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Problems and Theorems in Analysis I: Series. Integral Calculus. Theory of Functions
 9783642619830, 3642619835

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Classics in Mathematics George P61ya • Gabor Szego

Problems and Theorems in Analysis I

Springer

Berlin Heidelberg New York Barcelona Budapest Hong Kong London Milan Paris Santa Clara Singapore Tokyo

This picture shows G. P61ya (r.) and G. Szego (I.) delivering their original manuscript to Springer in Berlin in 1925 (courtesy of G. Alexanderson) . George P61ya Born in Budapest, December 13,1887, George P61ya initially studied law, then languages and literature in Budapest. He came to mathematics in order to understand philosophy, but the subject of his doctorate in 1912 was in probability theory and he promptly abandoned philosophy. After a year in GOHingen and a short stay in Paris, he received an appointment at the ETH in Zurich. His research was multifaceted, ranging from series, probability, number theory and combinatorics to astronomy and voting systems. Some of his deepest work was on entire functions. He also worked in conformal mappings, potential theory, boundary value problems, and isoperimetric problems in mathematical physics, as well as heuristics late in his career. When P61ya left Europe in 1940, he first went to Brown University, then two years later to Stanford, where he remained until his death on September 7, 1985.

Gabor Szego Born in Kunhegyes, Hungary, January 20, 1895, Szego studied in Budapest and Vienna, where he received his Ph. D. in 1918, after serving in the Austro-Hungarian army in the First World War. He became a privatdozent at the University of Berlin and in 1926 succeeded Knopp at the University of Konigsberg. It was during his time in Berlin that he and P61ya collaborated on their great joint work, the Problems and Theorems in Analysis. Szego's own research concentrated on orthogonal polynomials and Toeplitz matrices. With the deteriorating situation in Germany at that time, he moved in 1934 to Washington University, St. Louis, where he remained until 1938, when he moved to Stanford. As department head at Stanford, he arranged for P61ya to join the Stanford faculty in 1942. Szego remained at Stanford until his death on August 7, 1985.

George P61ya • Gabor Szego

Problems and Theorems in Analysis I Series. Integral Calculus. Theory of Functions Reprint of the 1978 Edition

Springer

George P6lya t Gabor Szego t Translator:

Dorothee Aeppli 1414 Chelmsfordn Street St.Paul,MN 55108 USA

Originally published as Vol. 193 of the

Grundlehren der mathematischen Wissenschaften

Mathematics Subject Classification (1991): 05-01,28-01,30-01,40-01

CIP data applied for Die Deutsche Bibliothek - CIP-Einheitsaufnahme P61ya, George: Problems and theorems in analysis I George P6lya; Gabor Szeg6'.- [Nachdr.].- Berlin; Heidelberg; New York; Barcelona; Budapest; Hong Kong; London; Milan; Paris; Santa Clara, Singapore; Tokyo: Springer (Classics in mathematics) 1. Series, integral calculus, theory offunctions.-Reprint [der Ausg.] Berlin, Springer, 1978.- 1998

ISBN-13: 978-3-540-63640-3 DOl: 10.1007/978-3-642-61983-0

e-ISBN-13: 978-3-642-61983-0

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted ouly under the provisions of the German Copyright Law of September 9, 1965,in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable for prosecution under the German Copyright Law. @ Springer-Verlag Berlin Heidelberg 1998 Softcover reprint of the hardcover 15t Edition 1998

The use of general descriptive names, registered names, trademarks etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. SPIN 10651015

4113143-543210- Printed on acid-free paper

G. P61ya G. Szego

Problems and Theorems in Analysis I Series Integral Calculus Theory of Functions Translation by D. Aeppli

Corrected Printing

Springer -Verlag Berlin Heidelberg New York 1978

George P6lya Gabor Szego Department of Mathematics, Stanford University Stanford, CA 94305/USA

Revised and enlarged translation of "Aufgaben und Lehrsatze aus der Analysis I", 4th ed., 1970; Heidelberger Taschenbucher, Band 73

AMS Subject Classifications (1970): 05-01, 28-01, 30-01, 40-01

ISBN-13: 978-3-540-63640-3 DOl: 10.1007/978-3-642-61983-0

e-ISBN-13: 978-3-642-61983-0

This work is subject to copyright. AU rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to "Verwertungsgesellschaft Wort", Munich.

© by Springer·Verlag Berlin Heidelberg 1972 Softeover reprint of the hardcover 15t Edition 1972 Printing and Bookbinding: BrOhlsche UniversiUitsdruckerei, Giessen

2141!314().5432

Preface to the English Edition The present English edition is not a mere translation of the German original. Many new problems have been added and there are also other changes, mostly minor. Yet all the alterations amount to less than ten percent of the text. We intended to keep intact the general plan and the original flavor of the work. Thus we have not introduced any essentially new subject matter, although the mathematical fashion has greatly changed since 1924. We have restricted ourselves to supplementing the topics originally chosen. Some of our problems first published in this work have given rise to extensive research. To include all such developments would have

O. denotes a quantity that divided by a.. remains bounded;

o(a.. ) a quantity that divided by lin tends to 0 as n _

00.

Such notation is used analogously in limit processes other than n _ 00. x _ a + 0 means x converges from the right (x _ a - 0 from the left) to a. exp (x) = e". e is the base of natural logarithms. Given 1J real numbers al' a l ..... a ... max (al. a2• .... an) denotes the largest (or one of the largest) and min (al • a l •..•• a.. ) the smallest (or one of the smallest) among the numbers al •. a l •..•• a... max f(x) and min f(x) have an analogous meaning for a real function defined on the interval a. b. provided f(x) assumes a maximum

Notation and Abbreviations

XIX

or a minimum on a, b. Otherwise we retain the same notation for the least upper and the greatest lower bound resp. (similarly in the case of a complex variable). sgn x stands for the signum function: sgnx= {

+1 for x > 0 Oforx=O -1 for x < 0

[x): denotes the largest integer that is not larger than x (x - 1 < [x] ~ x). Square brackets, however, are also used instead of ordinary parentheses if no misunderstanding is expected. (Their use in a very special sense is restricted to Part I, Chap. 1, § 5.) if is the conjugate to the complex number z. :For the determinant with general term a;.,,.., A,I4 = 1, 2, ... ,11, we use the abbreviated notation

Ia;.,..I~

or

Ia;.,..I;.,,..=1,2, ... ,n

or

IaU, aA2, ... , aAn I~.

A non-empty connected open set (containing only interior points) is called a region. The closure of a region (the union of the open set and of its boundary) is called a domain. (As this terminology is not the most frequently used, we shall sometimes overemphasize it and speak of "open region" and "closed domain".) A continuous curve is defined as the single-valued continuous image ofthe interval 0 ~ t ~ 1, i.e. the set of points z = x iy, where x = /p(t), y = lp(t), /p(t) and lp(t) both continuous on the interval 0 ~ t ~ 1. The curve is closed if /p(0) = /p(I), 11'(0) = 11'(1), without double points if /p(t 1 ) = /p(t 2 ), lp(tl) = 1p(t2 ), tl < t 2 , imply tl = 0, t2 = 1. A curve without double points is also called a simple curve. A not-closed, simple, continuous curve is often referred to as simple arc. A closed continuous curve without double points (Jordan curve) in a plane determines two regions of which it is the common boundary. Paths of integration of line, or complex, integrals are assumed to be continuous and rectifiable. (a, b) denotes the open interval a < x < b, [a, b) the half-open interval a ~ x < b, (a, b) the half-open interval a < x ~ b, [a, b) the closed interval a ~ x ~ b. When we need not distinguish between these four cases we use the term "interval a, b". "Iff" is used now and then as an abbreviation for "if and only if".

+

Problems Part One

Infinite Series and Infinite Sequences Chapter 1

Operations with Power Series

§ 1. Additive Number Theory, Combinatorial Problems, and Applications

*1. In how many different ways can you change one dollar? That is, in how many different ways can you pay 100 cents using five different kinds of coins, cents, nickels, dimes, quarters and half-dollars (worth 1, 0, 10, 25, and 50 cents, respectively) ? *2. Let n stand for a non-negative integer and let A.. denote the number of solutions of the Diophantine equation x

+ by + 10z + 200 + 50v =

n

in non-negative integers. Then the series

represents a rational function of C. Find it. *3. In how many ways can you put the necessary stamps in one 'ow on an airmail letter sent inside the U.S., using 2,4, 6, 8 cents stamps? The postage is 10 cents. (Different arrangements of the same values are regarded as different ways.) 4. We call B .. the number of all possible sums with value n (n a positive integer) whose terms are 1, 2, 3, or 4. (Two sums consisting of

2

Operations with Power Series

the same terms but in different order are regarded as different.} The series represents a rational function of C. Which one? 5. Someone owns a set of eight weights of 1, 1, 2, 5, 10, 10, 20, 50 grams respectively. In how many different ways can 78 grams be composed of such weights? (Replacing one weight by an other one of the same value counts as a different way.) 6. In how many different ways can one weigh 78 grams if the weights may be placed on both pans of the scales and the same weights are used as in problem 5 ? 7. We consider sums of the form £1

+ £2 + 2£3 + 5£4 + 10£5 + 10£6 + 20£7 + 50£8'

where £1' £2' ... , £8 assume the values 0 or 1. We call G.. the number of different sums with value n. Write the polynomial as a product. 8. Let £1' £2' ... , £8 assume the values -1, 0, 1. Modify problem 7 accordingly. Let D.. denote the number of different sums of value n Find the factorization of the following expression (function of C) 99

~ D ..e".

,,--99

9. Generalize the preceding examples by replacing the particular values of the coins, stamps and weights by aI' a2, ••. , al • 10. An assembly of p persons elects a committee consisting of n of its members. How many different committees can they choose? 11. There are p persons sharing n dollars. In how many ways can they distribute the money? 12. There are p persons sharing n dollars, each getting at least one dollar. In how many ways can they do it? 13. Consider the general homogeneous polynomial of degree n in the p variables xl> x 2 ' ••• , xp' How many terms does it have? 14. Any weight that is a positive integral multiple of a given unit can be weighed with the weights 1, 2, 4, 8, 16, ... on one pan of the scales, and this can be done in exactly one way. That is, any positive integer admits a unique representation in the binary system.

3

Pt. I. Chap. 1. No. 5-20

15. A set of weights 1, 3, 9, 27, 81, ... can be used to weigh any weight that is a positive integral multiple of a given unit if both pans of the scales are used, and this can be done in exactly one way. 16. Write (1

+ qC) (1 + qC2 ) (1 + q~) (1 + qC8 ) (1 + qC16 ) ••• = ao + alC + a2C2 + a3C3 + ....

Find the general formula for a". 17. Consider the expansion (1 - a) (1 - b) (1 - e) (1 - d) ... = 1- a- b

+ ab - e + ae + be -

abe - d

+ ....

What is the sign of the n-th term? 18. Prove the identity

(1

+ C+ C2 + C3 + ... + C9 ) (1 + CIO + C20 + .. , + C90 ) X (1 + C100 + C200 + ... + C900 ) ... = 1 ~ C'

18.1. The first and the third problem considered in the solution of 9 (concerned with A" and en respectively) are the extreme cases of a common generalization which, properly extended, includes also 18. Formulate such a generalization. 18.2. In a legislative assembly there are 2n + 1 seats and three parties. In how many different ways can the seats be distributed among the parties so that no party attains a majority against a coalition of the other two parties?

19. (1

+ C) (1 + C2 ) (1 + C3 ) (1 +~) ... =

(1 _ C) (1 _ C3)

(~ _

C5) (1 _ C7) •.••

20. Each positive integer can be decomposed into a sum of different positive integers in as many ways as it can be decomposed into a sum of equal or different odd positive integers. E.g. the decompositions of I) into sums with different terms are 6,

1

+ 5,

+ 4,

1

+ 2 + 3,.

+ 3,

1

+ 1 + 1 + 3,

1

+ 1 + 1 + 1 + 1 + 1.

2

with odd terms 1

+ 5,

3

4

Operations with Power Series

21. It is possible to write the positive integer n in 2,,-1 - 1 ways as a sum of smaller positive integers. Two sums that differ in the order of terms only are now regarded as different. E.g. only the seven following sums add up to 4: 1

+ 1 + 1 + 1,

1

+ 1 + 2,

2

+ 2,

+ 2 + 1, 2 + 1 + 1, 1

1

+ 3,

3

+ 1.

22. The total number of non-negative integral solutions of the following Diophantine equations is n 1:

+

x

+ 2y = n, 2x + 3y = nx + (n + 1) y = 1,

n - 1, (n

+ 4y = n + (n + 2) Y =

3x

+ 1) x

2; ... ,

O.

23. The total number N of non-negative integral solutions of the following Diophantine equations

+ 4y = n - 5, ... is smaller than n + 2; moreover the difference n + 2 - N is equal to the number of divisors of n + 2 (d. VIII, Chap. 1, § 5). x

+ 2y = n -

1,

2x

+ 3y = n -

3,

3x

24. Prove that the total number of non-negative integral solutions of the following Diophantine equations is n: x

+ 4y = 3n -

1,

4x

+ 9y =

5n - 4,

9x

+ 16y = 7n -

9, ...

25. The number of non-negative solutions of the Diophantine equation x + 2y + 3z = n is equal to the integer closest to

(n

-:-2 3)2 •

26. Let a, band n be positive integers, a and b relatively pr ime to each other. The number of non-negative integral solutions of the equation

is equal to

[:bJ or [:bJ + 1. [More may be less: to prove a more general

or a more precise theorem may be less trouble.] .

27. Let ai' a2 , ••• , at be positive integers without a common factor different from 1 and A" be the number of non-negative integral solutions

5

Pt. I. Chap. 1. No. 21-30

of Then we have .

1

An

lim - - =

n..,.~ n l - l

a l a 2 ••• al(l - 1)!

.

27.1 (continued). We suppose more: we assume that a j and aj are relatively prime whenever i =F i. Then we can assert more: An

=

P(n}

+ Qn.

where P(x} is a polynomial with rational coefficients of degree 1 - 1 and the sequence Qn is periodical with the period al a2 ... al :

27.2 (continued). In the particular case 26 where 1 = 2, a l

az = b


1, when n

= a,

a - b,

0 and generally A n+ab = An

+ 1.

28. The points in three-dimensional space whose Cartesian coordinates x, y, z are integers are called lattice points of this space. How many lattice points of the closed positive octant (x > 0, Y > 0, z > 0) lie on the plane x + y + z = n? How many lattice points of the open octant (x > O. y > 0, z > 0) are contained in this plane? 29. Let n be a positive integer. How many lattice points (Xl' x 2 , ••• , x p ) of the p-dimensional space lie in the "octahedron" 1Xl 1

+

x2 1

1

+

x8 1

1

+ ... +

1xp 1

:s;: n ?

30. Consider those lattice points in the closed cube -n < that satisfy the condition

-s :s;: x

X,

y, z :s;: n

+y +z
3. The number of positive integral solutions of x+y+z=n

that satisfy the additional conditions x

O.

+ 2P2q2 + SPsqs + ... + nP.. q.. = n2p.. q..

....... 00

78. The series a l

(al

...• q... ... be two sequences

-

~. (¥

+ fl

+ as + as + ... does not necessarily converge if a..) + (a 2

-

a..) + ...

+ (a.. _l

-

a.. )

is bounded as n -+ 00. If. however. the additional conditions

al >a2 >aS >···. are satisfied the series a 1

lima.. =O ....... 00

+ aa + a3 + ... must converge.

{J •

19

Pt. I, Chap. 2, No. 75-82

§ 2. More General Transformations of Sequences into Sequences 79. Consider the infinite matrix POO' POI' PO?' .,.

PlO' PU' P12' .,.

P2O> P21' P22'

.•.

Suppose that all the numbers PM are non-negative and that the sum in each row is convergent and equal to 1 (P ...

> 0; 1: PM = 1, for n, .=0

v = 0, 1, 2, ...). Let so' Sl' ••• , s..' '" form a bounded sequence. Defi.ne a new sequence to' t1 , t2 , ••• , t.. , ... by setting

t.. =

P..oso

+ P..1s1 + P..2S2 + ... + P...s. + ....

Show that t.. has a value between the upper and the lower bou'ld of the sequence so' S1' ••• , S .. ' ••• (whose terms are here supposed to be real). 80 (continued). The convergence of the sequence so" SI' S2' ••• to a limit S implies the convergence of the transformed sequence to' t1, t2 , ••• to the same limit s if and only if lim PM

"-+00

=0

for each fixed v. (This is the necessary and sufficient condition of the "regularity" of the transformation with matrix (P...); cf. 66.) 81. Assume that the series c1

+ 2c2 + 3cs + 4c4 + ... + nc.. + ...

converges. Then the series cn

+ 2c"+1 + 3c"+2 + 4n.. +s + ... =

t..

converges too and lim t.. = O.

"-+00

82. Let the power series

t(x) = ao + a1x + a2x2 + ... be convergent for X = 1 and assume 0 < ex t(ex) +f'( 0,

is convergent for b +bl+b.j'-+··· { o I divergent for

Then

/10

n = 0, 1,2, ... ;

It I


:E

0;

,,=0

b"tn converges for all values of t; .

a"

hmb=s.

Then a o + alt + and in addition

n-+oo

a2t2

+ ... + ant" + ... converges too for all values of t

ao + alt

.

lim

n

1-+00

+ a2t 2 + ... + ai" + ... 2 =s. bo + bIt + b2t + ... + bnt" + ...

(Cf. IV 72.)

95. If lim sn = s exists then n-+oo

. (

hm

1-+00

So

t2 t") -I + Sl -1't. + S2. -2' + ... + s" , + ... e = n.

S.

96. Assume that the sum ao

+ al + a2 + ... + an + ... = S

exists. Define

g(t)

=

ao

+ al iTt + a2 2!t + ... + an n!t + .... 2

n

Then

f

e-Ig(t) dt = s. o 97. The Bessel function of order 0 is defined as lo(x)

=

1

1 - i!l!

(X)2 1 (X)4 (_1)m(X)2m + .... 2 + 2! 2! 2 -'" + m! mT '2

Pt. I, Chap. 2, No. 92-97· Chap. 3, No. 98-102

23

We have

Chapter 3 The Structure of Real Sequences and Series

§ 1. The Structure of Infinite Sequences 98. Let the terms of the sequence aI' az, aa, ... satisfy the condition

m, n = 1, 2, 3, ... ; then the sequence

either converges to its lower bound or diverges properly to - 00. 99. Assume that the terms of the sequence aI' az, aa, ... -satisfy the condition Then an

lim - =

n-+oo

n




tn - 8n for all n.

Then the numbers t 1, t z, ... , tn' . " are everywhere dense between their lowest and highest limit points.

24

The Structure of Real Sequences and Series

103. Let "1' "2' ... , Vn , ••• be positive integers, "1 The set of limit points of the sequence




In-1sn-1'

lnsn

>

110. If the sequence

In-2 sn-2' ... ,

~\ ~2,

••• ,

~,

lnsn '"

>

l1S1'

tends to

+ 00

[107, 108.J and if A is

larger than its minimum [105J then there exists a subscript n (or several

25

Pt. I, Chap. 3, No. 103-112

subscripts n). n

> 1, so that the quotients L .. - L .. _ 2 --2-

L .. - L,,_1

---1

are

~

L" - L,,_a

L"

-3 - , .•. , n

A and the infinitely many quotients L,,+2 - L .. 2

L,,+1 - L .. --1--

L,,+S - L"

-3 - ' ...

---

are all > A. [The quantities in question can be interpreted as the slopes of certain connecting lines between the points with cartesian coordinates

(0, L o), (1, L 1 ), (2, L 2 ),

(m, Lm), ... ;

••• ,

Lo = O.

This interpretation leads to a geometric proof of the statement.] 111. Assume that the sequence l]> l2' la' ... , lm' ... satisfies the sole condition lim lm = +00. m-+oo

Let A be larger than II (A > ll)' Then there exists a subscript n, n such that all these inequalities hold simultaneously: 1"_/1+1

+ ... + 1"_1 + I" < A fI

-

~ 1"+1 -

p

=

>

1,

+ 1,,+2 + ... + 1,,+. , v

1, 2, ... , n;

11

=

1, 2, 3, ....

If A tends to infinity then so does n. 112. Let the sequence ll' l2' ... , lm' ... satisfy the two conditions

lim lm

m-+oo

m-+oo

=

0,

and assume II > A > O. Then there exists a subscript n, n that the inequalities 1"_/1+1

+ ... + 1"_1 + fA

I"

> A> -

1"+1

> 1, such

+ 1.. + 2 + ... + l,,+>, v

-

p = 1, 2, ... , n;

11

= 1, 2, 3, ...

hold simultaneously. If A tends to 0, then n tends to infinity.

§ 2. Convergence Exponent The convergence exponent of the sequence r1 , r2 , r a, ... , rm' ... where o < r1 < r2 < "', lim rm = 00, is defined as the number;' having the m .... oo

following property:

rIO'

+ riO' + riO' + ... + r;;;O' + ...

26

The Structure of Real Sequences and Series

converges for q > A. and diverges for q < A.. (For q = A. it may converge or diverge.) For q = 0 the series is divergent, therefore A. >0. If the series does not converge for any q then A. = 00. 113. Show that .

logm

lim sUP-I-

og 1'",

"'-+00

114. Let

Xl' X 2 ' X a'

••• , X""

=A..

be arbitrary real numbers, x'" =1= O.

•••

If there exists a positive distance IS such that IXI - xk I > IS, 1< k, I, k = 1, 2, 3, ... , then the convergence exponent of Ix1 1, Ix2 1, Ix3 1, ... , IX", I, ... is at most l.

115. Let {J be larger than the convergence exponent of the sequence

'1' '2' '" Then there exist infinitely many subscripts n for which the

n - 1 ineqUalities 1


0,

27

Pt. I, Chap. 3, No. 113-121

Then arbitrarily large values of nand r can be found for which the following inequalities hold simultaneously [111]: k = 0, 1, 2, 3, ...

(At the root of this fact, and of 122, lies the comparison of two power series

°

Suppose that Po > 0, PI > 0, ... , Pm > 0, and that P. =l= for at least one subscript i. Let e, e> 0, possibly e = 00, be the radius of convergence of the power series

Po The sequence

+ PIX + P2 X2 + ... + Pmx'" + ....

° °

Po' PIX, P2x2, ... , Pm xm , ...

converges to if < X < e. Therefore there exists [105J a maximum term whose value is denoted by ,u(x). I.e.

Pmx'"

~

,u(x),

m = 0, 1, 2, ...

The central subscript v(x) is the subscript of the maximum term, i.e. ,u(x) = P'(X)x·(X). If several of the terms Pmxm are equal to ,u(x) we call v(x) the largest of the corresponding subscripts. More details in IV, Chap. 1. 119. For an everywhere convergent power series in X which is not merely a polynomial the central subscript v(x) tends to CXJ with x. 120. The subscript of the maximum term increases as x increases. (One might consider this situation as somewhat unusual: in the course of successive changes the position of maximum importance is held by more and more capable individuals.) 121. The series

Po

+ PIX + P2X2 + ... + Pmxm + ...

with positive coefficients and finite radius of convergence e (Pm> 0, e > 0) is such that one term after the other, all terms in turn, become maximum term. Then ~ is the radius of convergence of the series (!

x x xm -+-+-+ ... +-+ .... Po Pi P2 Pm 1

2

28

The Structure of Real Sequences and Series

122. The dominance of the maximum term is more pronounced in an always convergent series than in one that does not always converge (it is strongest in a polynomial). More exactly: let the radius of convergence of the power series

tlo

+ a1x + ~X2 + ... + a".x'" + ...

be infinite and that of the power series bo + b1y

+ b2y2 + ... + b".y'" + .. . a". > 0, b". > 0, m = 0, I, 2, ...

be finite. Suppose The coefficients bo' b1 , bs' ••. be such that all the terms b".y'" become in turn maximum term [120J. Then a value y can be determined for certain arbitralily large positive such that for these corresponding values the respective series have the same central subscript. Let the common central subscript be n. Then all the following inequalities hold simultaneously:

x

ax"

by" b,.Y' -

-~ ~ -"- ~ 1

a,;;" -

00

1:

[Consider the maximum term of

".=0

a b'"

k = 0, I, 2, ...

,

z"'.]

".

123. If there are values x* to which no y corresponds in the sense of 122 then they are "rare". They have a finite logarithmic measure, i.e. the set of points log x*, x* exceptional value, may be covered by countably many intervals of finite total length.

§ 4. Subseries Let tl' t2, ts' ... , t.. , ... be integers, 0 < tl < t2 < ts < .... The series is called a subseries of the series

a, + a, + a, + ... + a, +. .. al+~+a3+"·+a.. +"·, J

I

I

"

124. From the harmonic series

~+.!..+~+ ... +~+ 1

2

3

n

...

remove all terms that contain the digit 9 in the decimal representation of the denominator. The resulting subseries is convergent. 125. If all the subseries of a series converge then the series is absolutely convergent. 126. Let k and 1 denote positive integers. Must the convergent series a 1 + as + a3 + ... be absolutely convergent if all its subseries of the

29

Pt. I, Chap. 3, No. 122-131

form

+ aHI + aH21 + aH31 + ...

a"

(subscripts in arithmetic progression) converge? 127. Let k and l be integers, k :?: 1, l :?: 2. Must the convergent series a 1 + all + as + ... be absolutely convergent if all its subseries of the form a"

+ a", + aid' + a"" + ...

(subscripts in geometric progression) are convergent? 128. Let q;(x) denote a polynomial assuming integral values for integral x, q;(x) = cox' + C1x'-1 + ... [VIII, Chap. 2]. Assume that the degree is l > 1 and that the coefficient Co of x' is positive (co> 0). The values q;(0). q;(1). q;(2), ... form a generalized arithmetic progression of order l; since Co > 0 only a finite number of terms of the progression can be negative. Must a convergent series a1 + as + as + ... converge absolutely if all its subseries whose subscripts form a generalized arithmetic progression a9'(O)

+ a9'(l) + a9'(2) + a9'(S) + ...

