk rk in t h a t domain, and

Zr Tzr o o

2. zk does not converge uniformly in {\z\ 1} In fact, th e

series is not a Cauchy sequence o f functions, because fo r every n

\ f k L,fkk= l

Z

1

Thus, fo r \ say there is no JV such tha t || =,,/ i fo r al lm > n > N, in fact, not even fo r m n .

3. e7 X^=1 zk k\ converges uniformly in a n y disk {|z| R} withR finite. Again, by comparison

Rk R

*T\and zr


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5. 1 Convergence 405

E X E R C I S E S

1. F o r what values o f x d o these series o f functions c o n v e rg e absolutely:

a) 2 2 V d) | jr+18)n=0 n=0

Nco s nx

b) 2 e> 2 n 1 X n 0

c) | e f | x< 2>n= 0 n 0

2. In which d o m ain s o f th ecomplex plane

d o th ese seriesconverge?

a) | h z- b) | il c | Vn 0 n 0 Z ) n 0

3. Wh i c h o f these series can be differentiated or integrated on their

domain o f c o n v e rg e n c e ?

a) Exercise 1 a) c) Exercise 1 d)

cos nx

b) Exercise 1 b) d) 2n 0 fl

4. Fin d th e p o w e r series expansion fo r these funct ions :

c) f e 1 dtJo

n m y

a) l + x 2 )

b)

P R O B L E M S

1. a) Find a p o w e r series expansion fo r sin x cos x

Hint : 2 si n x cos x sin 2x.)

b) F i nd p o w e r series expansions fo r sin2 x an d cos2 x

2. P r o v e Proposition 2.

3. Sh o w that

li m 2

log 2 l n l

C an y o u conclude

2 L_ii=i0g2?


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406 5 Ser ies of Funct ions

5.2 T he Fundamental T heorem o f Algebra

F o r th e r emainder o f this chapter we restrict attention exclusively to

complex-valued funct ions o f a complex variable. T he simplest class o f suchfunct ions ar e th e polynomial functions ; t h a t is, funct ions o f th e form

P(z) = az a^iz 1 axz a0 ,

(where th e at are complex numbers). W e shall always assume a 0; inthis case n is called th e degree o f P . I t is a basic fact o f mathemat i cs tha t

e v e r y polynomial h as a root ; tha t is, there is a n u m b e r c e C such tha t P(c) =

0. T he proof o f this fact consists in a systematic investigation o f th e analyticproperties o f polynomials. First, we recall de M oivre s theorem.

L e m m a . Every nonzero complex number has n distinct nth r o o t s .

Proof. L et c e C, c ^ 0. F o r this p u r p o s e , th e polar representation c re ism o s t convenient . A n n th root o f c is a n u m b e r w pe * s u c h t h a t p r an dem* ei8. tnat jSj n^

q js an integral multiple o f 2tt . L et

277 47T 2 n k 2tt(u 1)i

=

. i expO aj), . . . , a> exp(z a) ar e al l distinct an d have th e property co n = . These are called th e n th roo t s o f unity. Now, if p (r)1 a n d f 8 In,then (pe *)n

reie=

c. T h e n u m b e r s pe^ioi, pe cu,, , pe *co are t hen al ldistinct, an d ar e all n th roots o f c.

Now, we need tw o deeper facts depending on th e continuity properties o f

polynomials. T h e first is intuitively clear: t h a t \P(z)\ gets arbitrarily large asz - c o . T he second is th e cru cial fa ct fo r th e fundamenta l theorem: th e

place w here a polynomial h as a minimum modulus m u s t be a root.

L e m m a . L e t P(z)=

anz ax z a0 be a polynomial of degree n>0.(i) lim \P(z)\ = oo, that is, given any M > 0 there is a K>0 such that

|z|->CO

|P(z)| > M whenever \z\ > K.

(ii) If P(z0) 0, then z0 c a n n o t be a m i n i m u m point fo r \P\; that is, therear e z close to z0 such that \P(z)\ < |P(z0)|.


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5.2 The Fundamenta l Theorem of Algebra 40 9

(since ai ^ 0). N o w we consider th e case o f general degree n , assuming th e corollaryfo r polynomials o f degree n 1. By th e theorem, the re is a point c such t h a t

P(c) = 0. T h e n

P(z) = P(z) P(c) = 2 *(z c* = 2 *(* o zW-1 J

n-l/ n \

T he factor on th e right is a polynomial o f degree n 1, so th e induct ion assumption

applies: it can be wri t ten as a(z Zi) (z z_i) fo r suitable a ^ O , zu z i

Thus, writing c=

, we obtain

P( z ) = a ( z - z i ) - - - ( z - z ) (5-4)

This factorization is clearly unique, except fo r th e order o f th e z; s: a is th e

leading coefficient o f P and {zu . . . , z} are th e roots o f P . O f c o u r s e ,

Zjl, . . . , z need n o t be distinct; le t ru . . . , rs b e th e se t o f distinct r o o t s .I f

we le t m,. be th e n u mb e r o f occurrences o f th e root r{in th e list {zu . . . z j,

m t is called th e multiplicity o f th e r o o t r{ . W e ca n rewrite (5.4)as

P(z) = a(z rt mi (z rs m (5.5)

an d clearly m t + + m s = n , th e degree o f P .

Before concluding th is se ctio n we should remark on th e factorization o f

real polynomials. Rea l polynomials need not have real roots (viz.,z2 +

= 0), b u t their complex roots come in conjugate pairs. L et P(z)= az -\

+ ax z + a0 be a real polynomial. If P(r)= 0, then

P(f) = ani? n +---+a1r + a0=(anr +--- + a1 z + a0)~= PirY 0

so r is also a root o f P . Since

(z r)(z -r = z2 - ( r + r)z + rr

z2-2 Re(r)z + \r\2

th e polynomial has real coefficients. Thus,if we rea r ran g e th e r o o t s o f P

intoth e real roo t s

ru...,rk and th e conjugate pairs rk + l, r k + u . . . , r , , r , we

ca n rewrite th e factorization 5.5 into a product o f l inear and quadratic

real polynomials.

