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PROBLEMS AND PROPOSITIONS IN ANALYSIS
A
PURE AND APPLIED MATHEMATICS Program of Monographs, Textbooks and Lecture Notes Executive Editors
E arl J. Taft
Edwin Hewitt
Rutgers University
University ofWashington
New Brunswick, New Jersey
Seattle,Washington
Chainnan of the Editorial Board S. Kobayashi University of California, Berkeley Berkeley, California
Editorial Board Masanao Aoki
Paul J. Sally, Jr.
University of California, Los Angeles
University of Chicago
Glen E. Bredon
Jane Cronin Scanlon
Rutgers University
Rutgers University
Sigurdur Helgason
Mar-tin Schechter
Massachusetts Institute of Technology
Yeshiva University
G. Leitmann
Julius L. Shaneson
University of California, Berkeley
Rutgers University
W. S. Massey
Yale University
Olga Taussky Todd California Institute of Technology
Irving Reiner University of Illinois at Urbana-Champaign Contributions to Lecture Notes in Pure and Applied Mathematics are reproduced by direct photography of the author's typewritten manuscript. Potential authors are advised to sub mit preliminary manuscripts for review purposes. After acceptance, the author is respon sible for preparing the final manuscript in camera-ready form, suitable for direct reproduc tion. Marcel Dekker, Inc. will furnish instructions to authors and special typing paper. Sample pages are reviewed and returned with our suggestions to assure quality control and the most attractive rendering of your manuscript. The publisher will also be happy to supervise and assist in all stages of the preparation of your camera-ready manuscript.
LECTURE NOTES
IN PURE AND APPLIED MATHEMATICS 1. N. Jacobson, Exceptional Lie Algebras
2. L.-A. Linqahl and F. Poulsen, Thin Sets in Harmonic Analysis
3. I. Satake, Classification Theory of Semi-Simple Algebraic Groups
4. F. Hirzebruch, W. D. Newmann, and S. S. Koh, Differentiable Manifolds and Quadratic Forms
5. I. Chavel, Riemannian Symmetric Spac:es of Rank One
6. R.. B. Burckel, Characterization of C(X) Among Its Subalgebras
7. B. R. McDonald, A. R. Magid, and K. C. Smith, Ring Theory: Proceedings of the Oklahoma Conference 8. Y.-T. Siu, Techniques of Extension of Analytic Objects
9. S. R. Caradus, W. E. Pfaffenberger, and B. Yood, Calkin Algebras and Algebras of Operators on Banach Spaces
I 0. E. 0. Roxin, P.-T. Liu, and R. L. Sternberg, Differential Games and Control Theory II. M. Orzech and C. Small, The Brauer Group of Commutative Rings 12. ·s. Thomeier, Topology and Its Applications 13. J. M. Lo'pez and K. A. Ross, Sidon Sets
14. W. W. Comfort and S. Negrepontis, Continuous Pseudometrics
I 5. K. McKennon and J. M. Robertson, Locally Convex Spaces
16. M. Carmeli and S. Malin, Representations of the Rotation and Lorentz Groups: An Introduction 17. G. B. Seligman, Rational Methods in Lie Algebras
18. D. G. de Figueiredo, Functional Analysis: Proceedings of the Brazilian Mathematical Society Symposium
19. L. Cesari, R. Kannan, and J. D. Schuur, Nonlinear Functional Analysis and Differential Equations: Proceedings of the Michigan State University Conference
20. J. J. Schaffer, Geometry of Spheres in Normed Spaces 21. K. Yano and M. Kon, Anti-Invariant Submanifolds
22. W. V. Vasconcelos, The Rings of Dimension Two 23. R. E Chandler, Hausdorff Compactifications
24. S. P. Franklin and B. V. S. Thomas, Topology: Proceedings of the Memphis State University Conference 25. S. K. Jain, Ring Theory: Proceedings of the Ohio University Conference
26. B. R. McDonald and R. A. Morris, Ring Theory II: Proceedings of the Second Oklahoma Conference
27. R. B. Mura and A. Rhemtulla, Orderable Groups
� 8.
J. R. Graef, Stability of Dynamical Systems: Theory and Applications
29. H.-C. Wang, Homogeneous Banach Algebras
30. E. 0. Roxin, P.-T. Liu, and R. L. Sternberg, Differential Games and Control Theory II 31. R. D. Porter, Introduction to Fibre Bundles
32. M. Altman, Contractors and Contractor Directions Theory and Applications 33. J. S. Golan, Decomposition and Dimension in Module Categories
34. G. Fairweather, Finite Element Galerkin Methods for Differential Equations
35. J. D. Sally, N umbers of Generators of Ideals in Local Rings
36. S. S. Miller, Complex Analysis: Proceedings of the S.U.N . Y . Brockport Conference 37. R. Gordon, Representation Theory of Algebras: Proceedings of the Philadel hia Conference 38. M. Goto and F. D. Grosshans, Semisimple Lie Algebras
p
39. A. I. Ar ruda, N. C. A. da Costa, and R. Chuaqui, Mathematical Logic: Proceedings of the First Brazilian Conference
40. F. Van Oystaeyen, Ring Theory:
Proceedings of the 1977 Antwerp Conference
41. F. Van Oystaeyen and A. Verschoren, Reflectors and Localization: Application to Sheaf Theory
42. M. Satyanarayana, Positively Ordered Semigroups
43. D. L. Russell, Mathematics of Finite-Dimensional Control Systems
44. P.-T. Liu and E. Roxin, Differential Games and Control Theory III: Proceedings of the Third Kingston Conference, Part A
45. A. Geramita and J. Seberry, Orthogonal Designs: Quadratic Forms and Hadamard Matrices
46. J. Cigler, V. Losert, and P. Michor, Banach Modules and F unctors on Categories of Banach Spaces 47. P.-T. Liu and J. G. Sutinen, Control Theory in Mathematical Economics: Proceedings of the Third Kingston Conference, Part B 48. C. Byrnes, Partial Differential Equations and Geometry
49. G. Klambauer, Problems and Propositions in Analysis
PROBLEMS AND PROPOSITIONS IN ANALYSIS Gabriel Klambauer
Department of Mathematics University of Ottawa Ottawa, Ontario, Canada
MARCEL DEKKER, INC.
New York and Basel
Library of Congress Cataloging in Publication Data Klambauer, Gabriel, Problems and propositions in analysis ,
�Lecture notes in pure and applied mathematics Mathematical analysis--Problems, exercises, etc, I, Title . 1,
QA301,K53 ISBN 0-8247-6887-6
COPYRIGHT ©
1979
515 ,076
79-15854
49)
by MARCEL DEKKER, INC . ALL RIGHTS RESERVED
Neither this book nor any part may be reproduced or transmitted in any form or by any means , electronic or mechanical , including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER , INC . 270
Madison Avenue , New York , New York
Current printing ( last digit): 10
9
8
7
6
5
4
3
2
1
PRINTED IN THE UNITED STATES OF AMERICA
10016
To Helga Funk
PREFACE
Solving problems is an essential activity in the study of mathematics . Instructors pose problems to define scope and content of knowl edge expected of their students ; mathematical competitions and written qualifying exami nations are designed to test the participant ' s abi lity and ingenuity in solv ing unusual probl ems . Moreover, it is a famil iar fact of mathematical in struction that a s ingle good problem can awaken a dormant mind more readily than highly polished lectures do . This book contains problems with solutions and the reader i s invited to produce additional solut ions . To ensure a wide appeal I have concentrated on basic matters of real analys i s and have consulted problem sections in various mathematical j ournals and the col l ected works of some great mathematicians . Dr . John Abramowich aroused my interest to write this book and Dr . Ed ward L. Cohen has encouraged me throughout the proj ect ; I am grateful to both these personal friends . I am pleased to express my gratitude to Mrs. Wendy M. Coutts , my technical typis t , for her fine work and to the administration of the University of Ottawa for the generous support that I have enj oyed in connection with this and two other book writing pro j ects . My warmest thanks are due to my family . Gabriel Klambauer
v
CONTENTS
v
Preface Chapter 1 ARITHMET I C AN D COMB I NATORI CS Chapter 2 I NEQUAL I T I ES
(115
problems )
problems )
Chapter 3 SEQUENCES AN D SERI ES Chapter 4 REAL FUNCT IONS
L117
(115
(152
1
81
problems )
165
319
problems )
vii
PROBLEMS AND PROPOSITIONS IN ANALYSIS
CHAPTER 1 ARITHMET I C AND COMBI NATO R I C$
PROBLEM 1 . Let and B denote positive integers such that B. Sup pose, moreover, that and B expressed in the decimal system have more than half of their digits on the left-hand side in common. Show that PJA P/B p holds for p 2. Since xP - ly p-1 p-2y yp-1 pyp-1 for y x, we obtain, on setting xp and yP B, A A
A>
< _!_
_
3, 4, ...
=
Sol ution . X
X
-
+
+
X
+
. . .
>
PROBLEM 3. Show that, for n 1,2,3, . . .
,
Since
Sol u tion .
where
n(n - l)(n - k!2)···(n - k + 1) with k! 1 2. 3 (k - 1) k ' and, for 2 k n, (�) --\ k\ ( 1 - *)( 1 - *) . ( 1 - k � 1 ) "2--;;· 3,.=-: -.-k n we have, for n 2, •
�
�
=
.
�
.
�
•
•
�
• • •
3
A R I THM ETIC AND COMBI NATORICS
1{ + -n1 }n < 1 + 1 + -21 + -41 + ••• + 2n-11 < 1 + 1 = 3. --
�
=
PROBLEM 4. For n 3,4,5, . . . , show that < n/i1! < n ; We begin by showing that (n!) 2 nn for n 3,4,5, . . . Consider (n!) 2 [1·n][2(n - 1)][3(n - 2)] ••• [(n - 1)2][n •1]. But the first and the last factors in square brackets are equal and are less than the other factors in square brackets because, for n - k 1 and k we have (k + 1) (n - k) k(n - k) + (n - k) k · 1 + (n - k) n. Thus (n!) 2 nn which is equivalent with In < nln!. To verify that for n 2,3,4, . . . , we first note that n+1 1 + 1 }n+1 2 for n 1, 2, 3, . . . n : {� { by the first inequality in Problem 2. Thus r+1 � n+ l n + � { { n + 1 f\. --2- n + 1 {�f or n+1 for n 1, 2, 3, . . . (n + n1) n+1 < {� } 2 2 We now proceed by induction: if n +2-1}n n! < {-rn
1
Sol ution .
>
=
>
=
>
>
=
"i1+T
>
1
>
> 0
4
CHAPTER 1
holds, then n! (n + 1) _,_(n_+2n--
1
9
ARITHMETI C AND COMBINATORI CS
To evaluate the sum in Part (b) , we observe that
k (m +k!k) k l 1 n(j(j-k)- k)!l1 k 1 J, (k) (-1) (m + k l l1 n (j-k) 1 _ (-l) k (m + k) k l 1 (-l) k (m + k) (m + k - 1) ···(m + 1) (-m - 1) (-m - 2)···(-m - k + 1) (-m - k) (-m l) k l 1 . (-1) k (m�k}(j �k)
=
=
But
(-I)
n
J.
=
_
Thus
and so our sum equals
/! { (�) nj 11 + ({) n (j-1) 11 (-m + • • • + (�) ( -m - 1) j 11 } (n
�
(n -
(n
this equal s n . J
Observe that s 1nce form
1 1 l) 1
+ (�) n (j-2) 11 (-m 1) 211
- mj!- 1) j 11 - m - 1) - mj!- 2) ( -m-1) m - 1 j, . (m+k) (mm+k) (n
If n -
_
k
=
.
_
• • • (n
- m - j)
we can rewrite this identity in the
m - 1)(n - mj!- 2) ···(n - m - j)
PROBLEM
12.
Let
m j and
be pos itive integers and
Show the following results (due to Gauss) :
j m. �
Put
10
CHAPTER 1
(i)
(m, j ) = (m, m- j ) ;
(ii)
(m, j + l)
· (m- l , j + l) + xn- J - 1 (m- l , j ) ;
(iii)
(m, j +l)
(j , j ) + x (j + l , j ) + x 2 ( j +2 , j ) +
( iv) (v)
• • •
+ xm- j - l (m- l , j ) ;
(m, j ) is a polynomial in x; 1 - (m, 1) + (m, 2) - (m, 3) + • • • + ( - 1) m (m,m)
Sol ution .
{
�
1 - x) ( l - X 2 ) • • • ( 1
-
Xm- 1 )
if m is even , if m is odd .
Part (i) is clear from the fact that (m , j ) equals
To obtain Part (ii) we note that (m , j + l)
(m- l , j +l)
1 - xm 1 - xm-j - 1
=
(m- l , j +l)
= (m- l , j + l) + xm- j - 1 (m- l , J. ) .
rl + xm-j - l [
]
1 - xj + l 1 - xm- j - 1
To prove Part (iii) we make use of the result in Part (ii) and get (m, j + l)
(m- l , j +l) + xm- j - 1 (m- l , J. ) ,
(m- l , j + l)
(m- 2 , j +l) + xm-j - 2 (m- 2 , J. ) ,
( j + 2 , j +l )
( j + l , j +l) + x (j +l , j ) ,
(j + 1 ' j + 1 )
(j ' j ) .
Adding these equal ities termwise , we find (m, j +l)
• • •
(j , j ) + x (j + l , j ) +
+ xm-j -1 (m-. l , J. ) .
To verify Part ( iv) we observe that 1 - xm (m, 1) = --y-:-x-
+
• • •
+ Xm- 1
and so (m, l) is a polynomial in x for any pos itive integer m . Assuming that
AR I THMET I C AN D COMB INATORICS
11
(m, j ) is a polynomial in x for k � j , we get by Part (iii) that (m , j +l) is also a polynomial in x and so the claim follows by induct ion . W e introduce the notation f(x,m)
=
1 - (m , 1) + (m, 2) - (m, 3) + • • • + ( - 1 ) m (m,m)
to prove Part (v) . Since 1
1,
(m, 1)
(m- 1 , 1) + Xm- 1
(m, 2)
(m- 1 , 2) + xm- 2 (m- 1 , 1) ,
(m, 3)
(m- 1 , 3) + xm- 3 (m- 1 , 2) ,
(m- 1 ,m- 1 ) + x (m- 1 , m-2) ,
(m, m- 1)
(m- 1 , m- 1) ,
(m, m)
we get , upon multiplying these equalities successively by ± 1 and adding , f(x,m)
=
( 1 - xm- 1 ) - (m- 1 , 1 ) ( 1 - xm- 2 ) + (m- 1 , 2) ( 1 - xm- 3 )
+ ••• +
(-1)
m- 2 (m- 1 ,m- 2) ( 1 - x) .
But (1
xm- 2 ) (m- 1 , 1)
(1
xm- 1 ) (m- 2 , 1) ,
(1
xm- 3 ) (m- 1 , 2)
(1
xm- 1 ) (m- 2 , 2) ,
Therefore f(x,m)
{ 1 - (m- 2 , 1) + (m- 2 , 2) -
(1
xm- 1 )
(1
xm- 1 ) f(x,m- 2) .
Thus f(x , m)
(1
xm- 1 ) f(x,m-2) ,
f(x,m- 2)
(1
xm- 3 ) f (x,m- 4) ,
• • • + ( -l)
m- 2 (m- 2 ,m- 2)
}
CHAPTER 1
12
We first assume that rn is an even number. We get 3 f (x , rn) = ( 1 - xrn- 1 ) (1 - xrn - 3) • • • ( 1 - x ) f(x , 2) . But f(x, 2)
=
1
(2 , 1)
-
+
2 l - X 1 2 - � -
( 2 , 2)
-
This shows that
-
X.
when rn is even . Final ly, when rn is odd , But f(x, l) = 0, consequent ly f(x , rn) = 0 for any odd number rn .
PROBLEM 1 3 .
Show the following result (due t o Euler) : 2 x z) • • • ( 1 + xn z)
+
( 1 + xz) ( l
=
F (n) ,
where
2: n
F (n)
1 +
(1 -
k= 1
n
X ) (1
F (n+1) - F (n)
=
zxn+1 F (n) ,
that is , ( 1 + zxn+ 1 ) F (n) .
Therefore F (n)
(1
+
zxn ) F (n- 1) ,
F (n- 1)
(1
+
zxn- 1 ) F (n- 2) ,
F (3)
-
X
k (k+1) -2- Zk X
A straightforward calculation shows that
Sol ution .
F (n+1)
n- 1 ) • • • (1
X
n-k+1 ) ( 1 - x) ( 1 - i ) • • • ( 1 - xk) -
3 ( 1 + zx ) F (2) ,
•
ARITHMETIC AND COMBINATORI CS
13
F(2) (l + zx2)F(l) , F(l) 1 + zx. However, these equalities imply the desired result F(n) = (1 + xz) (l + x2 z) •••(l + xnz) . In a completely similar way one can show that (1 + xz) (l + x z) .. • (1 + x2n-1 z) n (l x2n) (l _ x2n-2)·· · (l x2n-2k+2) 1 + � (1 - x2)(1 - x4)•••(1 - x2k) k=l Remark .