(omitting the terms with negative subscripts) converge? 129. If the series a1 + all + as +' ... converges absolutely and if every subseries a, + a21 + aSI + "', l = 1, 2, 3, ... has the sum 0 then al = as = ... = o. 130. Consider the set of points determined by all the subseries of 222 9" 27

3"

2

+ + + .. , + 3" + ....

This set is perfect and nowhere dense (closed and dense in itself. but nowhere dense in the set of all real numbers). (We have to consider all the subseries. finite and infinite. including the "empty" subseries to which we attribute the sum 0.) 131. Let the terms of the convergent series

PI + P2 + Ps + ... + P.. + ... = s satisfy the inequalities

Pl:?:P2:?:PS:?: .. ·• o< P.. :=:;; P"+l + P.. +2 + P"+s + ... . Then it is possible to represent any number

(f

n = 1, 2, 3, ...

in the half-closed interval

30

The Structure of Real Sequences and Series

o< < (f

s by an infinite subseries:

+ Pt. + Pt. + ... = Find the series PI + P2 + Pa + ... that satisfies the conditions 1 PI =2' P.. = P.. +] + P"+2 + P"+3 + ... , n = 1, 2,3, ... , PI,

132.

(1.

and verify that in this case every (f mentioned in 131 can be represented by one infinite subseries only.

§ 5. Rearrangement of the Terms 132.1. By rearranging the factors of the infinite product

(1 + ~) (1 - !) (1 + !) (1 -

~) (1 + ~) ... =

Pl,l

we obtain the infinite product

(1 +~) 2 (1 +~) 4 ... (1 +~) 2p X(1-~)(1-~)"'(1--,1-)(1 +_1_) ... 3 5 2q + 1 . 2p + 2 P p,q =

in which blocks of P factors greater than 1 alternate with blocks of q factors smaller than 1. (Factors of the same kind remain in the "natural" order.) Show that

[II 202.]

pp,q=Vf·

132.2. By rearranging the terms of the infinite series

-21 - -31 + -41 - -51 + ... =

5 I,] = 1 - log 2

we obtain the infinite series 111

5p,q = 2' + '4

111 - 3 - 5'

+ 6' + ... + 2p

- ... -

2q

1

1

+ 1 + 2p + 2 + ...

in which blocks of P positive terms alternate with blocks of q negative terms. (Terms of the same kind remain in the "natural" order, steadily decreasing in absolute value.) Show that 1 p 5p,q - 5].1 =2 Iog -q' Let r1, r2 , r3 , ••• , s1> S2' S3' ••. be two sequences of steadily increasing natural numbers without common terms. Suppose furthermore that all positive integers appear in one or the other of the two sequences

31

Pt. I, Chap. 3, No. 132-136

(r",

< r"'+1' s.. < a"

s"+1' r",

Ee: s.. for m, n =

+ a" + a" + "',

1, 2, 3, ...). The two series

+ as, + as, + ...

as,

(the "reds" and the "blacks") are complementary subseries of the series a l + a2 + a3 + .... Let "1' "2' "3' ... be a sequence of integers such that each natural number 1,2,3, ... appears once and only once in it (a permutation of the natural numbers). The series

a."

+ a., + a•• + ... + a... + ...

is obtained from the series al

+ az + as + ... + a.. + ...

by rearrangement. We call special attention to the rearrangements where a, (m = 1, 2, 3, ... ) is the r",-th term before as well as after the re-

'" arrangement, i.e. which preserve the subseries a" + a" + a" + .... If before as well as after the rearrangement a, precedes a, and as '" '" precedes as for all number pairs m, n. In < n, we say that the rearrangement shifts the two complementary subseries relatively to each other (and it leaves each in its original order). 133. If one of the two to each other complementary subseries of a convergent series is convergent the other is convergent too. A rearrangement which only shifts the two subseries relatively to each other does not change the sum of such a series. 134. If one of the two to each other complementary subseries of a conditionally convergent series diverges to + 00 then the other diverges to - 00. Provided that all the terms of one of these two subseries are of the same sign it is possible to obtain an arbitrary sum for the whole series by shifting the two subseries relatively to each other. 135. It is not possible to accelerate by rearrangement the divergence of a divergent series with positive monotone decreasing terms. 136. By rearranging the series we can slow down arbitrarily the divergence of a divergent series with positive terms which tend to O. More explicitely: Assume

..

.

P.. > 0,

lim P,. = 0,

"-+00

lim

11-+00

(PI

+ pz + ... + P..) =

0< QI < Q2 < ... < Q.. < ''', Then there exists a rearranged series p" such that

P•, +' P., + ... + P',. < = Q,.

lim Q,. =

"-+00

00.

00.

+ P., + P" + ... + P.,. + ... for

n = 1,2,3 ...

32

Miscellaneous Problems

137. Assume that

+ a2 + as + ... + a.. + ... =

al

s is convergent, is divergent.

Let s' < s < sIt. By a rearrangement that leaves all the negative terms at their places the sum s' can be realized; by a rearrangement that leaves all the positive terms at their places the sum s" can be realized [136].

§ 6. Distribution of the Signs of the Terms

> 0, PI ~ P" ~ Ps ~ ... and that the series

138. Assume that P..

PI

+ P2 + Ps + ... + P.. + .. .

is divergent and the series

e)PI

+ e2P2 + eaPa + ... + e..p.. + ... ,

where e" is -lor +1, is convergent. If under these conditions a certain percentage of terms is positive then it is 50%. More precisely: lim inf

.. ·+e

~1+e2+

..

n

=:;: 0 < lim sup

"-+00

..

n

"-+00

139. Suppose

P.. >

e1P) where

e1 +£2+· .. + e

e" is -lor

0,

PI ~ P" ~ Pa ~ ... and that the series

+ e2P2 + eaPa + ... + e..p.. + ... ,

+1, is convergent. Then

+ e2 + ea + ... + e.. ) P.. =

lim (e1 ...... 00

O.

(Notice the two wellknown extreme cases

el

= e" = es = ...

and

e1

=

-e2 = es = -e, =

....)

Chapter 4

Miscellaneous Problems

§ 1. Enveloping Series We say that the series Ilo the relations

IA -

(ao + a1

+ a1 + ~ + ... envelops the number A if

+ a2 + ... + a..) I < Ia.. +1 I'

n = 0, 1, 2, ...

Pt. I, Chap. 3, No. 137 -139 • Chap. 4, No. 140-145

33

are satisfied. The enveloping series may be convergent or divergent; if it converges its sum is A. Assume that A, flo, ai' a2 , ••• are all real. If we have A - (ao + a l + a 2 + ... + all) = Olla,,+!, for all n = 0, 1, 2, ... and 0 < 0,. < 1.

the number A is enveloped by the series ao + al + a2 + ... , and in fact it lies between two consecutive partial sums. In this situation we say the series is enveloping A in the strict sense. G. A. Scott and G. N. Watson [Quart. J. pure appl. Math. (London) Vol. 47, p.312 (1917)] use the expression "arithmetically asymptotic" for a ciosely related concept. The terms of a strictly enveloping series have necessarily alternating signs. 140. Suppose that f(x) is a real function of the real variable x. If the functions \I'(x) \' \I"(x) \, ... are steadily and strictly decreasing in the interval [0, xJ, x > 0, then f(x) is enveloped in the strict sense by its Maclaurin series. 141. The functions e-%; log (1

+ x);

(1

+ x) -p,

are strictly enveloped by their Maclaurin series for x > 142 (continued). Prove the same for the functions l

p>o;

o.

cos x, sin x. 143 (continued). Prove the same for the functions

arc tan x,

1

(,,)2 + 2! 12! (,,)4 2" - ' "

Jo(X) = 1 - ffIT 2"

[141,142] .

144. Suppose that the terms of the series ao + al + as + ... are alternately positive and negative and that there exists a number A such that A - (a o + al + a2 + .. , + a,,)

assumes always the same sign as the next term, all+!' Then the series envelops A in the strict sense. 145. If the series ao + a) + as + "', all real, n = 0, 1, 2, ... , envelops the real number A and if in addition \all > Iaal > ItZa I > .. . 1 Obviously only the non-vanishing terms of the Maclaurin series are to be considered. E.g. the n-th partial sum of the Maclaurin series for COS" is

,,2

1 - -2!

,,4

+ 4!

-

...

"2,,

+ (-1)" -, (2n)!

n = 0, 1,2, ...

34

Miscellaneous Problems

then the terms al> a2 , a3 , '" have alternating signs and A is enveloped in the strict sense. 146. Let the function I(x) assume real values for real x, x > R > O.

If I(x) is enveloped fot x

>

R by the leal series a o + axl

+ xa~ + -~;,;- + ...

then the numbers aI' a2 , a3 , ••• have alternating signs and the series is strictly enveloping. 147. Suppose that the real valued function I(t) is infinitely often differentiable for t > 0 and that all its derivatives P')(t) (n = 0, 1, 2, ... ) have decreasing absolute values and converge to 0 for t -+ 00. Then the integral

J

I(t) cos xt dt o is, for real x, strictly enveloped by the series (0) (0) - f'---xz + 1'"-;4 -

fV (0)

---;G

fVII (0) + ---xs- ....

(Example: I(t) = e- t .) 148. The number : is enveloped by the series 31313131 4 - "8 - "8 16 16 - 32 - 32

+

4

+ +

+ ... ,

but not in the strict sense. 149 1• Plot the first seven terms of the series i

e

i 1 i 1 i 1 =1+-----+-+----··· 11 21 31 4! 51 61

successively as complex numbers and compute so the value of ei to three decimals. 150. Let ~ denote a ray with origin z = O. Assume that along ~ all the derivatives of the function I(z) attain the maximum of their absolute values at the origin and only there; i.e.

I/(n)(o)

I>

I/(n)(z) I

whenever z is on ~ and Iz I > O. Then: a} The function I(t) is enveloped for every z on series

~

by the Maclaurin

[140] . 1 In 149-155 the terms of the series are complex numbers; they are regarded as points in the Gaussian plane (complex plane) [III 1 et seqq.J.

35

Pt. I. Chap. 4. No. 146-155

b) the function F(z) = by the series 1(0)

j

e- I

I( : )dt is for every z on ~ enveloped

0

+1'(0) +f"(0) +/",(0) + ... Z

[147J;

z3

z8

•~ is the mirror image of ~ with respect to the x-axis; the integral is taken along the positive t-axis and converges under our assumptions on I(z). 151. The Maclaurin series of e-', log (1 + z) and (1 + z)-P, p > 0, envelop the respective functions for mz > 0, z =l= O. 152. Let z be restricted by the following conditions:

z =l= O. I"

00

Then the function e"2

,.

J e-"2 dt is enveloped by the series



1 1·3 1.3·5 -1z - --- + ... z3 +z5 Z7

(strictly enveloped if z is real). 153. Suppose that a" and btl are arbitrary complex numbers that have the same argument, i.e.

a

btl

z =l= 0 the two series ao + a1z

is real and positive. If at a certain point

"

+ a2z'l + ... + a..z" + ... ,

bo + blz

+ b2z2 + ... + b..z" + ...

envelop the values 91(z) and 1p(z) then the combined series

envelops the value 91(z) + 1p(z) for this z. (The same is true for enveloping in the strict sense if all the coefficients and z are real.) 154. If z lies in the sectors described in 152 the function z coth z is enveloped by its power series zcothz=z

e'

+ e-'

e' _ e -.

= 1

+B

(2z)2 _ 1 2!

B (2z~ 2 4!

+ B. (2Z)B + ... ;s

6!

(strictly enveloped if z is real). The coefficients B]. B 2, B3 • ••• are called the Bernoulli numbers. 155. The function w(z) = log r(l

+ z)

- (z

+ 1) log z + z -llog (2n)

[II 31]

36

Miscellaneous Problems

can, for

mz > 0, be written as an integral

f

co

w(z) = 2 o

arctan'!"" 2'"

e

Z

- 1

de.

[Cf. E. T. Whittaker and G. N. Watson, pp. 251-252.] Prove that the resulting (divergent) Stirling series Bl 1.2 ·z -

B. 3·4.

envelops the function w(z) if

B.

za + 5·6· z6 -

mz > 0 and -

•••

:

sin1

X

= sin x,

0 and that, moreover,

r 3"

lim 1 Ins sin,. x = 1.

11-+00

174. Assume that 0 < I(x) I(x) = x - a:l' for 0 < x




1.E,,(n) _

InIX+!

[VIII 81] .

1 we even have the inequality C(Ot lX

+ 1)1 < + 1 i=

2C(~) - 1 n

'

~

n-·, cf. VIII,

ft=l

n = 1, 2, 3, ...

[(! - [! J) is properly integrable over the interval [0, 1] if [107J; [! ] x'" is of bounded variation if > 1.J XIX

(X

tX

> 0,

57

Pt. II, Chap. 1, No. 42-48

46. Let 1'(n) =

I.(z) = l«,,-lZ + 2".-1,:2

Il)

+ y + z:S;; G.

+ ... + na. -1z" + ... ,

" = 1,2, ... , P

and

11 (z) Mz) ••• I/>(z)

=

Show that

= II'"

••• ;r"'/>-1-1 (1- x1 - x 2 - ' " - X1>-1 )",/>-1 />-'1

I~1-1~-1 1

1: a..z" .

.. -1

2

X dX1 dX2

... dX/>_l'

where the integral is taken over the domain described by the p inequalities xJ ~ 0, x2 ~ 0, ... , x/>_l ~ 0, xJ + x2 + ... + X/>_l :s;; 1 (p - 1 dimensional simplex). 88 (continued).

II ••• I ~.-lX;.-J ... X;~11-J (1 - Xl - x2 Xdx1 dx2

-

••• -

x/>_1)"'r 1

r(IX 1) r(IX2 ) '" r(IXp )

.. ·dx"_1=r( r IX1

87. Work out the ploduct

+ :;06

•••

I(x/» dX1 dX2 ••• dx/> =

when p is fixed and n increases to infinity. 88. Suppose that the 2m functions l(x 1 ), l(x 2 ),

....

I",(x),

9"(x1). 9"(x2),

••• ,

9"",(x)

~

(j

p. "

I(x) dX)/> ,

62

Inequalities

are properly integrable over the interval [a, b). Then we have

I j 11 (x) a

b

b

qJl (x)

f 11 (x) qJ2(X) dx

dx

f 11 (x) 'P,... (X) dx

a

"

"

"

" b

f I... (X) qJ2(X) dx

a

1

=

m!

a

b ~

b

IIJ (Xl) 11 (X2) •• , 11 (X... ) I

J j ... J Mx

a

a

j )

a

qJI (Xl) qJI (X2)

•• , qJl (X... )

12(x2.) .. , Mx... ) • qJ2(X1 ) qJ2(X2) '" qJ2(X... )

a

~~ i~'(;:)'i~(~~) ..... :i~'(~~) I~~(~~i ~~.(;:).::: ~~(~~)

[Compute in two different ways the product of the two matrices

II/!~ 11).=1.2........ • "qJ~ 111'=1.2 ........ ; ,,=1,2, ... ,'"

,,=1.2"0"'"

Chapter 2 Inequalities

§ 1. Inequalities Let aI' a2 , " ' , an be arbitrary real numbers. Their arithmetic mean is defined as the expression

~l(a)

m:(a) = If all the numbers aI' a2 , and harmonic means as

.•• ,

al

+ a2 + .. , + a,. n

.

an are positive we define their geometric

~(a)

=

lIn

1 '

-+-+ ... +a1 a2 an

63

Pt. II. Chap. 1 . Chap. 2

respectively. If m denotes the smallest and M the largest of the numbers Q i then

m
0 on Em. M]. Inthis case qJ(t) is strictly convex. If we have qJ"(t) ~ 0 only, then qJ(t) is convex. (A function can be convex in an interval where its second derivative does not exist at all points.) 73. The functions

r

(0 < k

< 1)

and

log t

and

tlog t

and

Ve s

are concave on any positive interval;

r

(k

< 0 or k> 1)

are convex on any positive interval; log (1

+ e')

+ t2

(e> 0)

are everywhere convex. 74. Assume that qJ(t) is a convex function defined on Em. MJ. that Pl' PS' ...• P.. are arbitrary positive numbers and that tl , ts' ..• , t.. are arbitrary points of the interval Em, M]. Then we have the inequality Pl tl + Psts + ... + P..t,,) ~ P1rp(tl ) + P2rp(tS ) + ... + P..rp(t.. ) • qJ PI + P2 + ... + P.. PI + P2 + ... + P.. (

75. Assume that I(x} and P(x} are two functions which are properly

integrable over [X1,X2 ] and that m ~l(x} ~ M, P(x) > 0 and

f P(x) dx>O. %.

%,

Let qJ(t) denote a convex function defined on the interval m Then we have j"P(X) f(x) dX)

qJ (

%,


[f(~)] d~

~'*...:'~----

-

I p(~) d~

'*1

I p(~) d~

-*1

~1

There is equality if and only if I(x) is a constant. 78. Prove the following generalization of the proposition on the arithmetic, geometric and harmonic means: Let PI> P2' ... , P.. , at, "2, ... , a.. denote arbitrary positive nuinbers, a. {: aj for at least one pair i {: j, . i, j = 1, 2, ... , n; then the inequalities

+ 1'2 + ... + 1' .. < 1'1 + p. + ... + 1'.. PI

~

a.

1>,log", +l>i10g"I+'"+1> ..log.... I>,+I>I+"·+P..

e


1 p.. -log.., + -log... '1-'·'+ -log.... 41

e

lin

".

p,

p.

al

4.

-+~+ ...

1>..

+A,.

+ ... + 1'.. I' , 1'1 +1'8 + ... +2 PI +1'2

1 or ,,< 1. Equality is attained only for a, = A.b., v = 1, 2, ... , n, or if " = 1. (What does the proposition mean in the case" = 2?) (Minkowski's inequality.) 91. The function j(x) is defined on [Xl' x2J, properly integrable and strictly positive. We introduce I

9R,,(/) = (}' [/(x)]"

dX)-;

Let g(x) be a function with the same properties as j(x). Then we have

9R,,(/

+ g) S;;

or

> 9R,,(/) + 9R,,(g) ,

according as " > 1 or " < 1. 92. Let a, A, b, B .be positive numbers, a < A, b < B. If the n numbers aI' a2 , " ' , a.. lie between a and A, and the n numbers bI' b2 , ••• , b" between band B we can prove that 1
0, but not identically zero. Let < a < b. If all the integrals occurring exist we find the inequalities

°

1-

(a : ~ :

lY
S2' S3' ••• are given, we have here a "perfect" scheme for computingI'. The- n-th step yields the lower bound S;;1/" and the upper bound s,,/s"+1 for 1', the next step improves both bounds and both bounds converge to the desired I' as n tends to 00. Observe that SIZ

zG/() + Sr2 + S3z3 + ... = .=1 ~ _z_ = _ _ z_ Y. - z G(z) 00

where G(z) =

jj (1 - Y.~)

.=1

and I' is the zero of G(z) nearest to the origin. Compare 197, III 342.) 95.4 (continued). Even if s,. is not given for all values of n but only for n = 1, 2, 4, 8, ... , 2"', ... , we can devise a scheme for computing y. Prove that 1

1

1

1

1 (1)2 o.

Then cp(x) vanishes identically. 143. The r-function r(s) - lim -

,,-+co

S

s(s

n nl

+ 1) ••• (s + n)

can be written as an integral [31]. Use this fact to prove that r(s) does not have any zeroes. [r(s + 1) = sr(s), 142]. We associate with each function that is defined on [0, 1] the polynomials

K" (x) =

~ j ( : ) ( : ) x· (1 -

>

x)" -v,

n = 0, 1, 2, ...

This polynomial is bounded on [0,1] from below by the greatest lower bound, and from above by the least upper bound, of j(x) and it coincides with j(x) at the endpoints. 144. Work out the polynomials K,,(x), n = 0, 1, 2, ... for j(x) = 1,

j(x)

=

x,

j(x)

= x2 ,

j(x) = ~.

145. Let x be any point on [0, 1] and

1=

1: (:) xV(1 -

x)"-V = EI

>=0

where EI refers to the subscripts for which I'll those for which I'll - nx I > n 3f', n > 1. Then 1

ElI O.

86

Various Types of Equidistribution

151. Suppose that L(r) is positive for r> 0, monotone increasing and that for r sufficiently large L(r) = (log 1')"" (log2 1')""

(log" 1')"''',

•••

IXl

>

O.

[log" x = log"_l (log x).] Then L(r) is slowly increasing. 152. If L(r) is slowly increasing then - 0 lim log L(r) . log r

r~oo

153. If N(r) denotes the counting function of the sequence 1'1' 1'2' rs ' ... , rn , ••• and if N(r)

N

rtL(r) ,

where L(r) is slowly increasing, 0 < A. < exponent of the sequence 1'1' r2 , 1'3' ••• , rn , A sequence r1 , 1'2' r8 ,

••• ,

00,

then A. is the convergence

•.•

r.. , ... of the type considered in 153 will

be called a regular sequence in the sequel, 154-159. Later on (e.g. IV

59-IV 65) sequences for which N(r)

A N

:'(r)

will also be termed regular.

If we take the term in this broader sense also the prime numbers 2. 3, 5,

7, 11, ... form a regular sequence and the propositions 153-159 remain valid without alteration. 154. The counting function of a regular sequence with convergence exponent A. satisfies the relation

r

N(cr) _

r-!.~ N(r) -

C> O.

A

c,

155. Let N(r) be the counting function of the regular sequence r1' r2 • rs •...• rn •... with convergence exponent A. and t(x) be a piecewise constant function on the interval (0, c], c > O. Then

156. The limit relation in 155 is also valid if t(x) denotes a properly integrable function on [0, c]. 157. Let N(r) denote the counting function of the regular sequence r1 , r2 , •••• rn , ••• with convergence exponent A. and let IX > O. Then 1

lim N()

'-+00

t'

L:

'n~"

(r~ )"'-A = J x",-A dx = -. ). r ex 1

0

A

87

Pt. II. Chap. 4. No. 151-161

158 (continued).

(1' )-"'-A =J X

-",-A

1

lim- ~ ~

'-+00 N(r)

',. > ,

l'

co

1

-

A

A dx=-. ex

159. Assume that N(r) is the counting function of the regular sequence 1']. 1'2' ••• , 1',., ••• with convergence exponent A, that I(x) is defined for x> 0 and properly integrable over every finite interval [a, b], o < a < b and that furthermore

and

I/(x) I < x"'-A

In

I/(x) I
0, < C1 < 1. The sequence

x,. = ana - [an"J is equidistributed on the interval [0, 1J. 176. Let a > 0, C1 > 1. The numbers

x,. = a (log n)" - [a (log n)"J



are equidistributed on [0, 1J. 177. For < a < 1, ~

°

sin 1"~ ~

the series

+ sin 2"~ + sin 3"~ + ... + sin n"~ + ... ~

~

~

is absolutely convergent if and only if e > 1. 178. Suppose that the square roots of the natural numbers 1, 2, 3, ... are written up one below the ,other in an infinite array. Examine the digits at the j-th decimal place (to the right of the decimal point), j > 1. Each digit 0, 1, 2, ... , 9 appears on 'the average equally often. More precisely: let vg(n) denote the number of those integers < n whose square roots show a g at the j-th decimal place. Then •

Pg{n)

1

lim -n- = -10 , g = 0, 1, 2, ... , 9. 179. Assume a> and x,. = a log n - [a log nJ, n = 1, 2, 3, ... and that the arbitrary function I(x) is defined and properly integrable

°

,.~~

91

Pt. II. Chap. 4. No. 173-182

over [0. 1]. Then the limit relation lim f(x l )

+ f(x 2 ) + ... + f(xn)

j I(x) K(x, ~) dx

=

n

n-+oo

0

holds provided that n increases to infinity in such a manner that xn ~~, o ~ < 1. The function K(x, ~) is given by

e

K(x,~)

=

j

log q rt-HI q- 1 log q

q-1

rt-
o. As t -+ 00 we obtain

§ 2. Modifications of the Method

210. Let

!X

and

f3 be

two real constants. Then the relation

+P B ( 1 ) -n!1 "+,,,yK" J e-xx" dx = A + -= + 0 -= 0 Vn Vn

holds where A =./r 21l

_1_

t'

(f3 B./r 21l

"'-"2 edt,

J

1

-00

_ /X2 + 2) e- ~ . 3

211. We denote by A a positive proper fraction and by x" the only positive root of the transcendental equation

1

X x, x" +-+-+ ... +-=AC 1! 2! n!

[V 42].

As n -+ 00 the root is given by x" = n

where

!X

and

+ V; + f3 +-0(1), !X

f3 satisfy the equations: 1

,/_ J e r 21l '"

00

t'

__

2

dt

= A,

212 (Continuation of 201). Let

!X

/X 2 + 2 f3=-3-·

denote a real constant. Then, for

n-+oo,

213. The functions 9'(x) , h(x) and I(x) = eh(x) are defined on the finite or infinite interval [a, bJ and satisfy the following conditions:

99

Pt. II, Chap. 6, No. 208-216

(1) tp(x) [I(x})" = tp(x) e"II(%) is absolutely integrable over [a, b], n = 0, 1,2, ... (2) The value of the function h(x} at a point Eof (a, b) is larger than its least upper bound in any closed interval to the left of E which does not contain E. Moreover there is a neighbourhood of E in which h"(x) exists and is bounded. Finally h' (E) > O. (3) tp(x) is continuous at x = E, tp(E) =1= o. Prove, for n -+ 00, the asymptotic formula H cdO~+l.

j

.

"tp(x) [I(x)]" dx

~ :~~

ePlI'(e) • n""'(~)-l •

e""(~),

where IX and {J stand for real constants. 214. Let E denote the only real root of the transcendental equation e1HE = 1. Then we have for n -+ 00 1 ~"+"'log"+11

J

I n.

0

ex" dx ~ nA B ,

where IX and {J are real constants and A=IXI+~_~

2 '

~

215. Suppose that n is odd and let-x" denote the only real root of the equation x x" 1 +-+-+ ... +-=0 I! 2! n! X

As n -+ 00

2

[V 74].

x" is asymptotically given by

x"

= En + IX log n + (J + 0(1),

where E is the only real root of the equation e1HE = 1 and IX and {J are given by

(J =

1

1 +~) +~ ~log (V~ 2n-~-.