P(z) = a(z rj 1 (z

rk m\z2 2 Rfi(rt+1)z + k+il2)

(z2-2Re(rt)z + |rr|2


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P R O B L E M S

4. L et w i . . . co be th e n nth roots o f unity. Sh o w that they are

arranged at n equidistant points a round th e uni t ci rc le . Sh o w t h a t th e sets

{cui, . oi}, {a>i, oil2, . . oj c1} ar e th e s a m e , if u> , is th e n e a r e s t such

point to 1.5. L et a> u . . . , w be th e n th roots o f unity. Choose k so t h a t kn 2 is

divisible by 4. Show that ikoju ikwn are th e n th r o o t s o f 1 .

6. Show that : (a) deg PQ = deg P + deg Q.

(b) deg(P + Q) = max(deg P, deg Q) if deg P deg Q.

(c) W h e n is th e equation in (b) not t rue?7. Given tw o polynomials P, Q show that the re is a polynomial R which

fa cto rs b o th P, Q an d is factored by a n y polynomial which factors both

P, Q. R is called th e greatest common divisor o f P and Q.8. Show that a real polynomial o f o d d degree h as a r ea l r o o t .9. Pr o v e tha t th e polynomial + zm a (m > 0) h as no m i nim u m m o d ulu s

a t z = 0.

10. F o r P(z) = 2S =o az a polynomial, le t

P (z)= Znanz 1n= 1

(a) Verifythat th e t rans fo rmat ion P^P ' is l inear and satisfies

( P Q Y = P Q + P Q

(by induc t ion on deg P) .

(P - * P is a complex analog o f differentiation)

(b) Pr o v e that r is a multiple r o o t o f P if a n d only if P(r) = 0 an d

P (r) = 0.

c) Define P =(P ) , P = (P ) , a n d so on. Then r is a root o f Po f at least multiplicity m if and only if P(r) = P (r) = = Pfm *> r) = 0.

5 .3 Cons t a n t Coefficient L i n ea r D i ff er en tia l Equations

N ow t h a t we k n o w th e factor izat ion th eo re m fo r polynomials we can return

to complete th e study o f constant coefficient equations in one u n k n o w n funct ion. L et L be a c o n s t a n t coefficient differential operator o f o rd er k ; tha t is,L is a mapping from functions to funct ions defined by

L(f) = fm + . f{i) ^ C (5-6)i 0


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5.3 C o nstan t Coefficient Li n ea r Differential Equations 411

Corresponding to L is th e polynomial

PLiX)

Xk kZai i> o

called th e characterist ic polynomial o f L. W e recall th e facts t h a t we alreadyk n o w a b o u t such differential operators.

Theorem 5 .4 . L e t L be given by 5.6). The collection S L) of solutions ofthe equation L f = 0 is an n-dimensional vector space of infinitely differ entiable

functions. If r is a root ofPL X) 0, then erx e S L).

N o w if all th e r o o t s o f th e characterist ic polynomial are distinct, we have n

solutions o f Lf 0, an d it is easily verified Problem t h a t they are inde

pendent. T h u s they sp an S L). T o examine th e case o f multiple roots,we m u s t ex ami n e more closely th e relationship between th e given differential

operator a n d it s characterist ic polynomial. I f P is a polynomial, we will le t

LP represent th e corresponding operator; t h a t is, fo r P X)

j= 0 a,-* , FPis defined by

i 0

Now, fro m w hat we already k n o w a b o u t these differential equations we ca n

guess t h a t th e factorization o f P will tell us all we w a n tto k n o w a b o u t LP .

In fact, we ca n fac tor th e corresponding operator accordingly as th e n e x t

l e m m a shows.

L e m m a 1. Lp+q LP Lq\ LPq LpLq .

Proof. O f c o u r s e , LPLQ is defined as th e composition o f operators : LPLQ) f)

LP LQ f)). T h e first equation is obvious . T h e second takes a little work . W e

will p r o v e it by induct ion on th e degree o f P . I f deg P

0, that is, P x)

a 0 , t hen

P Q = a 0 Q and LPQ f) a0LQ f) =LP LQ f)\ fo r a n y sufficiently differentiable

funct ion N o w s u p p o s e th e l e m m a is t rue fo r al l polynomials o f degree n L e t P

be a polynomial o f degree n 1. I f a is root o f P, we can wri te P X)

X - a S X , w h e r e S is a polynomial o f degree n Thus, by hypothesis,

Lsq LsLq. W e have left only to verify th e l e m m a fo r polynomials o f degree 1.T h a t is, we m u s t show that if R is a polynomial o f degree an d T is a n y polynomial,

t hen Lrt=LrLt F o r once this is verified, we take R X)

a , so that

P R S . T h e n

Lpq =LRSq =LrLSq = L rL s L q = L rs Lq =LpLq


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41 2 5 Series of Funct ions

So, le t R(X)

X-a, T(X) = 2?U b, X . Th e n

RT(x)=f(b,- abl+1)X,+1

abo

( 0

N o w we compute LRLT :

\ in m

IW =2(w,0) -^i,r( o / (

o i o

m

tibi-abi+l)f -ab0f1 0

T he lem m a is p r o v e n .

I t fol lows from th e lemma tha t if Q is a fa c to r o f P, then an y solutiono f LPQ f = 0 is a solution o f LP(f) = 0. N o w le t P be a given polynomial.W e c a n , by th e fac tor iza t ion theorem, write P as a product o f first-orderfactors.

P(X) ( X - a j - - - (X- as)m with m t + + ms =

degP

Be c a u se o f L e m m a 1 th e solutions corresponding to th e factors (X aj 1 1are in S(LP). T h u s we need to d isc ov er th e solutions o f th e differential

equation LP(f) = 0, where P(X) (X c m.

Consider, fo r example, th e differential operator corresponding to (X c 2.W e know one so lu t ion: e x we find another by th e technique o f v ar ia tio n o f

parameters. iX c 2 = X2 2 cX + c2. Tes t th e operator on y = zecx.

y= z ecx + zcecx y z ecx + 2z cee* + zc2ecx

T h e n

/ 2cy + c2y = z e = 0 or z = 0

T h u s z = x , an d th e second solution is xecx. W e ca n guess then th at th e

general situation is this.

L e m m a 2. The solutions of LiX_c m f = 0 are spanned by e , xecx, xm~V*.

