Problems and Propositions in Analysis (Lecture Notes in Pure and Applied Mathematics)

• • •

Sol u tion .

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A R I THME T I C AND COMB I NATO R I C$

63

Show that the reciprocal of every integer greater than 1 is the sum of a finite number of consecutive terms of the infinite series Ij=l j (j 1 + 1) . Since 23) j(j 1+ 1) j1 1 we see that b-1 1 1 1 j �a j (j + 1) Thus the problem is equivalent to that of finding positive integers a, b such that PROBLEM 90.

Solution .

]+1

(E .

a - b·

for fixed integer m 1. From (E.23) it is obvious that a solution is a = m - 1, b = m(m - 1) . Therefore, if m 1, 1 -m1 = m(m-1)-1 j=m-1L J J + 1) ' and we have what we set out to do. >

>

T

PROBLEM 9 1 .

y. X

Find all positive integers x and y with x y such that f

The only solutions are x = 2 and y Y= andx x = and y = 2. To prove that these are the only solutions, since x y , the prime factors of x and y must be the same. Write aP n and y n where p1 , p2 , . . . , Pn are primes and the a's and b's are positive integers. Now Sol ution .

4

4

CHAPTER 1

64

xY pal ly pa2 2y Pnany and y pbl 1x pb22 x Pnbn so that we have the list of equations aly b 1x, a2y b 2x, . . , any bnx. Suppose that x y, so that al b l , a2 b2 ' . . ' an bn ' and x is divisible by y, that is, xy kykyfor some integerk k. Thus thek-1 original equation may be written as (ky) y . Hence ky y , or k y . Now, since x y and x ky, then k 1 and we must have y 1. If y k 2, then we get 2 2 2-1 . If y 2k-1and k 2, then k 2k-l yk-1 and if y 2 and k 2, then k 2 y y Therefore y 2 and k 2, and x kyx, 4 is the only solution in positive integers if y. Similarly, if y the only solution is y 4 and x 2. =

=

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92. Show that the equation mnm nmn has no solutions in positive integers with m F n. Without loss of generality assume n m 1. If m 1, or if m 2 and n 3 or 4, we readily find that nmn mnm Suppose m 2 and n 4. Since f(x) xl/x 1s a dnecreas21ng functn1on form x e, 4 1/4 n1/n , or 4n n4 , or 2n n2 , or n2 2n , and nm mnn m Finally, suppose n m 3. Thenm ml/mn n1/n ' or mn nm , and again nm mn It follows that the equation mn nm has no solutions in positive integers if m F n. PROBLEM

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Sol ution .

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93. Find the product 3- 1 ) 3 - 2 3- 4) (1 3

PROBLEM (1

+

(1 +

) (1

+

+

- 8) • • • ( 1

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>

>

+

·

3-2n) .

>

ARI THMETI C AND COMB I NATO R I C$

If the given product is multiplied by 1 1 we have (1 - 3- 1) (1 + 3 - 1) (1 3- 2) (1 + 3 ) (1 3-2n) 3•

Sol ution .

-4 • • •

+

=

Thus

65

+

( 1 - 3- ) ( 1 + 3 ) 1 + 3-2n) (1 - 3-2n) (1 + 3- 2n) 1 - 3_2n+l -4 • • • (

4

1 - 3_ 2-1n+l i2 cl 3_2n+l ) . 1-3 =

PROBLEM

Show that + -n1 is not an integer for any n. Assume 1/2 + 1/3 + + 1/n t, an integer. Let 2m be the largest power of 2 less than or equal to n, and consider 94 .

• • •

Sol ution .

• • •

=

Every termm in this equation is divisiblemby a higher power of 2 than the term n!/2 . (To see this, observe that 2 mis the monly positive integer less than or equal to n which is divisible by 2 , so m2 "knocks out" more 2s from c n! than any of the other denominators.) Let n!/2 2 q, with q odd. Divide both sides of the equation by 2c . Then n!/2m2 c is the only odd term. This is a contradiction. =

CHAPTER 1

66

Show that any n 1 integers taken from 1, 2, . . . , 2n con­ tain a pair a and b such that a divides b. For any integer m, m 2kt, with t odd. Write each of the n 1kgiven integers in this way. If a is one of the given integers and a 2 t, then call t its "odd part". Since a 2n, t 2n, so t must be one of the n odd integers between and 2n. Then, since we have n 1 odd parts, some two of them must be equal. Hence for the given set of n 1 integers there exist a and b with a 2rt and b 2st. Then either r s and a di­ vides b or vice versa. Note that the set of n numbers n 1, n 2, . . . , 2n contains no pair such that one divides the other. +

PROBLEM 95 .