3
_
_
PROBLEM 14. nLet x and a be positive. Find the largest term in the ex pansion of (x a) , where n is a positive integer. Let the largest term be Tk = (nk) n-k ak . This term must not be less than the two neighbouring terms Tk- l and Tk+l ; thus Tk Tk-l and Tk Tk+l ' Whence and The first of these inequalities yields k (nx++ l)aa and from the second inequality we get k (nx++ l)aa _ l. e assume f'1rst that (nx++ 1)aa 1. s an 1. nteger. Then (nx++ aa - w1 be an integer also, and since k is an integer satisfying (nx++ l)aa 1 k (nx++ l)aa ' it can attain one of the two values +
Sol u tion .
X
�
�
�
�
1)
W
_
�
�
1
.
11
14
CHAPTER 1
k
=
(n + l) a x + a '
(n + 1) a x + a
k
1.
In this case there are two adj acent terms which are equal to each other but exceed all the remaining terms . Now consider the case when (n + 1) a x + a is not an integer . We then have (n + 1) a x + a
[(nx++l)aa] +
6
where 0 < e < 1 and the square brackets denoting the integer part of the number so enclosed; in other words , e denotes the fract ional part of the number (n + l ) a/ (x + a) . In this case the inequalities take the form �
k
]
[C n + 1) a + L x + a
6,
k
�
[Cnx++l)aa] - ( 1 -
e).
I t is clear that in this case there is only one value of k for which our in equalities are satisfied, namely, k =
(Cnx++l)aa] .
Hence, when (n + 1) a/ (x + a) is not an integer, there is only one largest term Tk .
PROBLEM 15 .
Let j and n be positive integers and put
Show that (n + 1) k+l - (n Solution .
n
I
p=l
We get
(p + l) k+l =
n
I
p=l
pk+l
pk-1
+
1) .
AR I THME T I C AND COMB I NATORICS
15
by summing p from 1 to n in the identity
) ( p + l) k+l = pk+l + (k+l) pk + (k+l) pk - 1 + • • • + (k+l k p + 1· 2 1 But n k+l l: (p + l) p=l
n k+l - 1 + (n + 1) k+l l: p p=l
and
(
k+l ) k+l-m
for m
=
1,2, . . . ,k.
Remark . With the help of the recursion formula in Prob lem 1 5 and the el ementary fact that s 1 = n (n + 1 ) / 2 we can easily see that
1 (n+1) ( 2n+l) , s 2 = 6? 1 2 (n+l) 2 s3 = �
= s 21 ,
1 n+1) (2n+1) ( 3n 2 +3n- 1) , s 4 = 3Q? C 1 2 (n+ 1 ) 2 (2n 2 +2n- 1) s s = un '
1 4 s 6 = � (n+ 1) ( 2n+1) ( 3n +6n 3 -3n+1) , 1 2 (n+l) 2 ( 3n 4 +6n 3 -n 2 -4n+2) s7 = � '
s 8 = g1on Cn+ 1) (2n+l) ( Sn 6 +lSn 5 +Sn4 -lSn 3 -n 2 +9n- 3) , 1 n 2 (n+l) 2 ( 2n 6 +6n 5 +n 4 - 8n 3 +n 2 +6n-3) , s 9 =za 1 (n+ 1) (2n+ 1) (3n 8 + 1 2n 7 +8n6 - 18n 5 -lOn 4 +24n 3 +2n 2 s 10 = 66? - 15n+S) .
Prove that i f k and m are positive integers , then
PROBLEM 1 6 . ( a)
m km - k (k - l) m + k (k1·2- l) (k - 2) m + • • • + (- l ) k- 1 k · l = 0
provided that k
>
m; if k
= m,
then
16
CHAPTER 1
(b) mm - m(m - 1) m m(�.; 1) (m - 2) m ••• (-1) m- 1 m m! . We have (x 1 ) m - xm mxm-1 m(m1•2- 1) xm-2 ••• mx Replacing x by x 1, we obtain (x 2) m - (x 1) m m(x 1 ) m-1 m(�.; 1) (x 1) m-2 ••• m(x 1) 1 . Subtracting the preceding equality from the last one, we find +
+
+
Sol ution . +
+
+
+
+ 1.
+
+
+
+
=
+
+
+
+
+
+
Analogously we obtain m(m - 1 )(m - 2)xm-3 p2xm-4 Using the method of mathematical induction, we can prove the following gen eral identity (x k) m - fcx k - 1) m k(� .; 1) (x k - 2) m - (-1) k xm m(m - 1) •••(m - k 1)xm-k pxm-k- 1 · • · , from which it is easy to obtain that for k m (x m) m - T(x m - 1) m ••• (-1) m xm m! . If k m, we get (x k) m - �1(x k - 1) m k(k1•2- 1) (x k - 2) m - ••• (-1) k xm Putting in the last two equalities x we get the required identities. +
+
=
+
+
+
+
=
I···
+
+
+
+
=
+
+
+
+
=
>
+
+
+
+
=
Remarks.
0,
It can be shown that there are exactly
+
=
0.
ARITHMETIC AND COMBINATORICS
17
m- digit numbers made up of and actual ly containing the digits 1 , 2 , 3 , . . . , k . Evidently, the identities in case k m and in case k > m which we proved in Probl em 1 6 are immediate consequences of this combinatorial result . In Prob lem 1 7 we shal l take up a proposition which easily lends itself to prove the comb inatorial result we invoked to interpret the identities discussed in Problem 1 6 . =
PROBLEM 1 7 . Principle of Incl usion and Excl usion . Suppos e that a set of N obj ects and a set of m properties a 1 , a2 , . . . , am are given . Some of the N obj ects may have none of the m properties and some may have one or more of these properties . We use the symbol N (a1. a . . . . ak ) to denote the numJ ber of obj ects which have at l east the properties a1. , a . , . . . , ak (and posJ sibly additional properties) . I f we wish to stress the fact that we are concerned with obj ects which lack a certain property, we prime the correspond ing a . For example , N (a 1 a 2 a4 ) denotes the number of obj ects which have the properties a 1 and a 2 and do not have the property a4 (the question of the remaining properties is left open) . In line with this convention , N (a i a z · · ·a�) denotes the number of obj ects with none of the properties a 1 , a2 , . . . , am . Prove the fol l owing relat ion
(E . 3)
The above sum is taken over all comb inations of the properties a 1 , a2 , , am (without regard to their order) . A summand involving an even number of properties enters with a plus s ign , and a summand involving an odd number of properties enters with a minus sign . Rel ation (E . 3) is refered to as the Principle of Inclusion and Exclus ion . This name reflects the fact that we exclude all obj ects which have at least one of the properties a 1 , a 2 , . . . , am' include all obj ects which have at least two of these properties , exclude all obj ects which have at least three of these properties , and so on . . . •
18
CHAPTER 1
To verify formula (E . 3) we use induction on the number of properties . In the case of a s ingle property a the formula is obvious ly true . Indeed , an obj ect either has this property a or does not have it . Therefore , Sol ution .
N (a ' )
=
N - N (a) .
Suppose formula (E . 3) is true for m - 1 properties , that i s , suppose
(E . 4)
We may use formula (E . 4) with any number of obj ect s . In particular , this formula holds for the s et of N (am) which have property am . If we replace N by N ( am) ' we obtain ••• +
(E . 5)
(to get (E . S) from (E . 4) one takes in each set corresponding to a summand in (E . 4) only those obj ects which have the property am) . Now we subtract (E . S) from (E . 4) . The di fference between the right-hand s ide of (E . S) and (E . 4) is j ust the right-hand s ide of ( E . 3) . The difference between the left hand s ide of (E . S) and the left-hand side of (E . 4) is (E . 6) But N (a i a z · · ·a� _ 1 ) represents the number of obj ects which do not have the properties a 1 , a2 , . . . , am_ 1 and poss ibly have the property am . However , N ( a i a z , . . . a�_ 1 am) represents the number of obj ects which do not have the propert ies a 1 , a2 , . . . , am_ 1 , but definitely have the property a . It follows m that the difference in (E . 6) is simply the number of obj ects which have none
ARITHME T I C AND COMB I NATO R I CS
19
of the properties a 1 , a2 , . . . , am- l ' am . In other words ,
This proves ( E . 3) for the case when the number of properties is m . Having proved the validity of (E . 3) for m = 1 and the fact that the validity of (E . 3) for m - 1 implies its validity for m , we conclude that ( E . 3) is true for any finite number of properties . Remarks . Recal l ing the comments at the end of the Solution of Problem 1 6 , we l ook at the fol l owing questions : How many positive integers of m dig its exist such that each digit is 1 , 2 , and 3? How many of these contain al l three digits 1 , 2 , and 3 at l east once? The answer to the first question is the number of permutations of three m obj ects (where repetitions are allowed) , m at a time , namely, 3 . To answer the second question we let
a 1 signify the absence of 1 ' s a2 signify the absence of 2 ' s a 3 signify the absence of 3 ' s and invoke the Principle of Inclusion and Exc lusion . Then
0; thus ( E . 3) gives
PROBLEM 1 8 . Let [ t ] denote the greatest integer less than or equal to t. If a is a positive integer , then [n/a] is the largest integer a such that aa � n . This definition is equivalent with saying that [n/a] = a, where n = aa + r with 0 � r < a . Thus [6/5] 1 , and [-6/5] = - 2 . Prove that for a and b greater than 0 ,
[[�J] = [:b].
20
and a
=
n
s. Then n
Let [n/a] a and [a/b] Sb + r 2 , 0 � r 2 < b Therefore
Sol u ti on .
=
=
Sab + ar 2 + r 1
and
However , r 2 is at most b at most a (b - 1) + a - 1
PROBLEM 19 . (i)
(iv)
(v)
r1
>
=
0
[n/p s+t ] . and b
>
1, then
[n/ab] .
I f m, n , and a are positive , then [mn/a] � m[n/a] .
If n = n 1 + n 2 + • • • + n t ' where n i ' for i sitive , then
1 , 2 , . . . , t , are po
For any real numbers a and b ,
[ 2a] + [ 2b ] � [a] + [ a + b ] + [b] .
( E . 7)
Parts (i) to (iii) are simple to verify . Part (iv) can be shown as fol lows : Letting [n i /a] = a i ' we have n i a i + ri , 0 � r i < a . Therefore Sol ution .
and [n/a]
a 1 + a 2 + • • • + a t + [ (r 1 + • • • + rt) /a] .
Hence
To see Part (v) we first note that [x + n] [x] + n for n an integer and x arb itrary . Thus both sides of ( E . 7) change by the same quantity if either a =
ARITHMETI C AND COMBINATORI C$
21
or b changes by an integer . It is thus sufficient to prove (E . 7) only for the case 0 s a < 1 , 0 s b < 1 . It then reads as fol lows [2a] + [ 2b ] � [a + b ] .
If [a + b) = 0 we have nothing to prove . If [a + b ] = 1 , then a + b � 1 , and hence at least one of the two numbers , say a , is � 1 / 2 , and thus [2a] + [2b ] � 1. PROBLEM 20 . If p is a positive prime, let Ep (m) denote the exponent of the h ighest power of the prime p that is a divisor of m. Show : If both n and the prime p are positive , the exponent of the highest power of p that divides n! is Ep (n ! )
[n/p]
=
Sol ution .
T
=
+
s [n/p 2 ] + • • • + [n/p ]
with [n/ps + l ]
0.
Consider the set T of integers from 1 to n , that i s ,
{1 , 2 , . . . , p , . . . , 2p , . . . , pk , . . . ,n} .
The last integer of the set that is divis ib le b y p is [n/p]p , and the coef ficient of p shows that there are [n/p] multiples of p in the set . Al l other integers of the set are prime to p. Hence Ep (n .' )
Ep (p• 2p • • • pk • • • [n/p]p) .
=
Now we take out one factor p from each of these multiples of p that are in the set T , thereby obtaining the factor p [n/p] . Therefore E p (n ! )
[n/p] + Ep ( 1 · 2 · · · [n/p]) .
=
But the last integer of the new set { 1 , 2 , . . . , [n/p]} that is a multiple of p is [ [n/p]/p] [n/p 2 ]p. We can , as before , remove the factor =
2 p [n/p ] from the product of the integers of the new set , showing that E p (n ! )
[n/p] + [n/p 2 ] + Ep ( 1 · 2·· · [n/p 2 ]) .
3
4
Likewise, we remove the factors p [n/p J, p [n/p J, . . . until we find that ps s n < p s+l , so that [ n/p s ] # 0 , while [ n/ps+ l ] 0 . Hence al l is proved . Remark .
If n
mk , then Ep { (mk) ! } � mE (k ! ) . P
22
CHAPTER 1
PROBLEM 2 1 . prime p so that
Let the pos itive integer n be written in the s cal e of the
+
• • •
+ a5 .
Show that n - (a0 + a 1 + ---'_:_.._, ---" p - 1
E p (n!) =
--
-
• • •
+ a5 ) ---=-
---
-
where E p (n!) is the exponent of the highest power of the prime p that is a divisor of n!.
+
Sol ution .
0 � ai
�
=
1
�
(1
···
1)
1•2
�
1•2
1)
} �
••• �
1,
>
_
CHAPTER 1
76
In the foregoing we restricted ourselves to the case n 1. If B 1, the inequality nq B(B 1•2- 1) c B becomes nq(B - l)c 2. If we suppose n 1, the last inequality takes on the form q(B - l)c 2 which is always true in the unique case B 2 and n q c 1. This unique case corresponds to the quotient 1•2•3 -4- . Thus the quotient 1•2•3•••(a a - l) (which corresponds to n 1) is a fraction when a is prime, and is an inte ger when a is not prime except in the unique case where a In this latter case we have the fraction 1•2•3>
Remarks . >
�
�
�
=
=
=
=
-4
= 4.
.
PROBLEM 112. Let m and n be positive integers. Show that { e:) (�n) } 2 (m+n n) is an integer. Since { (�) enn) } 2 (2m)! (2m)! (2n) ! (2n) ! m! n! (m+n n ) m! m! m! m! n! n! n! n! (m + n) ! m!(2m)m! ! m!(2m)n! ! n!(2n)n! ! (m(2n) n)! ! ' Sol ution .
+
A R I THME T I C AND COMB I NATORI CS
77
it will be sufficient to show that 2[2m'] 2[2n'] 3[m•] 3[n'] + [m' n'] (E.24) for all values of m', n' between and 1 (inclusively) . (a) For m' 1, n' 1, we have in fact 4 4 3 3 2; (b) For m' 1, n' 1, (E.24) takes on the form 4 2[2n'] 3 1 4, which is clear, since [2n'] is either 0 or 1; (c) For m' 1, n' 1, (E.24) takes on the form 2[2m'] 4 3 1 4, similar to case (b) ; (d) For m' 1, n' 1, we get 2[2m'] 2[2n'] [m' n']. Indeed, the larger of the two quantities 2m', 2n' is, by itself, larger than or equal to the arithmetic mean m' n' . We could also have used the result in Problem 25 to solve Problem 112. PROBLEM 113. Solve in positive integers the equation xY yx 1. Evidently this equation is satisfied for y whatever x is. To obtain the solutions in finite numbers, we note at once that x and y have to differ by little from each other and that their difference has to be an odd number. Suppose first that x y and let x y n. Then �
+
+
+
0
+
+
=
+
=
=
X
(1
_
1,
= �
X
X
3;
=
3,
= 1,
a value which, for n exceeds already the second member, and so much more so for n because the positive term increases with n more rapidly than the negative term. Moreover, for x >
1,
>
while
3,
1 l. Hence there is no solution for y n, and the only solutions in positive integers are y x arbitrary, y y X
X
0,
1,
X
2,
X
=
2,
3.
ARITHMET I C AND COMB I NATO R I CS
79
Solve the equat ion 3x
PROBLEM 114 .
54x - 135 .
We note that 54
Sol ution .
This shows that x = 3 and x = 4 are roots of the given equat ion . But these two roots are the only roots of the given equation s ince the curve y = 3x is strictly above the straight l ine y = 54x - 1 35 when x < 3 and when x > 4 and y = 3x is strict ly below the straight l ine y = 54x - 1 35 for al l x sat isfying the inequal ity 3 < x < 4 .
PROBLEM 1 15 .
Show that
is an integer for every integral value of x . Since
Sol ution .