Proof. W e have to show t ha t th e n a m e d funct ions ar e solutions. W e do tha t

by induct ion. T h e case m = 1 is already known (by L e m m a 1 . Th u s we m ay


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5 .4 Solutions in Series

I f now we are given a l inear differential equation w hich is not homogeneous,or has variable coefficients, we have a problem o f a m u c h d if fe re n t magnitude.In general, such problems cannot be solved explicitly. Thus, we m u st seek

ways to obtain approximate so lu tio n s . T h is is one o f th e places whereseries representations o f funct ions ar e usable . T he procedure o f series

approximation has tw o aspects. First, we m u s t e sta blish th e th eo re tic al

validity o f such a technique and, secondly (and this is essential from the

practical point o f view),we

needa

technique fo r effectively computing th eerror In this sect ion we shall describe this procedure, deferring these tw oessential points (which turn o u t to be th e same ) unt i l Section 5.7.

First, an example. Suppose we w a n t th e f un c tio n such t h a t

f (x)+gl(x)f (x)+g0(x)f(x) 0 /(0)

a0 f'(0)

a ,

w hereg0

andgx

are defined in aneighborhood

o f 0. We sh all assume

t h a t they are sufficiently differentiable. N o w ur i ni ti al c o n d it io n s

give us th e first tw o te rm s o f th e Taylor expansion o f / a t 0 :

f(x)

a0 a tx higher-order t e r m s (5.7)

O ur technique w ill be based on th e tacit assumption t h a t th e

higher-order t e r m s ar e computable, and knowing enough o f them will givea usable approximation to th e solution. Now, evaluating th e differential equation itself a t 0 gives us th e s ec o nd -o rd e r t e rm :

f i0)+g1(0)f'(Q)+go(0)fi0) 0

or

f (0)

-(g1(0)a1 go(0)ao)

so

f(x) a0 a tx i(aog0(0) + fli i 0))x2 higher-order t e r m s (5.8)

Differentiating th e differential equation will give an identity express-


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5.4 Solutions in Series 41 5

ing/' in t e r m s o f lower derivatives, so we m ay continue,

Fix ) g[ix)f'(x) (0)

10. Thus, to fivet e r m s th e

Taylor expansiono f th e

desired solution is

fix) 2x x2 \x3 h x4

A + higher-order t e r m s

Admittedly this is n o t very glamorous, b u t it is computableT he phrase


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5.4 Solutions in Series 41 9

do arise naturally, th e given functions usually are sums o f convergent p o w e rseries. I f this is th e c a s e , all o ther questions can be satisfactorily answered;t h a t is, th e solut ion also is th e su m o f a convergent p o w e r series w hos e co

efficients can be determined by th e ab o v e technique an d th e est imate on th eremainder can be effectively computed. L et us look a t ano the r i llu st ra tion.

Examples

10.

x3y x2y xy y ex (5.14)

y(0) 1 y iO) y i0) 1/6

L et th e solution be f(x) ^=0ax . Substituting in (5.14), weobta in

QO OO CO CO

n(n l)(n 2)anxn n(n

l )ax naxn anx

CO y

t-0

n=0 n 0 n=0 = 0

which gives these equations fo r th e coefficients :

ain(n l)(n

2) n(n

1) 1) fo r all n

or

a

n (n3-2n2-n l)(5.15)

Notice, t h a t we have n o t used th e initial conditions and fortunately

they conform to th e requirements (5.15). T h a t is, fo r this particular

equation, there is a unique solution independent o f a n yinitial con

ditions. Th i s does not contradict a n y previous results because

Picard ' s t heorems do n o t apply (since th e leading coefficient is not

invertible).

11.

y xy 2y

0 (5-16)

yiO) y iO)

0


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420 5 Series of Fu n c t io n s

Here Pica rd s theorem does apply, so we should get a unique solutionwith th e given i n it ia l cond i tions . Let/(x)

=0 ax be th e c an di

date. (5.16) becomes

nin l)flx -2 nanxn 2ax 0

n=0 n=0 n=0

or

a0 1 fl i 0

(n 2)(n+l)an+2-nan 2an 0 n> 0

or

(n

2)afl +2

(n 2) (n+l )

T h u s a2

1 a3 0, aA

0 and thu s all fu r ther coefficients are

zero. T he solution is/(x) x2.

In th e n e x t section, we shall fully develop th e theory o f p o w e r series. I tis m o s t advantageous (as we have already seen) to d o so in th e complexdomain .

EXERCISES

6. F i nd an approximate solut ion fo r

y x y

0 y(0)

0 y (0)

1

with an error o f at m o st 10 * in t he i nt er va l 1, 1].7. D o th e same fo r

y-x2y \ y (0 )=0 / ( 0 ) = 0 / ( 0 ) = 0

with an error o f \0~* in [ i J].8. Find a recursive fo rmula fo r th e coefficients o f th e solution, an d a

reasonable es t imate :

(a) / 2 v

0, v(0)

, / (0) 0.

(b) y 2/ x y

e\ y(0)

, / (0)

.

(c) y k + v 1, arbitrary initial condi t ions .

(d) / - /c 2 v = 0 .(e) x2 x y , y(0)

0.


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5.5 P ow er Series 42 1

P R O B L E M S

12. Why does n t Picard s theorem apply to Equation 5.14)?13. T h e

second-order equation xy y = 0 seems to have only onesolution by th e series method, b u t tw o independent solut ions by th e m e t h o do f separation o f variables. Explain that .

5.5 P o w e r Series

W e have already discussed at length th e p o w e r series expansion o f th e

exponential an d trigonometric functions, th e geometric series and some others.W e have also seen t h a t th e Taylor formula produces a p o w e r series expansionfo r suitable funct ions. W e have observed t h a t there is a certain disk corre

sponding to each p o w e r series, called th e disk o f c o n v e rg e n c e . T he series

converges inside t h a t disk an d diverges outs ide . W e shall recollect al l thisin formation as th e starting point o f ou r discussion o f complex p o w e r series.