Sol ution .

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0

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Remark .

Find the sum ••• (n n 1) ! Since (k k 1) ! 1 (k 1 ! . we see that (n n 1) !

+

PROBLEM 96.

+

+

·

Sol ution .

k!

+

+

1)

+

1 - (n 1 1) ! +

+

·

+

Find the sum 1•1! 2•2! 3 •3! ••• n • n! . Since k•k! (k - k!, we see that 1 · 1! 2•2! 3 · 3! ••• n•n! 2! 1! 3! - 2! 4! 3! • • • (n 1) ! - n! (n 1) ! +

PROBLEM 9 7 . Sol ution .

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1.

=

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1) !

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+

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+

ARI THME T I C AND COMB I NATO R I CS

67

PROBLEM Evaluate the sums Pn = + 2x + 3x2 + + nxn-l 98 .

• • •

1

and

Since n+l sn + + 2 we see that upon differentiation �dx sn and dxd (xPn) Thus n (n + -l)xx)n2 + nxn+l and + x - (n + 2xn + (2n23 + 2n - l)xn+l - n2xn+2 - x) Sol ution .

= 1

X

1 1

X

X

-

X

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p

1

-

(1

1

1)

(1

PROBLEM Evaluate the sum Hn = tan X + tan X Let Tn = cos �2 cos � cos -.2n Since sin 2t = n2 sin t cosn t, we see thatd {2n sin(x/2n)}Tn = -nsin x, thatn is, Tn = (sin x)/{2 -nsin(x/2 n)}. But -Hn = dx (log Tn) = cot x - 2 cot(x/2 ) and thus Hn = 2 cot(x/2 ) - cot x. PROBLEM Give an example of ten consecute integers all of which are not prime. 99 .

2

2

1

4

1

4

+

Sol uti on .

X

4

100 .

68

CHAPTER 1

The set of numbers 11! + 2, 11! + 3, 11! + 11 has the desired property. Sol ution .

PROBLEM 101. Show that _x +1_l + --2 2+-1 + + 2n2n__ 1 2n+l +1 1 Since 1 1 -2-2 , 1 we get 1 k 1 k+l2 1 2 +1 2 -1 for k 0, 1, 2, . . . n, or k k2 - 1 2k2 + 1 2k+l2k+l 1 for k 0, 1 , 2, . . . , n. Adding the obtained results, we arrive at the validity of our claim. • • •

X

x-:-r

_.:..

_

X

-

X

-

Sol ution .

-z::-r - Z+l =

X

X

l -

=

z

-

X

X

'

l

X

-

X

PROBLEM 102. Prove the identity + 2)a- 1 (n (n+ n)+ n)a -a 1 ' (n (n+ +l)al)a- 1 (n (n+ 2)a provided that a 0, 1, 1/2, . . . , l/2n. We have (1 + _ a -1 _1) (1 - _ 2a 1 -} ( 1 + _ 3a 1 ) (1 + (2n-1)a1 ) (1 - 2na1 - 1) f

Sol ution .

-

1

- 1

-

• • •

-

1

ARITHME T I C AND COMB I NATO R I CS

because

(aa(2a- 1)- (2a2)3a(4a - 1) ( -3a2)•••(2n - 1) (4a -- l)a(2na 1) • • • (2na- 2)- 1) n •2 •••(na 1) (a-1) (2a-l) l•a•3•a•S•a•••(2n-l)a• [(n+l)a-l][ (n+2)a-l] ••• [(n+n)a-l](a-1) (2a-l)•••(na-l) n l•a•3•a•S•a•••(2n-l)a•2 [ (n+l)a-l] [ (n+2)a-l ] [ (n+n)a-1 ] (n+2)a- 1 (n+n)(n+n)a a- 1 (n+l)a- 1 (n+2)a (n+l)a

-=--.1•2•3•4• 1;:-.: .--;2,-.-;;.3--'••5=--•••2n .-n--=-o:. (n + 1) (n + 2) • • • 2n . =

PROBLEM 103. Find the sum of n numbers of the form 1, 11, 111, , 1111, The sum in question is ..., - -=-1 + • • • + -10 -9--1 + 10 2 9 - 1 + .::.10 3....:... 9� 1 {10 IOn9 - 1 n} . Sol ution .