218. Assume that the function g(x) is monotone increasing for positive x and that

lim g(x} =

We define

%-++00

+ 00,

100

Functions of Large Numbers

If there is a positive number" such that g(!U') lim -

s-++oo g(x)

exists and is a continuous function of (¥ for 1 - " :s;; (¥

< 1 +" then

log a..



lim - - = 1.

"-+00 g(n)

The method of problem 201 to evaluate functions of large numbers can be generalized in the following way: We have to estimate an integral of the form

J rp(x) Mx) Mx) •• ·/.. (x) dx = J rp(x) .f b

b

1

a

(s)+II,(s)+ ... +II,,("l

dx

a

where the functions h1 (x), h2(X), ... , h.. (x) 3,(e positive on (a, b) and attain their maximum at the same interior point E. Then we approximah. h.(x) = h.(E)

+ i h;'(E) (x -

E)2

+ ...

by h.(E)

+ 1h;'(E) (x -

E)2

and the integral by

We have supposed that rp(E) :::j= 0, moreover h;(E) = 0, h:' (E) < 0 as condition for the maximum at the point E, and -h~(E) - h;(E) - ... - h~(E) = s. The method can be justified in many instances and it is capable of adaptation and refinement. 217.

. 1"I ." ." [Put i- and recall 59, 115.] 11m

"-+00

-11

n! 2211COS" ." ." df} (2ne' -.1) (2ne' - 2) (2ne' - 3) ••• (2ne' - n) I

= 2",.

f} =

217.1. Analogy to 201 suggests sufficient conditions under which for

n~

+00

JJrp(x, y) e""(s.,,l dx dy

N

rp(E, 'I) e"II(E.'1l

2"

n

1ft

2

Y". ,/,,,,, - "sy

where the partial derivatives of second order hsz, h"", and hsy are taken at the point (E, 'I). Give a full statement and a proof.

§ 3. Asymptotic Evaluation of Some Maxima 218. The function Y;(x - 1) (x - 2) •.. (x - n) a-",

101

Pt. II, Chap. 5, No. 217-223

where a > 1, has the maximum M .. on the interval approximated by M.. nl

1

N

J!2;

en, + 00). It can be

1

[16.]

---."",-,

(a _ 1).. +1/2'

219. The function x{x2

-

12) (x 2

-

~) •••

(x2 _ n2) a-x,

where a > I, has the maximum M .. on the interval (n, approximated by M ..

n!1

220. We define

Vi =

+ 00). It can be

N~(2Va)2"+1 2n

[17.]

a-1'

Qo(x),

V; (1 - ;) (1 - ;) ... (1 - =) = Q..(x),

n=

1, 2, 3, ...

The sequence of functions

... , is uniformly bounded for x

>

0 if a > 2; it is not uniformly bounded if

0< a< 2. 221. We define x = Po (x), n = I, 2, 3, ...

The sequence of functions P 1 {x) a-x, P2 (x) a-x,

... ,

is uniformly bounded for x > 0 if a ~ 3 + V8; it is not uniformly bounded if 0 < a < 3 + V8. 222. Assume that a > 0, 0 < # < 1 and that M .. is the maximum of e-(x+a%l') x" in the interval (0, + 00). We find lim

""00

)"-1' = -T n.

(M

e- a •

§ 4. Minimax and Maximin *223. The function I(x, y) is continuous in the rectangle

a

O. Then

]

[f(x, y)]" dy)-l dX]

"=

min max f(x, y) . "

Y

Part Three

Functions of One Complex Variable General Part Chapter 1

Complex Numbers and Number Sequences

§ 1. Regions and Curves. Working with Complex Variables The complex variable z is written in the form z= x

+ iy = "e'f

(x, y, ", {} real, " ~ 0, {} taken mod 2..,;).

We call

x = 8lz the real part of z, y = 3z the imaginary part of z, l'

= Izl the absolute value of z (also modulus),

{} = arg z the argument or amplitude of z.

z

The number = x - iy = "e- i6 is the conjugate of z. 1. The number z + z is real, z - z is purely imaginary, zz is real and not negative. 2. What sets of points in the z-plane are characterized by the conditions:

8lz> 0;

8lz~

0; a < 3z < b;

IX~

arg Z::;, {J;

8lz = 0; 1

IZ-Zol=R; Iz-zol PI > P2 .•. > P,. >

Po

+ P1z + P2 Z2 + '" + p,.z"

O.

cannot have a zero in the unit disc 1z 1 ::;; 1. 23. Suppose that all the coefficients Pl' P2' ...• P,. of the polynomial

Poz"

+ P1z"-1 + ... + P,._lZ + P,.

are positive. Then the zeros of this polynomial lie in the annulus IX < 1Z 1 < {3. where IX is the smallest. {3 the largest among the values

... ,

P,.

P,.-l·

24. Let ao• at. a2 • •••• a,. be digits (ill. the ordinary decimal notation. that is integers between 0 and 9 inclusively) n > 1. a,.~ 1. Then the zeros of the polynomial

ao + a1z + a2z2 + ... + a,.z" are either in the open left half-plane or in the open disk

Izl 0, m 2 > 0, ... , m" > 0, m 1 + m 2 + ....+ m" = 1 and Z

= m]z1

+ m2z2 + ... + m"z".

Then there are points z. on both sides of any straight line through z except when all the z;s lie on that straight line. We can interpret the numbers m 1, m 2 , ••• , m" as masses fixed at the points Z1' Z2' ••• , z". Then the point z defined in 30 is the center of gravity of this mass distribution. If we consider all such mass distributions at the points Z1' Z2' ••• , z" the corresponding centers of gravity cover the interior of a convex polygon, the smallest one containing the points ZI' Z2' ••• , z". The only exception arises when all the points are on a straight line. Then the centers of gravity fill out the interior of the smallest line segment that contains all the points Z1' Z2' ... , z". 31. The derivative P'(z) of P(z) cannot have any zeros outside the smallest convex polygon that contains all the zeros of P(z) (considered as points in the complex plane). Those zeros of P'(z) that are not zeros

109

Pt. III; Chap. 1. No. 26-36

of P(z) lie in the interior of the smallest convex polygon (the smallest line segment) that contains the zeros of P(z). 32. Let z1' Z2' " ' , z,. be arbitrary complex numbers, z,. 9= z. for all /-' 9= v, /-', v = 1. 2•...• n. We consider all the polynomials P(z) that vanish only at the points z1' Z2' ••• ,z,. (having there zeros of arbitrary order). The set of the zeros of the derivatives P'(z) of all these polynomials is everywhere dense in the smallest convex polygon that contains z1' Z2' ... , z,.. 33. Let P(z) denote a polynomial. The zeros of cP'(z) - P(z), c 9= 0 lie in the smallest (infinite) convex polygon that contains the rays parallel to the vector c starting from the zeros of P(z). A zero of cP' (z) - P(z) appears on the boundary of this region only in one of the following two cases: (a) the zero in question is also a zero of P(z); (b) the region in question degenerates into a ray. 34. Let el' e2' ... , ep be positive numbers, a1' a2 , ... , a p arbitrary complex numbers. and let the polynomials A (z) and B(z) of degree p and p - 1 resp. be related by B(z) A (z)

=~ +~ z - al

z - a2

+ ... +..-!L... z- a p

Suppose that the polynomial P(z) is a divisor of A(z) P"(z) + 2B(z) P'{z), i.e.

A(z) P"(z) + 2B(z) P'(z) = C(z) P(z), where C(z) denotes a polynomial. Then the zeros of P(z) lie in the smallest convex polygon that contains the numbers aI' a2 • ••• , ap ' 36. If a polynomial f(z) whose coefficients are all real has only-real zeros then this is true also for its derivative f'(z). If f(z) has complex zeros then they appear in pairs. the two zeros forming a pair are mirror images to each other with respect to the real axis; they are complex conjugates. We draw all those disks the "vertical" diameters of which are the line segments connecting the conjugate zeros of such pairs. If I' (z) has any complex zeros they lie in these disks. [Examine the imaginary part of j(~;

.]

§ 4. Sequences of Complex Numbers 38. Assume that the numbers z1' Z2' ••• , z,., ... are all in the sector -(X

< ~ . Then the series Zl -+ Z2 + ... + z,. + ... and 1zll + 1z21 + ... + 1z.. 1+ ...

~ arg z ~

(X. (X

are either both convergent or both divergent.

110

Complex Numbers and Number Sequences

37. Suppose that the numbers Zl' plane 9lz > 0 and that the two series

Z2' .•• ,

z.. ' ... are all in the half-

+ Z2 + ... + z.. + ... and zi + ~ + ... + z! + ... converge. Then IZI 12 + IZ21 2 + ... + IZ" 12 + ... converges too. ZI

38. There exist complex sequences

ZI' Z2' ••• ,

z.. ' ... for which all the

series

~+~+"'+z!+""

k = 1, 2, 3, ...

converge and all the series

+

+ ... +

+ "',

k = 1, 2, 3, ... IZI Ik Iz21 k Iz.. 11I diverge. 39. Let ZI' Z2' ••• , z.. ' ... be arbitrary complex numbers. If there exists a positive distance lJ such that Iz/ - zk I ; : : lJ for I < k, t, k = 1, 2, 3, ... the convergence exponent of the sequence IZ11, IZ21, IZ31, .. · is at most 2. [1114.] 40. The limit points of the complex numbers

+ 2'''' + 3'''' + .. , + n'''' ,

1''''

n

lX

real,

lX

~



!, but not if me ~ !, c =1= o.

49. Let u o' u 1 ' u 2 ' ••• , U.. ' ••• be arbitrary complex numbers. For what values of c does the existence of

. (

lim u" +c .

Uo + u t

"-+00

imply the existence of lim

"-+00

11

+ ... + U,,)

+1

U .. ?

SO. If the Dirichlet series

+ a22- s + as3- s + ... + a..n- s + ... converges for s = + iT, T real, > 0, then lim {l - W (a t + a t2 + a t S + ... + a"t" + ...) = o. '-+1-0 t 2 S a1 1-s

(I

(I,

(I

[192.]

§ 6. Rearrangement of Infinite Series 51. If every subseries of a series with complex terms converges the series converges absolutely. 52. Assume that the series 1zl 1+ 1z21 + ... + 1z.. 1+ ... diverges. Then there exists a direction 01 accumulation, that is a real number I¥

113

Pt. III, Chap. 1, No. 47-54

such that those t~rms of the series Zl. + Z2 + ... + z,. + ... that are contained in the sector €X - e < arg z < €X + e constitute an absolutely divergent sub series for any e > o. 53. If lim z,. = 0 and if the positive real axis is a direction of accu-

+ +

"~oo

mulation of the conditionally convergent series Zl Z2 Z3 ., .••• then there exists a subseries Z" + z'. + z'. + ... the real part of which diverges to + 00 and the imaginary part converges to a finite number. 54. If the series Zl + Z2 + Z3 + ... is convergent, but not absolutely convergent, any value represented by a point of a certain straight line can be obtained as the sum of the series rearranged in a suitable order. [Consider two complementary subseries shifted relatively to each other; 52, 53, I 133, I 134.]

Chapter 2 Mappings and Vector Fields If we associate each point Z of some domain ~ of the z-plane with a certain complex value w according to a given law then w is called a function of z. Two geometrical interpretations of the functional relation are particularly useful. One uses one plane, the other two planes. The value w belonging to the point z (or, if more expedient, w) can be thought of as a vector acting on the point z; in this way a vector field is defined in the domain ~. In the other interpretation, the value w associated with the point z in the z-plane is conceived as a point in another complex plane (w-plane). In this way the domain ~ is mapped onto a certain point set of the w-plane.

§ 1. The Cauchy-Riemann Differential Equations

Let u(x, y) and v(x, y) be two real functions of the two real variables x and y. Then w = u + iv is a function of the variable z = x + iy. The function w = u + iv of z = x + iy is called analytic in a certain open region if u and v are continuous as well as their first partial derivatives and satisfy the Cauchy-Riemann differential equations

114

Mappings and Vector Fields

Observe the combination 1 a( .) axa (U +.) W =-;- 8y U + w =

dftl dz •

55. Are the functions z, analytic ? 55.1. Assume that

Izl,

Z2.,

i

fez) is analytic, use the notation w = u + iv =f(z) = f(x + iy)

as above and use subscripts to denote partial derivatives in the usual way. Verify that 2

Uz

+ Vz2 =

2

U"

+ v"2 = Uz2 + u"2 = Vz2 + v"2 = UzV" -

U"Vz

dw 12 = Idi .

55.2 (continued). Prove that 55.3 (continued). Let tp(x, y) and 'I'(x, y} denote functions of the two real variables x and y having continuous derivatives of first order;. they can also be considered as functions of U and v where I'(z) =1= O. Verify that tpz'l'z

+ tp"'I',, =

(tp..'I'..

+ tpv'l'v)

I

I.

dW l2 dz

55A (continued). If tp(x, y) has continuous derivatives of the second order, also tpn

+ tp"" =

(tp....

+ tpvv) I!; 12 .

55.5 (continued). Assume that a, b, c and d are real constants, ad-bc=l,

and consider Then

2

2 + lPv lP.... + lPw 2 2= =

lP.. lPz

+ lP"

lPn

+ lP""

".

•. IJ

56. Find the analytic function of z that vanishes for z = 0 and has the real part U

= 1

+ 2X2 _

2".

+ (x~ + ,,1)2 •

57. We denote by a and b, a < b, two fixed real numbers, by z a variable point in the half-plane 3z > 0 and by w the variable angle

115

Pt. III, Chap. 2, No. 55-61

under which the interval [a, b] is seen from the point z. If possible find an analytic function the real part of which is w. 58. Show that for any analytic function f(z) = f(x + iy)

(a~B + ~:) If(x + iy) 12 =

+ iy) 12 • 59. Show that for any analytic function of z = x + iy ( a2

ax.

411'(x

a ) I . 12 _ + IJyB log (1 + f(x + ~y) ) 2

(1

+ iy) \2 + \/(x + iy) \2)2 •

4\/,(x

§ 2. Some Particular Elementary Mappings The Cauchy-Riemann differential equations express the fact that an analytic function brings about a conformal mapping of the z-plane into the w-plane. (Preservation of the angles including sense.) The import of the Cauchy-Riemann differential equations for vectorfields will be discussed later. Cf. § 3. 80. We consider an orthogonal coordinate system ~, 'YJ, C in three dimensional space. An arbitrary point (~, 'YJ, C) of the sphere ~2 + 'YJ2 + C2 = 1 (unit sphere) is projected from the point (0,0, 1) (north-pole of the sphere) into the plane C= 0 (equatorial plane). Let the projection be (x, y, 0). Express x + iy in terms of ~, 'YJ, C and ~, 'YJ, C in terms of x and y. (Stereographic projection.) 81 (continued). Let the point P on the plane C= 0 be the stereographic projection of the point P' on the unit sphere. A rotation through the angle :n; of the unit sphere around the ~-axis moves the point P' to the point P". This point P" is then projected stereographically into the point P'" of the ~, 'YJ-plane. Let the point P have the coordinates x, y, 0 and the point P'" the coordinates u, v, O. Express u + iv in terms of x +iy. We introduce on the unit sphere the geographic coordinates 0 and cp (longitude and latitude) whereby -:n;

< 0


'"' J o

sin (?) ;r"

dx = _1_r(~)sin(~ n-l

n

:t) 2

1.

153. We find, assuming p

f'"' e

n

-"Pcosc


0, 0
-2. The quadratic form of the infinitely many real variables Xl' x 2, X 3, •••

i i

A-I ,..=1

X}.X,..

1 + I' + IX

is bounded, i.e. there exists a constant M independent of n such that

i; i; xAx,.. I 0, regular and non-zero on its boundary and it has inside the half-plane the zeros al' a2 , ••• , a". and the poles b1 , b2 , ••• , b,., counted with correct multiplicity. If I(z) is regular at infinity (but also under weaker conditions on its behaviour at infinity) and if I(z) is regular and non-zero at z = x + iy, x > 0, we have

log I/(z) \ +

".

2

,.=1

log

1 +00 = n- flog -00

,. Iz+b I z _a!!..l- .=1 2 log i z _ b I •

IZ+ii i I ,.

P

_

I/Ci1J) Id arctan !L.1.. x

140

Cauchy's Theorem. The Argument Principle

178. The function f(z) is regular in the domain r~

Izl ~ R,

different from zero on the boundary of Si) and it has in Si) the zeros . d f'lnltIon .. al' a2 , ••• , am' a,. = r,.et6II, ft = 1, 2, ... , m. U smg the e iIJ ) 1 = -D) we have the formula log

If(ee

U(e,

:tr

1

~ ~~; r, f(z) are fixed.

b," log f(z) (1Z2

1)dZ T

+'R2

has to be computed along a path analogous

to the one in 175.]

§ 3. The Argument Principle 179. Prove 25 by examining the change of arctan

from -

ex>

to

+ ex> along the real axis.

~~:~ as x increases

We consider a closed, continuous, oriented curve in the z-plane that avoids the origin. If, starting from an arbitrary point, z describes the entire curve in the given direction (returning to its starting point) the argument of z changes continuously and its total variation is a multiple, 2nn, of 2n. The integer n is called the winding number of the curve. 180. Every ray from the origin intersects the curve in question at leas tin 1 times. In the sequel (181-194) L denotes a closed continuous curve without double points and Si) the closed interior of L. The function f(z) is assumed to be regular in Si), except possibly at finitely many poles, finite and nonzero on L. As z moves along L in the positive sense the point w = f(z) describes a certain closed continuous curve the winding number of which is equal to the number of zeros inside L minus the number of poles inside L. [The Argument Principle. Cf. Hille, Vol. I, p. 253.] The proposition remains true also when f(z) is only continuous and non-zero on L.

141

Pt. III, Chap. 4, No. 178-189

181. The functions p(z) and 'P(z) are regular in i), except possibly at finitely many poles, finite and different from 0 on the boundary L of i). We define I(z) = p(z) 'P(z). The winding number of the image under w = I(z) of L is equal to the sum of the winding numbers of the images of Lunder w = p(z) and w = 'P(z) respectively. 182. Prove the argument principle for a polynomial. 183. The argument principle implies: If p(z) is regular and non-zero in the domain i) the winding number of the image of L (the boundary curve of i)) generated by w = p(z) vanishes; i.e. the argument of p(z) is a single-valued function on L. Deduce the general statement of the principle from this particular case. 184. The real trigonometric polynomial

am cos ml}

+ b",sinml} + am+! cos (m + 1)1) + b",+! sin (m + 1) I} + ... + + a" cos nl} + b" sin nl}

has at least 2m and at most 2n zeros in the interval 0 < amine P(z) = (am - ibm) z'"

185. Let 0 < a l flo


0 which, consequently, is real. 197. A function (not necessarily schlicht) that maps the closed unit disk onto a domain contained in the open unit disk has exactly one fixed point. I.e. if /(z) is regular in the disk Iz I < 1 and if I/(z) I < 1 in Iz I ~ 1 then the equation /(z) - z = 0 has exactly one root in Iz I ~ 1.

198. The entire function r;~i assumes each value infinitely often in the half-strip -d

< 3z
2.5. The power series 1

+ -za + -a'z2 + -a'z2 + ... + -a'" + ... = Z"

F

(z)

defines an entire function which does not vanish on the boundary of the annulus and has exactly one zero inside the annulus, n = 1, 2, ... [Examine the maximum term on the circle Iz I = Ia 12.. , I 117.] 201 (continuation of 170). Let @) denote the set of all zeros of all the functions I.. (z), n = 1, 2, 3, '" in m. If the limit function I(z) does not vanish identically its zeros in mare identical with the limit points of @) in m. (The term "limit point" is used here to mean a point an arbitrary fixed neighborhood of which contains at least one zero of I.. (z) for all sufficiently large n.) 202. The functions

... ,

I.. (z) ,

are schlicht in the unit disk Iz I < 1 and converge in any smaller disk Iz I < l' < 1 uniformly to a not everywhere constant limit function I(z). Then the function I(z) is schlicht in the unit disk Iz I < 1. 203. Let gl (z), g2(Z), ... , g.. (z) , '" be entire functions which have real zeros only. If lim g.. (z) = g(z) "-+00

uniformly in any finite domain, the entire function g(z) can have only real zeros. 204. Assume a~

0,

d>O.

144

Cauchy's Theorem. The Argument Principle

The entire function

"

~ a.

cos (a

+ 'Pd) Z

has only real zeros. 205. Suppose that I(t) is a positive valued, never decreasing function, defined on the interval 0 < t < 1 and that its integral The entire function

1

JI(t) dt is finite.

o

1

JI(t) cos zt dt

o has real zeros only. [185.] 206. The domain 'l) contains the segment a < z ~ b of the real axis. The functions 11 (z), 12(z), ... , I,,(z), ... are regular in 'l), they assume real values for real z and they have no zeros on [a, b]. If these functions converge in 'l) uniformly to a not identically vanishing limit function I(z) then I(z) has no zero on the segment a ~ z < b. - This statement is lalse. 206.1. The analytic functions 11 (z), 12 (z) • ... , I,,(z) are regular and single-valued in the connected closed domain 'l); let c1• c2 • ••• , cn denote constants. If the function

c1/1(z)

+ c2/2(z) + ... + cJn(z)

does not vanish identically the number of its zeros in 'l) cannot exceed a certain upper bound which depends on 11 (z)./2{z), ... • In(z) and 'l) but does not depend on c1' c2' ...• cn' (206.2 is less general but more precise.) 206.2. Let AI' A2• .•• , Al denote real n umbers

Al < A2 < '" < Al and mI , m2 • ••• , ml positive integers, Let 11 (z), 12 (z), ... , In (z) stand for the functions

taken in this order, and N for the number of those zeros of the function

145

Pt. III. Chap. 4. No. 205-206.2

that are contained in the horizontal strip

cx 0 show that (Aj

~_~1.~_(1J ~.~ _ n 2n

+1< -

N ::::;;

().j -

-

"I) (/J. -.-x) 2n

+n

_ 1.

(206.1 is more general but less precise.)

Chapter 5

Sequences of Analytic Functions § 1. Lagrange's Series. Applications The power series

a1z

+ a2z2 + ... + a..z" + ... =

W

which converges not only for z = 0 and for which al =1= 0 establishes a conformal one to one mapping of a certain neighbourhood of z = 0 onto a certain neighbourhood of w = O. Consequently the relationship between z and w can also be represented by the expansion

blw

+ b2w2 + ... + b..w" + ... =.t,

albl = 1. To compute the second series from the first we set 1

al

+ a2 z + asz- + ... + a..z"-1 + ... = 9

q>(z).

The equation z

W = 9'(z)'

where q>(z) is regular in a neighbourhood of z = 0, q>(O) =1= 0, implies _

co

W"

z - 2: 1iI ,,=1·

[d. -

I

dx

[9'(X)]"]

,,-I

",=0

.

More generally, if I(z) is regular in a neighbourhood of z = 0, then (L)

I(z)

=

1(0)

+

i :; [d"-I j'(.~_rtx)t] .

.. =1

dx

",=0

146

Sequences of Analytic Functions

(Biirmann-Lagrange series, d. Hurwitz-Courant, p. 137; Whittaker and Watson, p. 129.) 207. We use the same notation and hypothesis as before and expand

i;

j(z) , = w" 1 - wrp (z) ,,=0 n!

[d" j(x) [q:(X)J"] dx"

. %=0

Derive this formula from Lagrange's formula or Lagrange's formula from this one by correctly using the generality of both formulas. [The existence of one formula for a certain I(z) implies immediately the other formula for another I(z).] 208. Prove the formula in 207 directly by expressing the coefficient of w" as a Cauchy integral. 209. Expand in ascending powers of w the solution of the transcendental equation that vanishes for w = O. 210 (continued). Expand e'" in powers of w = ze-' where IX is an arbitrary constant. 211. Expand in ascending powers of w the solution x of the trinomial equation 1- x wxfJ = 0

+

that becomes 1 for w = O. 212 (continued). Expand XX in powers of w, where IX denotes an arbitrary constant. (XX = Y is the solution of the trinomial equation 1 - 'lftX

+ wyPftX =

0.)

213 (continued). Note the cases p = 0, 1,2, -1,1 and derive 209, 210 by taking the limit in 211, 212. 214. Evaluate the sum of the power series 1

+ i; (n + ~)" w" ,,=1

n.

What is its radius of convergence? 215. Prove 156 with the help of the results of 214. 216. Let IX and p denote rational numbers. Then the series

1

e

e

+ (iX iii) w + ~ 2#) w2 + ... + ~ n~ W" + ...

represents an algebraic function of w.

147

Pt. III, Chap. 5, No. 207-219

217. Arrange the successive powers of the trinomial 1 + w regular triangular array 1 1

+ w

+ w 2 in a

+w2

1+2w+3w2 +2w3 +w'

1

+ 3w + 6w2 + 7w3 + 6w' + 3w5 + w6

The sum of the middle terms (in boldface) is 1

+ w + 3w2 + 7w3 + ... =

Vi -

1

2w - 3wl

218. Arrange the successive powers of the binomial 1 triangular array (Pascal's triangle)

.

+ w in a regular

1

l+w 1 +2w +w2

1+3w+3w2 +w3

1

+ 4w + 6w 2 + 4w3 + w'

Find the sum. of the middle terms (in boldface) and. more generally. the sum of any column. 219. Find the generating functions of the polynomials P,,(x). p~,IlI(x). L

-1

d" e-"x"+C
-1

(generalized Laguerre's polynomials). (Cf. VI 84, VI 98, VI 99. The generating function of the Legendre polynomials is the series Po(x)

+ Pl(x) w + P2 (x) w2 + ... + P,,(x) w" + ... 3x 1 = 1 +xw +---w + .... 2 2 -

2

148

Sequences of Analytic Functions

the sum of which has to be found as a function of x and w; similarly in the other two cases.) We define as usual LfF(z) Lf2F(z)

=

F(z

+ 1) -

F(z),

= .1 [LfF(z)] = F(z + 2)

.1 "F(z)

- 2F(z

+ n) - C) F(z + n - ... + (-1)" F(z)

. F(z

+ 1) + F(z) ,

1)

+ (;) F(z + n -

2)

220. Let s be a constant of sufficiently small modulus. Then the following formulas are valid for F(z) = esz : (1)

F(z) = F(O)

+

:!