T h e o r e m 5.6. L e t cn be a sequence of complex num bers. T here is a non-

negative number R called th e radius o f convergence of th e p o w e r series z )with these properties:

a) n 0 cn z diverges i f \z \>R. b) =0 cz converges absolutely and uniformly in any disk {zeC:

\z\ < r) with rR . T h e n the re is a t, \z\>t>R such t ha t {\c\t } is un

b o u n d e d . Since |c | |z | |c | t fo r al l n , we cannot have lim cz = 0 s o j r zdiverges.

b) L et r


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424 5 Series of F unc t i ons

th e hyperbolic cosine and hyperbolic sine, respectively. W ecan use

these expressions to define th e complex cosh and sinh :

ez e~ z . , ez e~zcosh z s inh z

Be c a u se o f (5.19) and (5.20), th e complex trigonometric functions a r e ,on th e imaginary axis, th e hyperbolic funct ions :

cosh z cos zz sinh z sin(z z) (5.21)

W e should also n o t e t h a t th e trigonometric identities imply th e hyperbolic ones. Since cos2(/z) sin2(/z) 1, it follows from (5.21) t h a t

cosh2 z sinh2 z 1 (5-22)

(see Exercise 10).

16. =1 (z / ) h as radius o f convergence 1. S o does th e series

=1nk (z /n ), fo r a n y integer k . W e shall see la ter tha t th e sums o fall these series can be given by closed expressions (such as z

(1-z)-1).

17. A polynomial funct ion in C is given by a p o w e r series. In

fact, writing th e polynomial p(z) =0 anz is th e same as givingit s p o w e r series expansion. W h a t is more interesting is t h a t an y

point in C ca n be chosen as th e c e n t e r o f a p o w e r series expansion fo r

p . L et z0e C and write

N

KZ) aniZ

Z0 Z0)n

0

Using th e binomial theorem this b eco mes

P(z)= (nW-z0)'zS-'' 5.23n

0 i 0 V /

A ll sums being finite, we m a y a r ran g e t e r m s a t will. T h u s we can

rewrite (5.23)as a sum o f

p o werso f z

z0:

Kz)= kr z-z0)n

0 m n /

which is t he d es ir ed expansion.


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5.5 P ow er Series 42 5

M o r e generally, an y series o f th e form =0 c(z

z0) will be called a

p o wer series expansion centered at z0 . C an we expand ez in a p o w e r seriescentered a t a point other than th e origin? T he answer is y e s (cf., Problem

15), and th e proof is like th e one ab o v e fo r polynomials, b u t th e question o fc o n v e rg e n c e in te rv en es a fte r th e analog o f Equation (5.23) above. I t is a

general fact t h a t fo r an y funct ion given by a p o w e r series, we m a y move th ecen t e r o f th e expansion to any other point in th e disk o f convergence. The

truly c o u r a g e o u s s tuden t should try to p r o v e this n o w ; it ca n be done . W ewill give a proof la ter which is simple and avoids convergence problems b u t

requires more sophisticated in formation a b o u t functions defined by p o w e rseries expansions.

Addition an d Multiplication of P o w e r Series

Suppose /, g are complex-valued funct ions defined by p o w e r series ex

pansions centered at a point z0 . Th en we can fin d s erie s expansions fo r th efunctions / g and/ also. Addit ion is eas y : if, say

f(z) an z

z0Y giz) bn z

z0)

then

(f+g)iz) yj(a b) (z -z0y

B ut to find th e series expansion fo r th e product requires a little more care.

Suppose t h a t z0 0 (this invo lves no loss o f generality). T o say tha t

/(z)

2 z is to sa y tha t in a certain disk A, / is th e l imit of th e polynomials2^=o az . Similarly, g is th e l imit o f th e polynomials ?=0 bnz . Thus,

fg is th e l imi t o fth e

sequenceo f

polynomials ( =0 z )(^=0 bz ). Now,we can multiply polynomials easily,

/ N \ I N \ N N

nz ) lKz afcmz +\n

0 / \n 0 / n 0 m 0

If we collect t e r m s in this expression to form a series o f p o we r s o f z we d o not

get a very aesthetic expression, b u t if we take some t e r m s from th e n e x t fe w

polynomials in th e sequence we obtain a reasonable expression.

g( nbm zk (5.24)k=0 \n+m=k IW e could hope t h a t fg is th e lim it o f this sequence o f polynomials. Th i s is a

reasonable hope; fo r even though we have modified th e original sequence


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42 6 5 Series of Funct ions

o f polynomials we have neither added nor deleted from th e series representedby t h a t sequence. In fact, by making careful use o f this fact, we can verifytha t fg is th e l imit o f 5.21).

Proposition 3. L et f z) =0 az , g z)

=0 bz and suppose r isless than th e radii of convergence of both series. Then

0 /+ 9)iz) =o ian b)z* uniformly and absolutely inA {zeC: \z\ / + g , q -+fg uniformly in A Problem 2.55).Since

Pn z) q z) 2 ia- + b*)zk= l

i) is p r o v e n .

ii) L et

rJLz)

k 0\t J k J

we w a n t to show tha t -+fg uniformly in A. W e know tha t pqn ->fg so it wouldseem w o r t h ou r while to compute q r. B u t that is e a s y,

Pnq rn I

2 bX

k 0\l + J k /t>n

Now, each t e rm on th e right is o f th e form aibjz +J with i> n or j>n. Thus,computing norms on A {z e C : \z\


28/52

5.5 Po we r Series 42 7

Now, we know that 2?=o \a,\r , Jj=0 j r> are finite. L e t M be a n u m b e r largerth an b oth . Given e > 0, there are

(1) i Vi > 0 such t h a t 2 Nu(2) N2 > 0 such tha t 2j=+i \bj\ r i < e i f n > N 2 ,(3) N3 > 0 such that \ \ p q - f g \ \ e if n>N2.

These assert ions follow from th e known c o n v e r g e n c e o f each case. Thus, ifn > max 7V~i, N2 , Ns),

2 |a, |r 21^1^ + 2 1 ^ 1 - 2 \bj

ll -n-/^ll< llpn?n-/^ll+ II.P4,.-rJ|

e-M M- 2 M + l)e

th e proposition is conc luded .

E X E R C I S E S

9. Verify, in th e w a y suggested in th e t ex t tha t cos2 z sin2 z = 1 is true

fo r al l complex n u m b e r s z , and thus cosh2 z sinh2 z 1 is always t r u e .

10 . F i nd a p o w e r series expansion fo r these func t ions :

(a) exp(z2) (e) e~ ^Z

(b) ez sin z (f) cosh z

cos z.