= 9

_

PROBLEM 104. Verify the identity (xn-1 + nl-1) + 2 (xn-2 + nl-2) + ••• + (n - l) (x + �) + n x x The sum considered may be written as 1 _ + _2_ + • • • + n x- lJ + [xn-1 + 2xn-2 + • • • + (n-l)x] + n. (_ Xn-1 Xn-2 The first bracketed expression equals Sol ution .

69

CHAPTER 1

70

n - nxn-l + 1]-"1n x + 2x2 + + (n-1) xn-1 = --"-x[(n-l)x -'� n x (x - 1) 2 x (see Problem The second bracketed expression is obtained from the first one by replacing x by 1/x. Hence we get the desired result. • • •

-[

]

-----;;:--

98) .

PROBLEM 105. Verify the identity --32 1 2 + --34 1 4 + + (2n) 31 - 2n 2n n+ 1 + --= -n +1 1 + -n +1 2 + + -2n1 · We have 1 = 2k1 . 21 = 4k1 ( 1 - 1 ) 3 (2k) - 2k (2k) - 1 k - (2k- -1) 1) - (2k2k(2k + 1) + - 1)2k) = l_h2\2k(2k = 2\1[ 1 - 2k1 - 2k1 + 1 ) Therefore nL 1 k=l (2k) 3 - 2k = l2{(1 + l3 + + 2n 1- 1) + (l3 + l5 + + 2n 1- 1) + --2n 1 1 + _!__2n) } +2n 1+ 1 - 2(l2 + l4 + + _!__2n) } 2n 1- 1) - 1 + --• • •

_

_

• • •

Sol ution .

2i(":1

Zi(":l

• • •

___

2i<+1

2i<+1

• • •

.

___

+

• • •

Hence

nL 1 + n_ = 1- 1 - -2n1 + -·2n-k=l (2k) 3 - 2k 2n + 1 = -n +1 1 + -n +1 2 + + in (see Problem 42). ---;;:___

__

AR I THMET I C AN D COMB I NATO R I CS

71

PROBLEM Prove the identity a +a 1 + (a + l)b(b + 1) + (a + l)(b c+ l) (c + 1) • • • + '(_a_+_""l )""" (b:-+--:- k1")-.-.-.--:(""-"k +--=-1"""") 1 - (a + 1) (b + 11) • • • (k + 1) Since _a a+_1 = l - 1 (a + 1)b(b + 1) 1 (a + l)1(b + 1) ' (a + l)(b +c l) (c + 1) (a + l)1(b + 1) - (a + l) (b 1 l) (c + 1) ' and, in general, (a + l) (b + kl)• · •(k + 1) (a + 1) (b + 11) • • • (j + 1) (a + l) (b + l)· 1·· (j + l) (k + 1) ' the desired result follows immediately. 106.

---

·

Sol ution .

a+l

'

a+l

-

+

PROBLEM Let be a continuous closed curve in a plane which does not cross itself and let be a point inside Show that there are points P1 and P2 on such that is the midpoint of the line segment P1P2 . We introduce a rectangular coordinate system in the plane containing with the point being at the origin, that is, having coordi­ nates We now map a point P on with coordinates (x,y) into a point P' with coordinates (-x,-y); this mapping takes into a curve with being congruent to The point is both in the interior of and and so the interior of and have nonempty intersection. the other hand, the region enclosed by the curve cannot be properly contained in the region enclosed by the curve because the areas of these two regions are the same. Therefore the curves and must intersect. But if P2 is a point common to and then, being a point on it must be the image of a 107.

C

Q Q

C

C.

Sol ution .

C ( 0 , 0) .

Q

C

C

C. C

C

C'

On

C'

C

C

C

C'

C'

Q

C'

C' ,

C'

C'

72

CHAPTE R 1

point P 1 on under the mapping (x,y) (-x,-y) and (0,0) is the midpoint of the segment P 1P2 . +

C

Q

PROBLEM 108. Show that every positive integral power of - 1 is of the form where m is a positive integer. Let a and b, a b, be two non-negative real numbers, and n a positive integer. Set p (a b) n 2 (a - b) n Then p2 - (a2 - b2 ) n [ (a b) n -4 (a - b) n ] 2 and it follows that 1:2

liD - � .