+ z(z (2)

F' (z)

=

LfF(z) -

+ z(z 2~ 1) Lf2F(0) + ... + 1) ···n(~ - n + 1) Lf"F(O) + ... ;

LfF(O)

! .1 F(z) + ~ .1 F(z) _ ... + 2

3

+(_1),,-1 ~Lf"F(z) + ... ; (3)

F(z)

=

F(O)

+ ;! F'(l) + z(z 2~ 21 F"(2) + ... +

+ z(z -

n)"-l F(")(n)

n!

+ ...

[210J;

~ z(z2 _ 12) (z2 _ 22) ••• [z2 _ (n _ 1\2] L.l 2,,-2 [F(-n + 2) _ F(-n)] (2n - I)! 2 ,,=1

+ .,:., --

[212, 216J.

221. The four formulas mentioned in 220 hold for any polynomial F(z). (In this case the series are obviously finite.) 222. The formulas (1), (2) given in 220 are also valid for any rational function F(z) if the real part of z is larger than the real part of any of the finite poles of F(z); formula (1) requires the additional condition that F(z) be regular for z = 0, 1, 2, 3, ... -Do the formulas (3) and (4) hold for rational but not entire functions?

149

Pt. III, Chap. 5, No. 220-229

In the sequel (223-226) we use the notation

+(~)ak+"-2

,1"ak = ak+" - (:)ak+"-1

-'"

+ (-1)"ak·

223. (1 - z)"

1:

where ak , k = 0, 224. Define

1:

ak~ =

k=-oo

,1"akz"+k

k=-oo

± 1, + 2, ... , denote arbitrary constants.

F(z)

=

a o + a1z

+ a2z2 + ... + a".7." + .. ,

and establish the relation

1:

t F(llt) = ao + ,1aot

+ ,12aot2 + .. , + ,1"aot" + ....

225. Define F(z) = a o + 2aiz

+ 2a2z2 + ... + 2a"z" + .,.

and show that 1

Vl+4t

F(1

+ 21 - V1 + 41) = a 2t

0

+ ,12a -1t + ,14a -2 t2 + ... +

+ ,12n a_"t"

226. Define F(z) = 2a 1z

+ ....

+ 2a2z2 + 2aszs + ... + 2a"z" + ... ,

and show that

~F(1+2t-V1+4t) _ t 21 -

227.

a1

_

a_I

if (1 + _.:i!-=-_.1_) = +

,,=1

n (n

1)

+(..12 _..12 )t LJ a o LJ a_ 2

sin :n:z . :n:z(l - z)

228. sin 1CZ is a single-valued function of w = z(l - z). The expansion of sin 1CZ in powers of w contains only positive coefficients (except the constant term) [227]. 229. Prove that [

d_"..;.(:n:_-_X)'--.-_"_-_l_c_os_x]

" dol'

0

> , . 0, P > 0, ex + P = 1, and put

!p(z) = exz

+ p-z1 .

The sequence of iterated functions

!p(z),

91 [!p(z)],

!p{!p[!p(z)]},

converges to +1, when 9!z> 0, to -1 when 9!z

9!z=O.


0 [12, II 220]; h( 0 in the upper halfplane 3z > 0 and that, besides, I(z) is regular at the point z = a of the

Pt. III, Chap. 6, No. 289-298

163

real axis and I(a) = h, h real. Then t' (a) is real and positive and we have the inequality

3_1_

> 3

b - /(z) =

_1-._

for 3z>

(a - z\ /,(u)

o.

294. Let the function I(z) be regular, have the zeros Zl' Z2' "., z.. and let I(z) be bounded, I/(z) 1 O. Then the stronger inequality 1I(z) 1 ~

1:1 - z .:, -+ z ".~z.. + z IIM

- I + 81

8

Z

.. -

8.

8

holds for all z with 9lz > O. We have equality either at every point or at no point of the right half-plane 9lz > O. 298. A function that is meromorphic in a closed disk and of constant absolute value on the boundary circle is a rational function; in fact it is, up to a constant factor, the product of linear fractional functions that map the disk in question either onto the interior or the exterior of the unit circle. 297. We assume that the function I(z) is regular and bounded in the disk Iz I < 1 and vanishes at the points Zl' Z2' Z3' ••. Then

(the sum of the distances of the zetos from the unit circle) is finite or else I(z) O. 298. We assume that the function I(z) is regular and bounded in the half-plane 9lz> 0 and vanishes at the points Zl' Z2' Z3' ••• outside the unit disk in the half-plane, i.e. Iz.. 1> 1, 9lz.. > 0, n = 1, 2, 3, ... Then the sum of the series

=

1

z. 1

1

91-+91-+91-+'" 81 8a

is finite or else I(z)

= O.

164

The Maximum Principle

§ 3. Hadamard's Three Circle Theorem 299. The sum of the absolute values of several analytic functions attains its maximum on the boundary. Here is a more detailed statement: The functions 11 (z), Mz), 13 (z), ... , I.. (z) are supposed to be regular and single-valued in the domain ~. Then the function

9'(z) = 1/1 (z)

1+ IMz) I + ... + 1/.. (z) I,

which is continuous in~, assumes its maximum on the boundary of ~. 300 (continued). The function 9'(z) assumes its maximum only on the boundary of ~ unless all the functions Mz), Mz), ... , I.. (z) are constants. 301. In three dimensional space, the n points Pl' P2 , ••• , p .. are given and P denotes a variable point. The function m(P) -- PP1 • pp2 ... P P .. T (PP7 is the distance between P and PI,) of the point P assumes its maximum in any domain on the boundary. (Generalization of 137.) 302. We assume that the functions Mz), Mz), ... , I.. (z) are regular and single-valued in the domain ~. Let P1' P2 , ••• , P.. denote positive numbers. The function

9'(z) = 1/1 (z) 11'1

+ IMz) II'· + ... + 1/.. (z) II'..

is continuous in ~. It reaches its maximum only on the boundary of ~ unless all the functions it(z), Mz), ... , I.. (z) are constants. 303. The function I(z) is supposed to be regular in the multiply connected closed domain ~ and I/(z) I single-valued in ~. [f(z) is not necessarily single-valued.] The absolute value I/(z) I attains its maximum at a boundary point of ~. The maximum cannot be attained at an inner point of ~ unless I(z) is a constant. 304. Let the function I(z) be regular in the disk Izl < R. Suppose

O 0 and to satisfy the following conditions: (1) there exist two constants A and B, A > 0, B > 0 such that in the entire half-plane

r

r

(2) we have for r

~

rs.

0

I/(ir) I ::;; 1,

I/( -ir) I ::;; 1;

I ...... 0 li m sup log !f(y) y ;;,.

(3)

,~+oo

Then I(z) is bounded by 1 in the entire half-plane: I/(z) I ::;; 1

for

9lz ~

o.

Proof: In treating 322 we extracted the desired conclusion from the maximum principle by introducing a variable parameter (the number e); now we introduce two parameters. Assume 1J > o. By virtue of condition (3) the function I/(r) e-'" I of the variable r converges to 0 as r -+ 00; it reaches its maximum F" at a certain point ro, ro > O. If ro = 0 the maximum is F" < 1 because of (2). We choose a fixed number A, 1

< A


O.

In the same way as in 322 we derive from (1), (2) and (3) that for

o S {} < ;

the absolute value I/(z) e-'1 Z I cannot be larger than 1 or F'1

whichever is larger. The same can be said for the sector - ~

< {} < O.

We claim that Ffl S 1: If Ffl were larger than 1 we would have I/(z) e- flz I O. Such a ray cuts the halfplane into two sectors, both with an angle smaller than 11:: Only this fact is essential [322, also 330]. 326. Let the function I(z) be regular in the half-plane ffiz > 0 and satisfy the following conditions: (1) there exist two constants A and B, A > 0, B > 0, such that in the entire half-plane

I/(z) I < AeB1z1 ; (2) I(z) is bounded on the imaginary axis,

I/( -ir) I < 1,

I/(ir) I < 1, (3) there exists an angle

1X, -

;


0 1/(±i1') I < Ce(n-l')'; (3) I(z) has the zeros 0, 1, 2, ... , n, ... Such a function vanishes identically. 329. Let w(x) be a positive function of the positive variable x that increases with x and tends to + 00 as x increases to + 00. A function I(z), regular in the half-plane ffiz ?: 0, that satisfies the inequality I/(z) I > e"'(/'!)/'/

for

ffiz ?: 0

does not exist.

330. Suppose that the function I(z) is regular at any finite point of the sector IX ro

for

in the above mentioned sector then the stronger inequality

I/(z) I < 1 holds in the entire sector. [Method of 325.] 332. The function g(z) is assumed to be an entire function, M(r) be the maximum of Ig(z) I on the circle Iz I = r. If lim log ~(r) = 0

Vy

,-+00

then g(z) cannot be bounded along any ray. [E.g. g(z) is not bounded along the negative real axis.] 333. Suppose that the function I(z) is not a constant and that it is regular in the half-strip @ defined by the inequalities

x>

0,

If there exist two constants A and a, A

I/(x and if

= ± ;)

0, 0 < a




@

(i.e. for x

=

0, - ;

< Y < ~- and for

x ~ 0,

then I(z) satisfies the strict inequality

I/(z) 1< 1 in the interior of @. [The comparison function is of the type eehz .] 334. Let w(x) have the same properties as in 329. Every function I(z) that is regular in the half-strip

x> 0,

z

must satisfy the inequality

I/(x

+ iy) f < e",(x)e

=

x

+ iy

X

at least at one point z = x + iy of the half-strip. 335. The assumptions of 278 are weakened insofar as (3) is satisfied in all but possibly finitely many boundary points Zl' zz, ... , z" of ~. An other assumption, however, is added, namely that there exists a positive number M' for which the inequality

I/(z) 1< M'

172

The Maximum Principle

holds everywhere in m. (Only the case M' > M is interesting.) This modification of the hypothesis does not change the conclusion of 278 that under those conditions If(z) I O. Then the point z = 00 is necessarily a boundary point of mand g(z) is not bounded in m. 339. Let Fl and F z be two continuous curves that have a common starting point, extend to 00 and enclose together with z = 00 a certain region m(e.g. two rays enclosing a sector). We assume that no point of the negative real axis belongs to m. The function fez) is supposed to be regular on Fv r z and in the enclosed region; in addition lim fez) = 0 as z tends to 00 along r 1 and r z. If fez) is bounded in mthen lim fez) = 0 as z goes to 00 along any path in m. .

[Examme

log z

A-+e -1og-f(z).] z

340. Let the curves r 1 and r z have the properties described in 339. Let fez) be bounded and regular in the region between r 1 and z and assume, in addition, that lim fez) = a as z tends to 00 along r 1 and lim fez) = b as z tends to 00 along r 2 • Then we have a = b. [Consider

r

a+ b)2 - (a- -2-b)2 .J (fez) - --2-

Solutions Part One

Infinite Series and Infinite Sequences *1. [Cf. HSI, pp. 238, 252-253, ex. 20.] AiOO = 292 [2]. *2. [For an intuitive solution see G. Polya:Amer. Math. Monthly Vol. 63, pp. 689-697 (1956). Cf. MD, Vol. I, p. 97, ex. 3.84.] 00

1: AnC" =

+ C + C2 + C3 + ... + cx + ...) (1 + t,5 + t,1O + C15 + ... + C5Y + ... ) (1 + C10 + C20 + t,30 + ... + C1O: + ... ) (1 + t,25 + C50 + C75 + ... + C25" + ... ) (1 + C50 + t,100 + t,150 + ... + t,50V + ...) (1

1

For the numerical computation of the coefficients An expand successively the functions (1 - t,)-I, (1 - C)-1 (1 - C5)-I, (1 - C)-1 (1 - t,5)-1 (1 -

C1O)-I,

(1 - t,)-1 (1 - C5)-1 (1 - t,10)-1 (1 - t,25)-I,

(1- t,)-1 (1- t,5)-1 (1-

C1O)-1

(1- C25 )-1 (1- C50 )-I.

It is convenient to dispose the coefficients needed for the computation of A 100 in a rectangular array.

174

Infinite Series and Infinite Sequen;;es

*3. B5 = 15 [4]. 4. The coefficient of (n in the expansion of

(( + (2 + (3 + (4)' is equal to the number of sums of value n with s terms of value 1, 2, 3, 4, where the order of the terms is taken into account. Therefore we have 1

=

+ 2;

Bi;" = 1 +

(( + (2 + C3 + (4) + (( + (2 + (3 + ~)2 + ...

"~l

1-C-C2 __ '3-C" For the numerical computation use the relation B .. = B"_l

+ B"_2 + B .. _ 3 + B"_4'

which follows from the definition of Bn or from the above equation. 5. 4 = C78 [7]. 6. 20 = D78 [8]. 99

7. 2; C,,(" 8.

=

(1

99

+ ()2 (1 + (2) (1 + (5) (1 + (10)2 (1 + (20) (1 + (50). + 1 + ()2 ((-2 + 1 + (2) ((-5 + 1 + (5) ((-10 + 1 + (10)2 ((-20 + 1 + (20) ((-50 + 1 + (50).

2; D"C" = ((-1 ,,~-99

9. [Cf. Euler: Introductio in Analysin infinitorum, Chap. 16, De Partitione Numerorum; Opera Omnia, Ser. 1, Vol. 8, pp.313-338. Leipzig and Berlin: B. G. Teubner 1922; also e.g. W. Ahrens: Mathematische Unterhaltungen und Spiele, 2nd Ed., Vol. 1, pp. 88-98, Vol. 2, p. 329. Leipzig: B. G. Teubner 1910, 1918.] The "change problem" [2]: 1 (1 -

a

C ')

(1 -

ca,) ••• (1 -

00

aj

C)

=

2; A,,(". ,,~O

An denotes the number of non-negative integral solutions of the Diophantine equation a1x 1

+ a 2x 2 + ... + a/x/ =

The "postage stamp problem" [4]: _ _ _ . __ 1 _._____ 1 -

Ca , - Ca , --

••• -

Cal

=

n.

1: B r"

.. ~O

"".

The "first weighing problem" [5] (all the weights on one pan): (1

+ ca') (1 + ca') ... (1 + cal) =

2; e"C. ,,~O

175

Pt. I, Solutions 3-13

The "second weighing problem" [6J (weights may be placed on both pans) : (C- a,

+1 + Ca,) (C- a• + 1 + ca') ... (C- al + 1 + Cal)

~

=

~

DnC".

11=-00

10. This problem is equivalent to the following: We have to weigh on one pan of the scales an object of n units with p different weights of one unit. According to 9 the number C" of the different possibilities is the coefficient of en in the expansion of

(1 + C)P = 1 + (;) C+ (~) C2 + ... + (~) C" + ... + CP, therefore: C

-(P) -

,,- n -

P!

n \ (n - P) ! •

In abstract terms: The number of different subsets of n elements contained in a set of p elements is (~). 11. This problem is equivalent to the following: Someone owns quarters minted in p different years. In how many different ways can he payout n quarters? According to 9 the number of different ways is An' the coefficient of Cn in the expansion of

-~-(1- C)p

=

1 +(-P) (-C) + ... + (-P)(-Ct + "', 1 n

thus

A = P(P+1) ... (p+n-1)=(p+n-1). n

1.2 ...

n

p-1

12. According to 11 this number is

(p +

(np-1 - P) - 1) = (np-1 - 1).

The result follows also directly from the expression (C + C2 + .. .)P. *13. Identical with 11. Another solution: Consider the P-fold series

1:

V1J"I.· .•

x~'x;·"· x;P = '''p =0,1 ,2,3, ...

(1 -

X1)-1

(1 -

X,z}-l ...

(1 -

Xp)-l

and identify the x;'s with C. Intuitive solution: Consider n + p - 1 places in a row. At the left hand end there are a certain number of places filled with Xl' then a place filled with a multiplication point, then a certain number of places filled with x 2, then a multiplication point and so on, as shown:

176

Infinite Series and Infinite Sequences

We have to choose

P-

1 among the n

+ P-

multiplication points which can be done in

1 places for the

P-

1

(n ; ~ ; 1) different ways,

by 10. ("Combination with repetition of P different elements taken n at a time" is the traditional term. In some of the cases that must be admitted the multiplication points are placed in an unorthodox way.) 14. According to the first weighing problem [9 extended to infinitely many weights we have to consider (1

+ C) (1 + C)2 (1 + ~1"4) (1 + ~1"8) ... =

1-c- 1-C' l-C 1-C· 1

=--= 1

Cf. 16, 17. 15.

1-Cl s 1-CB

1-CS

- _ . - - . - _ . - - ... 1-

C

1-C'

+C +C2 +CS + ....

(C- 1 + 1 + C) (C- 3 + 1 + CSl ... (C- 3" + 1 + C3") =

3

C- 1

=

9

3"+1

3,,+1

- 1 = C- N C C3" _ 1 C- 1

C-1 ~ C-3 ~ ..• C-3" C C3 - 1

C- N + C- N+1 + ... + CN- 1 + CN,

N =

3,,+1 - 1

2

1

.

16. a" = aF ", where F" is the number of digits 1 in the binary representation of n (its expansion in powers of 2). 17. [E. Catalan, Problem: Nouv. Corresp. Math. Vol. 6, p. 143 (1880). Solved by E. Cesaro: Nouv. Corresp. Math. Vol. 6, p. 276 (1880).] The series in question results from the expansion into a power series of

on setting C= 1. To determine the sign of a coefficient it is sufficient to examine the case where a = b = ... = 1. According to 16 the sign is given by (_1)F" where F" denotes the number of ones in the binary expansion of n. 18. 1 - CIO 1 - ClOO 1 - C1000 1 1 - CIO·t= CIOO ••• = 1 - C·

1="1".

This problem is not contained in 9. The result, however, is well known: Any positive integer admits a unique representation in the decimal notation. Cf. 14. 18.1. There are I kinds of coins and we have a limited number of each kind, PI coins of the first kind, each worth a1 cents, etc. In how many ways can we pay n cents?

177

Pt. I, Solutions 14-19

Given the positive integers

Pi' PZ' ... , PI' find E", the number of solutions of the equation alx l

in integers

Xl' X 2 ' •.. , X"

o
0, also X < 12 - X, Y - X < n - x - y ;£ x + y. Consequently the number of solutions in question is equal number of solutions of the inequalities ~-x p and !p(x)

=

1) .. ~~x -

xIx -

P+

1)

=(;),

'vVe have now

.t (:) G)

= p! (:I~p)!.~ (:

=:)

1p(x) =

1

2P

(XP )\

= (;) 2n - p = 2"1p(n) ,

__ n!_ ~'(_l).(n-p)_{ Oforn>p, ::P~ (_1).(n)(v) v p -p!(n-p)!.-:p v-p (-1tforn=p. 42. This is a special case of 40 with x case of 41 with !p{x)

=

(2x- n)2

=

We have 1p(n) = n.

IX

=

1/2. It is also a special

+ 4x2 = n 2 - 4(n - 1) x + 4x (x n 2 -2(n - 1) x + x(x - 1).

n2 - 4xn

1p(x) =

=

1)

185

Pt. I, Solutions 39- 47

43. Special case of 41 for l]1(x)

=

(2x - n)2 [42]. We obtain a"

* 2, a" 4 for n 2. 43.1. (1) First multiply both sides by

for n

=

=

0

=

~, then use 34 and 38. (2) Let y denote either the left hand, or the right hand, side of the desired identity and verify that, in both cases, y = 0 for x = 0 and

1 - e-~ dx=--x-

dy

44. Write I(x) = c,,(x -

Xl)

(x - x2 )

(z ! - x.)1 =

'"

(k -

(x - x,,). We have

x.) I.

45. [G. Darboux, Problem: Nouv. Annis Math. Ser. 2, Vol. 7, p. 138 (1868).] We deduce from 44

i

f(k)

k=O kl

1 = I(z~) e' = e'g(z) , dz

where g(z) is a polynomial with integral coefficients. 46. [E. Cesaro: Elementares Lehrbuch del' algebraischen Analysis und der Infinitesimalrechnung. Leipzig and Berlin: B. G. Teubner· 1904, p. 872.] The functions 1"+1 and I" are connected by the recursion formula

1"+1(z)

=

z[f..(z) (1- z)

Therefore the coefficients of I,,(z) = linked by the relations

+ (n + 1) I..(z)]. ai")z + a~")z2 + ... + a~)z"

11 = 1, 2, ... , n

This together with 11(z) determined by

are

e,,) = 0 . + 1 ·, aoe,,) = a"+l

= z concludes the proof. The value of 1,,(1) is 1"+1 (1)

=

(n

+ 1) 1..(1).

47. [Cf. N. H. Abel: Oeuvres, Vol. 2, Nouvelle edition. Christiania: Gr0ndahl & Son 1881, p. 14.] If g(x) is constant then the proposition is a consequence of 44. In assuming that the proposition holds for polynomials of a degree smaller than the degree of g(x) we write

g(z ~) y = g1(Z ~) [(z ~. = g1 (z

x1)yJ

~) (~~~) + ~(ti) z + ::~~) Z2 + ...) =

I (z ~)

1

~ z'

186

Infinite Series and Infinite Sequences

The given differential equation is soluble by successive quadratures because this is so for the equation

(z ~ - xo),..

= zy' -

xoY

= tp(z).

48. According to 44 both sides are equal to

1(1) z

+/(1) /(2) z2 +/(1) /(2) /(3) z3 + ... + /(1) /(2) ••• /(n g(1)

1) /(n) g(1) g(2) ••• g(n - 1)

g(1) g(2)

z"

+ ...



49. Apply 48 putting I(x) = (x - .)2, g(x) = 50. Comparing the coefficients of z" on both sides of the functional equation we get A,,(q" - 1) = A,,_lq", n = 1 2, 3, ... ; Ao = 1, hence x2 •

"("+1)

q-2-

A" =

2

1) (q - 1) ••• (q" - 1)

(q -

n = 1, 2, 3, ...

'

51. According to the functional equation (50] B,,(l- q")

=

B,,_lq,

n

=

1, 2, 3, ... ; Bo = 1,

therefore

B,,= (1

q" 2

- q) (1 - q ) ••• (1 - q")

n = 1. 2,3, ...

'

52. [Cf. R. Appell and E. Lacour: Principes de la theorie des fonctions elliptiques et applications. Paris: Gauthier-Villars 1897, p.398. For closely similar preceding work of Gauss see I.c. solution 55.] Calltp.. (z) the expression in question: 2 _ 1 + r"+1 z tp,,(q z) -tp,,(z)-+2il. qz

From this identity follows Cpq2~+1(1

I.e. C = •

(1 -

q

- t,,-2p) = Cp+1(l - t,,+2p+2), ,,' C,,=q,

r"+2 +2) (1 _l"+2 +4) ••• (1 _ P

P

2

4

Q

2

(1 - q ) (1 - q ) ••• (1 - q~"- p)

q4")

'P

= 0, 1, ... , n - 1,

'P

= 0, 1, ... , n - 1.

p'

q,

53. [Jacobi: Fundamenta nova theoriae functionum ellipticarum, §64,; Werke, Vol. 1. Berlin: G.Reimer 1881, p.234.J Take the limit n -+ 00 in 52. [181.] 54. [Euler: Commentationes arithmeticae, Vol. 1; Opera Omnia, Ser. 1, Vol. 2. Leipzig and Berlin: B. G. Teubner 1915, pp.249-250.] Special case of 53: q I q3/2, Z = _qI/2.

187

Pt. I. Solutions 48-59

55. [Gauss: Summatio quarundam serierum singularium, Opera, Vol. 2. Gottingen: Ges. d. Wiss. 1863, pp.9-45.J Special case of 53: replace q by ql/2 and z by ql/2, and apply 19 or the procedure of the third solution of 19. 56. [Jacobi, l.c. 53, § 66; Werke, Vol. 1, p.237.J Put z = -1 in 53 and use 19. 57. Setting _qnz = an we obtain

=

1 +G(z) -G(qz)

= 1 + 1:

n

-L:,. (1- qz) (1 -

n~11 -

q

q2 Z)

•••

(1 - qn-1z) (qn - 1)

+ a1 + a2(1 + a1) + a3 (1 + aI) (1 + a2) + a4 (1 + a1 ) (1 + a2 ) (1 + a3 ) + ... = (1 + a1 ) (1 + a2) + a3(1 + all (1 + a2) + a4 (1 + a1 ) (1 + a2) (1 + a3) + ... = (1 + a1 ) (1 + 2 ) (1 + a3 ) + a4 (1 + a1 ) (1 + a2 ) (1 + a3 ) + "', etc.

= 1

tl

58.

Do



= G(O)

q q2 q3 qn =-+-... +--1 - q 1 - q2 +--+ 1 - q3 1 _ qn + ... ;

applying 50 and 57 we find

G(z) - G(qz)

=

1: Anzn ,

G(qz) - G(q2Z)

n~l

G(tz) - G(q3Z)

=

=

2:

=

=

2:

Anqnzn,

n~l

A nq2H zn, ... ,

n~I

hence addition of the first m equations and taking the limit m -? ex:> yield = A G(z) - G(O) = 1: ~zn. n~11

-

q

59. Obviollsly G(l) = O. The functional equation in 57 implies with the help of complete induction that

C(q-n)

k (1 -- 1n) (1 -- qn) '" (1 - ~ k'-l) = -n, = k~11 1: -q-k - q q q q n

q-1

= a.

188

Infinite Series and Infinite Sequences

=

Introducing (1 - q) n

= q we get

y, 1 - L n

Let n ~ 00 for y fixed, and so q ~ 1: 00

1

J; - (1 - eY)k 1 O.

.=1

n-+oo

q > 0; i.e. n-+oo

==

N

(X> 0

because otherwise we would have

q log n -+ 00: contradiction.