(c) (g) s i n h z1 z

-dt

t

(d) f exp Z2


29/52


15 . Expand ez in a p o w e r series abou t a n y point z0 .

16 . Assuming that 1 +X2) 1 2 can be represented by a p o w e r series

centered at 0, find it . Find th e p o w e r series fo r arc cos x.

17 . Assuming tha t ta n x can be represented bya

p o w e rseries centered

at 0 an d using th e equation ta n x cos x sin x , find th e p o w e r series ex

pansion o f ta n x.

5.6 Complex Differentiation

A n easy property o f th e exponential function is

lim =1 (5-25)i 0 z

F o r

n=o n\

so

e l oo_n l / o o 7n-2\

-V-1+ ^r)N o w th e term in parenthesis is a convergent p o w e r series, so is continuous

a t 0. Thus, writing th e parenthesis as g(z) :

ez 1li m 1 li m zg(z)

12 - >0 z z -> 0

F r o m 5.25 an d the properties o f th e exponential it fo l lows tha t

exp(z0 z)

exp(z0) . N , . ez I

lim 2 ^ ^

exp(z0) li m

exp(z0)z 0 z z -> 0 z

T he s tudent o f c alc ulu s will recognize th e l imit on th e left as a difference

quotient an d th e entire equation as a replica o f th e behavior o f th e realex

ponential function. It might be a good idea to consider more generally such

a process of d i fferent ia tion on th e complex plane. This turns o u tto be a


30/52

5.6 Complex Differentiation 42 9

very significant idea, because there are m a n y beautiful an d useful ways to

represent functions which are so differentiable.

Definition 3. Let/be a complex-valued function d e fi ne d i n a neighborhoodofz0inC. /is differentiable at z0 if

,./ z)- / zo)

lim

z-zo Z Zq

exists. In th is case we write th e limit as / z0). If/ is defined in an o p e nse t U an d differentiable a t every point o f U, we say that is differentiableon U.

T he usua l algebraic facts on differentiation hold t r ue in th e complexdomain .

Proposition 4 .

i Suppose fi g are differentiable at z0 . Then so are f g and fg withthe derivatives given by

if+g) izo)=f izo)+9 izo)ifg) izo)

f izoMzo) +fiz0)9 iz0)

ii) Suppose f is differentiable at z0 and / z0) 0. Then 1// is differentiable at z0 and l/f) z0)

-f z0)/fiz0)2. hi) Suppose f is differentiable at z0 and g is differentiable at f z0). Then

g of is differentiable at z0 andig f) z0) g f zQ))f z0).

Proof. T h e s e propositions ar e so m u c h like th e corresponding propositions in

calculus tha t their proofs will be left to th e reader.

Examples

18. T he function z is clearly differentiable, and z z0) fo r all

z0 . A c o n s t a n t function is differentiable with derivat ive zero. Since

a n y polynomial is obta ined from z and constant funct ions by a suc

cession o f operations as described in Proposition 4 i), al l polynomials

are differentiable.

19. T he funct ion z is nowhere differentiable. F o r th e difference

quotient z

z^)/ z

z0 , fo r z j= z 0 , is a point on th e un i t circle

and as z r anges through a neighborhood o f z0 this difference quotienttakes on all values on the unit circle, so it could hardly converge .


31/52

430 5 Series of Functions

Sum of a Pow e r Series is Infinitely Differentiable

O u r introduction to th is s ec tio n was essentially a proof t h a t ez is everywhere differentiable. I t is in fact t ru e t h a t th e sum o f a p o w e r series isdifferentiable in it s disk o f convergence . W e now verify this basic fact.

T h e o r e m 5.7. L et /(z)

j= 0 anz have radius of convergence R . Thenf is differentiable at every point in the disk of radius R an d

f (z)= Y,nanz -1n= l

(5.26)

has th e same radius of convergence as =0 az .

Proof, lim sup(|a|)1/n

lim()1/n li m supGa,, )1' lim sup(|a|)1/n, so th e series

2 nanz 1 has th e same rad iu s o f c o n v e r g e n c e as th e given series. W e must showt h a t it represents t h e d e ri va ti v e off. Fix a z0 , \z0\ |z0|. Theseries 2 la l 1 c o n v e r g e s absolutely, so given e> 0 there is an N such that

2 >n lank 1 nanz0~ 1\ z0 th e las t t e rm t ends to zero. Thus, there is a 8 > 0such tha t if z z0 < 8, th e last t e rm is less than e. Thus, fo r \z\


32/52

5.6 Complex Differentiation 43 1

In particular, since / is given by a convergent p o w e r series, it also isdifferentiable, with derivative f (z) \)az ~2, an d so forth. W ethus obtain these results, which form th e complex version o f Taylor s theoremfo r sums o f convergent p o w e r series.

Corollary 1. L et f(z) Y^azn have radius of convergence R . Then f isinfinitely differentiable in {z: \z\ k

These corollaries are easily derived f rom th e theorem and the ir proofs are

left to th e student. Notice t h a t th e implication o f Corollary 2 is t h a t th e

coefficients o f a p o w e r series representation o f a function are uniquely an d

directly determined by th e funct ion . In particular, a function c a n n o t be

written as th e sum o f a p o w e r series in more t h a n one way. This observa

t ion allows us to easily verify th e identity

cos2

z + sin2 z 1 5-27

F o r th e function cos2 z + sin2 z is a polynomial in functions which are sums

o f p o w e r series and thus is th e sumo f a p o w e r series. Its coefficients can


33/52


be computed according to Corollary 3 just by letting z take on real values.B u t th e right-hand side o f (5.27) is th e Taylor expansion o f cos2 z + sin2 z,fo r real z , thus it m u s t be th e Taylor expansion fo r a ll z. Hence (5.27) is

always t r u e .

The Cauchy-Riemann Equation

It is o f value to c o m p a r e th e n o tio n o f complex differentiation with t h a to f differentiation o f funct ions defined on R2, since R2 C. Suppose t h a t

is a complex-valued funct ion d efin ed in a neighborhood o f z0 x0 + iy0 .