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Sol ution .

+

+

+

Take a

= 15,

b

= �

�

s and t positive integers, s t. Then

.

Moreover, p2 is a positive integer. Indeed, p (a b) n 2 (a - b) n an (n2) an-2b 2 (n4) an-4b4 [ 1 (-1) n ]bn . Since a b s and t positive integers, and s t, we see that p is the sum of positive integers if n is even; if n is odd, p is the sum of numbers of the form k/5, where k is a positive integer. Thus, p2 is a posi tive integer for any positive integer n. Finally, taking s 2 and t 1, we get +

+

= IS,

+

+

= �

=

an

+

• • •

+

+

�

.

=

PROBLEM 109. Show that the curve traced by a pointer which pulls taut inextensib1e string passing round a given ellipse is a confocal ellipse.

ARI THMET I C AND COMB I NATO R I CS

73

The claim is a consequence of the following theorem, due to Graves: If two tangents are drawn to an ellipse from any point of a confocal ellipse, the excess of the sum of these two tangents over the intercepted arc is constant. - proof of this theorem can be found in volume I, on page 166, of Course in Mathematical Analysis" by E. Goursat, Dover Publica­ tions, Inc., New York 1959. Sol ution .

A

"A

PROBLEM 110. Let a and n denote integers greater than 1. Show that the quotient n(n l) (n n2)•••(na - 1) a is a fraction if a is prime, and is an integer if a is not prime. To establish the claim we distinguish two cases, according as a is prime or is not. I. The integer a is prime. Denote by x the exponent of the highest pow­ er of a that is a divisor of the numerator of the quotient. This number x, as we know from Problem 20, is given by the formula x [ �J [ naa; 1 ] [ naa; 1 J [ y [ n :/ [ n :3 1 ] J J where the square bracket denotes the largest integer function. Clearly we have [ na ; 1 [ y J a J [ na ; 1 [ n :a 1 a J J [ na � 1 ] [ n :/ J a and so forth. Thus the formula given for x reduces to x - [ -na-a-- -1 J +

+

Sol ution .

+

=

_

+

_

=

=

=

_

'

_

+

• • •

CHAPTER 1

74

=

which gives x n - 1. Thus the factor a is contained n - 1 times in the numerator of the quo­ tient. But the factor a is contained n times in the denominator of the quo­ tient. Hence the quotient is a fraction, and the first part of the claim is verified. II. The integer a is not prime. To show that, in this case, the quotient is an integer, we shall prove that all prime factors of a appear in the nu­ merator at least as often as in the denominator. Let b be any prime factor of a and B be its exponent. We have a bB q, and b divides Bn times the denominator. In the numerator this factor b enters a number of times y given by the formula =

n- [ y] [ b/ ] But, clearly, -

[ nb)+; 1 ] [ n ;/] �

[ nb)+� 1

J

[ n -/

b J and so forth. Thus the formula giving y reduces to �

hence we canB write y nb - 1 q - 1 + nbB-2 - 1 + + nq - 1, �

• • •

ARI THMET I C AN D COMB I NATO R I CS

75

and, consequently, y nq bB - B. It is sufficient, in order to establish the second claim, to verify the inequality nq b B - 1 - B Bn under the assumption that a and n are integers larger than 1. To do this we distinguish two cases, according as B is equal to or larger than 1. If B 1, the inequality reduces to nq - n or q + -.n It is sufficient that q be larger than But this is so for if q is equal to 1 and at the same time B is then a would be prime, contrary to the hypothesis. If B 1, the inequality can be written as nq(bB - 1) (n + l) (b - l)B or, putting b + c (with c being an integer greater than zero), nq{ (l + c) B - 1 } (n + l)Bc. We expand + c) B- and we divide the two numbers by c; we get nq{ B + B(B1 • 2- l) c + nB + B or nqB + nq B (B - c + nB + B. But clearly we have nqB nB. Moreover, Since B and n are both larger than we also have nq B (B - c B. Thus the inequality is satisfied and the second claim is established as well. - 1 !)":"1

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1 �

1

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1,

1.