Also

the same argument as for Q.. can be

194

Infinite Series and Infinite Sequences

used. For

lX

=

0 lim np.. p;;l 11-+00

=

all the more lim np..

00,

11-+00

= 00,

conse-

quently }; P. diverges. Therefore the series }; np"q" is divergent. If .=1

lX

= 0 we conclude nq..

"

}; 1IP.q.
=1

i.e.

In the case

lX

,,=1

< KQ",

K independent of n,

.

K }; P.Q.

"


() replace the proposition by lim

ft.-+oo

P"Q" "

1: liP. q•

=lX+{J.

• =1

Apply 70: a..

=

P"Q.. - p .. - 1Q.. -1'

a"

p..

Q..

b..

=

1

np..q.. ,

-b.. =---+---_. ---+lX +{J = np" nq.. n 78. Example: a1 =

all = aa

= ... = 1.

s.

Now assume that a.. > a.. +!,

a" -+ 0 and al

+ ~ + '" + an -

nan

For a given m find n such that a.. K

> al


k the matrix is called triangular (d. 65, 66) or more specifically lower triangular. If Phi = 0 for l < k the matrix is termed upper triangular. 80. Contains 66 as a special case. Proof analogous. 81. With s" = nc" + (n + 1) c,,+! + ... we can write

Obviously lim s" = 0, which implies lim til = 0 [80]. "-+00

n-+oo

195

Pt. I, Solutions 78-85

82. [G. H. Hardy and J. E. Littlewood: Rend. Circ. Mat. Palermo Vol. 41, pp. 50-51 (1916); d. also T. Carleman: Ark. Mat. Astr. Fys. Vol. 15, No. 11 (1920).] Put

bo + b1 (1 - £x)

+ ... + b,,(1 -

£x)" = t".

It is known from analysis [Hurwitz-Courant, pp. 32-33; Hille, Vol. I, p. 128.] that for Iyl < 1 -£x

bo + b1 y

+ ... + b"y" + ... = ao + a1 (£x + y) + ... + a" (£x + y)" + ...

holds identically. Consequently 00

J; (1 - IX)

,,-/I

11=0

/I

00

1

Y J; b1y 1=0

(1 _ lX)"+1

=- ------ J; al (1X. + y) 1 - (x + y) 1=0 00

= (1 - £x)"+l

J; SI(£X 1=0

00

1

+ y)l •

The coefficient of y" on the left hand side is t" and on the right hand side (1

+ (n +1 1) £XS"+1 + (n +2 2) £x2S"+2 + ( n +3 3) £x3S"+3 + ...] = t,..

- £x) ,,+1

[

S"

J; P". = 1 (binomial formula). The present transformation has an

.=0

"upper triangular matrix" whereas the matrix considered in 65 should be termed "lower triangular". 83. Cf. the analogous propositions of 65 and 79. 84. Cf. the analogous propositions of 66 and 80. 85. In 84 put all

s,,=-,;-, ,. For given

11

and e, e> 0, choose n so that

We have IP.(t)


1. The proposition holds also if the radius of convergence is einstead of I, Q > O.

Th~

196

Infinite Series and Infinite Sequences

86. [N. H. Abel, l.c. 47, Vol. 1, p. 223.] According to 85 00

ao

+ alt + a2t 2 + ... + a..~'" + ... =

1: (a o + a l + ... + a,,)

,,-0

-

tIS

00

1: tIS

.. =0

ao + a l

.

-+lim "

+ ... + a.. 1

=s.

..... 00

87. [G. Frobenius: J. reine angew. Math. Vol. 89, pp. 262-264 (1880).] It follows from the hypothesis that n-1a" is bounded, therefore 00

L

I I
O.

[Solution 89. ] According to 75 we obtain

198

Infinite Series and Infinite Sequcnccs

93. According to solution 89 we have

OO(

lim (1 - t)3/2 1;

1->1--0

,,=1

IVnj -

2

[ ;.-]) [~/~~-]-2[V~] JI ~ r = lim - ..-----~-- < O. 2

The limit is - (V~- _- _1) V;, as ~. ~. 2 2

2

... 211+1

,,->oo_~.~.2

2.

4

6

'6

2n

... ~'!._-t_I ~ 2 1 / n -. 2n V"n

[II 202].

94. The statement 85 is true not only for t -+ 1 but also for t -+ Cf. 84. The sum of the series bo + bit

00.

+ bi!. + ... + b"t" + ... > O. a" = n. s~, b" =~. II.

increases to infinity as t -+ 00 because b,. 95. Application of 94:

Knopp, p. 471.] 96. We write 5" = ao + a l + ... + a", integration for the subtrahend we get [95]

5_ 1

=

[Borel's summation, O. Then using partial

97. In 96 put an = 0 for n odd a"

We have 5 =

=

35

m1

2m-l

(-1) 2'4'6 ... ~

(-V--~'==-)

r

1 - z :=-1

o

= V~ 2

for

n

=

2m.

. Similarly we obtain for

-1 ~ x

~1

e-I]o(xt) dt = ___ 1 ___ .

VI + ~2

98. [For a special case see M. Fekete: Math. Z. Vol. 17, p. 233 (1939).] It is sufficient to consider the case where the lower bound a is finite. a Assume e > 0 and .-.!!!. < IX e. Any number n can be written in the form m

+

n = qm + r where r is an integer, 0 :'S r < m-l. We define Then we have a,.

=

aqm +. < a,,. a,. aqm +,

+ am + ... + am + a. = qa + a, ,

-n = qm ---------+ I' < --

m

qa", qm

a" IXS-< (IX -II

+ a. = -a", ---- qm a, . -.- ++ I' m qm + I' II '

qm a. + e)----+-. qm+1' II

ao = O.

199

Pt. I, Solutions 93 - 99

*99. Since 2am - 1 < a2m < 2a m + 1 we have _ Ia2m 2m

(*)

am m

I < _1 • 2m

The series

a1 + (a2z_~) + (a, _ az) + (as _ at) + 1 4 2 8 4 1

a

... = lim ~ = w " ..... 00 2"

is convergent because

la1 1+

2-1

+

2-2

+ 2-3 + ...

is, on account of (*), a majorant series. Write the integer n in the binary system, i.e. n = 2'" + E12m- 1 + ... + Em' where

El ,

E2, ... , Em are 0 or 1; according to the hypothesis a 2", + E1a 2",-1 + ... + E",a 1 -

:s;; a"
O. According to the hypothesis there are terms of the sequence that are smaller than 'Yj. Let n be the smallest index for which l,. < 'Yj. Then we have n>m;

l,.

< ll,l" s. - The points (n, Ln) are to be enclosed in an infinite polygon convex· from above.

113. Set lim sUPJ~g rn = S. Then we have a) S ~ A.. This is obvious ",--;'00

for S log m

=


n, Pm> O. Then we have if

x>

"'-V"-- . .

P..

Pm

204

Infinite Series and Infinite Sequences

120. If for a certain x a term p"x" is larger than all preceding terms, i.e. if for a certain value of x all the inequalities

x·(p"x"-· - P.)

~

0, v = 0, 1, 2, .•. , n - 1,

hold, then this remains true also for any larger value of x. 121. Let m be arbitrary and choose x so that p",x'" is the maximum term. Then 1

P'" On the other hand P",«()e)'" is bounded for m quently

.

~r.-

e'"

~ PO' -+ 00,

0
l and collecting all the terms which belong to the same bIll we transform the subseries a" + ak+1 + a"'+21 + ... into the series

+ ++ ++ + ... b1

b2

bs

except for a finite number of

terms. 127. No [128]. 128. No. - Use b" of 126. Suppose that the functions !p(x) and c1>(x) assume only positive integral values and are strictly increasing: 0< !p(1) < !p(2} < "',0 < c1>(1} < c1>(2) < c1>(3) < ... ; !p(n), c1>(n) integers. Define a new series a1 + az + as + ... with the general term av

= 4>(m) _

bIll

4>(m _ 1)

forc1>(m -1) < 11 (m) ; a 1 = a 2 =

b

... = a(l) = 4>tl)

.

206

Infinite Series and, Infinite Sequences

The inequality qJ(t",) < «P(m) < qJ(t", + 1) determines the integer t", completely. Collecting the terms that belong to the same b", we transform the series

a'l'(l) +a'l'(2)

+ ...

into the series ! 1

t

- t

1P (mT -1P7':~ 1) b",.

If we set «P(x) = ~' we obtain a series a1 + as + as + ... which furnishes a counter example for the problems 126-128: If qJ(x) is a polynomial t - t

of degree > 2 or if qJ(x) =kJX the sequence lP(m)"'-IP(:-~

1)

is, beyond a

certain m, monotone decreasing [Knopp, p. 314]. If qJ(x) = k + lx we transform the contracted series by adding an absolutely convergent series into the series l-1(b 1 + b2 + ... + bn + ...) [126]. 129. [A. Haar.] Since the series SI = al + a21 + aSI + ... is of the same type as SI it is sufficient to show al = O. We denote the first m prime numbers by PI' P2' ... , Pm' The series

+ sp. + .. , + sPm) + (sp,p, + sp,p. + ...)

S1 -

(SPI

+ (_1)'" SpIP•...f.'" contains only a1 and the an's whose SUbscripts n are not multiples of one of the prime numbers PI' P2' .... Pm' Each of these an's appears exactly once [VIII 26]. I.e.

al < 1: Ian I, n=p",+1

The condition "absolutely convergent" is essential as can easily be shown by the example

5; ,1,(n)

[VIII, Chap. 1, § 5].

"=1 n

130. [G. Cantor, ct. E. Hewitt and K. Stromberg: Real and Abstract Analysis. Springer: New York 1965, pp. 70-71.} We obtain the set of points in question by removing from the closed interval [0, 1] the open middle third of the interval, then apply the same process to the remaining two intervals and so on indefinitely. (This set is often called the Cantor discontinuum or the Cantor ternary set.) 131. [Cf. S. Kakeya: Proc. Tokyo math.-phys. Soc. Ser. 2, Vol. 7, p. 250 (1914); T6hoku Sc. Rep. Vol. 3, p. 159 (1915).] Write

p" +Pn+l + ... +Pn+.=P".' lim p".=P n. n=1,2,3, ... ,v=0,1,2, ... , ' v~oo

207

Pt. I, Solutions 129-132.2

Assume that P'" is the first term for which P'" < (1. Either there exists a "1 such that p .."., < (1, PII"',+1 > (1. "1 ~ 0 or p'" :::;; (1. In the second case we have p'" = (1 because p'" > P.., - l ~ (1 (for n 1 = 1 this means PI = S ~ (1). i.e. (1 may be represented as an infinite subseries. In the first case we determine the first term P... with n 2 > n 1 + "1' p .."., + P". < (1. Either there exists a "2 with p .."., + P...... < (1, P""" + P"" •• +1 ~ (1, "2 ~ 0 or p .."., + p ... :::;; (1. In the second case we have p .."., + p ... = (1, because p"".' p ... ~ p ....., P... -l ~ (1 (n2 > n 1 +"1 + 1, because p"".' + P"'+',+1 = p""',+1 > (1) i.e. (1 may again be written as an infinite subseries. If this procedure never terminates (if the first case occurs at every step) then (1 = p ..... , + p ...,•• + p .... p• + .... 132. From the relations

+

+

P.. = P"+l P"+1 = we gather P..

=

2P"+1' thus

+ P.. +2 + P.. +s + .. . P"+2 + P"+s + .. . 111 P2 = ,., ...• P.. = 2'" .•. The

PI = "2'

representation by infinite binary fractions is unique. 132.1. [G. P6lya, Problem: Amer. Math. Monthly Vol. 51, pp. 533534 (1944). Solution Amer. Math. Monthly Vol. 53, pp. 279-282 (1946).] Define R -.!. ~ 2. ... (2n + 1) "-246

2n'

Then (1

+ !)(1 + ~)."(1 +2~)=R.. ,

(1 _~) ... (1 _ _ 1 ) ( 1 - ~) 3 5 2n + 1 1

R"

t'J

=....!...

R,,'

1

n"2 2n -"2

and so the product of the first m(p

by

+ q)

II 202

factors of p p•q is 1

R".p

Rmq

132.2. See 132.1. From log (1

t'J

(1-)2 q

+ x) =

log p p•q = Sp,q

x2

x -2"

x2 +"3 - ... follows

+A

where the infinite series A is absolutely convergent and so its sum is independent of P and q (of the rearrangement of its terms). Another

208

Infinite Series and Infinite Sequences

proof can be derived from the expression of Euler's constant given m the solution of II 18. 133. Insertion of appropriate vanishing terms into the two complementary subseries reduces the proposition to the termwise addition of two convergent series. *134. We assume that all the terms of the divergent subseries ar , + ar , + a" + ... are non-negative. Then the complementary subseries as, + as, + as, + ... will be such that to each positive e there corresponds an integer N so that

as... +as... +1 +···+as.. m > N. After this remark the proof essentially coincides with the well known usual proof for Riemann's theorem on the rearrangement of the terms of conditionally convergent series. [Knopp. pp. 318; cf. also W. Threlfall: Math. Z. Vol. 24. pp. 212-214 (1926).] 135. From PI;;;::: Ps ;;;::: Ps > ... , 0 < ml < ms < ms < ... follows that PI + P2 + ... + P.. ;;;::: .P..., + P..., 136. Determine the "red" subseries Pr,

+ P..., + ... + P..... . + Pr, + Pro + ... so that

Pr.. < min (2-", Q.. - Q.. -l)' n = 1, 2, 3•...• Qo = O. Then Pr, + Pr, + ... + Pr.. :s;: Q... the complete "red" subseries converges and Q.. - (Pr, + Pr, + ... + Pr.. ) increases beyond all bounds. The terms of the complementary subseries are successively accommodated where the relations

..

~

;=1

Pro
1; E~(P~ 7=1

,,-1

+ IX 1; v(P~ -

P~+1)

P~+1)

~=N+l

+ IXnp..

" where K is independent of n; therefore the right hand side tends to 00 in contradiction to the hypothesis. 139. [E. Lasker.] Set En = el + e2 + ... + e,. as in 138. The sequence (E)

has the property that between two terms with different signs there must be a vanishing term. We distinguish two cases: (1) Infinitely many terms of the sequence (E) vanish. (2) All but finitely many terms of the sequence (E) have the same sign. Suppose they are positive. In the first case choose the subscript M so that EM = 0 and that we have for M 0 for v > m the estimate (*) yields (E,. - E",) P,. < e, consequently E ..P,. < e

+ EmP,..

Since m is fixed and P.. converges to 0 we find for n sufficiently large E,.p,. X - 3i '

x'.

2" + cos x - 1 < 4!' I.e. cosx< 1- 21 Xl

143. arctan x -

=

x5

-"3 +"5 x3

(x

... +

J s

o

2 +1 (-x )" dx,,, (x) = i+x2 0

Xl

~"+1)

(-I)" 211 + 1 I

~J :If

0

144. Assume tlo > 0, hence al < 0, a2 > 0, A - tlo < 0, A - tlo -

~

>

x'

+ 4i' etc.

cos xl Vi-II

as
O. Summing up: the given series "=1 converges if and only if at least one of the following conditions is satisfied: a) !p(") = 0 for some positive integer n; b) laol < 1; c) ao = 1, a1 < -1; d) ao = -1, a 1 > O. 159. Special case of 158: m(x) T

convergent for

160.

J o 1

IX

1 ez1ogX-

>

=

'"

-

2- C

1, and IX

dx =

=

=

'"

",2

1

1 - -x - -2! -x 2

+ '"

log 2; log 4, log 8, ... are

>

1.

1: --,1 J1 x" ( log-1 ) .. dx. 00

n=on· o Substitute x.. +l = e- Y •

X

161.SetV1+V1+",+V1=t.. , then t!=1+t.. _1' t1 =1,

< tn' n = 2, 3, 4, ... For positive x we have x 2 < 1 + x if and only if x < !(1 + Vb), i.e. if x is smaller than the positive root of the equation x2 - x - 1 = O. Hence t! = 1 + tn - 1 < 1 + tn' t"_1 < t" < i(1 + Vb), n = 2, 3, 4, ... and lim tn = t exists, 0 < t < !(1 + V5), t2 = 1 + t, _ "-+00 i.e. t = i(1 + V5). We proceed similarly in the case of the continued

t,,_1

fraction where the recursion formula is U ..

= 1

+ u;_\' u1 =

1,

n

=

2, 3, 4, ...

214

Infinite Series and Infinite Sequences

182. [G. P6lya, Problem: Arch. Math. Phys. Ser. 3, Vol. 24, p.84 (1916). Solved by G. Szego: Arch. Math. Phys. Ser. 3, Vol. 25, pp. 8889 (1917).] If (for the sake of simplicity for v ~ 1) log log a" < v log 2,

a" < e2", then t" < -Ve2 + Ve4 + ... + Ve2" < e 1+,.V5 [181]. If, however, (f)" . - If a" < 1, then, of course, log log a" a" > ef1', P> 2, then t" > e --n-must be interpreted as 183. We prove

00.

by complete induction. Suppose that the corresponding relation is proved for the n quantities a2 , as, ... , a"+1' i.e. that

where

-Va2 + Vas + ... + Va" = Hence 2 t"+1 < a1 + t

t,

5)2 5)2 + s < (V-a1 + t + V-=< ( t" + 2 V. 2 +1 ~ 41

184. [Jacobi, I.c. 53, § 52, Corollarium: Werke, Vol. 1, pp. 200-201.] Write 1 - q = 110, 1 + q'" = a"" m = 1, 2, 4, 8, 16, ... , then the n + 1-th

partial product is

2-2-" 40

_

-(

4142444 8 '''42"

)2-'"

The product a1 a2 a4 a8 '" converges. Cf. also VIII 78. 185. Calling the sum in question F(x) we find F'(X) = F(x),

F(x) = const.· C. x

188. q/(x) = tp(x), 91(0) = 1, tp(X) = C = 1 + 11 + 2! L1tp(x)

= tp(x) , '1'(0) =

1, tp(x)

= ~=

1

Xl

+ ... + n!x" + .. .

+ ( ~) + (;) + ... + ( =) + .. .

215

Pt. I, Solutions 162 -169

for x

>

-1. We have

_ x"

() _

fJJ.. (x ) - n!' "P x 167.

log - x..

x .. +1

=

log

1
1 is obvious from the expansion 2(n 2n

log a.. =

=

+ P) +1

(

1

1

1

+ 3(2n + 1)2 + 5(2n + 1)' + ...

)

(1 + ~ ~ i) (1+ 3(2n 1+ 1)2 + 5(2n ~ 1)' + ...)

[solution 167]. This leads to log a..

=

log a..

=

1-p 1 +1" + 12n2

1- n

+ 0 (1) n3 '

thus log an +1

-

(n

- P+ i)3 + 0 (.;) , +\ )(n n

hence a.. increases for n larger than a certain subscript N if P < 1. If P< 0 this is true already for n > 1 as can easily be verified by expanding

(1 + ~

r

with help of the binomial formula.

.

169. We wnte a..

)"H (11++ ~)* -;- ; the fIrst . = (1 1 + -;;

[168]; the square of the second factor condition x

.

IS

1

factor decreases

- 1 x + 2x --1 + ( 1)' n + nn + 2

> t is therefore sufficient. Now expand

log a.. = 2n ( 2n

1

1

1

+1 + 3(2n + 1)3 + 5(2n + 1)5 + ...

)

1/X)3 1(X)5 ] + 2[X 2n + x + '3 \2n + x +"5 2n + x + ... _ 2n - 2n + 1

2x

1

+ 2n + ;\' + 12HZ +

0 (1) n3

'

The

216

Infinite Series and Infinite Sequences

Ga)

Since log an - log an+1 = ~!4~--~ + 0 the condition x > ~ is also necessary. 170. [Cf. problem No. 1098, Nouv. AnnIs Math. Ser. 2, Vol. 11, p. 480 (1872). Solved by C. Moreau; Nouv. AnnIs Math. Ser. 2, Vol. 13, p. 61 (1874).] The first inequality means

(1 + :r+1 2.

217

Pt. I. Solutions 170-176

173. [Proof based on a communication of E. Jacobsthal.] Cf. 174 for lim sin" x. We have

"-+00

x3



< sm x
VN+IX -61(VN+IX )3> VN+IX+1' c

c

C

C

1X

C ,n> N. Consequently for aIle < Va, liminfV;sin"x2:e n+1X i.e. > Vi If e > Vs choose"", so large that sinmx < v- C •• In a similar N+1 . t h C f'Irst d . c. c way as m caseI we conc U e smm+l x < V-==- , sIn",+2 < V~=- , N+2 N+3 et c. 174. The sequence v" is decreasing, v" > 0, therefore lim v" = v

thus sinnx > V.

,,~OO

"-+00

exists; v = I(v) implies v = O. Consequently it is sufficient to prove the proposition for small x. Let b' be fixed, b' > b. For sufficiently small x we have x - ax!' < I(x) < x - ax!' + b'xl and when n is sufficiently large, n> N(e),

1 -k-I1 _ a (l)k en -k=! > e(n + I) -k-l 1 ( 1)k (1)1 1 en -k=! _ a en -"=1 + b' en -k=! < e(n + 1)-k-1 en

or

depending on whether -1

e < [(k - l)a]k-l

-1

or

e> [(k - 1}a]k-1.

Cf. 173. The assumption on the sign of b is not essential. 175. [J. Ouspensky, Problem: Arch. Math. Phys. Ser. 3, Vol. 20, p. 83 (1913).] Convergence for s > 2, divergence for s < 2 [173]. 176. [Cf. E. Cesaro, Problem: Nouv. AnnIs Math. Ser. 3, Vol. 7, p. 400 (1888). Solved by Audibcrt: Nouv. AnnIs Math. Ser. 3, Vol. 11, p. 35*

218

Infinite Series and Infinite Sequences

(1892).] The inequalities

C-l

x>O;

x 0, and increasing in the second case, " .. < O. We have lim".. = " = 0 because ...... 00

e" - 1

" ~ log -uThe recursion formula

e"" - 1 =

for

" ~

o.

" ..e""+1, n = 1, 2, 3, ... yields

+"1 + "1"2 + ... + "1"2'" ".. _1 + "1"2 .. ·"..e.... +1 and lim "1"2 ... ".. e.... +1 = O. e'"

= 1

...... 00

177. [C. A. Laisant, Problem: Nouv. AnnIs Math. Ser. 2, Vol. 9, p. 144 (1870). Solved by H. Rumpen: Nouv. AnnIs Math. Ser. 2, Vol. 11, p. 232 (1872).] s =

! cos

qJ.

Notice that 4 cos3 qJ = 3 cos qJ

+ cos 3qJ.

178. [I. Schur, Problem: Arch. Math. Phys. Ser. 3, Vol. 27, p. 162 (1918). Cf. O. Szasz: Sber. Berlin Math. Ges. Vol. 21, pp. 25-29 (1922).] If e > 0 is so small that Iq I e < T then there exists a constant A independent of nand 11 such that

+

Ib;:p/


A

(Iql+ e)·,

11

= 0, 1, ... , n;

n = 0, 1, 2, ...

m we obtain

The sum of the last two terms is absolutely smaller than .=",+1

.=",+1

i.e. arbitrarily small with m- 1 • Choose m so large that these two terms are smaller than e. For fixed m choose n so that the first term becomes absolutely smaller than e. 179. [Special case of an important proposition in function theory by Vitali. Cf. E. Lindelof: Bull. Soc. Math. France Vol. 41, p. 171 (1913).J We show only that lim a" l = O. (Then form x- 1 / .. (x) - a..1 , etc.) Assume e

>

...... 00

x

0 and x so small that 0 < A 1 _

x

< e. Then we have

Iani I < X-I 1/.. (x) I + A 1 ~ x < X-I 1/.. {x) I + e.

219

Pt. I, Solutions 177-181.1

For fixed x choose n so large that Itn(x) I < ex.

180. We have Iak I < A k , k gent. Moreover

I

sn -

S

=

0, 1, 2, ... , therefore

ao I + Ia"l - all

I < Ia

nO -

1:

ak is conver-

k=O

+ ... + Ia"m -

am I + 2 1: A k • k=m+1

Assume e > 0. Choose m large enough to render the last term smaller than e. Having fixed m we select n so large that Ia"k - ak I < k = 0, 1, ... , m. Then we get

n 00

181. a) Since the infinite product

k=l

m:

(1 - q2k) converges for Iq I < 1

all its partial products lie between two positive numbers a and b, a Therefore C. as defined in 52 is bounded: Apply 180. b) Let y

1'




eY , v

=

k-1


(~)

o and note that the relation

1 - x" 1- x

dx

1

J~ dif>(~) =

o

=

F'(rx r(rx

+ 1) + 1)

+C

1

J if>(~) d~ =

if>(1) -

1- C

0

holds; i.e. the operations of taking the limit and computing the mean value can be interchanged [44]. 43. [Cf. Cesaro, Problem: Nouv. AnnIs Math. Ser. 3, Vol. 2, p. 239 (1883). ] n

lim 1_ ~

H"""*= n

.=

1

1

(1 - [_n-J~) = 1- J'[~J x dx v n • x 0

44. [G. L. Dirichlet: Werke, Vol. 2. Berlin: G. Reimer 1897, p. 97104; d. also G. P6lya, l.c. 41, p. 197 and Nachr. Akad. Wiss. G6ttingen 1917, pp. 149-159.] We are dealing with [VIII 4]

f' ([~J - [~ - ~J) dx 1

lim

=

n-+= •

X

1

X

= lim

n~ (~ - -~-) V V + ex

n-+='--

.=1

-n

1 = 1 -l+rx

1

+2 -

1 2--+-",

1

+ .. , =.

/' 1 _

o

x-'

~ dx. -,

t

we obtain 41. 45. [G. P6lya, l.c. 41, pp. 1£19-200.] We assume at first that Then If

(~

~ =

r[~J o 1

+ 1)

x

x'dx

=

i.; n f

n=l

C(~

>

1 n

I

n+l

=

~

(~+ l)x'dx= 1 +_1_ 2~+1

+ 1).