3x 5
sx 3
+
+
7x
=
3 (x - 2) (x - l) x(x + l) (x + 2) +
+
4 • S (x - l ) x (x
1)
+
lSx,
we need only to use the fact that the product of n consecutive integers is divisible by n , to see that the number (3x5
+
Sx 3
+
7x) I 15 = 51 x 5
+
1 3 3x
+
7 ISx
is an integer.
k
2:
PROBLEM 1 1 6 . 2,
Denote ( (n ! ) ! ) ! by n ( ! ) 3 , etc . , n ( ! ) 0
n . Show that for
k- 2 2 • • -1] ! (n ! ) [n- 1 ] ! [n ! - 1 ] ! [n ( ! ) - l ] ! • [n ( ! ) is an integer . From the Remark to Problem 2 2 we know that the product of any n consecutive integers is divisible by n ! . Now n ( ! ) k is a product of Sol ution .
CHAPTER l
80
n(!) k-1 consecutive integers. We divide these numbers into groups of n con secutive integers. Then we have [n - 1]![n! - 1]![n(!) 2 - 1]!•••[n(!) k-2 - 1]!, k ;:: 2, groups, since n(! /-2 [n( ! ) k-3 - l]![n(!) k-2 - 1] ! n(n - l) ![n! - l]! •••[n (!) k-2 - 1]!. Thus, n(!) k is divisible by (n., ) [n-1]! [n! -1] ! • • • [n ( ! ) k-2 -1 ! . =
J
PROBLEM 117. Sum the series Sol ution .
Let S be the desired sum. Multiplying the identity
by (-1) r- 1r and sum from r 1 to r 2n. the left-hand side we obtain (S - 2n) (2n 2)/(2n 1), and on the right-hand side (2n2n1 1) - 2n - 2n, the binomial .coefficients cancelling in pairs. Solving for S we find that S n/(n 1) . =
+
+
=
On
+
=
+
CHAPTER 2 I NEQUAL I T I E S
PROBLEM 1 . Suppose that a real -valued function g , defined on a nonempty set T of real numbers , satisfies for arbitrary elements t 1 , t 2 , t 1 f t 2 of T the inequal ity
Then the more general inequal ity
holds , where the t 1. ' s are arbitrary elements of T but t 1. f t . for at least J one pair i , j . Sol uti on .
The proof is carried out in two steps . Step 1 . Assume the validity of the claim for n = m and prove its val id ity for n = 2m. We have t 2m- l
m +
log s 1 + + log sn 1 + exp ------------n
I NEQUAL I T I ES
85
or
For b ].. a ].. = s l..
(i = 1 , 2 , . . . ,n)
this is the desired inequality .
PROBLEM 5 . Let a 1 , a2 , . . . , an form for i = 1 , 2 , . . . ,n) . Show that � �
1 n
an
arithmetic progress ion (a i
>
0
n la a • • • a 1 2 n
In particular n +-. r-;-: < r l � n vn vn 2 _
Sol ution .
We have by Prob lem 2
a l + a2 + • • • + an n /a a • • • a � ----n n 1 2 Since a 1 , a 2 , . . . , an form an arithmetic progression , the term on the right hand side of the last inequal ity equals ( a1 + an ) /2 . To prove the rest of the claim , consider
But
(because ak = a 1 + (k- l ) d and hence
and
an-k+l
an - (k- l ) d
86
CHAPTER 2
indeed, in any arithmetic progression , whose common difference is not zero , the product of two terms equidistant from the extreme terms is the greater the closer these terms are to the middle term) . Thus
PROBLEM 6 .
Let a > 1 and n b e a positive integer . Verify that
)
n; l a . Sol ution .
Let a
s2 .
It is required to prove that
or, which is the same , ---,,.---- �
s 2n - 1 s2 - 1
n sn- 1
But 5 2 (n-l) + 5 2 (n- 2) + • • • + 5 2 + 1
=
because 2 + 4 + • • • + (2n-2)
PROBLEM 7 .
Let x i >
0
n (n-1) .
for i = 1 , 2 , . . . ,n. Show that
){_l_
(x l + x2 + • • • + xn x + _l_ + ••• + l n l x2
Sol ution .
}
�
n2 .
By Problem 2 , �
n n ix1 • • • xn '
1 > ... + xn - n
and the result follows . Remarks .
I f we carry out the multipl ication
�
1 ••• 1 X 1 Xn '
I N EQUAL I T I ES
87
. t h e sum o f th e fol lowLng . n 2 terms : we ob taLn 1, xl x2 '
-
X
-
x2 xl '
-
1,
-
n xl
'
X
n x2 '
...
1.
'
But (xi /xk ) + (xk/x i ) � 2, since t val idity of the claim in Problem
7.
+
1/t � 2 for t > 0. This again shows the
An interesting analogue from integrat ion is the fol lowing result : Let f be a cont inuous strict ly pos it ive valued function on the interval [a,b] . 2 1Then I = Jtba f(x) dx .Jtba f(x) dx � (b - a) . Indeed, I where
=
f(x) dx dy jjr.J f(y) dx dy /.1 f(y) ' f(x)
S
=
S
S is the square [a,b]x[a,b] . Therefore 1 �r [ f C x) f C y) ] dx dy f.T f2 C x) + f2 C y) dx dy � I zJ� f(y) f(x) S 2 f(x) f(y) _ -
+
_ -
dx dy
because of the trivial inequal ity 2AB � A2 + B 2 . Hence I � (b - a) 2 .
8.
PROBLEM Let f be a real-valued function defined on an interval (a,b) . Then f is said to be convex if for each x 1 , x 2 in (a,b) we have (E . 1 ) regardless of how the posit ive numbers q 1 and q2 = 1 - q 1 are chosen . The funct ion f is said to be concave if the inequality in (E . l ) is reversed . Prove the fol l owing Inequali t y of Jensen : I f f is a convex funct ion on (a,b) , then
88
CHAPTE R 2
(q l ' · · · ' qn >
O; ql + • • • + �
=
(E . 2)
l)
holds for any points x 1 , . . . , xn of the interval (a,b) .
We note that in case n = 2 , we are back to the definition of convexity . We therefore assume that the inequal ity in question is true for n � 2 and show that it wil l also be true for n + 1 . In other words , we pick n + 1 points in (a,b) , namely , x 1 , . . . , xn , xn+ l ' and we sel ect n + 1 positive numbers q 1 , . . . , qn ' �+ l such that q 1 + • • • + � + �+l = 1 , and we seek to establish that Sol ution .
(E . 3)
+ ••• +
To this end we replace in the left-hand side of the above inequality the sum �xn + �+l xn+l by the sum
In this way we can use inequal ity (E . 2) and see that the expression on the left-hand s ide of (E . 3) is smaller than or equal to
We now only have to apply to the values of the function in the last expres sion the basic inequal ity (E . l) in order to obtain (E . 3) . Hence (E . 2) is proved completely .
PROBLEM 9 .
show that
Sol ution .
have
I f x i > 0 , q i > 0 for i
1 , 2 , . . . , n , and q 1 + • • • + qn
Since x i > 0 for all i we may set y i
1,
log xi . Then we wil l
I N EQUAL I T I E S
89
But f(t) et is convex on the entire real l ine and we may appeal to Problem 8 to write
where the summat ion on i is from 1 to n .
PROBLEM 1 0 .
a,
S,
be positive and
Let
... ... a
'
'
+ + S
a,
+ a = 1 . Show that 52a + • • • +
where A =
n
L a , B i=l i
Sol ution .
n 2: b , i=l i
From Problem
(a;rtBi t • • • Cs9a
and so
n 2: i=l
:s;
9 we see that
a
a (a;rC:/ • • • Cs9
n 2: s . . i=l l
s
A+ a l.
:s;
s T
b l.
+
• • •
+ a -ss. l
+ b . + • • • + a -ss. a + + ••• + a 1 n 2: i=l
a l.
Cl -
A
S
s T l
l
90
CHAPTE R 2
Thus l AaB""" s___s _a _
PROBLEM 1 1 .
where a 1. . J
>
J l (a�b �
• •
+
1 , 2 , . . . ,n. Moreover , it is given that
0 for
akn
=
1
1,
for k = 1 , 2 , . . . , n . Prove that
Since log t is concave and a il
Sol ution .
log y i
2:
+
a il log x 1
a i2 log x2
+
...
+
+
a i2
+
...
+
a 1n .
1'
a 1n . log Xn
1 , 2 , . . . ,n. Hence
for i
n l: log y i i=l
2:
PROBLEM 1 2 . that xy
< _
k
1 xk
n ( log x 1. ) l: a il i=l
Let x
1 + k'
>
0,
y >
+
...
0, k
>
+
n . ( log xn ) l: a 1n i=l
1, k'
>
1 , and 1 /k
n l: log i=l
+
1/k '
X. 1
•
1 . Show
yk '
Sol ution . In Probl em 9 consider the special case n 1/k, q2 = 1/k ' , x 1 = xk , and x 2 yk '
2 . Then put q 1
=
PROBLEM 1 3 . pl
+
Show that
p2
Let x i
+ ••• +
>
0 and y i
Pn = P .
>
0 for i
1 , 2 , . . . , n . Moreover , put
I NEQUAL I T I E S
91
0� 1 • • • x�nr /P pl xl + + pnxn In particular nix1x2 • • • xn xl + x2 +n • • • + n p2 1. In Problem take qi = p i/P and all is clear. Moreover, note that the last claim has already been established in Problem 2. : 1, and 1/k + 1/k' = 1. Show that n a . b. ( n ak.) l/k( n bk'. ) l/k' (E. 4) i=l i=l 1 i=l For the special case when k k ' 2, we get the We first assume that nI a.k nI b.k' 1 (E.S) i=l i=l and observe that the inequality to be proved will be of the form nI a . b. 1. i=l In Problem 12 we put successively x = ai ' y = b i for i = 1,2, . . . ,n and then add up all inequalities obtained in this way. By (E.S) we get what we have set out to do. The general case can be reduced to the foregoing special case if we take in place of the numbers ai ' bi the numbers b. a! a. and b! b.k' ) 1/k' for which condition (E.S) is fulfilled. By what has been shown Lni=l aib i 1 holds and this is equivalent to (E.4) . >
0
>
0
>
L:
1 .1
L:
0 we have t + t � 2 because (t - 1) � 0 . Setting x for k = 1 , 2 , . . . , n and x > 0 , we therefore get Sol ution .
t
10
b
k
X
n 1 � = n k -k k X 1 + (x + X ) k=1 k=O
I
PROBLEM 43.
I
::;
l+2n
Let a,b > 0, a + b
1
1 , and
q
> 0 . Show that
I NEQUAL IT I E S
113
The funct ion
Sol ution .
f(x)
=
q (x + .!. X)
for q
0
>
is convex for 0 < x < 1 because x -2 ) 2 + 2qx -3 (x + x - 1 ) q- 1
1) (x + x - 1 ) q- 2 (1
q(q
f" (x)
+ X -4 for q
>
1 + 4x 2
>
0
0 and 0 < x < 1 . Consequently, for a,b > 0 and a + b
f (a)
; f(b)
PROBLEM 44 . Show that
�
; b)
f (a
Let x , y
>
1,
f(�)
F
0 with x
y and m and n be pos itive integers .
Xmyn + Xnym < Xm+n + ym+n Consider xm+n - xmyn - xn ym + ym+n
Sol ution .
Xm
ym
xn
yn
and
=
=
(xm - ym) (xn - yn) . But
(x - y) (xm- 1 + xm-2 y +
•••
+ xym-2 + ym- 1 )
(x - y) (xn- 1 + xn-2y +
•••
+ xyn-2 + yn- 1 ) .
PROBLEM 4 5 .
Let x
0 but x
>
F
1 and n be a pos it ive integer . Show that
l.
x2n- l + x < x2n + Note in particul ar that Xn- 1 Sol ution .
1 +Xn
for x ' 1 , x
Since (x2n- l - l ) (x - 1)
we have that 2n- l + 1 x2n - x - x
>
0.
>
0 , and n a posit ive integer. X 2n- 2 + X 2n-3 +
•••
+ X + 1
>
0,
114
CHAPTER 2
Let a > b > 0 and n be a posit ive integer larger than 1 .
PROBLEM 4 6 . Show that
Let , for n > 1 , x � 1 ,
Sol ution .
x l/n - (x - 1 ) 1/n .
f(x)
Then , for x > 1 , x l/n-1
nf' (x)
_
(x
thus f (x) decreases for x > 1 . Since f(l} = 1 and f(x) < 1 for X
1 /n
-
Letting x
(�y /n
1
1.
a/b and not ing that a/b > 1 , we get - 1
PROBLEM 4 7 .
0 and a � b . Show that
(� : �) b+x > (�) x .
Sol ution .
f(x) = Then
Let , for x � 0 ,
(� : �) b+x .
b - a + log a+ x) f(x) . f ' (x) = (-a + X X b +-
The sign of the derivative is the same as the sign of the funct ion b - a + log "i)+X· a + x g (x) = -a + x
Since g' (x) = g (x)
>
(a - b) 2 (a + x) 2 (b + x)
48.
>
0
0,
Since (x - y)(xn-1 + xn-2y + + xyn-2 + yn-1 ) n yn and putting x �an + kn , y 1Vbn + kn we get, since x � a and y � b, k � and a b Sol ution .
X
• • •
=
=
0,
>
>
0,
But an-l + an-2b + + abn-2 + bn-l is positive and so the last inequality yields • • •
=
If n 2 and a," b, and k are arbitrary real numbers, then
Remark .
holds. PROBLEM Compare the magnitudes of v'n+ 1 and ( vn + 1) . We note that log > 10 g ( vn) v'n +l 49.
( Iii)
� rn
Sol ution .
and
( vn)
But
v'n+l
> ( v'il"+"T) rn
++
< ( v'il"+"T) rn
++
f(x) log x =
X
rn
rn
rn < rn
log
v'il"+"T
v'il"+"T
log
v'i1+l
v'il"+"T
116
CHAPTER 2
is increasing for 0 and decreasing for (vn)
and
1>
C li1+1)
PROBLEM 50. X
X
log
�
17
( li1+1) rn
ln+ l
e . Moreover < e < Hence for n
li1
= 1 , 2 , 3, 4 , 5 , 6
for n � 7 .
Using the elementary inequal ity
X
-
x>
for
1
0,
show that n
n
:E p l og p . � :E p . log q i i=l i i=l l.
for p i
>
0 , ql..
l.
>
0 (i = 1 , 2 , . . ,n) and .
:E q . i=l i
Sol ution .
(since q i
l.
n
n
:E p .
i=l
>
Since p i /q i
>
0 , we get that
0) . Summing over i , we obtain
n n p. � � p . log � � (pi - qi ) . q i i=l i=l l. But n � q. i=l l. and so p. n � p . log � qi � 0 i=l l. or
18.
I NEQUAL I T I ES
117
n (p log p - p log q ) i=l i i i i or n p log P nL p log q . i=l i i i=l i i There is equality if and only if pi qi (i l, . . . ,n) . L
� 0
L
�
Remark .
PROBLEM 51. Show that if a1 a2 a3 k k I: a. I: b. for k 1,2, . . . ,n, i=l i=l �
then
�
1
�
�
�
an and � 0
1
(E. 15)
n a.2 I:n b� . (E .16) i=l i=l After multiplication by ak - ak+l ' (E.l5) becomes for k 1, . . . ,n, (E .17) where an+l = Summing both sides of (E.l7) from k 1 to k n, we get nI a.2 n a. b . (E .18) i=l i=l By the Cauchy-Schwarz Inequality (see Problem 14), (E.l8) yields (i=lnI a.2)2 (i=lnI a.b. )2 (i=lnI a.z)(i=lnI b2.) . I:
�
1
1
Sol ution .
0.
� L
1
1
�
1
1 .
1
1
�
1
1
PROBLEM 5 2. For a positive integer n, let P(n) be the proposition P(n) : f(n) i + �n - + �. + v2 + < + 1 1_
••
If
rn
CHAPTER 2
118
Verify this propos it ion . Since f(l)
Sol ution .
1 < 2 , P (l) is true . Suppose now that P (n) is
val id . Then v'n + 1 + f(n)
f (n + 1)
!:, vn + 1 +
0 , it follows that g (x) > 0 for all positive x # 1 . Since f ' (x) has the same sign as g (x) , we see that f(x) is strict ly increas ing. The verification is complete when we note that f(O) n - m < 0 and f ( l) 0 . =
=
=
=
=
I N EQUAL I T I ES
>
Assuming that m n are posit ive real numbers and x nonnegative , we have shown that Remarks .
I
119
1 is
(m - n) - (m + n) xn + (m + n) xm - (m - n) xm+n
1 . It is easy to see that this
and 2m( l - xm+n ) 1 - xm
--"-----�
> (m + n) (1 +
xn ) .
Along the same l ines of reasoning we can establish that (m - n) + (m + n)xn - (m + n)xm - (m - n) xm+n
1 whenever m > n are positive
> (m + n) ( 1 + xm)
and
must hold.