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···

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1)

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1,

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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 . >

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=

=

=

=

-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

+

+

=

+

<

+

<

+

� �

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=

+

=

+

=

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+

+

+

Remark .

+

0,

Sol ution .

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or, dividing by yY , (1 y Y - yn But (1 .!!.y) Y is less than en which in turn is less than 3n ; hence it can not surpass 3n . If we let y 1, we have n 1, x 2. +

.!!.)

+

=

=

=

CHAPTE R 1

78

For y

=

2, 1 2

z·

= 3,

This equation is satisfied for n hence x For n the first mem­ ber becomes negative, and so much more for n Hence we have the follow­ ing two solutions: y y and there are no others with x y. Let y x; putting y x + n, xx+n - (x + n) x and, on dividing by xx , n + �) x x· But x, after the preceding discussion, can not be smaller than however, for x the first member becomes =

=

1,

>

1,

X

2,

X

=

3. 3.

2

3,

>

>

=

=

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 +

<

• • •

m

+

g (t 2m- l )

+

2

81

g (t 2m)

2

+

CHAPTER 2

82

+ 2m (since, by assumption, not all of the elements t 1 , t2 , . . . , t 2m are equal to one another, they can be grouped so that, for example, t 1 t 2) . Thus the claim is valid when n is a power of 2. Step 2. Let n 2 and n not be a power of 2, that is, let 2m-l n 2m . Then + + tn + s 1 + s 2 + + s ) n+p • • •

F

>

<

• • •

• • •

(here t 1 , t 2 , . . . , tn are not all equal to one another) , by what we have already established in Step 1 . Put +t sp n n Then + tn p -t l-+ -t 2-n+---and g (:l + + t� : ;1 + + the other hand g(t 1 ) + + g(tn) + g(s 1 ) + + g(sp) n+p n n+p From the last inequality we obtain + t g (t ) + + g ( t ) n n) ___l__ n ____n_ • • •

• • •

• • •

• • •

On

• • •

• • •

<

____

PROBLEM 2. Show that, for xi with i 1 ,2, . . . ,n, >

0

<

IN EQUAL I T I E S ..., ---:-

-'n:.::.._ 1 ----:;--,-1 + - + . . . x l x2

--,-

__

+

1 xn

83 �

+

+

x 1 x2 + • • • xn n lx x · · · x � ----1 2 n n

holds w ith equal ity being obtained only in the case x 1 = x2 = • • • = xn . In other words , the harmonic mean of n posit ive numbers is less than or equal to their geometric mean which in turn is less than or equal to their arith­ met ic mean . Let g (t)

Sol ution .

it is clear that

(

t g l since (t 1 1'

: t 2)

- log

+

<

- log (1

+

t) . From the graph of the function g

; g (t 2 )

g (t l )

t 2 ) /2 is the midpoint of the segment [t 1 , t 2 ] . Hence , by Prob lem

(+ + + ) tn t1 -=-----__ _c::_ n

1

<

log (1 _

+ + t1)

n

or n or

+ Putt ing 1

+

ti

n ix • • • x 1 n

+

n

••• n

x i we get <

xl

+ +

X

n

n

Obvious ly, if we assume the poss ibil ity x l n ix • • • x n 1

tn

�

xl

+ + n

x2

xn , then we wi ll get

X

n

Final ly, replacing x i by 1/xi in the foregoing inequality we obtain that the harmonic mean is less than or equal to the geometric mean for the same set of positive numbers x 1 , . . . , xn .

CHAPTER 2

84

PROBLEM 3 .

(

x l + x2

Verify that , for x i

: • • . + xnt

s

>

x + x + � � n

0

with i

1 , 2 , . . . ,n,

• • • + Xnk

when k i s a positive integer. Sol ution .

Let g (t) = t k . Then

The des ired result now fo llows from Problem 1 .

PROBLEM 4 .

Sol u tion .

Let a i

>

0,

bi

>

0 for i = 1 , 2 , . . . , n . Prove that

Let g (t) = log (1 + et ) . Then

)

t By Problem 1 we therefore get (changing the notation from e to exp (t) )

.. . . ' 'n

(

t + t2 . log 1 ' exp l <

or

log (1 + exp (t 1 ) ) +

• n•• + log ( 1 + exp (tn) )

•••

1 1 + t2 + + tn 1 + exp --'------__c;;_ n Putting exp (t) = s , that is, t = log s , we obtain

>

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 �

• •

<)·

•

For i = 1 , 2 , . . . , n , let

0 and x .