If/ is differentiable as a funct ion o f tw o real variables, then th e differential

dfix0,y0) is defined and is a complex- valued lin ea r f un c tio n on R2. Iffis also complex differentiable, then

u r- fiz fizo e /,0-.f (z0) lim (5.28)

z - > z o Z Zq

exists. L e t z ->-z0 along th e h o riz o nt al line. T h e n (5.28) specializes to

t \ i;fix>yo)-fixo>yo) Sff (z0) h m (x0 , v0 (5.29)x- xo X Xq O X

If we le t z - z0 along a vertical, we also h av e

t \ v f ixo ,y - f ixo ,y0 ld ff (z0) h m T x , y0) (5.30)y - j -o Cv

J o Sy

T hus th e right-hand sides o f (5.29) an d (5.30) are th e same. In conclusion,a complex differentiable funct ion m u s t satisfy (when considered as a functiono f tw o real variables) this relation

(DM ) (5.31)Th i s is called th e Cauchy-Riemann equation. M o r e precisely, th e Cauchy-R i e m a n n equations are found by writing /= u + iv an d splitting into realan d imaginary parts. L et us record this important fact .

T h e o r e m 5.8 . L et f be a complex differentiable function in a domain D.

Split f into real an d imaginary parts and consider f as a function of tw o real


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5.6 Complex Differentiation 433

dyjidx i dydu dv

dx dy

du

~dv

dx

variables z = x iy Then these partial differential equations hold in D :

5.32)

5.33)

Proof. Equation 5.32) was o bs erv ed a bo ve . Equation 5.33) follows from

5.32) a n d th e identities

dfdu

.dv Bf8u

,8vd~x~~ ~x ~dx ~ ~ ~~ ~ Hy

Notice t h a t w h e n / is complex differentiable, it s differential is given by

dfizo , ir is))

^ z0)r

-jt z0>

f z0)r + if iz0)s f izo)ir is)

Thus th e differential o f a complex differentiable function is a complex l inear

complex-valued function.W e shall show, via th e techniques o f th e n e x t few chapters, tha t a complex

differentiable function can be written as th e sum o f a convergent p o w e r

series. Thus, just by virtue o f th e differential being complex linear, th e

function has derivatives of a ll orders and is th e sum of it s p o w e r series.

P R O B L E M S

18 . P r o v e Proposition 4.

19 . P r o v e Corollaries and 2 o f Th e o r e m 5.7.

20 . Show t h a t if is an infinitely differentiable funct ion on an interval

e), and t here is an M > 0 such tha t

| /w x |


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434 J Series of Funct ions

22. W rite th e Cauchy-Riemann equations in polar coordinates. Hint:Different iate along th e r a y and circle through a point.

23. Suppose ,...,/, ar e given by convergent p o w e r series in a disk A.

If F is a polynomial in k variables such t h a t

F Mz),.. . ,fk z))=0 5.34

fo r real z , t hen 5.34 is t rue fo r all z in A.

24. Compute th e limits o f these quotients as z > 0 :

co s z

daar c tan z

z

bsin h z

sin z

ccos z

eco s z

TT72

si n tanz f n = 0 , 1,2, 3, 4

25. Suppose is a differentiable complex- va lued funct ion o f tw o rea lvariables in a d o m a i n D . Show that/is a complex differentiable if an d onlyif th e differentia l df z0 is complex l inea r fo r all z0 e D .

26. Suppose that is twice differentiable in D, a n d is complex differ

en tia b le . Show also that / is complex differentiable. Supposing that

/ ) 0, show that /is a quadratic polynomial in z.

27 . If f=u iv is complex differentiable in D a n d twice differentiable,t hen

82u d2u d2v 82 v

dx2 dy dx2 dy

5.7 Different ia l Equations with Analytic Coefficients

A funct ion which ca n berepresented

as th e sum o f aconvergent

p o wer

series at a point a e C will be said to be analytic a t a. W e now return to th e

study o f l in ea r d i ffe ren ti a l equations in o r d e r to answer some o f th e questions

posed in Section 5.5. W e can use th e in fo rm a t io n in Section 5 .6 to d o this

an d to provide th e sought-for est imates. In particular, we shall verify th e

following fact.


36/52

5 .7 Differential Equations with Analytic Coefficients 43 5

Proposition 5 . Suppose h, g 0 , fc_l5 gk are analytic at 0. Then thesolution of the differential equation

ym g,yw + h(x) = 0 y(0) = a0,..., ^ (O) ak^i 0

is also analytic at 0; that is, in some disk centered at 0, it is th e sum of a con

vergent p o w e r ser ies whose coefficients ca n be recursively calculated from th e

differential equation.

W e already know, from Section 5.5, h ow to compute th e Taylor coefficientso f th e solution; ou r business here is to show t h a t th e resulting series does in

fact converge . This, o f c o u r s e , involves producing the k ind o f est imate

required by T h e o r e m 5.7.

Suppose t hen t h a t h,g0, . . . ,gk- i are analytic in th e interval |x | R . Th e n

oo oo

h(x) = 2 x g,(x)

2 a* xn=0 n=0

an d fo r some positive num ber M, \a \ < M R ~ , |a |


37/52


coefficients :

m(m \) (m k cm

or

k l m- k

+ 2 2 aa (m + i (k+a)) m /c + a cm +,_,*+) =01 = 0 a = 0

1 k l m-k

2 2 + - * + ))m k) i=o a=o

(m (k + a))a cm+i-(t+1,) (5.36)

By th e restraints on i and a we have m + i (k + a ) < m + k k m 1, soth e highest subscript o f c on th e right is m 1. Thus, given c0 , . . . , ck-i we cansolve Equations (5.36) successively. W e no w try to find an estimate. F o r this

p u r p o s e assume M > 1, an d le t

^,P i?-l -1 Pt-l ,^,^y/J,0


38/52

5 .7 Differential Equations with Analytic Coefficients 437

W e might perhaps have m a d e a better estimate by more clever subst i tu t ions ; bu tou r a b o v e estimates were sufficient fo r th e results desired. In a n y particular casewe cou ld usually be clever and obtain even better estimates.

Examples

20. y exy + x y 0, y(0) 0, /(0) 1 .