PROBLEM 54 . show that
I f a 1.
al a2 ----- + + 1 + x 1 1 + x2
;::
. . .
0,
l: 1. a 1.
an __ + _ 1 + Xn �
1 , and
1 + xl
0
al
�
X. 1
x2
1 a2
� 1 for i
X
n
1 ,2 , . . . ,n,
an
>
Sol ution . Assume without loss of general ity that a i 0 for proposed inequal ity fol lows from Jensen ' s Inequal ity (see Prob lem is convex on an interval I , then for al l y i in I , l: ai f(y i ) � f (l:
all i . The 8) : I f f ai y i ) with
CHAPTER 2
1 20
equal ity if and only if y 1 inequal ity , let and
Yn · To apply this to the proposed
Yz
f (y) =
1
ey
---
1
>
+
,
assuming for the moment that xi 0 for all i . Since that - oo y i $ 0 and since , for y < 0 ,
1/4 for al l x and y in the unit in then , in part icular,
g (l) I < 1/4 ,
l f (O)
g (O) I < 1/4,
+
and
l f ( l)
+
g (O) I < 1 /4 .
But then by the triangle inequality 1 1 - f(l) - g ( l) I �
11-
�
>
PROBLEM 62 .
l f (l)
- l g ( l)
1
+
+
+
g(l) I g (O) I - 1 - g (O) - f (O) I f (O) I
Lf ·
Let t > - a
l f ( l)
if
0. 0
Show that < a < 1
(E . 19)
and t a - at
�
Sol ution .
t yiel ds f' (t) f ' (t) > O < 0
0
1 - a
if a > 1 .
(E . 20)
Different iating the funct ion f (t) = t a - at with respect to a (t a 1 - 1) . Clearly, if 0 < a < 1 , then for 0 < t < 1 , for t > 1 , for t 1 ,
I NEQUAL I T I ES
1 25
and f assumes its largest value at t
M
> 0,
i
1,2,
• . .
i,j
� an and , by homogeneity ,
)
n 2 L a 1. i=l
(E .
,n- l . Then 1 ,2, . . . ,n
and (a.1 - a . ) 2 L J l �i<j �n Inserting this in (E . n n < n L a.2 i=l 1 2 or � ni=l a i
>
35)
>
M2 •
n.
we get
(I ) i=l
2 (i - j ) L l�i<j �n
a 1.
2
n 2 � n L ai , i=l
1 , a contradiction to our normal ization . Hence
as asserted.
PROBLEM 82 .
Let
x
and y be non-zero real numbers . Show that
35)
1 41
I N EQUAL I T I ES
lx - y l
�
ic l x l
Sol ution .
lx i
+ lyl)
1 1�1
- lfl-1 ·
We have
i J¥ - lfl- 1 � l x l l fxr - -&rl
l � lx - yl + l ( y
+ lxl
l k - lfr-1
(E . 36) x l )y l � l x - y l + l l y l - lxl l
1; 1
� 2 lx - yl . Similarly by adding and subtracting x/ l y l we obtain (E . 37) The desired result fol l ows from (E . 36) and (E . 37) .
PROBLEM 8 3 . Show that
Sol ution .
Let x and a be real numbers and n be a nonnegative integer .
(x + na) 2 that
We have for A = ( x - a) 2 and B
l/ (n+l ) < nA + B = x2 + na2 An/ (n+l ) B - n + 1 from Prob lem 2 .
PROBLEM 84 . Given an arbitrary finite set of n pairs of positive num bers { (a i ,b ) : i = 1 , 2 , . . . , n } , show that i TT [xa 1. + ( 1 - x) b 1. ] � max
n
i=l
for all x if
E
� Tr 1
i=l
a. , 1
TI
i=l
b.
1
t �
[0 , 1 ] , with equal ity attained only at x = 0 or x
1 , if and only
CHAPT E R 2
1 42
Let
Sol ution .
TI [xa
n
f (x)
i=l
i + ( 1 - x) b ) ,
g (x)
log f (x) .
Then g ' (x) =
n
a . - b 1. 1 + ( 1 - x) b i ' xa i=l i 'i' L
0
g" (x) =
0, 0.
- 1=1 ) n
(
a. - b.
xa 1.
/ ( 1 - \) b 1.
)2
·
Since g" (x) < for all x E [ 1 ] , the maximum of g (x) [hence of f (x) ] is attained at x or x = 1 if and only if g ' (O) and g ' ( l) do not differ in sign , that is , g ' ( O) g ' ( l) � Since TI a ' i
n
f (O)
g'
(0)
0
i=l =
f ( l)
n a. - b . I 1 a. 1 i=l 1
TI b i ,
n
i=l and
g ' (1)
-
n a. b 1. 1 I b 1. i=l
the assertion is proved .
PROBLEM 85 . Show that if m and n are positive integers , the smaller of the numbers n rrn and m ,;n cannot exceed 3 13 . Let f (x) = x 1/x . E lementary calculus shows that f (x) + 0 as x + O , f (x} + l as x + oo , f is increasing in [O , e ] , and f is decreasing in [ e , oo) . Hence Sol ution .
C
= sup { f (k) : k = 1 , 2 , 3 , . . . } = max{f(2) , f ( 3) } .
C
1 Since 3 2 > 2 3 , f ( 3) > f ( 2 ) , whence = 3 1/3 . Then f (m) s 3 1 3 for all pos i. 1 /n ,n 1 /m } tive integers m. I f m s n , then m 1/n s m 1/m s 3 1 / 3 , whence m1n{m 1/3 . s 3 x l /x PROBLEM 86 . Show that if a � 2 and x > 0 , then a + a equal ity holding if, and only if, a = 2 and x = 1 . Sol ution .
Let f (x , a)
(ax + a 1/x) /ax+l/x for a
�
s
2 and x
ax+l/x ,
>
0.
We
I N EQUAL I T I ES
1 43
observe that f (x , a) = f(l /x, a) and hence the problem wi ll b e solved if we show that f(x, a) � 1 for x � 1 with equal ity if and only if a 2 and x = 1 . Next we note that f(x, a) is a strictly decreasing function of a for each x and, hence, that it is enough to show that f(x, 2) < 1 for x > 1 , it being cl ear that f(l , 2) = 1 . We will obtain this latter inequal ity by verifying that for x > 1 . It is easily s een that F ' (x)
x l/x (log 2) g ( l/x) . l/x 2 2 x- l/x ( log 2) ( 1 + l/x - 2 ) = 2 -
=
The proof will be complete if we can show that g (l/x) < g" (t) = 2 - 2 t ( log 2) 2 > 2 - 2 ( log 2) 2 >
0
for x > 1 . Now
0
for 0 < t < 1 , so that g (t) is strictly convex for these t . Since g (t) is continuous and g (O) g ( l) = 0 , it fol l ows that g (t) < 0 for 0 < t < 1 or that g ( l /x) < 0 for x > 1 .
Show that if x i � 2 1 � 1. � 1 , then L� =l PROBLEM 8 7 .
sol ution .
and ( 1 2
+
x1)
-x 1
-1
+ 2
-X .
0
for i
· 1/ (1 + x ) 1 , 2 , . . . ,n and L ni=l i
The result is obvious for n = 1 ; while for n = 2 , x 1 , x 2 � - 1 � 1 imply x x � 1 and x , x > 0 . But then + (1 + x ) 1 2 1 2 2
-x2
and the latter is no greater than 1 by Problem 86 . The proof proceeds by induction . L ni=l ( 1 + xi ) - 1 (l + z) - 1 +
n
I
i=3
( 1 + x i) -
1
�
�
1 implies that
1,
-1 where z i s defined by ( 1 + z) - 1 = (1 + x 1 ) - 1 + ( 1 + x2 ) z � 0 . By the induct ive assumption , 2-z +
n
I
i=3
2
- x1.
�
1.
It remains to show that
�
1 , and hence
0
1 44
CHAPTER 2
2 -x1 + 2 -x2 � 2 -z or 2 -y1 + 2 -y2 � 1, where yi = x . - z with i = 1, 2. This, however,-1 follows from-1the case for n = 2 upon verifying that y 1 ,y2 0 and (1 + y 1) + (1 + y2) � 1. PROBLEM If x and y are positive, show that xy + yx 1. The inequality is trivially true if either x or y � 1. Let 0 x, y 1, and put y = kx. Because of the symmetry we consider only 0 k � 1. Now f(x) = xkx + (kx) = (x ) k + k x ak + ka, min(xx) and kx k. But F(k) ak + ka has a unique where a = exp(-1/e) minimum at k0 = 1 - e 0 and is increasing for k k0 . Since F(O) 1, = 2a 1, f(x) 1. PROBLEM Let 0 ai 1, i l, . . . ,n and put �ni=l ai Show that n a. i=l - a. - n equality occuring only if all the ai are equal. b./b. + b./b . . Then Let b. 1 - ai ' B B . b � b.b+ b.� 2, nI B - n + n-1I nI B . . n + 2 n-1I (n - i) n2 ' i=l i=l i=l j=i+l whence, after dividing by B, nI (...!.._ l) - n (n - B) ' B i=l b. which is the required inequality for ai 1. Equality holds only if every Bij 2, only if all the bi are equal, only if all the ai are equal. 1
�
88.
Sol ution .
0 and x 1. Show the following inequalities due to Karamata: (i) xlog- x1 (E. 39) (ii) Xlog- x1 + xX1/3l/3 ' (E. 40) We start with the expansion log l1 +- tt 2 (t + -t31 3 + -t51 5 + -t71 7 + ) ' which holds if - 1 t 1 . To prove Part (i) , we put
>
y
PROBLEM 2x 2+ 1
>
�
97.
Let x
0
0.
for x > we see that g is an increasing function. Since, moreover, limx -r oo g(x) we obtain that g(x) for x > If in the second inequality of (E.42) , for x > 1, we replace x by 1/(x - 1), and for x 1 we replace x by x/(1 - x) , we obtain (E.39) of Problem Replacing 1 + (1/x) by x/y with x > y > in the first inequality of (E.42), we get (E.41) of Problem --------=-2 < 0
0,
0
0,
0.
(
0,
0
0,
0. Show that 1 - e -x-y The foregoing inequality is equivalent with __:1;:__ 1 12 ' 1 - e -x_ + 1 _ e -y - 1 -< ..!.x + ..!.y + � that is, f(x) + f(y) � 1, where f(x) 1 - 1 e-x x1 X Since f(x) 1/2, the desired result follows. Sol ution .
_
�
n·
CHAPTER 2
1 52
PROBLEM 101 . satisfy A + B � E
Let A, B , C , D, E , and F be nonnegative real numbers which and
(E. 42)
C + D � F.
Show that Since one may interchange C and D in (E . 42) , another valid inequal ity is
Multiplying the inequalities in (E . 42) , we obtain AC + BD + x (AD + BC) � EF. But 2 (AD • BC) 2 � AD + BC (by Prob lem 2) , and so Sol ution .
(v'AC +
/iio) 2
PROBLEM 102 . O �
AC + BD
=
+
UAD • BC � AC + BD + (AD + BC) � EF .
Show that , for x
log X < .!_ . - 2 x2 - 1
>
0 and x
#
1,
X
Consider the function
Sol ution .
x2 - 1 2 log x - x-·
f(x) Since f' (x)
=
2
X
-
-
1 -
� X
=
-
(�) 2 � 0 '
the function f is decreasing and, consequently, x2 - 1 2 l og x - -x-
1,
>
0
for 0
3 sin x.
Since 2
form X
f(x) But f ' (x)
- 2
+
(E . 43)
> 0 for al l x, we may write (E . 43) in the
cos x
> O.
3x
+ COS X
(1 - cos x) 2 2 + COS X
=
> 0,
Show that , for x
�
0
and so f(x) is seen to increase as x increases . Moreover , f(O) the desired result .
PROBLEM 104 . 2 s in x
+
Sol ution .
0. Hence
Show that , for 0 < x < n/2 ,
tan x
> 3x.
Since 2x 3 - 3x2 X
for < 0
+
=
1
(x - 1) 2 (2x
> -Y, and
X
+
r 1,
for x < -Y,.
1) , we see that (E . 44)
As 0 < cos t < 1 for 0 < t < n/2 , we get by (E . 44) : 2 cos 3 t - 3 cos 2 t
+
>0
1
for 0 < t < n/2 ,
or +
2 cos t
sec 2 t - 3
>
for 0 < t < n/2 .
0
(E . 45)
Integrating the inequal ity (E . 45) between the l imits 0 and x we obtain 2 s in x + tan x - 3x
>0
for 0 < x < n/2 ,
which is the desired result . The result in Prob lem 104 i s due to Huygens . Another way of obtaining the result in Problem 104 is to set Remarks .
f(t)
2 sin t
+
tan t
3t
and to note that , for 0 < t < n/2 ,
for 0 < t < n/ 2
CHAPTER 2
1 54
f' (t)
2 cos t + sec 2 t - 3 (sec 2 t) (cos t - 1) 2 (2 cos t + 1) > 0 .
Integration of f' over the interval (O , x) , where 0 < x < rr/ 2 , gives the de s ired result . In an ent ire ly s imilar manner we can show that 2 s inh u + tanh u > 3u
for u > 0 .
In Problem 9 5 , Part ( ii) , we proved that log x < 1 + x 1/3 X - 1 - X + X l/3
for x > 0 and x � 1 .
Putting x = e 3u , the foregoing inequal ity becomes 3u e 3u - 1
--=....:.:._ :....: _ ,:;
1 + eu 3u e + eu
(E . 46)
which is invariant under the transformat ion u (E . 46) can b e reworked to 2 s inh u + tanh u � 3u.
PROBLEM 105 . Show that
+ -u. Assuming that u > 0 ,
Let x > 0 , x � 1 and suppose that n is a pos it ive integer.
1 > 2n X - 1 . X + n xn - 1 X ---
Sol ution .
The inequal ity in quest ion trans forms into
n + l + l ) ( xn - 1 _,(�x____ �"---"-,:-__-'--) > 2n xn . X - 1
(E . 4 7)
The identity ( Xn+l + 1) (xn - 1) X - 1
n-2 + n 1 (xn+l + l) (x - + x (xk + x-k )
together with
• • •
+ l)
1 55
I N EQUAL I T I E S
impl ies (E . 4 7) .
PROBLEM 1 0 6 . Let a be a fixed real number such that 0 � a < 1 and k be a positive integer sat isfying the condition k > ( 3 + a) / ( 1 - a) . Show that • • •
+
for any pos itive integer n . Since, for a , b
Sol ution .
a
1
+
>
0 and a � b ,
4 a+b'
1
b"
we get
• • •
+
+
(--1 nk - 1
n (k - l ) n (k
>
k
+
1) n
+ -
4 1) - 1
4 (k - 1) + 1 - 1/n
n (k - 1) . k + 1
>
The stated inequal ity holds for every pos it ive integer n if 2 (k - 1) k + 1
>
1
PROBLEM 107 .
+
a,
for 0 � a
that i s ,
>
1.
1) is a convex funct ion , we obtain
CHAPTER 2
1 56
a + b r l 2 l
s
1 r r 2C I a I + I b I ) ,
whence
For r 1 , the inequal ity (E . 48) becomes I a + b l s l a l + l b l . Now , let 0 s r < 1 . I f a and b have opposite signs , the result is evi dently true . Otherwise, let t b/a with a F 0. Then (E. 48) becomes =
=
(0 For r
s
r
a t a a a 1 1 t ae
(t
1 + na if n � 2 and a > 0 or 0 > a > - 1 . Thus
,
,
K - x"- 1 p + m xmp- 1
We also see that
I"
>
1 +
K - xm- 1 p_� ---"' m xmp-l
2 v'ri'"+1
-
2
(see Problem 26 of Chapter 2) . Hence xn > 2 clil+T - vn) - 2 > - 2 . But any decreasing sequence which is bounded below is convergent .
PROBLEM 6 .
Show that
15 = 1 + 1 + 1 1 3 --2 3 3• 7 3 • 7 • 4 7 + 3 • 7 • 4 7 • 2207 + • • • (each factor in the denominator is equal t o the square of the preceding fac tor diminished by 2) . Sol ution .
Yo 2
A
---
We put
1 + 1 + 1 =--- + Yo Yo Y l Yo Y l Y2
where y0 = 3 , y 1
--
7 , y 2 = 4 7 , . . . , and, in general ,
and determine A . Noting that 1 1 =y2y 3 + y 2 + -.
..