+ _1__ + .. ' 3~+ 1

1.

241

Pt. II, Solutions 43-50

The total variation [d. solution 9] of (1(1) =

[! ] xo< =

I(x) is

+ 0») + (1(1- 0) -/(1 + 0») + (/(1- 0) -/(i + 0») + ... 1(1-0< - 2-0-1 1 2 I> -1 )"'1>-1 ' '] +r'+"'+"1>_1:;;;"

thus

1

a..

n

("1 )",]-1 ("2--;;- )",.-1 ••. ("1>-1 )"'1>-1-1 -n-

1>-1

~ ~ •.. ~ --;;.]+ •• + ...•1>-1:;;;.. .

. (1 _

"1 _ "2 _ ... _

"']+""+"'+"'1>-1 =

n

n

n

"1>-1 )"'1>-1 . n

Cf.23. 66. We use the same notation as in solution 65. According to solution 31 we have for t --? 1 - 0 k

=

1, 2, ... , p.

Introducing 00

F(z) = ~ n""+"'·+···+"'I>-lz.. .. =1

we obtain F(t)

(t)

t--..l -0

F(t)

lim

=

t)-(""+""+"'+"'I» ;

r(~1) r(~2) ... r(~I» r(~1

+ ~2 + ... + ~I»

.

On the other hand this limit is, according to I 85 and 65, equal to the integral in question. The p-fold Dirichlet-Jordan integral can be easily computed with the help of this relation, d. E. T. Whittaker and G. N. Watson, p. 258.

247

Pt. II, Solutions 64-69

67. The term in question is ~!

~~

... ~ I.,nf••,,··· I.p"

1 ;;;;"-", and so our proof is not suitable for specifying the cases in which equality is attained. For an essential remark on such cases see 109. 70. [J. L. W. V. Jensen: Acta Math. Vol. 30, p.175 (1906).] The proof is analogous to Cauchy's proof for the inequality between the arithmetic, geometric and harmonic means given in the footnote on p.64 (which deals with the case qJ(t) = log t). First the statement is proved for n = 'i' (k integer), then it is extended to arbitrary n. 71. [J. L. W. V. Jensen, l.c. 70.] Using a similar notation as in 48 we get for each n qJ

/ .. ( 1

+ 12.. + ." + I,.,.) < fP(f1,.) + fP(f2,.) + ... + fP(f,.,.) n = n

Let n increase to infinity and notice 124, 110. 72. Let t1, tz be two arbitrary points on [m, M], t1


m > 0; further1 a.

more replace a. by -, v = 1, 2, ... , n. 79. In 77 put: tp(t} = -log t, and t log t resp., M > m >.0; then 1

replace I(x) by f(x) • 80. First proof:

Second proof: If ). and p denote real variables the quadratic form ().al + pb l )2 + (Aa 2 + pb2)2 + ... + ().a" + pb,,)2 = A).2

+ 2B)'p + Cp2 ?:: o.

Provided that ).2 + p2 > 0 the case of equality presents itself only then when there is a particular set of )., p for which ).a. + pb. = 0, v = 1, 2, ... , n. Therefore AC - B2 is positive or 0 as asserted.

81. By taking the limit in 80: Writing I." = I

g... = g (x]

(I

l"gl"

'

+ vXs -: ~!) , we obtain [80]

(Xl +

V XI :

~) ,

/,

2 2 + J2"g2" + ... + .."g.... ) 2 $; 121.. +. f,22" + ... + /,2"". g11l + g2" + ... + g"2...

n

-

n

n

Let n increase to infinity. It is also possible to adapt both methods used in 80 to the present problem; as to the first method cf. 68.

250

Integration

81.1. See 81.3. 81.2. See 81.4. 81.3. Let E stand for

" :E , set

.=1

Eb. = B, ... ,

Ea. = A,

Et. = L,

and use 78:

a. b. (t),

P(O) = @(t),

P(1)

=

~(f).

Let M denote the maximum of t(x) on [Xl' x2] and c) the length of a subinterval of [xl' x2] in which t(x) > M - e. Then we have for t > 0 1

(M - e)

C2 ~ ~)' < P(t)::;; M,

251

Pt. II. Solutions 81.1- 87

i.e. !P(oo) = lim !P(t) = M; !P(- 00) is found in a similar way. This

''''00

proposition contains therefore a new proof for 69. M. [H. Minkowski; cf. e.g. Hardy. Littlewood and P6lya. l.c. 69, p. 21.] First proof: We assume 0 S;; t S;; 1 and define

,.

... :2:: b". (If b. < bp ' v < /-t, we interchange b. and bp: b! + b! = b! + b; and apbp + a.b. :2:: apb. + a.bp.)

254

Integration

We may also assume that not all the a7 's are equal. nOl all the b/s. i. e.

a,.b l - alb,. = (a,. - al)b l

>

2 the numbers u l • u 2 • equations If n

a;

= u.ai

+ v.a! •

b;

••••

+ al(bl -

>

b,.)

O.

U,._l' VI' v2 • •••• V"_l are defined by the

= u bi 7

+ v b! • 7

'JI

=

2. 3•... , n - 1.

We find u. > 0, v. > 0 and a.b. > u7 al bl + v.a,.b,. [80]; u. = 0 if and only if a. = a.+ l = ... = a,., b. = b.+ l = ... = b,. and v. = 1. In a similar way v. = 0 implies u. = 1, etc. If u. > O. v. > 0 then a.b. > u.al bl + v.a,.b,.. Thus the expression in question is

< (pai + qa~) (pbi + qb~) (pal bl + qa,.b,.) 2 where 1 + u 2 + Us + ... + U"_l = p, v2 + va + ... + V"_l + 1 = q. The inequality becomes an equality if and only if the u.'s and v.'s are o or 1. P. q. are integers, a l = a2 = ... = ap • ap + l = ap +2 = '" = a,.. bl = b2 = ... = bp • bp +1 = bp +2 = ... = b,.. The last expression is = 1 +pq( a,.bl - alb,.

pal bl

+ qa,. b,.

)2 0 and sufficiently small, I(X)2 >/(;)2 whenever

Ix -

~I < 6,

and therefore ;+a

b

f I(X)2 dx:;::: f

"

;-a

I(X)2 dx :;::: 61(~)2 > O.

110. Assume that e, fJ are given, e, fJ > 0 and that 6 is such that Itp(Yl) - tp(Ys) I < e whenever IYl - Y21 < 6. Since I(x) is integrable, a subdivision of [a, b] can be found for which the total length of the subintervals where the oscillation of I(x) is :;::: 6 is < fJ. On the other subintervals the oscillation of tp[f(x)] is at most e. 111. [Cf. C. Caratheodory: Vorlesungen fiber reelle Funktionen. Leipzig and Berlin: B. G. Teubner 1918, pp. 379-380.] Let I(x) be defined as in 99 and G(x) as in 98 and

1 for Y = 0

tp(y) = {

o

for'y

~

O.

Then tp[f(x)] = G(x). 112. Assume that I(x) is non-increasing. We obtain

j "'/(C) dC > I(x) j C"dC =

"

2"

f

C"/(C) dC

"

2"

< I(x) f C" dC = "

r+ ~

(1

x"+1 I(x) 1 - a

"

f~r 0 < x
-1. The integral over [00,00), w > 0, is convergent for (:J < 0 if and only if £x < -1. If (:J > 0 and n an integer

(:J

262

Integration

the two following integrals can be compared as n -+ 00: ("+1)..

f

X' Icos x IJ dx

....

(nn;)'"

N

0

The second integral lies between

..

f 1cos x 1(·...)1 dx

o

..

f

and

Icos X1(%+ ....)1 dx.

.

f 1cos x 1[("+1)"1" dx,

0

"

which increase like n -"2 [202J. For the integral to converge we must have

~ < -1. Combining all these results we find that the integral in

IX -

question is convergent if and only if either -1

< IX < ~ -

114.1. Let

1.

~,as,

IX


€p(x

+ ~) -

€p(x) > €p(x) - €p(x - ~) >

--n-'

q>(x) - G

Let ~ converge to 0 and n increase to infinity in such a way that x ± n ~ remains in (a, b). Then the continuity of €p(x) is established. Assume

~ > 0 and replace ~ by .! in (*) : 1J

~

q>(x) - q> (x - :

-

6)

m

q>(x) _ q>(x _ 6)

> = --6 --.

I

-6 1J

Since €p(x) is continuous: q>(x-t 6) - q>(x) ~ q>(x

6

-

>

=

+ 6')

- q>(x)

6'

0

,

q>(x) - q>(x - 6) ~ q>(x) - q>(x - 6)

6'

-

.1\ a > 0, 0 ~ xn < 1, a independent of n.

Let

~

denote an accumulation point of the xn's and m be such that

270

Integration

Im(~) < a. Determine a neighbourhood of ~ in which Im(x) < a, thus I,,(x) < a for n ~ m. There are infinitely many x,,'sin this neighbourhood: contradiction. 127. [Cf. G. P6lya, Problem: Arch. Math. Phys. Ser. 3, Vol. 28, p.J 74 (1920).] The limit function is monotone too, say monotone increasing. Subdivide the interval of convergence of the sequence I,,(x), n = 1, 2, 3, ... , into subintervals [x._ I , x.], '11= 1, 2, ... , N, so small that I(x.) -/(x._ I ) < e, moreover choose n so large that 1/.. (x,) -/(x.) I < e for every v. Then we have for X'_I < x < x.

l(x._ I )

-

e < 1.. (x._ 1) < I,,(x) < I.. (x.)

< I(x,) + e

thus 1/.. (x) - I(x) 1< 2e; we have used the hypothesis that I.. (x) is increasing. 128. Obvious. 129. Assume a < x < b. (It is obvious what has to be changed in the proof below to accommodate the cases x = a and x = b.) x+.

b

JP,,(t) I(t) dt -/(x) = J P.. (t)

a

x-.

[f(t) -/(x)] dt

a

%-8

2 max I/(t) a:;i;I:;ob

I


ko

Writing

f" e-k,l rp(t) dt =

J(k) = [«P(x) e-("-ko)"]; The relations e-"X = y,

«P(x)

we

obtain

for

0

«pC log ~)

+ (k -

co

ko) f «P(x) e-(k-ko)" dx. o

= V'(Y), '1'(0) = J(ko) = 0,

define V'(y) as a continuous function on the interval [0, 1], furthermore I

fV'(y) y.. -l dy = 0, n= 1, 2, ... o Hence [138] V'(Y) = 0, «P(x) = 0, rp(x) = O. 143. [M. Lerch, communicated by M. Plancherel.] If So were a zero of the F-function So + m, m = 1, 2, ...• would be zeros too [functional equation]. Let m be so large that + m) is positive, put s = So + m + 1 = (1 + it, (1 > 1. The equation

meso

j

o

e-""xu - 1 cos (t log x) dx =

1e-""x'-l dx

m

0

=

mr~) = n

0, n = 1,2,3, ...

would imply [142] that xu - 1 cos (t log x) = 0: contradiction. 144. For I(x) = 1, x, x 2 d. 140. For I(x) = e":

K,,(x) =

i e~ (:Z) x'(1 -

,=0

x)"-'

=

(e~ X + 1- X)" = [1 + (e~ - I)X

r.

275

Pt. II, Solutions 141-148

145. We have [140]

1; (v -

nx)2 (:) x'(l - x)"-' = nx(l - x),

,=0

thus

~

n

n2 L;1 1 < nx(l - x) ::::;: 4" . 146. [Cf. S. Bernstein: Communic. Soc. Math. Charkow Ser. 2,

Vol. 13, pp.1-2 (1912).] We define e,,(x) = max I/(x)

-/(:)1

for

1

all v for which

en(x)

< en'

I~n - xl < n - '. lim e,,(x) = 0 =

lim e"

"-+00

11-+00

uniformly in x, i.e.

O. Moreover

I(x) - K,,(x)

= ,~[/(X) - I( :)] (:) x'(1 - x)"-'.

According to 145 we have the inequality

I/(x) - K,,(x) 1< e" L;1

+ 2M L;II < en + ~ n- 1 ,

where I/(x) 1 < M on [0, 1]. 147. [Cf. J. Franel: Math. Ann. Vol. 52, pp.529-531 (1899).] Obvious for 0 < r < r l . For rm < r < rm+l the right hand side is equal to

ml(r) - 1[f(r2 ) -f(rl )]

-

2[f(ra) - l(r2)]

_

... -

(m -1) [f(rm) - ICrm- l )]

- m[f(r) - I(rm)]. But this exactly equals the expression on the left hand side. Indeed, the formula we have proved is the formula for "partial integration" :

,

f ICt) dN(t) =

o

,

N(r) ICr) -

f N(t) rCt) dt.

0

148. If r"-k-I < rn _ k = ... = r" (possibly k = 0 or l = 0; ro = 0) then

=

r"+1

= ... = r"+l < r,,+l+l lim ~= O.

n-+oo

If rm

< r < rm+l we find N(r) = m and m+~ _ _ 1_= ~< N(r)

rm+1

rm+1

Analogously in the second case.

rm+1


1 and choose m so large that 1 < c < 2m • Then 1
n~l! I(x~)dx, (r,..) 1 - ----

N(r) ~ 1 r - N(r)

-

1

,,-1

,..=1

~

',..~'

(r,..) -

"

n ~--

1r - n

-1

J1xfJ .!_ 1

II

(~)

dx

.

279

Pt. II, Solutions 159-163

l3-)

The integral! dx is a continuous function of fJ [131]. Similar o arguments apply if f(x) is increasing. Replace f(x) by -f(x). 161. Let 0 < IX < A < fJ. By making use of 1116 we establish similarly as in 160 the existence of arbitrarily large n for which 1

N(-) Yn

1: f (r) ~ 00

k=l

Yn

1 ( 1)

1)

-:::;,ff x--; dx+ f f xP dx. 00

0

(

}

IX and fJ sufficiently close to A [131]. 162. [For 162-166 d. H. Weyl: Nachr. Akad. Wiss. G6ttingen 1914, pp. 235-236; Math. Ann. Vol. 77, pp. 313-315 (1916).] If f(x) = 1 on the subinterval [IX, fJ] of [0, 1] and f(x) = 0 otherwise the equation

Choose

(*) on p. 88 leads to the condition lim ~.. ~p~ n

n.-+CCI

= fJ -

IX.

Now suppose

that the condition is satisfied. To begin we note that it does not matter whether the subinterval IX, fJ is open, halfopen or closed. The relation (*) holds for any function that is constant (=1= 0) on a subinterval and vanishes outside this subinterval, therefore (*) holds also for any linear combination c1f} (x) + c2f2(x) + ... + cd/(x) of such functions f.(x) , c. constant, i.e. for any piecewise constant function. If f(x) is properly integrable there exist piecewise constant functions [102, with a = 0, b = 1], tp(x) and 'l'(x) such that

+ tp(x2 ) + ... + tp(xn) /(x}) + /(x2 ) + ... + /(xn ) -=----=-----< ---'=--~--n n

tp(x 1 )

1

1

The first and the last expression converge to f tp(x) dx and f 'l'(x) dx o 0 resp. and both are arbitrarily close to •

1

f

o

f(x) d~. The weaker conditions

vn(O, fl)

0


OX>

(

T

_~_ 1-"

(1 + x)' -

1-

IXo¥)

----~;;---- dx.

Special case of 201 : a

= -1, b =00, tp(x) =1, h(x) = ~= 1

209. Substitute e-1t' (1 e-1t: exp (e-1iXt ~)

(1

+ x)'"

-

IX

.- - -, 1 - ",x

0,

+ x)for x. There results

1= exp {e-1iXt! [x -

(1

+ x) log (1 + x)] } dx.

291

Pt. II, Solutions 206-210

Special case of 201 :

a = -1, b =

lP(x)

00,

=

~

=

210. We put 'fJ

=

1, h(x) 0, n

=

n- H ., 0 < e

=

x - (1

+ x) log (1 + x) ,

1

e-10f,t-;'-· 1 < (I'

The integral

III

question

becomes [205]

=

V;n [1

n

+O(!)r -jP"-\e- X (l +x)tdx

Y

= ;n [1 + 0 (!)r"-iY"-\e-

X

°

+ x)t dx + (V ne- l ,,2e).

(l

-'I

Since the function e- x (1 + x) increases for x < 0 the integral over [-1, -'fJ] is smaller than [e'l(l - 'fJ)t. We now expand the integrand on the remaining interval

e- X (l

-~+~-~(lHx)-4

+ x) =

e

-n~(lHx)-4

The factor e 4 comes therefore

2

3

4

is of the form 1



0< 0

+ O(n-

= O(x) < 1.

1 +4 6 ).

The integral be-

We have x'

/-3 ",,-i+lln- 1 -n~

J

Since

e

2

=

1

+ n ~ + O(n-1+6,). n~

dx is of order n- l the O-term of e

3

yields a contri-

-'I

bution of O(n-1+6,). Hence

Y;n

[1

+O(n-

1

+4 6

-'I

1 o:+pn-t =--:-

1 )r- y,,-1 e-"~ (1 +

J

V2n _,,'

x'

e- Z (l

3

+~)dX

3Vn

+ O(n-1+66) =

nX;)dx

+ O(n-1+6,)

.

"+lIn-1 1

V2n

X dx

J

x'

e-~(l+

-00

+ O(n--1+6,)

3

x _)

3Vn

292

Integration

211. [Cf. A. de Moivre: The Doctrine of Chances, 2nd Ed. London 1738, pp. 41-42.] We have K (x) "

= ~ IX e-Xx" dx = n!o

1_e- x (1 + -=- + x + ... + x"). 2

I!

2!

n!

Therefore x" is the only positive root of the transcendental equation K,,(x) = 1 - A. According to 210 we find for arbitrary n-free (X and {J

+ (XV -n + (J) =

K,,(n

A

B + V;-+ 0 ( V;-1 ) ,

where A and B have the same meaning as in 210. Determine (X and {J so that A

=

1 - A and B

=

+ (X V; + (J) must converge + (X V; + (J) > c > 0 for

O. Then x" - (n

to zero. If, on the contrary, we had x" - (n infinitely many n we could conclude that

B' + (X V-n + (J + c) = A + V;+ 0 ( V;-1 ) , where B' depends on (X and {J + c as B depends on (X and {J. In particular a , 1 -"2 1 -2" . B = B + - --- ce = - - ce > O. Smce A = 1 - A the last in-

1 - A = K,,(x,,)

>

K,,(n

l

V2n

(XI

V2n

equality is impossible. In a similar way we show that

x" - (n

+ (X V; + (J)
I(x", y,,) when 1Xl

-

8

x,,1 is sufficiently small. Hence tp(x1)

and we can interchange Xl and *224. Let

~

tp(x,,) -

8

x" in the foregoing argument.

m;x(mzin/(x, y)) = I(x l , Y1)'

rnzin (m:x/(x, y)) = l(x2, Y2)·

In view of the first operation (the inner one, written on the right)

296

Integration

*225. max I(x, y) = 1, min I(x, y) = { x

y

1 - (y - 1)2 when

'\I

>

- -

2

,

1 - (3 - y)2 when y :s;: 2.

min max I(x, y) = 1, max min I(x, y) = 0 x

y

y

"

*226. By using 198 complete the following outline: The quantity whose limit is desired is ,.'

'" [ [qJ(x) -n dx

] -1/..

'" min qJ(x) .

Part Three

Functions of One Complex Variable

z

1. Z + Z = 2x, Z = 2iy, zz = r2. 2. The open right half-plane; the closed right half-plane; the open horizontal strip bounded by the lines y = a and y = b parallel to the x-axis; the closed sector between the two rays which form the angles IX and {3 resp. with the positive x-axis; the imaginary axis; the circle with center Zo and radius R; the open disk and the closed disk resp. with center Zo and radius R; the closed annulus between the two circles with radii

Rand R' and centred at the origin; the circle with center at Z = ~ and . R ra dlUS2"' 3. The ellipse, and the domain bounded by the ellipse, with foci Z = a and Z = b and the semimajor axis k if Ia - b I :::;; k. (If Ia - b I = k the ellipse degenerates into a segment.) If k < Ia - b I no point z satisfies the condition. 4. Let Zl and Z2 denote the two roots of the equation Z2 + az + b = O. The region in question is the interior of the curve IZ - zlll Z - z21 = R2 with "foci" Zl and Z2' The curve is the locus of all points for which the product of the distances to Zl and Z2 is constant, equal to R2. It consists of two pieces for R < [Zl ~

R

=

[Zl ~

z2I

Z2 [

and of one piece for R

> ~i.!~[ . If

the curve is called lemniscate.

5. The condition in question is equivalent to

I

Z -

a 12 ~ 11 -

az 12

or to

(1

-I a 12) (I

Z

12 - 1) ~ O.

The first set is the open disk IZ I < 1; the second set is the unit circle IZ I = 1; the third set is the exterior IZ I > 1 of the closed unit disk. The value of the expression in question is ! a 1-1 for Z = 00, thus Z = 00 belongs to the third set.

298

Functions of One Complex Variable

8. The condition in question is equivalent to

la-zIZ:la+zIZ

or to

a

-m(a+fi)z:O.

Since a + is real and positive the condition means mz ~ 0. The first set is the open right half-plane; the third set is the open left half-plane; the second set is the imaginary axis. (The value of the expression in question is for z = 00 equal to -1; z = 00 belongs to the second set.)

+ i Cor z < C. 17. We have

Izo I" =


~

-Iall Mn-I_la2IMn-2

- cIMM"-l -

c2M2~-2

- ... -Ianl -'" - c"M" > O.

*21. From 20. Substitute for ck

respectively and add a little remark to 20 in the last case. 22. [G. Enestrom: Ofvers. K. Vetensk. Akad. Forh. 1893, pp. 405415; T6hoku Math. J. Vol. 18, pp. 34-36 (1920); S. Kakeya: T6hoku Math. J. Vol. 2, pp.140-142 (1912); A. Hurwitz: T6hoku Math. J. Vol. 4, p. 8f! (1913).] We find for Izl < 1, z =l= 1,

1(1- z) (Po

+ Plz + P2 Z2 + ... + Pn z") I

= Ipo - (Po - PI) z - (PI - P2) Z2 - ... - (P"-l - P,,) z" - p"Zn+11 > Po - I(Po ~ PI) z + (P1 - P2) Z2 + ... + p"z"+11

>

Po - (Po - PI

+ PI -

P2

+ ... + P,,) =

0,

because (Po - PI) Z, (PI - P2) Z2, ... , PnZ"+1 cannot have all the same argument. (Unless z? 0, in which case the proposition is trivial.) A weaker statement, < instead of

0 (in

the second case we first multiply by z"), so that 22 can be applied. 24. We call the polynomial in question I(z). For have

m-z1

?- 0, therefore

If(z) I > la I z" I = I ,,+

an_II _ an_ z

I

2 _

IZ l2

> m(a" + a"z_~) -1:1

mz >

Izl >

1, we

a,,_~ _ ... _ ~

Izl3

2 -

0,

1:1

Izl"

3 -'"

9

? 1 - Iz I2 -IZI'

The last expression is ~ 0 if Iz I ~ r, where r is the positive root of the

V

equation r2 - r = 9, r = ! ((1 + 37). 3 < r < 4. The polynomial corresponding to the number 109, 9 + Z2, has the roots + 3i.

302

Functions of One Complex Variable

25. [Ch. Hermite and Ch. Biehler; d. Laguerre: Oeuvres, Vol. 1. Paris: Gauthier-Villars 1898, p. 109.J We write

+ iV(z) =

P(z) = U(z)

let x be a root of V(x)

U(x)

+ iV(x) =

=

ao(z -

0 or U(x)

Zl)

(z -

=

O. Hence

U(x) - iV(x) or U(x)

Z2) •••

(z - z,,), ao =F 0;

+ iV(x) = -

[U(x) - iV(x)J,

that is

ao(x -

Zl)

(x -

Z2) •••

(x - z,,) = +ao(x - i l ) (x - Z2) ... (x - i,,).

Such an equation, however, is possible only for real x. Assume 3x > 0; then Ix - z.1 < Ix - i.1 for 11 = 1, 2, ... , n. By the same token xcannot lie in the open lower halfplane.

26. [Cf. 1. Schur: J. reine angew. Math. Vol. 147, p.230 (1917).] Put P(z) = ao(z - zl )(z - Z2) •.. (z - Z,,), a o =F O. Let x be a root of P(z) + P*(z) = o. Then IP(x) I = IP*(x) I; since P*(z) = o(l - Zl z) (1 - Z2Z) ••• (1 - z"z) we have

a

n" Ix - z.1 = .=1 n" 11 - i.x I·

.=1

Such an equation can hold only for ~ x I = 1. Assume Ix I < 1: then [5J I z.1 < 11 - I for all 11, thus the first product is smaller than the second. By the same token it is impossible that Ixl > 1. We argue analogously in the case of P(z) + yP*(z), II'I = 1.

x-

z.x

27. [M. Fekete.J Define I' = AP(a) + p.P(b), 0 < A < 1, A + p. = 1. If all the zeros of P(z) - I' = ao(z -'- Zl) (z - Z2) ···(z - z.. ) were outside the domain in question, the inequality a -z"

-- 0,

304

Functions of One Complex Variable

thus for C=F z.

m• = A1 IC -

Zl

A.IC - z.12 12 + .... +.. A Ie Z2

12 + ~ I'"., -

z.. 12

Approximate m. by the rational numbers

,

P;,

11

11

= 1, 2, ... , n.

= 1, 2, ... , n,

PI + P2 + ... + P.. = P, and remember that the roots of algebraic equations are continuous functions of the coefficients. The derivative of the polynomial

n. (z -

.=1

z.)i'. has a zero arbitrarily close to C. Since C

lies inside or on the border of at least one of the triangles determined by three of the points z. it is sufficient for this problem to know that the zeros of a polynomial of degree 2 are continuous functions of the coefficients, which is obvious. [Remark due to A. and R. Brauer.J

33. [M. Fujiwara: T6hoku Math. J. Vol. 9, pp.102-108 (1916); T. Takagi: Proc. Phys. Math. Soc. Japan Ser. 3, Vol. 3, pp. 175-179 (1921).J Set P(z) = ao(z - Zl) (z - Z2) ••• (z - z.. ) and let z denote a point at which P(z) - cP'(z) = 0, P(z) =F O. Hence (1)

P'(Z)

1 c

1

1

1

1

---=-+-+···+----=0. P(z)

Z -

Zl

Z -

Z2

Z -

z..