W e will find a polynomial which approximates th e s olu tio n to within10~5 on th e interval [0.01, 0.01]. L et 2 cx be th e supposed solut ion. By substituting th e series we obta in

n(n l)cx -2 lt- M l ^x 1) + cx +1 0 5.

n 2 \n 0 n \] \ i / o

38)

T he initial condit ions give c0

0, q 1. Equation (5.38) becomes

-1cm

-

m(m 1)

T h u s

c2

-\cx

\

lm-2 1 \

| - m-l-i>m-i-i cm_3

c3

\Qc2 c t c0) 0

c4

-J1ji3c3 2c2 Iq cx)

5L

Cs

-2Xo 4c4 3c3 c2 iq C2)

TJo

an d so for th . T he question is not really w ha t th e coefficients are

(that is to be left to a machine) b u t h o w m a n y coefficients need to

be computed. T he coefficients a p p e a r to be bounded (we could in

fact show t h a t they m u s t be, cf . Pro b l em 29). L e t s try to prove t h a t

|c j < K fo r a ll n by induct ion. W e have

1 m_2 1|c J ^

^73T) fS 7T( 0|Cm-1-il |Cm-31

m m yso long as m > 4. T h u s we m a y tak e fo r K a b o u n d for th e first

f o u r terms, tha t is, k 1/2. Then th e d iffe ren ce b e tween th e solution


39/52


and th e k th partial su m o f it s Taylor expansion is dominated by

|c||x| k Z k L 1 |X |

fo r |x| < 1. T he interval we are concerned with is |x| < 10_1 so our

bound on th e error is

i.i_.l i i o2 10 9 18

Th i s is less than 105

if k

6, thus computing also c6) th e so lu tio ndiffers from

lv2 -L V4 J- i_ V 1 y6X 2X T 34 X T J J 0 A 3 60A

by at m o s t 10-5 fo r all values o f x in [-0.01, 0.01].

21. Suppose we needed tha t good an est imate in th e interval

[1 , 1].I t is easy to see t h a t

justknowing t h a t th e coefficients are

bounded is n o t good enough. W e h a v e to k n o w tha t \c\ < Kr fo r

some r < 1 an d some K . L e t s try r \. T h a t is, we attempt to

verify by induct ion tha t \c\ < 2~ fo r all n. Now, using th e equationdefining {c},

\1 y m

~ 1 ~ K

K \|Cml Sm m 1) I ik i\ 2m-i-i^2n,-3J

K 1 /m-22; \

e2 + 4 K< 7

m z

Th i s is less t h a n 2 ~ m K as soon a s m 2 e2 + 4), o r m 26. Thusth e induct ion step proceeds as soon as m > 26 ; we need only choose

K so t h a t th e inequality holds fo r al l m < 26 K 2 will do). Th u s

\cn\ < 1/2 fo r all n. T he desired solution differs in th e interval

1, 1] from its k th -o rde r Taylor polynomials by

c ( Dn>k

1 1y

2 -1 t o2 ^2^ lc*l ^ ~fc-i on ok-2


40/52

5 .7 Differential Equations with Analytic Coefficients 439

This is less than 10 5 if k is 19. Thus we need to compute 19 t e r m so f th e Taylor expansion to find th e desi red approximation.

22 . Compute th e solution o f

+ (l^y' + exy 0 y(0) 1 y'(0) 0

to an accu racy o f 103

in th e interval i ].L et f(x) =0 cnx be th e solution. W e have these equations

fo r th e solution :

c0= 1, Ci 0

-1c

n(n

1) \i=n 2n- 2 J \

n l-i)c-i-i+ -.cn-2-i)i=o =oi /

(5.39)

W e will show by inductions t h a t th e {c} are bounded. Suppose t h a t

Ic.l < K fo r all n =o /K / - x . \ X | m m

1)

m(m -1)1 j 1 / m(m

e

1)+ e

K1

2 m(m 1).

3. Thus we ca n take A as a b o u n d fo r th e first f o u r

t e r m s . W e have from (5.39) c2= -\,c3= 0, so we m a y take K=\.

Then th e estimate o f th e remainder af ter k t e r m s in th e interv al

[-i, H is

k l W < ^

^k n>k^

Z.

Th i s is less than 1 03

when k 10, so we need 11 coefficients. W e

compute

c4

h c =20 c6 T 2 0 e tc .


41/52


Up to six te rm s (giving at m o s t an error o f 1/64), ou r solution is

x^ x^ x^ 13x6

~ T+

L2+

20+720+

E X E R C I S E S

13. Find a p o w e r series expansion for th e general solut ion in a neighborhood o f 0 fo r this equation,

1

x2) / 2 x / + k(k + 1 v 0

14. F i nd a p o w e r series expansion for th e general solut ion o f

yh

Ixy + 2k v 0

15 . Find th e p o w e r series fo r th e funct ion y =/ (x) such t h a t

(a) y + e-y

x2, v(0)

l, v (0)=0

(b) (/)2

v , y(0)=0

(c) (y )2=yy , y(0)=0, y (0) 1

(d) + 2x^2=016 . H o w m a n y t e r m s o f th e p o w e r series for th e solut ion y do we need:

(a) fo r an a c c u r a c y o f 103 in th e in te rva l i, in Exercise 3(a)?

(b) fo r an a c c u r a c y o f 105 in t h e i nt er v al 10 , 10) in Exercise 3(a) ?

(c) fo r an a c c u r a c y o f 1 0 a in th e in te rv al 0.1, 0.1) in Exercise

3(b)?17 . F o r w h a t k are th e solut ion o f Equations (5.1), (5.2) polynomials?

P R O B L E M S

28 . Generalizing th e argument in th e text p r o v e this t h e o r e m :

Theorem. Ifh ,g0, ...,gk-i are analytic in an interval R,R) about th e

origin, then a n y solution of th e differential equation

/k> + k Zglyi hr 0

can be expressed as th e sum of a convergent p o w e r ser ies in a neighborhoodof th e origin.

29 . Suppose tha t g , h are convergent p o w e r series in some disk (|z| R)

with R>\. Sh o w t h a t t he s olu tio n o f th e linear differential equation

y + gy + hy 0 (5.40)


42/52

5.8 Infinitely Flat Funct ions 441

is th e sum o f a convergent p o w e r series w ith b o un d ed coefficients.

30 . If th e p o w e r series g(x)

~2 ax h(x)=^bx bo t h have infiniteradius o f convergence , then so does th e series expansion o f th e solution

o f (5.40).