'
CHAPTER 3
1 70
and , l etting n -+ "" , we see that A
From yn
lim n -+ "" Yo Y 1 2 Yn- 1
Yn Yn-1
2 we get
This in turn yields
and so
Remark . A more interesting way of solving the prob lem is to observe that yn can be represented by the expression
+ 1n ' x2
where x is any of the roots A = (3 + 15) /2 or B = 0 . Thus 1 + x4 , -2- Y 2
X
and the series
(3 - IS) /2 of x 2 - 3x + 1
1 +x8 -4 , X
1 + 1 + 1 + Yo Y o Y l Y o Y l Y 2 --
---
to be evaluated becomes 8 + ____"""2�-_....::.:.._--;4:--___-::8:- + (l + x ) (1 : x ) (l + x )
•• • J .
But the series inside the square brackets tends to 1 or x 2 , according as x = A or x = B ; this is clear from the fact that (see Problem 106 of Chapter 1)
1 71
SEQUENCES AND SERI ES
a b c � + (1 + a) ( 1 + b) + ( 1 + a) (1 + b) ( l + c)
.....,.--; 1 + b).,-:- 7=k(1 �-=----;+ • • • + -:=+ a) (�---(1 + c) · · · (1 + k)�
= 1 - (1 and A
+
1 and 0
>
1 a) ( l + b) ( l + c) • • • (1
x2k-l and a" nlim-+ «> x2k exist. We now show that a' a". In xn+l = c/2 + xn2 /2 we let n -+ we first let n «> with n even, and then with n odd, and we obtain a' c/2 + (a") 2 /2, a"
- 3 , and so a' = a" must hold . Indeed , if we set a" = a ' - 2 in a" = �2
(a ')2 2
+
'
we would get (a ' ) 2
+
2a'
+
(4
+
c)
0
which has no real root for c > - 3 . For c = - 3 both factors ( a ' - a") and ( a ' + a" + 2 ) vanish because then a ' a" = - 1 . Thus in all cases a ' = a" . I f we denote this common value by a we see that a = 1 - � because the limit of the negative-valued sequence (xn ) � =l cannot be (strictly) positive . Final ly, for c < - 3 the sewuence (xn ) �= l does not converge ; for example , for c = -4 we obtain the sequence - 2 , 0 , -2 , 0 , - 2 , 0 , . . . and this sequence has no limit .
PROBLEM 2 0 . Setting
ax with a > 0 .
Let x b e real and consider the equat ion x
investigate the l imit L (x) We observe that it is sufficient to suppose x to be pos it ive ; if x < 0 , we may take x 1 = ax in place of x. First we show that l imn + oo xn = oo if a > e l/e = 1 . 44466 . . Indeed , since log x � x/e , where e is the base of the natural logarithms , we have for the considered case log x � x log a or x < ax ; that is , x < x 1 . Thus we see that Sol ution .
.
SEQUENCES AND S E R I ES
1 87
We now show that the difference
is larger than a certain pos it ive number . We cons ider the function g (x) = ax - x and form its derivative g ' (x) = ax log a - 1 .
>
We see that , if a e , the derivative is posit ive and the funct ion g has its least value at x = 0 ; but we have g (O) = 1 , and it fol l ows from this that g (x) l . I f e l/e < a < e , the derivative vanishes for
>
x
=
log log a l og a
_
and we obtain ax
>
_
x
log log a > 1 + log a
Denoting by A the positive number ( 1 + log log a) / ( log a) , we have that g (x) A, that is , all differences xk + l - xk are larger than A. We thus see that xn becomes arbitrarily large as n oo . We now take up the case 1 < a < e 1/ e The equation ax = x has two real roots in this cas e , one being between 1 and e and the other between e and oo We denote the first root by a and the second by S . The funct ion x 1 - x = aX - x = g (x) is positive for x = 0 and x = oo ; it only remains to study its values for al l x 0 . We form the derivative g ' (x) X = a log a - 1 and denote its posit ive root by x0 ; we evidently get that
+
>
1 log a· As 1 < a < e l/e , we have 0 < log a < 1/e, and so 1/ ( log a) or
x0 l og a
> l.
>
> e; we have
Thus the function g decreas es in the interval from 0 t o x0 attaining its minimal value at x0 and then increases for al l x x0 . By subst ituting x0 for x in g (x) , we get
CHAPTER 3
1 88
1 - x0 log a log a
--��------
< 0;
thus the equat ion g (x) 0 has two s imple roots , one between 0 and x0 and the other between x 0 and oo I t is easy to see that the number e is between the real roots of the equation g (x) = 0 . Indeed , g (e)
=
ae - e
= (a - l) S - o < o ( log S - 1)
0
o
because a s - S = 0 and (a 0 - a0 ) /o = at (log a) for some t such 0 < t < o by the Mean Value Theorem of differential calculus ; but log a = (log S) /S and at 1 in the considered case. Thus we see that
>
xk+ l - xk
>
> o ( log
s -
1) , oo for x
and so, however smal l o is , we have l imn -+ oo xn As a 1 , it fol lows that
n 1 + k 1 + 1 , because P , k + pn +k +1 = Pn , k + 1 0) , Pn ,k + Pn n1 1 1 1 1 1 1 1 2 = a , that is , a is again representable by an infinite subseries . I f this process never terminates (that i s , if the first case always occurs) , then
PROBLEM 32 .
Find the series p 1 + p 2 + • • • + pn + • • • such that n = 1,2,3, . . .
and note that in this case each a mentioned in Problem 3 1 can only be repre sented by infinite subseries . From p n
Sol ution .
Pn+2 + Pn+3 +
Pn+l
2 Pn+ l " Hence
we get pn p1
=
Pn+l + Pn+2 + Pn+3 + • • • and
1
2'
1 p 2 = 4'
1 pn = 2n
The representation by non-terminating dyadic fract ions is unique .
I f the series a 1 + a 2 + a 3 + • • • is absolutely convergent and every subseries
PROBLEM 33.
SEQUE N CES AN D S E R I E S
205
k = 1,2,3, . . . • • • = 0.
has sum 0 , then a 1
Sol ution . Since the series s k ak + a 2k + a 3k + • • • is of the s ame type as s 1 = a 1 + a 2 + a 3 + • • • , it is sufficient to show that a 1 = 0 . Let p 1 + p 2 , pm be the first m prime numbers . Then =
+ 5
P2
+
•••
+
•••
+
(- l) ms p p • • • p 1 2 m
+ 5
Pm
)
contains only a 1 and not thos e terms an whose index n is not divisib le by the primes p 1 , p2 , . . . , pm ' in fact , it contains every such an only once (see Prob lem 1 7 of Chapter 1) . This means
and so a1
0.
PROBLEM 34 .
where
9
Stirl in g ' s Formula :
is between 0 and 1 .
Sol uti on .
Then
Verify
Let
I f n i s a positive integer , then
CHAPTER 3
206
tn
1•2•3 • • • n
or n+� 1•2• 3 • • • n = � tn n
(E . 6)
We cal culate t n . We get 3 2 5 3 7 4 log t n = 2 log T + 2 log 2 + 2 l og 3 +
2n -1 1 og n + -2 J1:1
·
But 1 1 n = 2 � 1 2 log il=l + 2 + 3 + ••• T (2n - 1 ) 3 5 ( 2n - 1 ) 5 and so 2n --1 log il=l n -2
=
1 + u , n- l
where 1 1 • + 5 -----=:____ + 4 .. ( 2n - l ) �
Thus
}
.
+ •••
(E . 7)
In order to calculate Sn- l ' we study the series
We have un- 1 < 1 3
1=-----={ (2n 1 l) 2 + ---= ( 2n - 1) 3 _
+
•••
}
=
.!. }
� { -1- 12 n - 1 n
·
We obtain , by addition , 1 5 n- l < 1 2 ( 1 -
�) .
Thus Sn- l tends to a l imit S , smal ler than 1 / 1 2 . The terms being pos itive , we have
sn - 1 < s .
(E . 8)
207
SEQUENCES AND SE R I ES
We also obtain 1
< 12 n or (E . 9) Using the inequal ities (E . S) and (E . 9) , we can put s n- 1
=
_
s - 1 28 _ n'
is between 0 and 1 . Inequality (E . ?) gives
where
n - 1 + S - 128 n " Therefore 1 tn
-n + 128 n 1-S e e
holds . Substitution in (E . 6) gives n
n + 21 -n + 8 l2 n C n e
=
(E. 10)
where C is the constant e l-S to be determined. Let
"2il"-=1 2n
It is easy to see that f(n)
=
l 2 4n--- (( 1 • 2 • 3 --n 1•2•3
n) 2n
2
)
"
2
Now, by (E . lO) , ( 1 · 2 · 3 • • • n) 2 1 • 2 • 3 • • • 2n where
8',
f(n)
=
e C (2n)
the same as
=
8,
2 4612- n8 ' c 4 e
- 2n + � 6n
.!. - -
2n + 2 2n + 248 n e
is between 0 and 1 . Thus
CHAPTER 3
208
where
and
e
are between
e•
and
0
z c l im f(n) = 4. n + oo
1.
Hence
But , by the Formula of Wal l is (see Problem 8) , z· 71
l im f(n) n + oo
=
Thus C = 1:2-IT. Hence , final ly n
1
!
e
-n + 2 n 12Tin nn e To obtain
Remark .
log
1
_n_ n -
l_ L - 1 2n - 1 _
__
+
with
0 < e
227 PROBLEM 56. What is the smallest amount that may be invested at inter est rate i, compounded anually, in2 order that one may withdraw dollar at the end of the first year, . . . , n dollars at the end of the n-th year, in perpetuity? 2 dollars n years from now at rate i The present value of n per year is n2 i) -n . Thus the required sum is n2 n I n=l i) Since - x) -l L�=O n it follows by differentiation that x(l - x) -2 n=lI n xn , (x x2 - x) n=lI n 2 n all series being convergent for x Taking x = i) , the required sum is found to be i) (2 i) SEQUEN CES AND S E R I E S
1
Sol ution .
(1
(1
+
+
(1
=
X
'
+
-1
(1
+
-3
) (1
1 A
repetit ion of this argument gives then
3
�
(n + 2) a
+ •••
Vl
+ (n - 1) ll+!l,
where a = 2 1-n . Remarks .
In Probl em 5 7 we showed that
,/1 + 2 V1 + 311 + • • •
= 3.
This formul a can easily b e conj ectured along the fol l owing l ines : since n (n + 2) = nil + (n + l) (n + 3) and letting n (n + 2) = f (n) , we see that f (n)
n il + f (n + 1) = n Vl + (n + 1) 11 + f (n + 2)
that is ,
�
-- -
:r� �� �� � � � =
• •�. n (n + 2) = n �1 + � c n +�l � + ::3 )::11�+�· � ) , l +�cn + 2 l �Vl ::+ c::n::::
Putt ing n = 1 , we have + •••
3.
a
SEQUEN CES AN D S E R I ES
229
(rr
In similar manner, since n(n + 3) nln + + + l) (n + and supposing that g(n) = n(n + 3) , we have g(n) = nln + + g(n + 1) = n"n + + (n + l)ln + 6 + g(n + 2) and so forth; we may conjecture that (taking n 1) : + �6 + a
5
5
4)
5
•••
>
4.
PROBLEM Let a1 b 1 be given. We form the numbers a2 al +2 b l and b 2 lalb l ' a3 = a2 +2 b 2 and b 3 la2b 2 , an + bn and b n+l 2 58.
> 0
Show that the sequences (an) n=l and (bn) n=l tend to a common limit L(a1 ,b1 ) and prove that L(a1 ,b 1) 2G7f ' where G 17f/2 f 2 2 dx 2 2 al cos X + b 1 sin X We observe that a1 a2 b2 b 1 and that in general "'
"'
=
0
Sol ution .
>
>
>
and hence (an) :=l is monotonely decreasing and bounded and (bn):=l is mono tonely increasing and bounded because the an 's and the bn 's are, in fact, the consecutive arithmetic and geometric means of the initially given num-
CHAPTER 3
2 30
bers a1 and b 1 with a1 b 1 Indeed, it is evident that a1 a2 and b 2 b 1 (because a1 b 1 To see that a2 b 2 we might refer to Problem 2 of Chapter 2 or note directly that al + b l v'a b = (� - £1) 2 for a b . 1 1 2 ll 2 In the same way we can show that an an+l bn+l bn Moreover, it is easy to see that al an bn b l . The sequences (an) :=l and (bn) :=l are therefore convergent; let = lim an and But an+l = an +2 bn and so, for n we get >
>
>
>
>
>
>
'f
0
>
>
>
0) .
>
-
>
>
0.
>
a
� oo ,
a
hence = We denote this common limit by L Let s.
G
We put 2a1 sin t sin x = as t changes from to rr/ 2, so grows x from to rr/ 2. Differentiation gives 2--t � ) b sin b (a (a ) + 1 1 1 1 --�----------� ---- cos t dt. cos x dx 2a1 [(a + b ) ---1 1 + (a1 - b 1 ) s 1n. 2 t]-2 0
0
231
SEQUEN CES AN D S E R I E S
But
2 - (a1 - b 1) 2 sin 2 t b ) ca + ,l 1 1 COS (a + b ) + (a - b ) sin2 t cos t, 1 1 1 1 and thus 2 dx 2a1 (a(al ++ bb l)) -+ (a(al -- bb l)) sinsin 2 tt l l l l where dt T X
T,
the other hand 2 �ai cos 2 x + b 12 sin 2 al (a(al ++ bb l)) +- (a(al -- bb l)) sinsin2 tt l l l l and thus dt dx 2 cos + Putting
On
X
=
X
we get /2 ' G J f 2 dx 2 . 2 J''' f 2 dt 2 2 a; cos t + b 2 sin t ai cos + b l S1n By repeated application of this transformation, we get /2 rr G f --;;:�;:c;:: o;:::;2;:dx:::::: ;:: ;:;;::;:;:::;2;:= (n 1,2,3, . . . , �a s x + b� s in x where an and bn are defined by the recursive formula �
0
0
X
X
0
)
CHAPTER 3
232
an bn ./an- lbn- 1' we know already, these two sequences converge to the common value L(a1 ,b 1) L. It is easy to see that 11 G 2b11 ; n n passage to the limit as n gives or The foregoing formula L(a1 ,b 1) 11I 2G and its derivation is due to Gauss. We consider an application of this formula and compute the integral 2 dx ! 11I2 1 / 1 I -;==:::::;2;:::: = -:===::::;:2=dx====.::;2 ;: G 0 ,;1 + COS x 0 ,; 2 COS X + S1n X Here a1 and b 1 1; the numerical sequences (an) :=l and (bn) :=l converge rapidly to L in this case and a5 and b5 are approximately equal to 1.198154. Hence we may put L 1.198154 and obtain the approximate value G 2111 1.3110138 . . . The integral 1112 ! dx G (a b 0 -f-:;:a;2=c=o=s ;:;:2 x:i:::: +=:: b::;2;:s=: i=n::;2;:; x ::: may be changed into a complete elliptic integral of the first kind by setting /2 G � f -f�===::;2;:=:=;2== === l a - b S 1. n2 x and can be computed with the help of tables (see Jahnke, E., Emde, F. , and Losch, F. (1960), Tables of Higher Functions, McGraw-Hill Book Co. , Inc., New York, N.Y.) . As
=
za
O)
2 33
SEQUENCES AND SERI ES
We now consider the complete e l l iptic integral of the first kind K (k) for every value of the module k it may be obtained from G if we set and
a = 1
� = k' .
b
In order to apply to this integral the method of Gauss , we first compute 1 + k' ...J� = -2- , 2
1 +
1 - kI , = � thus
or
By repeated application of this formula we obtain
where the sequence of numbers (kn ) , as can be verified by induction is given by 1 - ..f/1 - kn2 - 1
kn s o that 0
2 , then S i s dense in no interval . =
The set T of express ions of the form
Sol ution .
t
= ±
Vk
±
• • •
± 1i
2. It will suffice to show that S contains none of its own lim !. it points . Evident ly M = sup S satisfies M (k + M) 2, whence M = Y,{ l + ( 1 ).. ).. + 4k) 2} . Now k > 2 impl ies M < k, whence m inf S = (k - M) 2 > 0 . Therefore ).. the greatest s in S less than lk is less than (k - m) 2 , and the least s in ).. S greater than lk is greater than (k + m) 2 • It can be seen that if s has a k deleted neighborhood disjoint from S , then each of (k + s) !:.:2 and (k - s) 2 has such a neighborhood. It fol lows by induction that each s in S has a neighborhood containing no other point of S . PROBLEM 7 3 . converges .
Let x0
249
SEQUENCES AN D SER I ES
It is clear that 1/2
Sol ution .
trary n, k
;::
s
0,
xn
s
1 for all n . Moreover , for arbi
and so I Xn+l+k - Xn+l I
By induction on n we get
But (4/9) n + 0 as n + oo and the sequence (an ) n=O is a Cauchy sequence and hence converges ; its limit a clearly satisfies the equation a2 + a = 1 and a cannot be negative ; thus a = ( 15 - 1) / 2 . Remark .