C

Introducing m1

1

=-1Z--1 - Zl 2 '

we can write (1) as

(2) The first term pn the right hand side of the equation represents the center of gravity of a certain mass distribution at the points zl' Z2' ••• , z.. , that means a point inside the smallest convex polygon containing all the points z•. The second term represents a vector parallel to the vector c. Hence the statement follows. -Cf. V 114.

305

Pt. III. Solutions 33 - 36

34. [T. J. Stieltjes: Acta Math. Vol. 6, pp. 321-326 (1885); G. P6lya: C. R. Acad. Sci. (Paris), Vol. 155, p. 767 1-769 (1912).] Let %v, V = 1, 2, ... , n denote the zeros of P(z) and assume A (z.) =l= O. Then p' (z.) =l= 0 because otherwise the differential equation for P(z) would imply that pIt (z.) = 0 and repeated differentiation would show that P(z) is identically zero. The equation

__~ ___ +_ 1_+".+===+===+.,,+

1 3. -

3"

implies [31] that z. lies in the interior of the smallest convex polygon that contains the points zl' Z2' .'" Z._l' z.+1' "., Zn' aI' a2 , ••• , ap (on the line segment that contains all these points). Consider now the smallest convex polygon that encloses Z1' Z2' ... , Zn' a1' a2, .•. , ap ' Only t.he a:s and no z. different from the zeros of A (z) can lie on the polygon.

35. [L. J. W. V. Jensen: Acta Math. Vol. 36, p. 190 (1913); J. v. Sz. Nagy: Jber. deutsch. Math. Verein. Vol. 31, p. 239-240 (1922).] We denote the zeros of /(z) by z1' Z2' ••• , zn and assume

1 + --1 + ". + --1 = ----3 -

31

3 -

32

3 -

-

z =l= z, z =l= z.' v

0,

3,.

=

1, 2, "., n.

Because of the pairwise symmetry of the zeros we have

. (1

~3

v=1

3-3

"

1) =0. + 3-Z Ii

The formula Z= X

. + zy,

Zo

=

Xo

+ zYo, .

~

,-S

(_1_ + _1_) = 3 -

30

3 -

Zo

2 J'~

I (3- - (x30)-

Y

X O)2 (3 -

Iy2 '

30) 2

shows that the above equation can not hold when z is outside all the circles described. 36. Writing zn

=

xn

+ iy..

we find IZn

x

1

t.

~ I < 1, i.e. if and only if

49. [I. Schur: Math. Ann. Vol. 74, pp. 453-456 (1913.)] The case e = -k, k positive integer, can be excluded from the start; example: U o + u l + ... + u" n ) u" = ( k _ 1 ' u" + e n+ 1 = 0 for k > 2, u" = log (n + 1)

310

Functions of One Complex Variable

for k = 1 [169]. The case e = 0 is obvious. We put un

= Zn'

Un

=

wn

+e

U

0

+ U 1 + ... + U 'n

and, as on p. 111,

+1

n

Multiplying the relation (n

+ 1) zn -

=

nZn_l

(n

+ 1) wn -

nWn _ 1

+ ewn ,

n

=

1, 2, 3, ...

byr(;(~ : ~) 1) and adding the first n equations we obtain r(n + c + 2) r(n + 1)

w.. - T(e

_ r(n + c + 1) r(n + 1)

(n

+ 2) Wo =

+ 1) zn

Y' r(v

e "-'

.=1

n+1

n

+c+

z-e

1 n

me >

ano With

me =

r(t-; ~_ ~fl n- c- I

-1. Assume

U = U = U ='" = _1_ o 1 2 1+c'

+ c + 1) + 1) z.

r("

+ 1) z. -

_

T(e

VZ,_I]

+ 2) Zo,

r(n+1) " - I r (v+c+1) "" Z r(n + c + 2)"';;" r(v + 1) .' .=0

For fixed v one finds an. '" - e exists if and only if

+ c + 1) + 1) [(v

r(v

,,-1

_

i.e.

w " -

;, F(v

'~I

me >

-1 and put

then Z = 1 W = "

[1155]. Thus !~~ an.

'"

_1_

l+c'

+ anI + an2 + ... + a"" =

1

thus

1

+ c.

y we have

I A l' Ir(nr(n+ +c +1) 1) 1< n,

I F(n + 1) I < B -1'-1 Ir(n + c + 2) n ,

where A and B are constants, independent of n. Hence ,,-I

lanol

+ IanI I + ... + lan,n-II < lei ABn- 1' - I J:

.=0

-+

lei AB of1 x1'dx =

v1'

I lAB _c__ 1

[1122].

+y

The desired necessary and sufficient condition is therefore SO. Set an = cx.. + ip.. , cx,,' P.. real. The relation

me >

-1.

311

Pt. III, Solutions 50-55.1

(binomial series) and a generalisation of the proof of 175 imply lim (a 1 + a2 + '" + a,,) n- a = 0, i.e.

"'-+00

lim (IX1

n.,.,oo

+ IX2 + ... + IX,,) n- a =

lim (PI

ft.-i"OO

+ P2 + ... + P,,) n- a = o.

Now I 92 can be applied to both power series

IXlt

+ IX2t2 + ... + IX"t" + "',

hence lim (1 -

1.-+1-0

=

lim (1 -

1.-+1-0

PIt + P2t2

+ ... + p"t" + "',

W(IX1t + IX2f + ... + IX"t" + ...) W(PIt + P2t2 + ... + p"t" + ...) = O.

51. Whenever the four subseries consisting of the terms in the four quadrants (ffiz > 0, 3z 2: 0, etc.) converge, the series converges absolutely. 52. By successive bisection: Assume that all the terms z,,' z,,' ... lie in the sector -81 ~ arg Z < -82 and that \Z" \ + \Z" \ + ... diverges.

CoItstruct the two subseries with terms in -81 < arg Z < {}l ;

{}B

{}l ;

{}B

and in

< arg Z < -82 resp. At least one of the two is divergent.

53. Choose a finite number of terms zm = xm lj- iYm from each of the successive sectors

(- ;. ;).(- :' :) .... , (-;'2~)'''' The different sectors should contribute different terms so that the points

z'h' Z'h+l' .... Z'h+k

that correspond to (- ; . ;) are in this

sector and that 1


+ 2ixYl2 =

IX2 - y2

+ 2ixyl2,

i.e. x2 - y2

= t.

They define an equilateral hyperbola which intersects the real axis at the points z =

± V~ . The rotation becomes ~-1 Z2

:rr.

arg --2- =±2 when 91 ~ Z

=

1, i.e.

,2 = cos 2{}.

/z -V~ liz +V~/ = ~ .

These points lie on the lemniscate

317

Pt. III, Solutions 79-89

1

82. The auxiliary function, onto the upper half-plane

,=

00, z

=

=

i to ,

3' >

= -

z+-

~ maps the region in question

0 [81]. The origin z = 0 corresponds to

= ± 1 to , = + 1. The function w = :2 half-plane 3' > 0, note 75, into the w-plane cut

0 and z

transforms the upper open along the non-negative real axis. The point' = 00 corresponds to w = 0, ,= 0 to w = 00, ,= ±1 to w = 1. The mapping function having the required properties is therefore given by W

2)2 =

= (- - 1 Z

The images of z =

83. arg w =

R

+-

(1

+ Z2)2 • 4Z2

Z

± 1 are both at w =

1 but on different sides of the cut.

2~ (arg z - ex), i.e. 0 < arg w < 2n. "

,.,-IX

84. The first auxiliary function, = (e- i " z)Il-e< maps the circular sector onto the upper half of the disk < 1. The second auxiliary

1'1

function s

3s >

C +~-

= - ___ C 2

[81] maps the disk onto the upper half-plane

O. Apply 78. 85. Follows from u - iv

= uX : [rp(x, y) + i1jJ(x, y)] = ~:[rp(x, t oy

y)

+ i1jJ(x, y)]

by separation of real and imaginary parts. 86. They are the images in the z-plane of the lines 'lftl = const. and 31 = const. parallel to the axes in the I-plane under the conformal mapping I = I(z). 87. Follows from 85 by virtue of the Cauchy-Riemann differential equation

~:

+ ~.~

=

O. Also the function 1jJ(x, y)

differential equation. 88. With u cos T + v sin T becomes

= '1ft fiz L

eiT ds

satisfies Laplace's

= 'Ift(u - iv) eiT the integral in question

= '1ft fiz

dz= 'Ift[f(Z2) - I(Zl)]

L

where dz denotes the directed line element with modulus ds and argument T. 89. u sin T - v cos T = Z'\'(u - iv) eiT ; d. 88.

318

Functions of One Complex Variable

90. The third equation is identical with the second Cauchy-Riemann differential equation. We get the first two by differentiation, keeping the first Cauchy-Riemann differential equation in mind: lOp

au

Ov

au

8u

l8p f!8y

8u 8y

Ov 8y

Ov 8x

8v 8y

- = - u8x --v= - u8x- - v8y' f! 8x 8x --= -u--v-= -u--v-.

91. The vector

w= -.!.. ei8 forms

the angle l} with the positive real

I'

axis, its modulus is -.!... The functions in question are up to a constant I'

I(z)

=

log z, 9'(x, y)

=

log r

=

log

V + y2,

tp(x, y) = {} = arctan L. x

XS

The level lines are concentric circles around the origin, the stream lines are rays perpendicular to these circles. 92. 9's - 9'!

=

logrs -logr!

=

tp' - tp

log I'z, 1'1

=

231;,

'"

1 ( ,

- "') 4n ----=--.

-'-

qiz - qil

2 log 1'2 1'1

93. The amplitude of

w is

equal to {}

+; ,

the modulus is : .

Furthermore we have (up to an additive constant)

= -

i log z,

tp (x, y)

= -

I(z)

9'(x, y) log r

=

= {} =

arctan Lx ,

-log VX2

+ y2.

The level and stream lines are the stream and level lines resp. of 91. The potential 9' is infinitely multi valued. 94. According to 93 the field of force is described (up to a real constant factor) by W

2i = ZI-=t'

t h us I() z

=

'1 z - l ~ og z + l ' tp

=

IZ-ll

log z +

1 ' -

9'

=

-l arg z z+ 1.

The level lines are circles through the points z -" -1 and z = +1, the stream lines are circles too, namely the ones with respect to which the points z = -1 and z = +1 are mirror images of each other (circles of Apollonius).

319

Pt. III, Solutions 90-97

95. We are looking for the points z for which [93]

i'-2

iAI

iAn

--------···---=0, Z -

Z -

zl

z2

Z -

z,.

i.e. Al A2 A,. - + -Z-- + " ' + -Z-- = 0; Z - zl z2 z,.

the positive numbers AI' ~, ... , A.,. are proportional to the intensities of the currents. Cf. 31, in particular the first solution given. 96. The vector field is generated by an analytic function I(z) for which fRl = const. on the given ellipses. The function [80]

z = kZ

+ k~'

2kZ

= z-

4

VZ2 -

maps the region bounded by the two ellipses onto the annulus'l < The semi-axes and the radii are related by

provided that a l

>

IZ 1< '2'

a2 and the positive roots are chosen. The problem is

! defines a vector field in the Z-plane for which

now reduced to 91 : w =

the concentric circles around the origin Z = 0 are level lines, i.e.

fR

J~

I I

= const. Consequently the same is true for Z = const.

J

dZ ~dzdz Z -

-J V

dz z2-4

along the given ellipses in the i-plane, i.e. w= -

V

1 , zl-4

I(z) = log (z -

VZ2 -

4) •

The stream lines are confocal hyperbolas, the level lines confocal ellipses with foci - 2 and 2. The relations between the potentials are

The capacity is 1

97. Put [93] i

w = -""""i; "PI = -log a, "P2 = -log b, fill =

(X,

fIl2 =

p.

320

Functions of One Complex Variable

The resistance is (the sign is not important) equal to

fJ-1X logb-loga·

98. According to 85 the unit circle Iz I = 1 is a stream line. If the constant value of the conjugate potential along the unit circle and on the real axis inside the vector field is assumed to be zero the function I = I(z) transforms the unit circle Iz I = 1 into a segment of the real axis. In view of 79 set

Since

w= 1 for z =

00

I(z) = k (z

+ !) + ko,

we have k

=

1, i.e.

w=1-~. Z2

w

99. The stagnation points are z = ± 1; assumes the same values at each pair of points that are symmetric with respect to the origin. Therefore the resultant total pressure on the pillar vanishes [d. 90]. The pressure is minimal or maximal when 11 - :21 is maximal or minimal resp., i.e. for z = +i and z = ± 1 resp. Rotation of all the vectors through 90° generates a field of force which admits the following interpretation: A homogeneous electrostatic field is disturbed by a circular, insulated wire perpendicular to the direction of the field. (The most simple example of electrostatic influence.) 100. [G. Kirchhoff: Vorlesungen tiber Mechanik, 4th Ed. 1897, pp. 303-307; A. Sommerfeld: Mechanics of Deformable Bodies. New York: Academic Press 1950, pp.215-217.] Supplementary continuity condition: the boundary of the wake (the stagnant water) stretches to infinity where Iwi = 1; since Iw I is constant on the entire boundary it has to be equal to 1. The direction of is known along the boundary segments AB, AD (barrier) and the magnitude of wis known along theboundary lines BC, DC (along the wake), is completely known at the four points A, B, C, D. Noticing th"at either the direction or the magnitude of is constant on the respective parts of the boundary we find a half-circle in the w-plane as image of the boundary of the field of flow. We fix the constant contained in I [po 123] so that 1= 0 corresponds to the stagnation point z = o. Then the left and right "banks" of the positive real axis of the I-plane correspond to the streamlines ABC and Abc, respectively. The I-plane cut along the positive real axis corresponds to the whole field of flow; it is not possible that only a subregion

w

w

w

321

Pt. III, Solutions 98-101

of the I-plane so cut should correspond to the field of flow because W = ::

C'V

i as z -+ 00, thus

I C'V iz. The same point, but associated with

the left or right bank of the cut in the I"plane, corresponds to the two points Band D, respectively, because of symmetry. Note that a dilatation of the I-plane with retained, causes the same dilatation of the z-plane because dl = w dz. Now we dilate the I-plane so that I = 1 becomes the image of z = ±l. In this way we attribute a numerical value to l. - If a one to one relationship can be established between the corresponding parts of the z-, and I-planes we can find out [d. 188J whether the interior is to the left or to the right as one moves in the direction given by ABCDA (see the last line given in the table). In the following diagrams the points in the different planes corresponding to A are also called A; B, C, D are used similarly.

w

w-

A B C D

z

iii

0 1 00

0 1 -i

-1

-1

w 0 1 i -1

right

left

region lies to the left

I 0 1 00

1 left

c

CD

o

C Velocity plane (iV-plane)

A

A 8 e=en ------0---0--o

8

(w-plane)

Potential plane (f-plane)

101. According to 82 we have 4w 2

1= (f+-W2)2'

i.e.

1 w

1

+ V1 - I VI

Vl-

where I becomes 1 as I = O. Observe the continuous change of the value of w on the two banks of the cut in the I-plane. Hence

z= z= x

J

dl o w

= 2 Vi + VI V1 - I

+ iy = 2 VI + ~ - i [I

+ arcsin VI.

F ~-

log

(VI + VI -

1)].

The first formula for z is to be used when 0 < I < 1, the second when I > 1 (positive roots) ; it furnishes the boundary of the wake:

322

Functions of One Complex Variable

z = l = 2 + ; for I = 1; X

2

N

VI, y

N

-I, thus the width of the wake

VfYT

is 2x N 4 at a great distance from the barrier. 102. If the pressure is P = c - t 1W 12 at the point z [90] it is in particular equal to PI = C - t on the boundary and therefore everywhere in the wake. The total pressure desired is +1

J

-I

=

(P - PI) dz =

J (1 o I

+1

J

-I

1

t

(1 - 1W 12) dz

dJ

=

w2) -

W

J 4 V1 0 I

= J (1 0

w2) dz

V-

--

I d 1= 'Te.

103. Put z = rei/}. Since the sign of the angular velocity is positive, {} is increasing, i.e. z describes the circle in the positive sense. We have

~~ = iz and the velocity vector in question is dJ(z) _ dJ(z) dz dD dz dD -

.

1'( )

~z

z.

104. Let w denote the angle through which the vector w (radius vector) has to be rotated in the positive sense to fall into the direction of the vector iz/, (z) (tangential vector, 103). Then the distance in question is given by iz!'(z) 7(z) -ffiz!, (z) J(z) I iz!,(z) I = !z!'(-;rr-'

c>< .

I/(z) 1 SIll W

I/(z)

=

;;j

1

J(z)

105. The amplitude of the vector in question is 3 log I(z). The angular velocity is therefore, with z = rei/}, ~ Q< I

dD~ og

I() z

=

106. The curvature is

Q< d logJ(z) dz ~ dz dD

~

=

=

Q0.

The proposition can also be easily proved by elementary geometric reasoning. 112. h(cp) = ffiaeiq> = lal cos (cp - ex). 113. (1) The velocity vector [103J iz/,(z) forms the angle cp with the positive real axis: cp = arg z/' (z) = Slog z/' (z). (2) h(m) = r

[104.]; note the sign.

ffizj'(z) j(z) Izj'(z) I -

+ ;

324

Functions of One Complex Variable

114. iO

z =3Iog~l+z

cp

115. For :

.=1

r + If'~~o) r + ... + Itt:~Zo) r 1'2

2,.

= 2171:

1'2"

1"

f

I/(zo

o

+ re''') 12 dO.
~

=

R •

142. If f(z) =F 0 everywhere inside the level line the absolute value \ f(z) \ attains its maximum and its minimum on the boundary according to 135 and 138. This implies that \f(z) \ must be constant in the interior,

329

Pt. III, Solutions 140-146

and so !(z) = const. Geometric interpretation: Since there is no peak inside a closed level line on the modular graph there must be at least one pit unless the modular surface is a horizontal plane. 143. At least one zero must lie inside any closed line [142] along which the absolute value of the polynomial !z - Zl) (z - Z2) ••• (z - z,,) is constant. There are only n zeros. 144. The theorem is not valid for !(z) = const. Thus we may assume !(zo) =F o. If there is a saddle point on the circle Iz I = r the projection of at least one of the sectors with points above the saddle point mentioned in 132 protrudes into the interior of Izl < r. Therefore Zo cannot be a saddle point, i.e. !:(zo) =F O. We put !(z) = wand consider the image of the circle Izl = r in the w-plane. The point farthest from w = 0 is Wo = !(zo), the curve has a definite tangent at Wo because /,(zo) =F O. The tangent, that is the vector izol'(zo) [103J, is perpendicular to the vector Wo = !(zo) (obvious for geometric reasons). Hence iZf~~;o) is purely imaginary. In the neighbourhood of Wo the side of the image curve turned towards the o~igin corresponds, according to the hypothesis, to the side of the circle. turned towards the origin. Therefore iZ~~:;o) must be positive imaginary.

145. [Cf. A. Pringsheim: Sber. bayer. Akad. Wiss. 1920, p.145; 1921, p. 255.] .. a(w' - w·- I )

21;

,=1

a(w



.-1

+w

"

= 2 1; J ,=1

w t - w- t t -I

W

+w

.

n

.

= 2m tan n~ 2;7n.

146. [A. Pringsheim, I.c. 145.] We define

C(k) = •

I

-

k

~ 1 (.(-1 + .(=~z. + ... + .()

for k = 0, 1, 2, .. , ,

k

+1 1 ('(~f Z;l + ~~iZ.-2 + ... + Z.-!l.(+l)

fork = -2,-3, ... ;

then Cik)(Zl - zo}

+ C~k)(Z2 -

Zl)

+ ... + C~)(z" -

~+l

Z"_I)

_ zk+l

= -~r- =

O.

The total length of L is called t, R is the largest, r the smallest distance of L from the origin z = 0; assume R > 1. The quantity

is the length of an inscribed polygon, thus < t. The points Zl' Z2' ... , z" can be chosen so that for a given 15. 15 > 0, and sufficiently large n,

330

Functions of One Complex Variable

Iz. -

Z._11


0 n

J x'-le- x dx + wSi f ro

~

o

eisO-roeiO

~

e2

df} -

0

ro

J x'-le- ix dx =

O.

0

Let w converge to + 00. Then the first integral converges to r(s) and the third to the integral in question. The modulus of the second integral is


a,

the rectilinear path of integration by the semicircle over the segment (a - iT, a + iT) to the right or to the left according as lX < 0 or lX > o.

333

Pt. III, Solutions 153-157

In the first case the integral does not change and is absolutely smaller 1 e-

e-

than 231 T2 nT = 2T ; in the second case it decreases by

IX

(residue at

the pole s = 0) and the new integral is absolutely < 2~ (T e":- a)1 nT. Now let T increase to + 00. 156. [The case A = 1 + e-1 is due to H. Weyl.] We have t"

+ 1,

for n < t < n

lI(t) = n! --

r

thus (the interchange of summation and integration can be justified by various arguments)

u !! +00 +00 sin-e2

! f I

:

,,=-00 1=0

J e lX, -31: 0 the integral along each of the L..'s has the same value. By letting lX increase to + 00 we successively extend E(z) analytically over the entire plane. 159. The integral along L = Lo has the same value as the integral in = in = e- r the contributions of the along L .. , lX ~ o. Since horizontal parts of L .. cancel each other. The vertical segment of L .. supplies

mz

r+

1

2n

r-

J" e...+iy d"y.

-"

This value is independent of lX, hence =1, as can be seen for 160. (1) Let z be outside ~. We have [169] E(z) =

lX

-+ -

00.

-..!..+..!..~ [(-c +~)e·CdC. g gB2m, C-z L

Evaluate the integral not along L but along L', an inner parallel curve to L at the distance 15 (boundary of the region ffiz> 15, -31: + 15 -

[II 217].

n!

Explanation: The principal parts of both integrals stem from an arc centred at z

=

2n whose length is of order

V;. Along it the argument of

dz is close to ; and the argument of In(z) is nearly O. [43.]

162. The number of zeros in the disk Iz I < r is:

=

1 2ni

rf. j'(z) ':f j(z) dz

1 = 2n

1'1='

f

2"

0

j'(z) 1 z j(z) df) = 2n

f

2"

j'(z)

ffiz j(z) df).

'19

z = re' .

0

163. The function w(C~ ~ ;(Z) is a polynomial of degree n - 1 in z. Furthermore v = 1. 2..... n.

164. Apply Cauchy's integral formula to both sides. The proposition is now identical with V 97. 165. Let e > O. Consider that region of the z-plane where all the

I n; I>

inequalities z -

e. n

=

O. ±1. ±2•... are satisfied (the riddled

plane). There exists a constant K depending on e such that in the entire riddled plane

336

Functions of One Complex Variable

It is sufficient to check this in the region - ~ ::;: x

o < e < i; . The integral ~

< ~, Iz I > e,

1 rf. F(C) dC 2:ni l' sin eC ~- Z)2

along the circle

ICI = (n + ~); converges to 0 as n ~ 00 because

IF(C) (sin eC)-ll < C]( along the path of integration. Compute the sum of the residues. [Hurwitz-Courant, pp. 118-123.J 166. We substitute in 165 G(z

+ ~) for F(z) and then z -

(_l)"G(n+J~) 1; e e

~ for z:

00

d (G(Z) )

- dz cos ez = -::~-oo

(ez - (n

+!) n)2

.

We now combine the terms with the subscripts nand - n - 1:

.!-. (G(Z)

) _

dz cos ez -

£ (-1)" (n + i~) ( e + i)

and integrate G(z) _

G

,,~O

£

[ez - (n

e

+ [ez + (n(!) + i) :n]2

(1

1)

+ !) :n) - - e - ez-(n+!):n+ez+(n+!):n'

1 n G (n

COS(F--n~o(-)

:n]2

The constant of integration has to be 0 because there must be an odd function on both sides. 167. The functions I(z) log z and :i'1(z) are regular in the domain described in the diagram.

168. Since Ilog z -

inl
I/(z) I. 198. [Example of a general theorem of G. Julia: Ann. Sci. Ecole Norm. Sup. (Paris) Ser. 3, Vol. 36, pp. 104-108 (1919).] Let R,. be the rectangle with the four corners n ± t ± id, n integer, d fixed, d > 0 and V';Tite z = x + iy = reiD. Because of 1155 we have on the boundary of R" as n -+ 00 log IF(z) I N (x - t) log r - y{} - x,

mz

mz

so that the minimum of IF(z) I on R" tends to 00 as n -+ 00. On the other hand we find Isin ;;r;z I > c on the boundary of R,. where c is independent

345

Pt. III. Solutions 193-200

of nand c >

o.

Hence the minimum of

1= Isin nz1lr(1

_1 1r(z)j

n

converges on the boundary of R_ .. to arbitrary a and sufficiently large n

+

00

_ z)

I

as n -+ 00. Thus we have for

Ir~z)1 > lal 1

on the boundary of R_ .. whereas at the center of R_ .. r(z) = O. Apply 1

1M to fez) = r(z) , 9'(z) = -a. 189. Integrating by parts twice we obtain

zF(z) = /(0) -/(1) cos z + = /(0) - /(1) cos z

! [1'(1) sin z - / t"(t) sin zt dt]

+ 9'(z) .