J

>

+

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. :<;;

p

X

:<;;

9

Sol ution .

Hol der ' s Inequali ty :

PROBLEM 14. Let ai and bi for i = 1, 2, n and suppose that k 1, k' > 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:

<

L:

1

Cauchy-Schwarz Inequa l i ty .

Sol ution . 1

1

1 1

1

1

1

1

J

:<>

92

CHAPTER 2

Let a i

PROBLEM 1 5 . Minkowski ' s Inequal i t y : . . . ,n and suppose that k > 1 . Show that

) l/k

(a.1 + b 1. ) k

:>

(

(

)

n l/k L ak. + i=l 1

0 and b i > 0 for i = 1 , 2 ,

>

)

n l/k L b k. . i=l 1

(E . 6)

We observe that

Sol ution .

Applying to the l ast two sums inequality (E . 4) in Probl em 1 4 , we get (since 1/k + 1/k I 1) :

n

I

(a 1. + b i ) k

i=l

+

(I

(

)

)

l/ k n 1 1/k 1 bk1. ca 1. + b 1. ) ( k -l) k I i=l i=l n

[( I n

i=l

) l/k + ( nI bk. )l/k] ( nI

ak1.

i=l

1

i=l

) l/k 1

c a1. + b 1. ) k

and finally arrive at inequal ity (E . 6) by division using the last factor as divisor.

PROBLEM 1 6 .

Let

An

n

with a . > 0 for j J

+ an 1 , 2 , . . . , n . Show that , for p > 1 ,

m p- l m L Anp -< _L_ L p - 1 n=l An an . n=l By Probl em 12 ,

Sol ution .

Anp- 1An- 1

:>

(p - l ) Ap + Ap __:.:n __::cn:.._-.=.1 p

_ _ _

_

I N EQUAL I T I ES

But

93

Anp - p - 1 Anp-l an = Anp - __E__ p - 1 [nAn - (n - l)An-1 ]Anp-l __E__

s

p - nAnP} . AnP(l - 2L) p - 1 np -- 11 { (p - l)Anp An-p 1 } = p -1 -1{Cn - l)An-1 +

+

PROBLEM 17. Verify the a1 , a2 , , an positive,

Inequali t y of Hardy and Landau :

For p 1 and >

. • •

Using Holder's Inequality (see Problem 14) we get from the result in Problem 16 the following: mL\ A < __E__ m\ Ap-l a - � m\ aP)1/p( m Ap )1/p n=l n p - 1n=l n n p - 1 n=l n n=l n where p' = p/(p - 1). Dividing by the last factor on the right-hand side (which is certainly positive) and raising the result to the power p, we get the desired inequality. Sol ution .

L

•

L

L

<

PROBLEM 18. Prove the If an for n 1,2, . . . , then n L la1 a2 • • •an e:E an n=l n=l provided that E�=l an converges. From Problem 17 we see that l/p a2l/p ••• anl/p)p ( a 1 n n=lI provided that E:=l an converges; the " <" turned into because of the limit process involved (as m If we now let p and note that > 0

Inequali t y of Carleman :

3,

co

s

Sol ution . co

+

+

� co ) .

+

s

� co ,

" s"

CHAPTER 2

94

+ a2 + • • • + an

n /a a • • • a � a l n 1 2

n

------

(see Prob lem 2) , we obtain the desired result since (1

+ _p-l_1) P

+

PROBLEM 1 9 .

e

as

Prove

p

+

"' ·

Hadamard ' s Inequali ty :

For any determinant

ann with real elements a ik we have the est imate A2 �

TI

n

n 2 L a . i=l k=l ik

Sol ution . We commence with some observations . In the Cauchy-Schwarz Inequal ity (see Problem 14) equal ity occurs if and only if a 1 /b 1 = a 2 /b 2 = • • • = an/bn . From the theory of determinants we recal l the following fact :

if h if h

f.

j, j,

where �k denotes the cofactor of the el ement �k · Final ly, a continuous real -valued function on a closed bounded set in Euclidean space assumes its largest value on this set . We now return to our problem and let the el ements a ik vary, but keep the sums of the squares n 2 L a k=l ik

c 2l.