Note that the sequence 2 + Xn xn+ l = 1.+X""" n
(n
0, where e2u 1 + 1/x. Thus max log1 2 - 1 0.4426950 . . . and nlim-+ co F (n). By expanding log(l + x) in a Maclaurin series, log(l + x) + }"1 X 41 x4 + we have 1 2 + cl ) I -1 n -F(n) 2n where O(xn) with xn > 0 signifies a quantity that divided by xn remains bounded; it follows that i3min n1-+imco F (n) 1 a.
=
a.
3
0
=
3
• • •
_
•
= z·
PROBLEM 95. If r > 1 is an integer and x is real, define f(x) k=O�"' rj=lL-1 [x r+k+ljrk ] . where the brackets denote the greatest integer function. Show that f(x) [x] if X 0 [x + 1] if X 0. By Problem 27 of Chapter 1,
>
�
>
PROBLEM Show that the integer nearest to n!/e is a multiple of n - 1. The error made in stopping the expansion of e-l with the n term (-l) /n! is less than 1/(n + 1) !. Hence the integer nearest to n!/e is Pn n!{l - 1� + }! - + (-l) n n� } . The divisib�lity property in question can be verified as follows: n nPn-1 + (-1) n (n - l)Pn- 1 + Pn-1 + (-l) n (n l)Pn-l + (n - l)Pn-Z + (-1) n-1 + (-1) n 96.
Sol ution .
=
•••
p
PROBLEM Prove that a necessary and sufficient condition for the rationality of R \/a + 3 /a + where a is a positive integer, is that a N(N + l) (N + the product of 97.
=
· · · ,
2) ,
CHAPTER 3
276
three consecutive integers. In that case find R. Define R1 = 3 /a, Rn = 3 /a Rn-l · Now R2 R1 , and R� - R�-l (Rn) is monotone increasing, Moreover, �-l R3k_ 2 , so that by induction R1 1 + /a, and �-l 1 + 3 /a implies that R� a + 1 3;a (1 3 ;a) 3 , so that by induction (R ) is bounded. It follows that (R ) converges to a limit R. But then R3 nR - a 0 . If R is rational and anintegral, then R is integral, and a = (R - l)R(R + 1), the product of three consecutive integers. Hence the condition is necessary. It is also sufficient, since R N + 1 satisfies the equation R3 - R - N(N l) (N 2) = 0 , and, as it is the only real root, it is the value of the radical. Cardan's Formula yields the explicit expression R = { a/2 ..va2/4 - 1/27 }1/3 {a/2 - ...V a2 /4 - 1/27 } 1/3 The result in Problem can be generalized as follows: a necessary and sufficient condition for the rationality of R = "a + n la + where a is a positive integer, is that a N(Nn-l - 1) . >
+
Sol ution .
e x ---1 log --X
>
-
0,
X > 0;
X
0 , and increasing in the second case , un < 0 . We have l imn + co un = u 0 because =
u
>
eu---1 log -u
for u
> 0
u
Sol ution .
M =
b 1 ( f C x) dx . (b - a) 2 } a Let
sup I f ' (x) I a�x�b
.
REAL FUNCT IONS
323
Then , by the Mean Value Theorem (see Problem 1} , f (x)
f ' (t) (x
a)
f (x) = f ' (x) (x - b)
0
D
D.
U
D
PROBLEM 22. subset E of the real line R1 is said to have if, for any £ there is a countable set {lk };=l of intervals such that k=l I k E and where I k l denotes the length of the interval Ik . Clearly, any countable set {tn : n = 2, . . . } hasn measure zero, the point tn can be enclosed in an interval of length £/2 , and A
measure zero
> 0,
:o
U
!
1,
Show that if E n=l En ' where each En has measure zero, then E has measure zero. Given £ the set E 1 can be enclosed in the union k=l I l,k ' where U
Sol ution .
>
0,
U
and in general, the set Em can be enclosed in
344
CHAPTER 4
k=l Im, k ' U
where
The intervals Im, k with m, k 1,2, ... then satisfy m,k=l Im,k m=l m and m,k=lI I Im,k l 2 2 2m u
:::>
u
E ,;; 2
E
E,
E
+ - +
. . .
E
+ - +
E
•
23. The consists of all numbers in the interval [0,1] that admit a ternary development in which the digit 1 does not appear. Show that has measure zero and is an uncountable set. The set can be constructed by deleting the open middle third of the interval [0,1], then deleting the open middle thirds of each of then intervals [0,1/3] and [2/3,1], nand so on. If Fn denotes the union of the 2 closed intervals of length l/3 which remain at the n-th stage, then n=l Fn . Fn (and therefore contains no interval of length larger than l/3n . The n sum of the lengths of the intervals that compose Fn is (2/3) , which is less than if n is taken sufficiently large. Hence has measure zero. Finally, each number x in (0,1] has a unique non-terminating binary development PROBLEM
Cantor Set
C
C
C
Sol ution .
c
n
C)
E
C
If y = 2x then O.y1y2y3 . . . is the ternary development with yi 1 of some between x and y, extended by mapping 0 pointi y ofi ' This correspondence onto itself, defines a one-to-one map of [0,1] onto a (proper) subset of It follows that is uncountable; it has cardinality c (the power of the continuum) . �
C.
C.
C
345
REAL FUNCTIONS
PROBLEM 24. Let C be the Cantor set (see Problem 23) . Show that any subinterval [a,b] of [0,1] contains an interval (a' ,b') free of points of C such that its length satisfies the inequality b' - a' � �1 (b - a) . If [a,b] is free of points of C, then the claim is trivially true. If not, then let V be one of those closed intervals of length 3-n that remained at the n-th step of the removal of the open middle thirds such that V [a,b] but with n being as small as possible. If V [a,b], then the claim is true, for at the (n 1)-th step we remove the open middle third of V. In the other case, that is, in the case when V is a proper subset of [a,b], there are adjoining V (possibly on the left and on the right) intervals of the same length as V which are free of points of C. If one of these (two) intervals has a subinterval of length � (l/5) (b - a) in common with (a,b), the claim is once again fulfilled. If this, however, is not the case, then the length of V is � (3/5) (b - a) . Since the middle third (a' ,b') of V is free of points of C and b' - a' � (l/5) (b - a) holds, the claim follows in general. Sol ution .
=
c
+
PROBLEM 25. Let C be the Cantor set (see Problem 23) . We arrange the complementary intervals into groups as follows: The first group contains the interval (1/3,2/3) , the second the two intervals (1/9,2/9) and (7/9,8/9), the third group the four intervals (1/27,2/27), (7/27,8/27) , (19/27,20/27) n-l and (25/27,26/27) , etc. The n-th group contains then 2 intervals. We define the g as follows: g 1/2 for (1/3,2/3) 1/4 for (1/9,2/9) 3/4 for (7/9,8/9) In the four intervals of the third group wen-1 set the function g consecutively equal to 1/8, 3/8, 5/8, and 7/8. In the 2 intervals of the n-th group we set g consecutively equal to 1 3 5 2 2 2 The function g is in this way defined on the open set [0,1] - C and is Can tor Function x
(x)
E
x E
x E
n'
n'
n'
CHAPTER 4
346
seen to be constant on each component interval and is also seen to be non decreasing on [0,1] - C. We extend the domain of definition of g to all of [0,1] by putting g(O) 0, g(l) 1, and g(x0) Sup{g(x) : X E [0,1] - C, X < x0 }. We see that g is now defined on all of [0,1] and is nondecreasing throughout [0 ' 1]. Show that the Cantor function g is continuous on [0,1]. The claim follows from the fact that the set of values of the function g on the set [0,1] - C is dense in [0,1], that is, every sub interval of [0,1] contains at least one point of the set of values of the function g on [0,1] - C. Indeed, since g is a nondecreasing function, any point of discontinuity of g must be a simple jump; hence, if g is discontin uous at x0 , then at least one of the intervals =
=
=
Sol ution .
where g(x0 - 0) denotes the limit from the left and g(x0 + 0) the limit from the right of g(x) at x x0 , must be free of values of g. =
PROBLEM Let P(x) denote a statement concerning the point x of the interval [a,b]. We say that P(x) holds on [a,b] if P(x) holds for every point of [a,b] except for a subset of points of [a,b] having measure zero (see Problem The Cantor function g (see Problem is an example of a continuous, nondecreasing function on [0,1] having derivative 0 almost everywhere on [0,1]. Moreover, g(O) 0 and g(l) 1. Give an example of a strictly increasing continuous function on the interval [0,1] whose derivative is zero almost everywhere on [0,1]. Let g denote the Cantor function and let us set g(x) 0 for x < 0 and g(x) 1 for x 1. If we have the interval [a,b], we define the corresponding Cantor function in a similar way, namely, we set it equal to 26.
almost everywhere
22) .
25)
=
=
Sol ution .
=
=
>
347
REAL FUNCTI ONS
for a x b. Now, let 1 1 , 1 2 , ... be the intervals [0,1], [0,1/2], [1/2,1], [0,1/4], ... [0,1/8], and let gn be the Cantor function corresponding to In Then f(x) n=lI is continuous and strictly increasing on [0,1]; moreover, f' (x) 0 almost everywhere on [0,1]. �
�
'
PROBLEM 27. Prove the following characterization of the Riemann inte gral, due to Lebesgue: bounded function f on [a,b] is Riemann integrable if and only if the set of discontinuities of f has measure zero. Let E {t [a,b]: w(t) 1/n}, where w(t) = w(f;t) with w(f;t) as in Problem n21. Clearly, f is continuous at t if and only if w(t) = 0. The set En is closed (by Problem 21) , and the set of points where f is discontinuous equals n= l En . Now assume that f is integrable, so that for any £ 0 there is a partition P£ such that U(f,P£ ) - L(f,P£) < £ (see Problem 11). If A D
E
Sol ution .
�
D
uro
>
then the set of intervals can be split into two groups, where the intervals in the first group have nonempty intersection with Em , and those of the second group do not meet Em (m fixed). Then where the prime (resp. double prime) indicates summation over intervals of the first group (resp. second group) . On the primed intervals, Mk - mk 1/m, so that �
CHAPTE R 4
348
and E'(tk - tk_ 1 ) This Ern can be enclosed in finitely many intervals of length less than and since is arbitrary, we see that Ern has measure zero. It now follows from Problem 0022 that has measure zero. Conversely, assume that = Un= 1 n has measure zero and, hence, that En has measure zero for all n. Choose K so that 1/K The set E� [a,b] - Ek is open; therefore, by Problem 20, < IDE .
IDE ,
E
D
E
D
< E.
where the intervals (Ik) �=l are open and disjoint. We now show that k=ll: J rk l b - a. Suppose, on the contrary, that E�=l J rk J b - a. Since EK has rneas ure zero, we can find (Jk)�=l such that b - a2 and The intervals (Jk)�=l and (Ik)�c l together cover [a,b], and hence b - 2 b _ a. This contradiction shows that A b - a. We now see that, for some n, A
=
=
I
-_
IX
$
$
I.
+
35 1
REAL FUNCT I ONS
Thus f1 (x) exists everywhere. But at the points of E it is not contin uous. Indeed, if x0 E, then in every neighborhood of x0 there is a point, and therefore also an endpoint, of one of the removed intervals, and we know that at such an endpoint the oscillation of the function f is equal to But the set E is not of measure zero and so f1 cannot be Riemann integrable according to the result in Problem E
2.
27.
PROBLEM Show the following theorem: If f is differentiable in [a,b] then the derivative f1 is Riemann integrable over [a,b] if and only if there exists a Riemann integrable function g in [a,b] such that f(x) f(a) + lax g(t) dt. We give the verification in three steps. I. Let g be Riemann integrable and assume that m g(x) M for all x [a,b]. Define, for x [a,b], G(x) j(x g(t) dt, and assume that G is differentiable on [a,b]. Then we have m G 1 (x) M for all x [a,b]. Define K(x) G(x) - m(x - a) . Then K is differentiable and K1 (x) G1 (x) - m. We also have K(x) J[x g(t) dt - m(x - a) j[x (g(t) - m)dt. Since g(t) m we see that K is monotonically nondecreasing. Hence K1 (x) 0, or G 1 (x) m for all x [a,b]. In a similar fashion one proves G1 (x) M for all x [a,b]. II. Under the same conditions as in step I we have that G1 is Riemann integrable over [a,b]. 29 .
Sol ution .
E
E
=
�
�
�
�
E
Proof.
=
=
=
�
� �
�
�
E
E
CHAPTER 4
352
Step I is applicable on every closed suinterval of [a,b]. From this it follows that the oscillation of is not larger than the oscillation of g. According to the result in Problem 11 we may therefore conclude that is Riemann integrable over [a,b]. III. Proof of the theorem : It is clear that the given condition is nec essary; take g = f'. To prove the sufficiency we write f(x) - f(a) = 1x g(t) dt. Then it is clear that f' is the derivative of a function of the form 1x g(t) dt. According to step II we obtain that f' is Riemann integrable over [a,b], completing the solution. Clearly we have ix f' (t) dt = 1x g(t) dt for all x [a,b]. However, it is easy to construct examples in which f' g. Proof.
G'
G'
Remark .
E
#
PROBLEM 30. A function f is said to be if for any 0 there is some o 0 such that
absol utel y continuous on
E >
>
[a,b]
for every finite, pairwise disjoint sequence ... of open intervals of [a,b] for which n .E (bk k=l Show that the Cantor function g defined in Problem is not absolutely continuous on [0,1]. '
- �)
< 0.
25
353
REAL FUNCT I O N S
First we extend the domain of definition of g as follows: we set g(x) = for x < and g(x) = for x Next, we enclose the Cantor set C defined in Problem in a union Sol ution .
0
0
1
1.
>
23
or pairwise disjoint open intervals such that E;=l (bk - ak) is arbitrarily small. It is easily seen that k=l [g(bk) - g(�)] and n [g(b ) - g(�)] k=l k for sufficiently large n, while n (b - a ) k=l k k is arbitrarily small. E
1
E
�
1
2
E
PROBLEM Give an example of a differentiable function which takes rational numbers into rational numbers but whose derivative takes rational numbers into irrational numbers. Let (E. 14) f(x) n=O g(n!x) (n!) where g(y) is the periodic function of period defined on by g(y) = y(l - 4y ). The function g vanishes at all integers and has a continuous derivative which is unity at all integers. For any rational x, the series (E.l4) has at most finitely many non-zero terms and they are rational. The formal derivative of (E.l4) converges uniformly and absolutely and therefore converges to the derivative of f. For any rational x, the deriva31 .
Sol ution .
\ L
2'
1
2
[ - 1/2 , 1/ 2 ]
354
CHAPTER 4
tive of the series (E.l4) is the same as the series for e, n=OL n\ ' save for at most finitely many terms which are rational. Thus for rational x, the derivative of f is e plus some rational number. But e is irrational (see Remarks to Problem 13 of Chapter 3) . PROBLEM 32. Give an example of an everywhere continuous but nowhere differentiable function. For x let T(x) n=ll: x. Putting qn = 2-n , pn 2n2 , and h' 2h, we obtain T(x + h') - T(x) n=l 2 (sin pnh) cos pn (x + h) . For any 0 there exists some n0 , and for this n0 an h0 0, such that and, if h h0 , For h' 2h we therefore get TCx + h') - T(x) T is thus uniformly continuousI on0the entire number line. For h' 0 and holding x fixed, we define Q(h ') T(x + h')h' - T(x) Considering Q(h') only for the special values h = s pm ' where m and s = ±3, we obtain: (a) The first m - terms of the series for Q(h') in absolute value have a sum smaller than - oo
Sol ution .
< oo ,
E
>
J l
J
!
5
m are zero. Indeed, pnh TIUn ' where un which is an integer because n2 - m2 - 2 > and hence sin pnh (c) The m-th term of the mentioned series is where STI . Sl. n cm pm --STI 4 cos (pmx + STI) whose sign for the four mentioned s-values is determined by the variable of the cosine term. Since we can assign to this variable four consecutive values at distance TI/2 from each other (whatever pmx may be), it is possible to make the cosine term both � 1/1:2 and -1/1:2 for suitable values of s. Moreover, for these values of s, 2 -m+l � 2m2 -m-3 ' 1/1:2 1 1 m 2 I cm I � 4 Pm qm 3� 3TI
2 and there m which Q(h') -2 . The desired result therefore follows. The foregoing function, due to Lebesgue, provides an example of a function for which at every point the right and the left upper Dini derivatives are +oo and the right and left lower Dini derivatives are
5
�
s
>
1
>
36 2
CHAPTER 4
But nthe functions gm form a monotonically decreasing sequence; since gm (x) fmm (x) f.(x), where j for each value of x is a number (in general dependent on x) larger than or equal to m, namely one of the numbers m, m back in . . . , nm , it follows from fm (x) + 0 that gm (x) + 0. We areovertherefore the special+ case already settled and the integral of gm [a,b] tends to zero as m oo. the basis of the inequality just proved we can see that Jm must sink below any positive number, hence specifically below thus, for sufficiently large m, we obtain Jm < But the Jm 's are nonnegative and may be picked as small as we please; hence Jm + 0 as m + oo and the proof of the theorem of Osgood is complete. The foregoing proof of Osgood's theorem is due to Riesz. =
�
-
J
+ 1,
On
0
2o.