We draw circles of radius e around all the zeros of the periodic function /(0) - /(1) cos z, where e > 0 and 2e smaller than the distance between any two zeros. In the z-plane from which all the e-disks have been lifted

_'PX~) cos z

converges to 0 as z -+ 00. [To begin with prove this for the strip -:n~ mz~ :n.] With the exception of finitely many zeros zF(z) has, therefore, in any disk the same number of zeros as the function /(0) -/(1) cos z [1M], that is 1. This zero is necessarily real in the case 1/(1) I> 1/(0) I where the disk is cut in half by the real axis; the non-real zeros of the functions that assume real values for real z appear in pairs [solution 196]. 200. The term z"a-'" assumes its role as maximum term of the series

1(0)

2.. - 1 and abandons it on the circle Iz \ = \a \2"+1 on the circle Iz I = Ia 1 [I 117]. To study the dominance of the maximum term in between those two circles notice the formula

F(z) - z"a-'" z ~'-z"-a--""" = -a2-. . +-1

z

z

z

+ a-2"-+1 . -a2-"+-S + -a2-"+

z

J •

z

-a2-. . +-S • a-2-"+-5

+ ...

a 2.. - 1 a 2"- a 2.. - S a 2"- 1 a2n - S a2.. - 5 +-+_._+_._._+ .... z z z z z z J

346

Functions of One Complex Variable

On the circle Iz I= Ia 12.. the corresponding terms of the two subseries on the right hand side have the same absolute value. thus

I,!(z) - z"a-"'I < 2(-~ +~._1_ +~._1_._1_ z"a-'" I lal lal lals lal lal s lali 2

1

+ ...)

21al8

< Tiii 1 - Ia 1-3 = Ia IS - 1 < 1. because the only positive root of the equation z3 - 2Z2 - 1 = 0 is smaller than 2.0 [18]. Hence F(z) has in the disk Izl < lal 2.. the same number of zeros as z"a-..•• namely n. The disk Iz I < Ia 12.. - 2 contains. by virtue of the same proof. n - 1 zeros. -Cf. V 178. 201. [A. Hurwitz: Math. Ann. Vol. 33. pp. 246-266 (1889).] The closed disk D has a as its center. lies completely in mand contains no other zeros of I(z) than possibly a. We have I/(z) I > 1/.. (z) -/(z) Ion the boundary of D when n is sufficiently large. Apply 194.I,,(z) -/(z) = tp(z). More generally: Each subdomain of on the boundary of which there are no zeros of I(z) contains exactly the same number of zeros of I.. (z) as of I(z) if n is sufficiently large. Important for the applications! 202. The limit function is regular in the unit disk Iz 1 < 1 [170]. Assume that l(zl) = l(z2) for ZI =l= Z2' 1zll < 1. 1z21 < 1; consider the sequence I.. (z) -1.. (z1)' n = 1. 2•... which converges to I(z) -/(zl). In a disk that has its center at Z2' lies completely inside the unit circle and does not contain zl.I.. (z) -I.. (z) would have to vanish for sufficiently large n [201]: contradiction. 203. [170.201.] 204. In the case where a and d are integers the proposition is proved in the same way as proposition 185 because the zeros of the polynomial aolf + a 1lf+ d + a 2t'+2d + ... + a..t'+nd lie in the disk Iz I ~ 1 [23]. If a and d are rational z has to be replaced by a suitable multiple of z. If a and d are irrational approximate these two constants by rational numbers and apply 203. 205. [G. P6lya: Math. Z. Vol. 2. p. 354 (1918).] We have [II 21]

m

1

,,-1

JI(t) cos zt dt = ..lim I o ...,.00

.=1

~ 1(~) cos ~ z n

n

n

[185.203].

208. Counter-example

I ..(z) =

Z2

1 +-;-.

n

= 1. 2. 3•... ;

:1): Izl~ 2;

a = -1.

b=

+1.

208.1. [G. P6lya. Problem: Jber. deutsch. Math. Verein. Vol. 34. 2. Abt.. p. 97 (1925). Solved by R. Jungen: Jber. deutsch. Math. Verein.

347

Pt. III, Solutions 201-208

Vol. 40, 2. Abt., pp.6-7 (1931).] We may assume without loss of generality that Mz), fs(z), ... , f,,(z) are linearly independent and 1c1 12

+ 1c2 12 + ... + 1c,,12 =

1

so that the set of coefficients cl' c2 ' ••• , c" can be conceived as a point c on the surface 6 of the unit sphere in 2n dimensions. If there were no finite upper bound of the nature stated, there would exist an infinite sequence of points c', c", ... such that the linear combination corresponding to c(m) has not less than m zeros in ~. This sequence has at least one limit point c(co) on 6, yet the linear combination corresponding to c(co) has only a finite number of zeros in ~. Hence contradiction to the last remark of solution 201. 208.2. [G. P6lya, Problem: Jber. deutsch. Math. Verein. Vol. 34, 2. Abt., p.97 (1925). Solved by Nikola Obreschkoff: Jber. deutsch. Math. Verein. Vol. 37, 2. Abt., pp. 82-84 (1928).] Apply the argument principle to a rectangle with corners

-a

+ ilX,

a

+ ilX,

a + ifJ,

-a

+ ifJ

where a is sufficiently large. Use V 75 and solution 180 in considering the horizontal sides. , dw 207. From z - w91(z) = 0 follows 1 - w91 (z) = 91(z) dz' Thus Lagrange's formula (L) (p.145) for fez) implies by differentiation with respect to w /,(z)

IP~ =

1 - WqJ'(z)

1:

w"-l

,,=1 (n -

[d"-l/,(X) IP(X) [1P(X))"-1]

i)!

dx"-1

:1:=0 '

i.e. the formula to be proved for the function t'(z) 91(z). The family of admissible functions f(z)- is identical with the family of functions t'(z) 91(z), where t'(z) is the deIivative of an admissible function, because 91(0) =l= O. Thus 207 leads to Lagrange's formula (L), p. 145, by integration. 208. ~ [d"j(X) [IP(X)]"] = -.!.., ,(. j(C) [1P(Cl]" dC, n! dx" %=0 2:/tI- ':J' C" C

1: w:n.

,,=0

[d"j(X) [:(X) dx

J"]

= .. =0

~ 2m

,(. j(C) dC

':J'

C

i

,,=0

(WIP(C) )" ,

C

integrated along a circle around the center C= 0 and for w so small that 1C1 > 1w91(C) 1 along the path of integration. Then the path of integration encloses the same number of zeros of C - w91(C) as of C [194], i.e. exactly one. Denoting this single zero by z we further find 1 ,(. 2ni'Y

j(C) dC

C- wlP(C) =

j(z)

1 - wlP'(z) •

348

Functions of One Complex Variable

209. [L. Euler: De serie Lambertiana, Opera Omnia, Ser. 1, Vol. 6. Leipzig and Berlin: B. G. Teubner 1921, p. 354.] In (L), p.145, set ljI(z) = e', j(z) = Z, 210.

32w3

2w2

w" + 2T +31 + ... +-n-!+ .... Introduce in (L), p. 145, ljI(z) = e', j(z) = eat', at. _ 1 + "'";, IX(IX + n}"-1 w." e Z= W

n"-l

n!

,,=1

211. We write x = 1

+ z, ljI(z) =

+ z)P, j(z) = 1 + z; (L), p. 145,

(1

yields

P

"

= 1 + z = 1 + 2: (n .: 1) wn . 00

x

,,=1

212. [Cf.l.e. 209, p. 350. ] Set x = 1 + z, ljI(z) = (1 (L), p. 145, implies

y=

xat

=

(1

+ z)P, I(z) = (1 + z)";

+ z)at = 1 + i (IX +n pn- 1- 1) IXW" • n ,,=1

213. We obtain for {1 = 0, fJ = 1 the binomial series, for fJ = 2 = (1- JI~)at 2w

1

+ lX

i (IX +

2n n - 1

"=1

for f1 = -1 essentially the same series; for f1 =

)2 (1Y;-;--2 1+~ +; Put x

=

1

at

= 1

1) w" . n'

i

(IX + ~ + lXn~' n _ 1 00

1) "

:.

+ ~ , w = ; , lX = af1; fix~, w, a and let f1 increase to +

The equation in 211 becomes w=

~ (1 + ~

t

P c-J

00.

~e-~ .

214. By setting ljI(z) = e', I(z) = eotl and applying 207 we obtain 00

(n

+ IX)" w"

,,~--n-!--- =

eat. 1- we'

=

eat' 1- z'

where z has the same meaning as in 209. The radius of convergence is

IX)"

. (n + (n + 1)! = hm -----n~oo n! (n + 1 + IXt+1

215. We are dealing with

~

J

n+1

"'" "=0"

t" e-J.I dt = ~ n! "'" ,,=0

J 1

0

(n

= e-1 •

+ IXt e-J.(n+ A *(r) - M, thus

A (r) > ~ao - M

+11 a.. I r" .

If a.. is different from 0 we conclude

lim inf ~g A(r) ,~~

> n.

logr-

There are arbitrarily large n's for which a.. =F O.

356

Functions of One Complex Variable

238. [E. Landau: Arch. Math. Phys. Ser. 3, Vol. 11, pp.32-34, (1907); d. F. Schottky: J. reine angew. Math. Vol. 117, pp.225-253

(1897).] It is sufficient to prove the inequality Ima1 1 R

! L1(1); for


0,0 < 21J < lX 0; put I(z) = l(eilJ) = U(e, {}) + iV(e, {}), then

The upper bound for 1(0) is reached if Vee, {}) = 0 for 0 < {} < 2n, hence I(z) 1(0) [230]. m. [134.] 274. The function I(z) = :~:~ is regular for Iz I < 1. It is also

=

regular on the unit circle, where I/(z) I = 1 unless lp(z) = O. If Zo is a zero of lp(z) it is also a zero of 91(z) and with the same multiplicity [otherwise z = Zo were a zero or a pole of I(z) which is impossible: at other points of the unit circle, arbitrarily close to zo, we have I/(z) I = 1]. We drop the common factors of 91(z) and lp(z) and so obtain the regular function I(z) which is different from zero in the closed disk Iz I ~.1 and whose modulus is equal to 1 on the unit circle, I/(z) I = 1 for Iz I = 1, therefore [138] I(z) c, Ic I = 1. Since 91(0) and 11'(0) are real and positive we have c = 1. 275. The absolute value I/(z) I is a real continuous function in ~; it assumes therefore its maximum in ~. This is impossible at an inner point if I(z) is not a constant [134]. 276. [Cf. E. Lindelof: Acta Soc. Sc. Fennicae, Vol. 46, No.4, p.6

=

(1915).] A rotation through 2nv around a point C maps the domain onto n

and the set 58 onto the set 58., 'V = 0, 1, 2, ... , n - 1, 580 = 58, ~o =~. The intersection 3 (largest common subdomain) of the domains ~o' ~l' ••• , ~.. _) contains Cas an inner point. Those inner points of 3 that can be connected with ~ by a continuous curve in the interior of 3 form a region 3*. The boundary of 3* consists, according to the hypothesis and the construction, of certain points of the sets 580 , 58), ... , 58.. _). The absolute value of the function I(C + (z - C) £0-') is < A at all the boundary points of 3* [275] and < a at those boundary points of 3* that belong to 58•. The absolute value of the function the domain

I[C

~.

+ (z -

C)] I[C

+ (z -

C) £0-1] ... I[C

+ (z -

C) £0-"+1]

is therefore not larger than aA"-l at all the boundary points of 3*, consequently [275] also at the inner point z = C.

369

Pt. III, Solutions 272- 281

2n. [Cf. E. Lindelof, l.c. 278. J Assume IX < n without loss of generality because we can consider f(#), with (J suitably chosen, instead of f(z). Draw a circle around an arbitrary point of the ray arg z = !(IX - 6) that is tangent to the ray arg z = IX. The chord cut by this circle from the real axis is always seen from the center under the same angle. Apply 276 where Cl) is identified with the portion of the disk in the upper halfplane and \8 with the chord on the real axis; thus lim fez) = 0 along arg z = !(IX - 6). By modification of the conclusion we show that the convergence lim fez) = 0 is uniform in the sector 0 < arg z < !(IX - 6). Repeat the argument for the rays 3

arg z = 4" (IX - e),

7

8 (IX - 6),

15 16

(IX - 6), ...

278. [Cf. E. Lindelof, l.c. 276. J Let R denote the least upper bound of If(z) I in ffi. There exists at least one point P in ffi or on the boundary of ffi so that in the intersection of ffi with a sufficiently small disk around P the least upper bound of If(z) I is equal to R. If there is no such point P in ffi then If(z) I < R in ffi, but then there exists a boundary point P with the required property. According to condition (3) we have R ~ M, i.e. If(z) 1< Min ffi. In case there exists at least one point P, z = zo' of the described type, then If(zo) I = R. Along a sufficiently small circle around Zo we have If(z) I < R, thus, according to 134, fez) = constant. 279. [Po Fatou: Acta Math. Vol. 30, p. 395 (1905).J Put w = e2ni/". If n is sufficiently large then lim fez) f(wz) f(w 2 z) .•• f(w"-J z) = 0,

r->1-0

the convergence is uniform in the unit disk, 0 ~ {} < 2n. The function fez) f(wz) f(w 2z) .. ·f(w"- l z) vanishes identically according to 278. [This proposition is not an immediate consequence of 275.J 280. [R. A. Schwarz: Gesammelte mathema1ische Abhandlungen, Vol. 2. Berlin: Springer 1890, pp. 110-l11.J Apply 278 to the function f(z)

z

which is regular in the disk Iz I < 1.

281. [E. Linde16f: Acta Soc. Sc. Fennicae, Vol. 35, No.7 (1908). Concerning problems 282-289 d. P. Koebe: Math. Z. Vol. 6, p.52 (1920), where also ample bibliography is provided.J We denote by C = tp-1(W) the inverse function of w = tp(C). The function F(C) = tp-1{f[q>(C) J}

370

Functions of One Complex Variable

satisfies the conditions of 280. Hence IF(C) I ::;: e for ICI ::;: e; there is equality only if F(C) = e''''C, ()(. real. The inequality states that the points F(C) lie in the disk IC I ::;: e, that means that the points tp[F(C)) = l[tp(C)] lie in the domain 9; z = tp(C) represents an arbitrary value in t. In the extreme case we have tp(e''''C) = l[tp(C)), i.e. I(z) = 'I' [e''''tp-l (z)], where C = tp-l(Z) is the inverse function of z = tp(C), ()(. is real. This is the most general function that maps 9l univalently onto @) and z = Zo onto w = Wo [IV 88]. 282. [C. Caratheodory: Math. Ann. Vol. 72, p.107 (1912).] Apply 281 to the following special case: 9l: the disk Iz I < 1; @): the disk Iwl < 1; Zo = 0, Wo = 1(0),

(") '1'''

tp(C) = C,

C+ Wo

=l+iiioC·

The subdomain t is the disk Iz I < e, t is the image of ICI < e under the function w = tp(C), thus 9 is also a disk. The points of 9 satisfy the relation

IW -

-ICI (1 - !Woll ) < Wo I 11 + iii~CI =

The inequality becomes an equality only if I(z)

.'" _+ ~o , in which case 11 =~ 1

+ woe""::

1 - IWol 1

e 1 - IWo I (I • = tp[e''''tp-l(z)] = tp(e''''z) =

+ woe''''zl = 1 -lwollzl, i.e.

arg z = arg Wo - ()(. + :rt. 283. Apply 281 to the following special case: 9l: disk Iz I < R; @): half-plane 9lw < A (R); Zo = 0, Wo = 1(0), 9lwo = A (0),

tp(C)

=

RC, tp(C)

= Wo + [iii;:= ~A(R)] C = Wo + [wo + Wo -

2A(R)] I ~ C.

is the disk Iz I < eR = r, 9 is the image of ICI < e under the mapping w = tp(C). The points of 9 satisfy the inequality C (I 9lw = 9lwo + [wo + Wo - 2A(R)] 9l 1 _ C< 9lwo - 2 [9lwo - A(R)] T+ (I

t

1-(1

= 1

2(1

+ (l 9lwo + 1 + (I A (R).

It is an equality only if I(z)

= tp[e''''tp-l(z)] = 'I' (e"" ~).

284. The following relation [solution 283] holds Iwl::;: IWol

+ [2A(R) -

Wo - wo] 1~

(I

= M(O)

+ 1 ~ (I [A(R)

which is a weaker statement than 236. 2BS. Apply 283 to log I(z) :

9l log I(z) = log I/(z) I ::;: log M(r).

- A(O)],

Pt. III, Solutions

282- 289

371

286. Proposition 285 implies 2!~

Iln(z) 12 < 1/,,(0) 1 Hizi For

Iz 1< ~

the exponent is 2 ~

~

287. Use 281: ffi: the disk

Zo

=

0,

Wo

=

1(0)

>

C.

Wo

1

'P(C) =

Wo

1+C 1 - C•

e, s: the disk whose boundary circle intersects the real

axis orthogonally at the points circle is

1; 6: the half-plane ffiw> 0;

0,

9'(C) = t: the disk 1z 1 :::;;

1:1> 1, consequently Iln(z) 12 < 1In (0) I.

Izi
0; its absolute ciently close to any boundary point [6]. Apply 278.-A different proof is based on 177. In fact, both methods go beyond the particular case of the half-plane mz> 0; both can be easily adapted to a generalized proposition which relates to 294 as 281 relates to Schwarz's lemma 280. 296. The function t(z) is assumed to be meromorphic with the zeros aI' a2 , ••• , am and the poles b1 , b2 , ••• , b.. (counted with proper multiplicity) in the disk 1z 1 < 1 and It(z) 1 = c > 0 for 1z 1 = 1. The function t(z)

n

m 1-az

1'=]

--I'

ap

Z -

n ___ .. z-b

v

.=1

1 - b.z

= z +Z L.: 1-~ Iz. + IX '=1/1 + : I 2z.z. 00

(

IX

• -.

00

.=1

(1

1

~ zL.:m-

+ IX)

00

.=1

z.

converges. 299. [T. Carleman; cf. also P. Csillag: Mat. phys. lap. Vol. 26, pp. 74-80 (1917).] Let Zo be an inner point of '1) and put It.{zo) 1 = e.t.(Zo)' '11= 1, 2, ... , n. (In the case t.{zo) = 0 we choose e. = 1.) The function F(z) = edl{Z)

+ e2t 2 (Z) + '" + e..I.. {z)

is regular and single-valued in '1); F(z) assumes its largest absolute value at a boundary point ZI of '1). Hence

IF(Zl) 12 IF(zo) I = J: ,..

Obviously

J: 1I. (ZI) IP' >

0=

1/,..(zo) IP"',

J: 1I. (zo) IP', •

i.e. fI/(Z1) > fI/(zo); in fact fI/(Z1) > rp(zo) if at least one I,..(z) is not a constant or, in the other case, if at least one I.(z) does not vanish identically. Since il is closed and fI/(z) continuous there exists a point in il where the function fI/(z) assumes its maximum: it cannot be an inner point except in the particular case mentioned in the problem.

376

Functions of One Complex Variable

303. Since 'l) is closed the function Ij(z) I, which is single-valued and continuous, attains its maximum in 'l). Proposition 134 shows that this cannot happen at an inner point of 'l) except in the case where j(z) is a constant. 304. [J. Hadamard: Bull. Soc. Math. France Vol. 24, p. 186 (1896); O. Blumenthal: Jber. deutsch. Math. Verein. Vol. 16, p.108 (1907); G. Faber: Math. Ann. Vol. 63, p. 549 (1907). Hurwitz-Courant, pp. 429430; E. Hille, Vol. II, pp. 410-411.] The function ZXj(z) is not singlevalued in the annulus r 1 < Izl s r 3 , its modulus however is. Hence the maximum of 1ZXj(z) 1is either 1 M(r1 ) or raM(rs) [303]. Choose (X so that (*)

1M(r1) = r;M(rs)'

Considering a specific point on the circle 1z 1= r2 we see that ~M(r2) S

1M(r1)

= r;M(rs)'

We introduce the value (X from (*). (The condition that j(z) be regular and Ij(z) 1single-valued on the punctured disk 0 < Izl < R is sufficient.) 305. The maximum of ZXj(z).is reached at a point of the circle 1z I = r2 , i.e. in the interior of the annulus r1 :::; 1z 1< r3, only if ZXj(z) is a constant. 306. Put j(z) = ao + a1z + a2z2 + ... + a"z" + .... The integral 12 (r) becomes

12(r) =Iao 12

+ la112 r2 + la212 r4 + '" + la,,1 2 r2" + ... =

""

~ p"r",

where p" > 0 and at least two p,.'s are non-zero [II 123]. 307. Assume that j(z) is not a constant and that none of the zeros Z1' Z2' ... , z,. of j(z) in the disk 1z 1< r coincides with z = 0 (for simplicity's sake assume also j(O) = 1). Then we have [120] log &(r) = nlogr -loglz11-log IZ21-'" -log Iz"l. Hence the graph of log &(r), as a function of log r, consists in a sequence of straight pieces with monotone increasing slopes. The change of slope' for log r = log ro is caused by the appearance of additional zeros on the circle Iz I = roo The increase in slope is equal to the number of such zeros counted with appropriate multiplicity. 308. [G. H. Hardy: Proc. Lond. Math. Soc. Ser. 2, Vol. 14, p. 270 (1915).] Suppose 0 < r1 < r2 < r3 < R. Define the functions e(&), F(z) by the relations

e(tt) j(r2eiD )

= Ij(r2eiD ) I,

0

< tt
< n 1: 1/(1'awi"·) II>· .=1

As n -t 00 this inequality becomes 11>(1'1) :::; 11>(1'a)'

Assume 0 < 1'1 < 1'2 < 1'3 < R, IX real. The functions

'"

'"

'"

zl> l(w 1z), zl> l(w2z), ... , zl> I(w"z)

are regular in the annulus 1'1:::; Izl < 1'3' however, only their moduli are necessarily single-valued. All the same [303, 302] we may conclude that the sum of the p-th power of their absolute values assumes its maximum on the boundary of the annulus. Applying the same arguments as in 304, 308 and taking the limit we establish the behaviour of 11>(1') with regard to convexity. -Cf. II 83 for the limit cases p = 0 and p = 00. 311. We may assume that the center of ~ is at the origin. Apply 230. The proposition states in other words: A harmonic function that is regular in a closed disk and that vanishes on the bounding circle vanishes identically. 312. We denote by u(x, y), z = x + iy, a harmonic function that is regular in the disk (x - xo)2 + (y - YO)2 < 1'2. The value of u(x, y) at the center is 1 2,.

J

u(xo, Yo) = 2 u(xo no

+ l' cos {), Yo + l' sin {)

d{}

[118],

378

Functions of One Complex Variable

consequently

IU(Xo. Yo) 1
0 or u(xo, Yo) :$; O. The integrand must vanish identically. i.e. u(x. y) cannot change sign on the given circle (u(x. y) possibly becomes 0 at some points). 313. Suppose that the point xo. Yo at which the maximum is reached is an interior point of 'il. Choose r so small that the disk with radius l' and center xo. Yo lies in the interior of 'il. The equation 1 2,. -2-

f

1fo

[u(xo. Yo) - u(xo +

l'

cos {}. Yo +

l'

cos {}, Yo+ r sin (}) = 0, 0
A) is best established first outside and then inside the circles Iz - n I = i, n = 0, 1, 2, ... [solution 165].

381

Pt. III, Solutions 323 - 332

Second solution: 178 and the fact that the terms on the left hand side are positive imply

"(11-') Ii - n2 ~ 2n1J"(1e2 - n21) [2 log C + 2(n -

~

p-l

1

where C and C' denote constants. The left hand side is hand side

N

y) e] de N

+ C,•

log n, the right

n - " log n: contradiction. This method can be generalized n

[F. and R. Nevanlinna, l.c. 177].



329. Assume e > 0; the function q:>(z) = plane 9h > O. We have

Iq:>(z) I ~1

for

if r is so large that w(r)

mz = 0 >

_ 6-

is regular in the half-

[f(z)t

and Iq:>(z) I ~1 for

Iz I = r,

! .hence Iq:>(z) I ~ 1 in the entire half-plane

»lz;;::: 0; and finally, as e -+ 0, le'l < 1: contradiction. (Borderline case of 290: The region % becomes a halfplane.) .IJ 0, h = ee 2, 0 < (1 < «5, (1({J - IX) < n. Apply the reasoning of 322 to the function F(z) = I(z) exp (_

(hz)fJ~IJ< -a) .

There results the conclusion: IF(z) I < 1 in the entire sector. Let e converge to O. - The proposition could also be reduced to theorem 322 with the help of a function that maps the sector IX ~ arg z ~ {J onto the sector -y

n < arg z < y, y ="2 -

~(/l

-

2

IX)

'

and leaves z = 0 and z=oo

unchanged.

331. In the special case

IX

= - ; , (J

= ; the statement is weaker

than 325. The proof involves the function f .fJ+IJ w 2 , ••• , W n . Determine a regular and analytic function q;v(z) in the half-plane 3z > 0 so that n91q;.(z) = w. [57J and put cf>(z)

=

a .( ~y,(Z)+Q>'(Z)+"'+Q>n(z).

The function t(z) cf>(Z)-l is regular and bounded in the interior, < 1 on the boundary, of ~ with the possible exception of 2n boundary points [335]. 337. The function l(eU ) is single-valued, regular and bounded in the half-plane 91u < O. We have It(ell ) I < 1 at all the boundary points of this half-plane except at the boundary point u = 00. [335.] 338. If z = 00 were not a boundary point we would have [135 suffices] necessarily Ig(z) I ::;;; k in ffi: contradiction. He1\ce z = 00 is a boundary point of ffi. If g(z) were bounded in 91 we could use 335 (only excluded point z = (0) which would lead to Ig(z) I < k in ffi. 339. There exists, according to the hypothesis, a constant M, M > 0, such that It(z) 1< M in the region ffi bounded by 1 and 2• Choose R > 1 and so large that It(z) I < e on 1 and 2 outside the circle Iz I= R. Take the branch of log z that is real for positive z. This branch is regular and single-valued in 91, where Ilog z I < log Iz I + n provided that Iz I > R. The inequality

r

1M (log R

+ n) + e (log z + n)1

r

> M (log R

r

r

+ n) + e (log Iz\ + n)

384

Functions of One Complex Variable

holds and for Iz I ~ R both addends on the right hand side are positive. Thus we obtain for Izl = R, z in m,

I

M(log

If

R+

I

log z

n)

+ e(log z + n) I(z)