(i

1, . . . , n)

2 is the largest value which the funct ion A2 of the elements fixed . I f Amax a ik can assume under these n conditions (such a maximum exists because we are clearly dealing with a continuous function on a bounded closed set) , then

INEQUAL I T I E S

95

the el ements of Amax in every row must be proportional to the corresponding cofactors . Indeed, for a fixed h A = a.nl Ahl + • • • + a.nn·A.nn '

where �k is the cofactor of ahk ' and by the Cauchy-Schwarz Inequal ity n

n

k= l

k= l

2 . I ahk I

�k

n ch2 • I Ah2 k ; k= l

if. � k is not proportional to Ahk ' then A2 can certainly not assume its max­ imal value because then the inequal ity sign holds while by changing the n values �k (k = 1 , . , n) under keeping fixed c� and � k the square of the determinant can be made equal to the right-hand side expression . Mult iplying Amax with itself we get .

1T

.

n

c 2. i=l 1 because , for h

r j, 0

� l Aj l + • • • + �n Ajn and therefore , for h

r j,

0

as a]. n

A.·

Jn

Thus for the initial determinant A we certainly have A2

::;

1T

n

c 2. i=l 1

1T

n

n

I a2 i=l k=l ik "

Equal ity holds if and only if, for h

r

j , � l aj l + . . + �n ajn .

0.

Remarks . Hadamard ' s Inequal ity has the following geometrical interpre­ tation : The volume of a paral lelepiped obtained by n vectors of given length in an n- dimens ional space is the largest when the vectors are mutual ly per­ pendicular. It is clear that if all elements a1. ]- of the determinant A are bounded in absolute value by the constant M, then

CHAPTER 2

96

PROBLEM 20. T (x)

A trigonometric polynomial

1 a + n ca cos kx I k 2 0 k=l

+

is an expression of the form

b k sin kx) .

If I an I + l bn I > 0, then the number n is cal led the order of T . It can shown that a trigonometric polynomial of order n cannot have more than real roots in [o, 21!) , even if each multiple root is counted the number t imes it occurs (see e . g . , A Zygmund, Trigonometric Series , vol . 2 , p . Cambridge University Press , 1959) . Prove the Inequality of Bernstein : If T (x)

n 1 ca cos kx 2 a0 + k=l k

I

+

be 2n of 2,

b k sin kx)

and I T(x) I <; M for a l l x E [0 , 21! ] , then the derivative T ' satis fies I T ' (x) I nM for all x E [0 , 21!] . Moreover, T (x) = sin nx shows that the result is the best possible. <;

Sol ution .

Suppos e , on the contrary,

sup I T ' (x) l = n K, 0 $ X $ 21T where K > M. Since T' is cont inuous , it attains its bounds and so for some c, T ' ( c) = ± n K. Let us suppose that T ' (c) = nK . Since nK is a maximum value of T ' , T " (c) = 0 . Define S (x) = K sin n (x-c) - T (x) . Then R (x) = S ' (x) = nK cos n (x- c) - T ' (x) and S and R both have order n . Consider the points (1 .,; k .,; 2n) .

Then

I N EQUAL I T I ES

97

Each of the 2n intervals

then contains a root of S , say S (y i )

=

0, where

u i < Y i < ui+l with 0 $ i $ 2n - 1 . Clearly y 2n- l < y0 + 2 � . Put y2n Roll e ' s Theorem, there is a root x1. of R ins ide each interval (y1. , y . 1 ) , 1+ where 0 $ i $ 2n - 1 . Evidently x 2n- l < x0 + 2� . Now R (c) = nK - T ' (c) = 0. Since the trigonometric polynomial R of order n has at most 2n real roots , it fol lows that , for some k , c = xk (mod 2�) . But R (c) = -T" (c) = 0 . Therefore c (and so xk ) is a double root (at least) of R. Therefore the x i with 0 $ i $ 2n - 1 provide at least 2n + 1 real roots of R. This is only poss ible if R = 0 and so S is a constant function . But S (u0 ) > 0 and S (u 1 ) < 0 and we have a contradiction .

PROBLEM 2 1 . I f P is an al gebraic polynomial of degree n and I P (x) I $ M for all x E ( - 1 , 1 ) , show that nM I P ' (x) I $ ---

for al l x

\[;7

E

(- 1 , 1) .