F.
Remark .
PROBLEM Let f be a continuous and increasing function on the inter val [a,b]. Show that f has a finite derivative almost everywhere on [a,b], that is, at every point of [a,b] with the possible exception of the points of a set having measure zero. The proof will be based on a lemma. Let g be a continuous function on an interval [a,b], and let E be the set of points x interior to this interval such that there exists a point t lying to the right of x and satisfying g(t) > g(x) . Then the set E is either empty or it decomposes into countably many disjoint open intervals (ak ,bk) and (E for each of these intervals. We first note that the set E is open, since if t x0 and g(t) > g(x0 ), then by virtue of the continuity, the relations t > x, g(t) > g(x) 36 .
Sol ution . LEMMA .
. 18)
Proof.
>
36 3
REAL FUNCT I ONS
remain valid when x varies in a neighborhood of the point x0 . Thus E, if not empty, decomposes into countably many disjoint open intervals (ak ,bk) (see Problem 20) ; the points a bk do not belong to the set E. Let x0 be a point between ak and bk ; we shallk ' show that (E. 19) (E.l8) will follow by letting x0 tend to ak . To verify (E.l9) , let x1 be a point at which the function g assumes its largest value on the interval [x0 ,b]. The point x1 cannot belong to E, for there exists no t such that x1 t b and g(t) g(x1 ) . Since the part of [x0 ,b] to the left of bk be longs entirely to E, we necessarily have bk x1 b. We cannot have g(bk) g(x1 ) since bk does not belong to E. Thus g(bk) = g(x1 ) and so g(x0) g(bk) . This completes the proof of the Now let f be a continuous and increasing function on the interval [a,b]. To examine the differentiability of f, we shall compare its Dini derivatives at a point x: D , D_ , that is, the right-hand limit superior and inferior and the left-hand limit superior and inferior, respectively, of the ratio f(t)t -·x - f(x) as a function of t, at the point t = x. The values are admitted. Then function f is differentiable at x if all the four Dini derivatives have the same finite value there, and then f' (x) = D+ D As an immediate consequence of the definition we have D D in the case of an increasing function the Dini derivatives are of course non negative. Our problem consists in showing that for the increasing function f we have almost everywhere (ii)
s
D+
>
X
>
> c,
>
=
LEMMA
This yields, by addition,
(E.20) here we used the fact that the total increase of an increasing function on an interval cannot be less than the sum of the increases on disjoint sub intervals. The inequality (E.20) shows that, for sufficiently large C, the total length of the intervals (� ,bk) will be as small as we please. That is, the set E00 is of measure zero. The statement (ii) is proved by analogous reasoning which is repeated alternately in two different forms. Let c and C be two positive numbers, c C. We first show that the set EcC of points x for which C and c, is of measure zero.
D
LEMMA
D
=
LEMMA
+
hence Taking account of (E.21), it follows that k,mI (bkm - akm c kL (bk - ak). If I 5 1 I and 1 52 1 denote the total length of the systems 51 {(� ,bk)} and 5 2 {(akm' bkm)}, respectively, it follows that ) s
c
=
=
Repeating the two steps alternately, we obtain a sequence 1 , 52 , of systems of intervals, each imbedded in the preceding, and we5have for n 1, 2, . It follows that =
.
.
Thus the set EcC can be covered by a system of intervals of total length as small as we please and EcC is seen to have measure zero. Now we form the union E* of all the sets EcC corresponding to pairs c, C of positive rational numbers (c < C) . As a union of countably many sets of
CHAPTER 4
366
measure zero (see Problem E* itself is of measure+zero. If at a point we can interpolate between and D two rational numbers, x we have D
22) ,
< D+ ,
D
then x is a point of the set E and consequently of E*. Thus the points x where (ii) does not hold, formcCa set of measure zero. This completes the so lution. The foregoing proof is due to F. Riesz. Remark .
PROBLEM We assume that the upper and lower integrals j(b f(t) dt and Jfab f(t) dt of a bounded function on a bounded closed interval [a,b] have been defined (see Problem We also assume the following elementary properties of upper and lower integrals: (a) j(b f(t) dt J:b f(t) dt for each bounded function f. (b) For a c b, �b f(t) dt �c f(t) dt �b f(t) dt 37.
1 1) .
�
=
>
=
�
=
�
>
>
O)
1
o)
>
1
-
=
=
+
=
�
CHAPTER 4
368
Therefore (t) ::;; fax f(y) dy + [f(x + O) + a./ 2] (t - x) - �x f(y) dy - [f(x + 0) - a./2](t - x) - a.(t - x) ha. (x) ::;; 0. This contradicts the fact that x is the greatest lower bound. Of course the dual result on left-hand limits is an immediate consequence by consideration of f(-x) . Clearly the result includes the prop ositions that continuous functions are Riemann integrable and that monotone functions are Riemann integrable. Given the "continuous almost everywhere" characterization of Riemann integrability (see Problem 27), we have that bounded functions with right-hand limits at every point of an interval are continuous almost everywhere on the interval. \�
Remarks .
PROBLEM 38. Let (�) := O be the sequence of polynomials defined by (E.22) �+1 Ct) = � Ct) + 21 {t - �2 Ct) } , n 0. Show that in the interval [0,1] the sequence (qn) := O is increasing and that for any> £ > 0, there exists a positive integer n0 (depending only on £) such that n no implies - � Ct) I < £ for every t in [0,1], that is, n=O converges uniformly to on [0,1]. To prove the claim it is enough to show that, for all t in [0,1], we have 0 � (t) 2 +2/tnit' (E. 23) for (E.23) implies that 0 � (t) ::;; 2/n. We prove (E.23) by induction on n. It is true for n 0. If n 0 it follows from the inductive assumption (E.23) that ;:>:
l rt
(0 ) 00
It
'll
Sol ution .
::; It -
::;; _::...__ _:: ::;;
It -
;:>:
REAL FUNCT I ON S $ It -
0 C�n (t) and hence 0 qn (t) s
369
$ It, s
It,
and therefore from (E.22) we have
'ln+l (t) 0, and from (E.23) - +1 (t) 2 2/t + nit { 1 - �} 2/t � 1 } = 2/t 2 + nit 2 + (n + 2 + (n + To verify that (qn) :=O is an increasing sequence, we note that
so that
It -
It
'In
?::
$
It
$
but we know already that 0
$
It -
1)
qn (t)
It
1)
It
$
rt.
PROBLEM 39. Show that for any > 0 there exists a polynomial p such that p (x) I x I I for all x in [ -1, 1]. By Problem 38, for any 0 there exists a polynomial q such that for all t in [0,1]. jq(t) Replacing t by x2 and noting that x V x2 , the desired result follows. Consider the function jx - ci on an arbitrary closed bounded interval [a,b]. We choose a number d such that the interval [c - d,c d] includes the interval [a,b]. By the substitution s = -d£
-
I
< £
Sol ution .
£
>
It! < £
=
Remarks .
+
X -
C
CHAPTE R 4
370
the interval c - d x c + d goes over into the interval -1 s 1. From Problem we know that for any 0 there exists a polynomial p(s) such that I P Cs) - lsl l for -1 s 1; this implies that �
�
�
39
�
E >
�
�
on the interval [c - d,c d] and, a fortiori, on the interval [a,b]. Thus the function P(x) d•p (---x d----c) , which is a polynomial in the variable x, approximates the function l x - c l on [a,b] with accuracy Hence the polynomial Q(x) �[P(x) + x - c] approximates the function Lc (x) �{(x - c) lx - cj } on the interval [a,b] with accuracy +
=
E.
=
+
=
E/ 2 :
Note that
�
0 for x c, x - c for x c; Lc is a polygonal function whose graph has an angle at the basic point x c. Suppose now that g is any polygonal function on [a,b] whose graph has angles at the basic points a a0 < a1 • • • < an b. Then g is a linear combination of the Lc . Indeed, let g0 (x) g(a) c0 Lao (x) + c1 Lal (x) + • • • + n-1 Lan-1 (x) , and define the constants by the equations 0 ,1, . . . ,n. �
=
=
=
=
+
0, there exists a polynomial P such that for all x [ a,b ] i f(x) - P(x) I < and we say that f admits uniform approximation by P. If f is continuous on a closed bounded interval [ a,b ] , then f is uniformly continuous on it and hence admits uniform approximation with an arbitrarily small error by a polygonal function on [ a,b ] . In fact, f is uniformly continuous on a closed bounded interval if and only if f is such that j f(x) - g(x) l < £ for all x [a,b ] with g a polygonal function. But we know from the Remarks following Problem that polygonal functions on a closed bounded interval can be approximated uniformly with an arbitrarily small error by polynomials. In the foregoing, the proof of the Approximation Theorem of Weierstrass was reduced to the problem of approximating the function l x l by polynomials, a procedure due to Lebesgue. For emphasis we add that f is, of course, assumed to be a real-valued function. Approximation Theorem of Weierstrass :
£
£
E
Sol ution .
E
39
Remarks .
H.
PROBLEM 41. Let f be a continuous function on the set of1 real numbers 1R . Show that if f can be approximated uniformly throughout R by polynomials, then f is itself a polynomial. If polynomials Pn approach f uniformly, then for some n we have Sol ution .
CHAPTER 4
372
IPi (x) - f(x) I 1 for all i � n and all x in R1 . Hence for all i n, IPi (x) - Pn (x) < 2 for all real x, so that P i - Pn is a bounded polynomial, and hence is constant. Taking limits as i f - Pn is constant, and hence f Pn is a polynomial.
0 such that lf(x) > d/2 for all x in the interval x1 x1 + o, by the continuity of f on [a,b]. Thus Jlb lf(x) l dx > %·2o = do > In the same way we can see that if f # 0, then f2 (x) dx > 0. Thus Remarks .
E
I
s
o s x
o.
b I f(x) I dx > 0. Ja PROBLEM Let f be a continuous function on [-�,�J and suppose that f: f(t) cossin ntnt dt 0 for n 0 ' 1, 2, ... Show that f 0. Suppose not, that is, suppose lf(t0) I > 0, say f(t0) = > 0. Then by continuity there are two positive numbers and o such that f(t) > for all t in the interval I, where I = [t0 - o,t0 + o]. It will be enough to show that there is a sequence (Tn) of trigonometric polynomials such that (i) Tn (t) 0 for t I, (ii) Tn (t) tends uniformly to in every interval inside I, (iii) the Tn are uniformly bounded outside I. For then the integral 43 .
a
Sol ution .
E
�
E
+ "'
J
E
CHAPTER 4
374
may be split into two, extended respectively over and over the rest of By (i), the first integral exceeds \J\ tminJ Tn (t) , and so, by (ii) , tends to + oo with n. The second integral is bounded, in view of (iii). Thus f(t) Tn (t) dt = 0 is not possible for Tn with large n and we have reached a contradiction.n we set> Tn (t) = [x(t) ] , x(t) = 1 cos (t - t 0) - cos then x(t) in x(t) in J, \x(t) 1 outside Moreover, conditions (i), (ii) , and (iii) are satisfied. I
(- rr , rr ) .
E
f- rr rr
If
I,
1
\
�
+ I.
o,
0 by a- polynomial £ < lx1 xz l = a � x � b. There follow the inequalities =
=
+
• • •
+
X
= 1
\ xz - k!O ak [h(x2) Jk \ < £,
i x1 - x2 1 < 2 £. This contradiction completes the proof.
#
=
377
REAL FUN CT IONS
It is clear that the result in Problem remains valid if we replace the interval [0,1] by any closed bounded interval [a , S ] . It can be seen therefore that if f is real-valued and continuous on the interval [-n/ 2 , n/ 2 ] , then it is possible tok approximate f uniformly on this interval by linear combinations of (sin x) with k 0,1,2, . . . ; however, f cannot bek approximated uniformly on this interval by linear combinations of (cos x) with k 0 , 1 , 2 , . . 45
Remarks .
=
=
.
PROBLEM Let f be a function with at least k derivatives. Given that for some real number r, r r (k) 0 lim and lim x f(x) x (x) 0, f co x-+ co x-+ show that r (j) lim x f x-+ co (x) 0, 0 k. For each integer j, 1 j k, expand f(x + j) in a Taylor polynomial about the point x: .k-1 f(x j) f(x) j f' (x) + �. 2! f"(x) (k - 1) ! f (k-1) (x) kll f (k) . ) where x x j. This may be considered as a system of linear equations in the unknowns f(x) , f' (x), . . . , f (k-1) (x) . The matrix of coefficients has for its i-th row 1, i (k - 1) ! . From the corresponding determinant we may factor out the denominators common to the elements of each column and will have 1! 2 ! 1 (k - 1)! times the familiar Vandermond determinant. Hence the determinant of coeffi47.
=
�
0, an = �n ){0n cos ax cos nx dx -nl io0n [cos(a + n)x + cos(a - n)x] dx (-l) n """"'"a2-=2:.-a=-.: n,.2- . sinn an ' we see that the Fourier cosine series expansion of the function f yields -n2 --sincos axan = -2a1 + n=lI aa2sin- nnx2 (-n n) . Setting x = 0 we get n sin1 an = an1 + 2 n=lI (an)(-l)2 - (nn)an 2 and putting an t yields -t1 + n=lI (-l) n (_ t -l_nn + _1 t +_nn) '. here t is arbitrary real number, but not an integral multiple of n. We now write { Ck+l)n/2 sin x dx. = }o{ "" sin x dx = k=OI lkn/2 For k = 2m we consider the substitution x = mn t and for k 2m - 1 we consider the substitution x = mn - t. This leads to 1(2m+l)n/2 sinx x dx (-l) m lon/2 mn . + t dt mn/2 �
�
.!
an
I
X
X
+
�
CHAPTER 4
384
and
[ 2mTT/ 2 J (2m-l)TT/2
sin x X
dx
(-l) m- 1 ln0TT/2 mTTsin- tt dt.
It follows that I = la TT/2 s1n. t t dt + m=l"'L la TT/2 (-l) m (--t -1 mTT + --t +1 mTT) sin t dt. But the series m (1-mTT + 1-mTT) sin t (-l) I t t + m=l converges uniformly in the interval 0 t TT/ 2 because it is majorized by the convergent series 1 � -2-1-1 m=l m and so can be integrated term-wise. We therefore have I = )0{ TT/2 sin t fl lt + m=lI (-l) m (t -1-mTT + t +1-mTT) } dt. But we already know that O
O
s
n
L
s
4
where t is arbitrary, but not a multiple of TT; we may therefore conclude that I = lo0TT/2 sin t -s1n. l-t dt = la0TT/2 dt = (This elegant calculation of the integral I is due to I. Lobatshewski.) z· 1T
N.
PROBLEM 53. Show that, if x is a positive integer, 1 e = 1 + IT + x2 + • • • + 8 (x) , (E. 34) where lies between 1/2 and 1/3 and is decreasing as x increases from 0 to 2
X
S (x)
"' ·
X
2T
XX
iT
385
REAL FUNCT I ON S
From we get S(x) + -ex -xx!- - x! { + ITx + ••• + �x } . Using integration by parts, we note that x! and (E . 34)
Sol ution .
=
1
x
1
2
X
1
X.
X
XX = + -
we obtain Let w be defined by we-w te -t for all t 0 with w(t) when 0 < t and w(t) when t. The func tion w is given explicitly by w(t) ts (t), where s is defined by slog- s t for t > 0; in other words, s is the inverse function of (log t)/(t - Hence >
� 1
� 1
� 1
1 �
=
1
=
1) .
and substituting t w(u), u w(t) we have - !ol (ue -u) x w' (u) du. Thus =
=
0
CHAPTER 4
386
= 1 ){ 1 (te -t ) (1 w') dt. Using integration by parts, after having multiplied and divided by (1 - t)/t and set t {l w' (t)}, we get 1 1 w1 (1) - fr l (te-t) wi (t) dt. From the definition of the function w we get w ' (t) = t _1 w-_w ' t w' w1 (t) "l""=t wi (t) = l � t ( (l -t t) 3 (1 -w w) ) we obtain w1 -1, and since slog- s1 t ' we see that w1 (1) -4/3. Substituting this later value into we obtain x + xe2 0
S (x)
"l""=t
S (x)
x
-
=
+
x 2
2
=
X
�
+ -
.!____:__!