5.8 Infinitely Fla t Functions

N o t al l functions are susceptible to th e kind o f Taylor series analysiswhich we have been doing. A first requirement is t h a t th e function h av e

derivatives o f all order; even tha t h o wev e r is insufficient. Another glance a t

Theorem 5.6 will remind th e r eader tha t there is a behavior requirement on

these successive derivatives in order tha t th e given function be th e sum o f it s

Taylor expansion. W e shall show by example tha t there are infinitelydifferentiable functions which are not sums o f p o w e r series. First, we shall

make th e notion o f analyticity precise.

Definition 4 . L e t /b e a complex- valued function defined in an o p e n se t U.

L et a e U. /is analytic a t a if there is a bal l {z: \z

a\ < r} centered a t a

such t h a t / is th e sum o fa

convergent p o w e rseries in th is b all. is

analyticin U if / is analytic a t every point o f U.

W e have deliberately stated this definition without reference to th e domain

o f definition o f th e function; it applies equally well to functions o f a real or

complex variable. T he only functions which we k n o w to be analyticare

th e polynomials and ez. F o r example, i f / is th e sum o f a convergent p o w e r

series a t th e origin, we d o n o t yet k n o w t h a t we can expand in a series o f

powers o f (z

a) with a an y o ther point in th e disk o f convergence. W e

shall see in th e n e x t chapter t h a t this is th e case. W e have already seen t h a tan analytic function h as derivatives o f al l orders (is C and

now we will

produce a function which is not analytic. T h e clue tothis function is

given by th e following fact, which follows from l Hospital srule.

Proposition 6 . lim P(t)e~

0, for any polynomial Pt. 0O

Proof. Problem 31.

T he function we h av e in mind (Figure 5.1) is defined by

expx > 0

/ (5.41)

0 x ^ O


43/52


44/52

5 .8 Infinitely Flat Functions 44 3

F o r each n there is a polynomial P such tha t

ff n x P (- j exp j fo r x > 0 (5.43)

This ca n be verified by induction. Assuming (5.43), we compute

0 x>0

Thus is also infinitely differentiable a t 0. B ut it is certainly not analytic.It s Taylor expansion is =0 0

x which converges to o(x) only fo r x 0 if a


45/52


Figure 5.2

Theorem 5.9. L et [a, b~] be a given interval an d Uu . Un a finite collection of open intervals covering [a, b~ T he re e x is t C functions p t such that

i 0 < pix) < l fo r all x e R , all i, ii Pi x) 0 i fx$Ut ,

iii) pt x) = l, fo r all x e [a, U].

Proof. L et [/ ,

at , bi), an d take rt = Tib{ . T h e n t x)

2 Tt x) > 0 ifx e a, b). L et

ti x)x 6 a, , bi)

Pix) = l.33 . Using Theorem 5.9 it can be shown tha t a n y cont inuous function is

th e l imit of C funct ions. Let/e C [0, 1] an d e > 0. F ind a C function

such that | | g\\ < e.Here ' s h ow to d o it. First pick an integer N> 0 such t h a t |/(x) f y \

< e/2 if \x y \< l jN . N ow cover th e interval [0,1] by th e intervals


46/52

5.9 Summary 44 5

U0, UN, where

/ i l /+1\

u -(n- n)and le t px,...,pN be th e corresponding functions o f Theorem 5.9. L et

^-l/^HF o r a n y x , x is in only tw o o f th e intervals {U,}, say UT, U,+1. Th e n

l / * -* * l=M * **>- ) + Pr+1 /-, ^e e

0 such t h a t

(0 IIAIIiV

YjPk R .

b) cnz converges absolutely in {|z| < /} fo r r < R .

c) P=[limsup |C|)1/ ]-1.

If/(z)

anz , g z)

YJbnzn in th e disk {\z\ < r}, then

f(z) g z)=

t an+ bn)z

n 0

fiz)giz)= abm zk=0 \n +m=k 1in tha t same disk.


48/52


49/52


F U R T H E R R E A D I N G

I. I. Hirshman, Infinite Series, Holt, Rinehar t an d Winston, N ew York,1962.

H . Cartan, Elementary Theory of Analytic Funct ions of O ne or SeveralComplex Variables, Addison-Wesley, Reading, Mass., 1963.

This t e x t develops th e subject o f complex analysis from the point o f viewo f p o w e r series. I t also contains a complete discussion o f th e theorem onexistence o f solutions o f analytic differential equations.

Fur the r material can be found in

T. A . B ak and J. Lichtenberg, Mathematics fo r Scientists, W . A . Benjamin,Inc., N ew York, 1966.

Kreider, Kuller, Ostberg, Perkins, A n Introduction to Li n ea r Analysis,Addison-Wesley, Reading, Mass., 1966.

W . Fulks, Advanced Calculus, John Wiley and Sons, N ew York, 1961.M . R . Spiegel, Applied Differential Equations, Prentice-Hall, Englewood

Cliffs, N . J., 1958.

MISCELLANEOUS P R O B L E M S

34. Find as e q u e n c e {/}

o f con t inuousnonnegative

real-valued funct ions

def ined on th e interval 0,1 such tha t f x ) = 2 = i / * ) exists fo r al lx e [0, 1], b u t /is n o t continuous.

35 . F o r |z| 1, define

oo 7k

l n z = 2 r l k

Sh o w tha t fo r all such z, z

exp ln l z .36 . Show tha t th e series

i z-n)2

c o n v e r g e s to a complex differentiable function in th e d o m a i n

C - { 1 , 2 , ...}.37. Show tha t

exp zlim

1+

z/m)m. Hint: Computeth e

p o w e rm oo

series expansion o f 1 zjm)m.)38. Show tha t a real polynomial o f odd degree always h as a real r o o t .39 . Let oj exp 277f //j). Show tha t

z 1 ojz \ a>2z) - wnz)


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5.9 Summary 451

T o p r o v e this we must observe t h a t Legendre's equation m a y be written as

( ( l - x 2 ) / ) ' + k (k + l )v 0

Th u s

\\ f - f - ^ T T ) /. i [(1 *'>/]'/ dx

by integration by parts. N o w do th e s a m e , interchanging m an d n.61 . L e t P be a polynomial o f degree d. Show that /(z)

ep(2 is complexdifferentiable . Show that/(n)(z)e ,

7-advanced calculus book - chapter five

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