+
(E . 35)
0
l=W"
+
3 ;
(0+)
=
(E . 35) ,
To see that to
S (x)
is decreasing from
1 it is sufficient to show that w i (t) t __..:.:.w_ ( 1 - t) (w - 1 )-;:;S (oo)
=
3'
3
�
3"
�
0,
that is, (E . 36)
387
REAL FUNCT IONS
But
w ts and t = log s and so (E.36) is seen to be equivalent to t slog- s1 - s1 ss l/31/3 for s > However, the latter inequality is true (see Problem of Chapter 2) and we have the desired result. The result in Problem was posed as a question by Ramanujan. The Solution given above is due to Karamata. We draw attention to the fact that the result in Problem can easily be deduced from the result in Problem Indeed, since (by Problem n n-1 k ;- = �! (nn/n!)en , 1/3 < en < 1/2, k=O we have m-1L mx/x! - em-1 x=O s:I
=
Sol ution .
>
not dependent on 8 such that +n for all 8 whenever r M. Since = for r = (note that in this case we have 3P/3 8 = 3Q/3 8 the interior integral of I 2 leads to the value for r R. In case R + oo, this value tends to -n uniformly in 8 . Hence we obtain lim I 2 = - 2nn. R+oo We see therefore that the integral I 2 is negative for sufficiently large R. Thus, the equality I 1 I 2 cannot be fulfilled and this completes the proof. > 0, l < E
E l au; a e
0
>
au; a e
=
0
0
0) ,
au; ae
=
PROBLEM polygon is said to be convex if it contains the line seg ments connecting any two of its points. Prove the following proposition: Any convex polygon (in the complex plane) which contains all the zeros of a polynomial P(z) also contains all the zeros of the derivative P' (z) , regard less of whether z is real or complex. Let P be a polynomial of degree n and have the zeros z 1 , z 2 , zn . Then A
59 .
K
Sol ution .
Thus
P' (z) n z -1 z k=l k Let be the smallest convex polygon containing z 1 , z 2 , . . . , zn . If z is a zero of P' and coincides with one of the zk , then there is nothing to prove; if z is a zero of P' but different from all zk ' then n 1 k=l z - zk holds and it is sufficient to show that the foregoing equation cannot be sat isfied for any point z outside of We do this now. Let z , z , . . . , z > 0, be arbitrary points of the complex plane, m 1 1 n 2 m2 0, . , mn > m1 m2 + • • • + mn = 1 and �
-
-
L
K
l:
--
=
o
K.
>
.
.
0,
+
395
REAL FUN CTI ONS
Interpreting the numbers m1 , m2 , . . . , mn as masses fixed at the points z 1 , z2 , . . . , zn ' the point z defined by z = m1 z 1 m2 z 2 + • • • + mn zn is the cen ter of gravity of this mass distribution. If we consider all such mass dis tributions at the points z 1 , z2 , . . . , zn the corresponding centers of gravity cover the interior of a convex polygon, the smallest one containing the points z l , z 2 , . . . ' zn . The equation nI 1 = o k=l z - zk implies z - zn 0. + ••• + z - zn Thus z m1 z 1 m2 z 2 + • • • mnzn ' 1' where the k-th "mass" mk is proportional to 1 k 1,2, . . . ,n. Hence if z were outside the smallest convex polygon K that contains the zk 's, there could be no equilibrium. From Rolle's Theorem (see Solution to Problem 5) we know that any interval on the real line which contains all the zeros of a real-valued polynomial P also contains all the zeros of the derivative P'; the result in Problem 59 generalizes this fact. Any convex polygon which contains all the zeros of a polynomial P also contains all the zeros of its derivatives. +
--
----..,-2
l
=
+
l
+
Remarks .
PROBLEM Let f(x) be a complex valued function for isfying f(x + y) f(x) f(y) 60 .
- 00
0, there ex ists a o > 0 such that for all x and y in E with lx - < o we have lf(x) - f(y) l sequence < £. (x ) : of real numbers is said to be a if n =l for any £ > 0 there is an integer n0 such that lxn - xn , l < £ if n � n0 and n' � no . Show: If a real valued function f is uniformly continuous on a set E of real numbers and if (xn) :=l is any Cauchy sequence of elements in E, then {f(xn)}:=l is also a Cauchy sequence. Conversely, if a real -valued function f, defined on a bounded set E of real numbers transforms Cauchy sequences of elements of E into Cauchy sequences, then f is uniformly continuous on E. Let f be uniformly continuous on E and let (xn) :=l be a Cauchy sequence of elements in E. Given £ > 0, there exists o > 0 such that if x' and x" are in E and I x' - x" I < then lfCx') - , f (x") I < £. But, associated with the number o > 0, there exists an index n0 such that if m,n � no then l xm n < o; therefore, lfCxn) - f(xm) I < £, which shows that {f(xn)}:=l is a Cauchy sequence. 61 .
A
uni formly con tinuous on
Yl
Cauchy sequence
A
Sol ution .
o
-
X I
398
CHAPT E R 4
the other hand, assume that the function f is not uniformly contin uous on a bounded set E. The negation of uniform continuity may be thus ex pressed: there exists an 0 such that for any 0 there exist points x�,x� in E such that lx� - x�l however, lfCx�) - f(x�) Since E is bounded, it is possible to extract a convergent subsequence of (x�) n=l ' namely (x�k\-- 1 ; it is not implied, however, that the limit x0 of (x�k) k=l belongs to E '· at any rate, (x�k )�- l is a Cauchy sequence. In the same way, there exists a Cauchy sequence (x�k) k=l having the same limit x0 , as it follows from the inequality lx"nk - x0 I � since both l x"nk - x'nk I and lx'nk - x0 1 may be arbitrarily small. Therefore, the sequence (xk)�= l ' where x 1 x� 1 , x2 = x� 1 , ... , x2k-l = x�k , x2k = x"nk , . . . , obtained by "mixing" the two convergent subsequences (x�k)�=l and (x�k)�=l ' is again a Cauchy sequence. thus, By hypothesis, we deduce that {f(x )}�=l is also a Cauchy sequence; that if k,k' k0 , · then I f(xk) given any 0, there exists an index kk0 such f(xk , ) I However, this is not the case, since for any k0 we have l fCx2k0_ 1 ) - f(x2k0) I = lfCx'nk0) - f(x"nk0 This shows that f must be uniformly continuous on the bounded set E. On
£
>
a >
I
£.
CX>
CX>
-CX>
=
£
>
1 ,
w (x) ,
1
:s;
(E . 61)
x,
where w(x) is defined by the equation we -w
=
xe -X
w (x)
�
1
for all x
>
0,
(E . 62)
with when 0 < X :'> 1 :s;
when 1
:s; 1
(E . 63)
X.
The function w (x) is explicitly given by w (x) = xs (x) ,
X
>
0,
( E . 64)
where s i s the inverse function of x-:-1'
log x
that is,
s-:-T log s -
x,
_
X
>
0;
(E . 65)
hence F (x)
1
0 < X :'> 1 ,
s (x) ,
1
:s;
(E . 66)
x.
Indeed , the expans ion of the funct ion a ( A) defined by ae-a = A ,
for A
:s;
1/e,
into a Burmann- Lagrange series being (see, for exampl e , A. Hurwitz & R. Courant : Vorlesungenen uber allgemeine Funktionentheorie und e l l iptische Funtionen , 4th Ed. Berl in-Gott ingen-Heidelberg - New York : Springer 1964 ; pages 1 3 8 , 141 , and 142 are relevant) a ( A)
_ nn-1 _ An , n n=l ! \"' L..
IAI
n 1 - X2 1 - X
•
n- l ( 1 + X2 )
. .
•
n-1 (1 + x 2 )
-+ oo .
REAL FUNCTIONS
423
Thus , for l x l < 1, l imn + oo P (x) n
PROBLEM 84 .
{
. 1 1m t + oo
Compute
1 2t 2t t + t2 + 1 2 + t2 + 22 +
( Cm+l)h ./h
But
and
+
""
f(x) dx
2t_ _ --:::2 t + nn
+
••
h[f(h) + f(2h) +
$
•
l""
f(x) dx $ h
J(""
f(x) dx
Ia""
f(x) dx
�
=
nL
-
f (nh) $
PROBLEM 85 .
1
""
• • •
+
•
•
•
1 /t . Then + f(mh) ]
$
}. {mh ./ f (x) h
f(x) dx .
2 arc tan h
�
and the desired l imit equals
�.
Show that for all real numbers x we have
j e" ( l 2 - 6x + x 2 ) - ( 1 2 Sol ution .
•
Let f(x) = 2/ (1 + x2 ) and h
Sol ution .
and, as m
1 / (1 - x) .
+
6x + x 2 )
I $
1 5 !xi 60 J x l e .
Consider the identity
1 +x +x2 + ex - (1 + x + x 2 /2) = 3 ! 3 ! 4 5! X Differentiating twice the above identity, we get
dx
CHAPTER 4
424 =
=
=
�
5.
+
60
+
1 1( 60 1
• • •
+
• • •
+
•••
+
(n
+
(n
+
+
(n
l) (n + 2) xn + + 5) ! 3) (n
60
+
(n
1 4) (n
+
+
3) (n + 4) (n
• • •
n X 5) n ! +
5)
+
Xn + n!
-
... . )
But 60 l x ln - l x ln ...,.(n .- -=-..,...: ..,. . .: :. .,. .,. .=-3) (n 4) (n 5) n ! n! -
-
< --
+
+
+
and so x 60 e ( l 2 - 6 x
J
+
x2 )
; (12
X
+
6x + x2 )
I�
which yields the desired result . The result in Problem 85 can be generalized to :
Remark .
elxl for any real number x .
PROBLEM 86 .
(1
+ X
Show that the product
l -) ( _ -1 -1 ( -1 a - 1 1 - 2a - -1) ( 1 3a - 1) 1 - 4a - 1) 1 (2n - l) a - 1) ( 1 - 2na l - 1) +
( +
-
-
+ oo ,
By Problem 102 of Chapter +
+
(
+
•
provided that a f 0 , 1 , 1/2 , 1/3 , 1/4, . . .
1, 1 -) ( -1-) ( -1 -) ( -1 ) (1 a - 1 1 - 2a - 1 1 3a - 1 1 - 4a - 1 1 (2n - l) a 1) ( 1 - 2na 1 - 1) X
• •
l
tends to the limit 2 l/a as n Sol ution .
X
1
-
X
• • •
REAL FUNCT I ON S
425
(n (n l)al)a- 1"(n (n 2)a2)a- 1 (n (n n) n)a a- 1 +
+
+
+
+
+
--;:;---Jt . . . { { - n *)a Jt { 1 n --)a n l
1
- l
1
(1
+
1
(1
+
2
1
But the foregoing expression tends to the reciprocal of exp � - al fcl _1 d_x _x � exp(- -a1 log 2) = 2 - 1/ a as n + "' because nI log (1 - 1 ) lim nI 1 1 . lim n (1 �)a n +oo k=l n n+oo k=l n (1 �)a n Indeed, [log(l x) - x[ x2 for [x[ i· Hence, letting �,n = (l -+ 1-nk) a and Bn 1 we see that nI log(l B ) - nI B B • nI 2 B --k ,n n k=l k,n n I n k=l -K,n n I k=l whenever --k ,n Bn 1/2 . By taking n large enough, we get that I --k ,n ['Bn 1/2 for k = 1 , 2 , . . . ,n and that 2 B Bn • k=lnI Ak,n n differs from by as little as we please. s a particular case of the result in Problem note that for a = we have 1 3 5 7 9 11 • • = . 1 The limit of +
=
l
+
+
�
+
�
n'
+
I A.
I
A.
A
�
A�
�
A.
0
Remarks .
A
86
2
2"2"6"6"1o"1o
12
�
CHAPTER 4
426
(n 2 + 1) (n 2 + 2) (n 2 1) (n 2 - 2)
(n 2 + n) (n 2 - n)
.!. 1 + .!. n n 1 1 - .!. n .!. n 1
2 n 2 n
1 n 1 n
+ - -
n -1 1 +n n 1 nn n1
is
1 Ia 1x dx � exp 1 - fa x dx � 1
exp
= e
as n � oo. The verification is s imilar to the method used in the Solution of Problem 86 .
Find the value of
PROBLEM 87.
I n=O
S
(-l) n (p+� - l ) rn ,
\r\ < 1
as the solution of a differential equation . Let un denote the (n + l) st term of S ; then
Sol ution .
u _n_ = -r p un- 1
+
n - 1 n
and n=O
I
nun
-r
I
nun
-r
I
(p + n - l ) un-1
I
(p + n) u . n
n=O
or n=O
n=O
But
I nu n=O n
dS rdr
and so (E. 71) becomes dS dr
+
____£___
1 + r S = 0,
(E . 71)
427
REAL FUN CT I ONS
whence log S + p log (l + r) + C If r
0 , we have S = 1 and C
S = C ( l + r) - P .
or
0
1 ; therefore
1
s
PROBLEM 8 8 .
Express the infinite series
as a definite integral , and find its value . From Problem 8 7 we get
Sol ution .
l + 21 X
1
+
1 • 3 • • • (2n - 1) n 1· 3 2 1 •3•5 3 2 • 4 X + 2 • 4 • 6 X + • • • + 2 • 4 • • • (2n) x + • • •
and so
• ••• - 1) Xn- 1 + • • • + 1 23 · 4 • • •(2n(2n) Thus
=
10
I1="X"
1 1 dx xll="X"
PROBLEM 89 .
2
f rr /2 tan t dt
Show that , for a
a- 1 - 2 a- l + 3 a- l lim 1 na n + oo Sol ution .
n-1
I .!. n k=1
2
0
Let f(x)
) c - 1 l - 1 f (� n
...
>
2 log 2 .
0,
+ (-l) n- 1 n a- 1
0.
xa- 1 Then , for m = [n/ 2 ] , m
�n I k=1
n-1 ) f(� .!. I f (�n) n n k =1
+
•••
428
CHAPTER 4
which tends to Remark .
� l f(x) dx - )"0 1 f(x) dx
For a s imilar quest ion see Prob lem 43 in Chapter 3 .
PROBLEM 90 .
{
Show that
-x1 1·m + X 1 X + 1- 0
X
X2 - --1 + X2
X �X (1
___£____
_
+
X2
+
1
For l x l
Sol ution .
!+X
0 as n + "'
p=l p=l n n = n Thus, the sequence (Sn) :=l ' being bounded below and above by sequences that converge to�l/2, must itself converge to 1/2. Therefore Tn exp(Sn) con verges to e +
l:
_.E._
l:
4
=
4
->-
.!_
"' ·
2•
PROBLEM For n = 1,2,3, ... , let g be continuous functions on a closed, bounded interval [a,b] such that n 1. If f is a continuous function on [a,b], show that all terms of the numerical sequence ab gn (x) f(x) dx, n = 1,2,3, . . . are situated between the smallest and the largest values of f on [a,b]. Clearly, 99 .
Sol ution .
�
�
whenever m f(x) M for all x [a,b]. Alternately, since f is continuous, there is a point t in [a,b] such that (see Problem 17) ib gn (x) f(x) dx = f(t) ib gn (x) dx. E
PROBLEM 100. Let a1 , a2 , . . . , ap be positive. Show that n � n ,;a; • • • n ,;a- n ap . p n-rlimoo +
+
+
REAL FUNCT I ONS
437
In Problem 77 of Chapter 3 we took up the case p 2 . The proof of the present claim is completely analogous to the proof given in the special case p = 2 . Sol ution .
=
PROBLEM 101 .
Show that
1 (4n 4n+ 3)+ (2n4 + 1) 211 h�
Sol ution .
�
>
� 2n (2n + 1) . 11 l !z � · l 4n + 1
I n the Solution o f Problem 9 6 we noted that , for k an integer
larger than 1 , 11 1 2
2 • 4 • • • (2n)
> 1 • 3 • • • (2n - 1)
sink x dx
=
k
�1
�
11/2
s ink-2 x dx;
hence
{ 11/2 }" s ink x dx 0
(k - 1 ) (k - 3) • • • 4 . 2 k (k - 2) • • • 5 • 3 • 1
if k is odd,
(k - l) (k - 3) 3 • 1 11 k (k - 2) • • • 4 . 2 "2
if k is even .
Since (sin x - 1) 2 � 0 , we have the inequal ity 1 � 2 sin x - sin 2 x . Multi· by s 1n · 2n- l x an d by s1n · th 1s · 2n x, an d 1ntegrat1ng · b etween · p 1 y1ng " 1nequa1 1ty 0 and 11/ 2 yiel ds the desired resul t . ·
PROBLEM 102 .
Show that
1 1 1 1•3 1 • 3 · 5 -1 1 2n + 2 - + 2 · -2n + 4 - + -2 • 4 · -2n + 6 - + --· 2 • 4 • 6 2n + 8 + -2 • 4 • 6 • • • (2n) 3 • 5 • 7 • • • (2n + 1) Sol ution .
1
·
\
We first note that , for x \