Sol ution . This is cl early the algebraic equivalent of Bernstein ' s Inequal ity (see Problem 20) and i s obtained by putting T ( 8) = P (cos 8 ) and noting that T ' (B) - P ' (cos e ) s in e . =

Remark .

- 1 and 1 .

A

The bound for P ' (x) given in Problem 2 1 fai ls at the endpoints better result , due to Markoff, will be taken up in Problem 24 .

PROBLEM 2 2 . Tn (x)

=

Setting cos 8

cos ne

=

=

x, the express ion

cos (n arc cos x)

are polynomials in x of degree n , called Chebyshew polynomi al s . The leading coefficient of Tn (x) is equal to 2n-l . The first five polynomials of this kind are

98

CHAPTER 2

T 1 (x)

x,

r 2 (x)

2x 2 - 1 '

T 3 (x)

4x 3

3x,

r4 (x)

Bx4

si + 1 '

r 5 (x) = 16x 5 - 20x 3 + Sx. The roots of Tn are al l real and dist inct and l ie in the interior of the in­ terval (- 1 , 1) . The roots of Tn are - 1) 11 cos (2k 2n

(k = 1 ' 2 , . . . , n) .

Let x 1 , x 2 , . . . , xn be arbitrary distinct real or complex numbers . Set with a0 -F

o,

1 f (x) f' (xk ) x - xk

Every polynomial P o f degree n- 1 may b e represented by means of its values at the points x 1 , x2 , . . . , xn as follows

this is the Lagrange interpolation formul a . The polynomials fk are cal led the basis pol ynomials of the interpolation . Prove the fol lowing result : Let X

k

- 1) 11 = cos (2k 2n

for k = 1 , 2 , . . . ,n

be the roots of the Chebyshew polynomial Tn . If Q is a polynomial of degree less than or equal to n- 1 , then Q (x)

I NEQUAL I T I E S

99

Sol ution .

Since Tn (x)

T; (x)

n

__ __ __ __

�

cos (n arc cos x) , we get

s in (n arc cos x) .

From (2k - 1) 1T 2n

arc cos xk we obtain

s in (n arc cos xk )

s in (2k 2- 1 ) 11

and so

-� f .1 . - xk To prove the equal ity in question , we note that on both s ides we have poly­ nomials of degree � n - 1 and so it is sufficient to show that they agree for n values xk . As x + xk , Tn (x) x - xk

--- +

T'n (xk)

Al so, for x = xk ' every term on the right-hand s ide except the k-th vanishes because, for i = 1 , 2 , . . . , n and k # 1 , 0 as the x i are roots of Tn . Thus the right-hand s ide express ion is a Lagrange interpolation polynomial for Q .

PROBLEM 2 3 .

Let Q be a polynomial o f degree � n - 1 and for x

Show that in [- 1 , 1 ] we then have the estimate I Q C x) I � n . Sol ution .

With the notation of Problem 22 , if -x 1

=

xn � x

�

E

[- 1 , 1 ]

x1 ,

CHAPTER 2

1 00

Hence the assertion is true for xn � x � x 1 . For the remaining points of [- 1 , 1 ] we apply the Lagrange interpolat ion formula to the polynomial Q found in Problem 22 : 1 n Q (x) = L ( - 1 ) k- l n k=l Since e ither x I Q (x) I

<

- _!_n <

But

n 1 - xk Q (xk )

x-:-x-·

Tn (x) k

xn or x > x 1 , the numbers x - xk have the same sign . Hence

I ��� -

k=l x - xk

and so

Therefore I Q (x) I

�

� � T� (x) I ·

But , s ince x = cos 6 , T� (x)

n s in n6 sin 6 '

which gives I Tn' (x) l � n 2 because ! s in n6 1 � n i s in 6 1 for eas ily be verified by induction) .

PROBLEM 2 4 . gree � n, then

Prove

Markoff ' s Inequal i t y :

2 I P ' (x) I � n ! sup I P (x) I L l �x�l for all x

E

[-1 , 1] .

I f

- co <

6

< co

(as can

I f P is any polynomial of de­

I N EQUAL I T I E S

1 01

If

Sol ution .

sup I P (x) I - l::;;x ::;; l

M,

=

take Q (x)

�

=

Mn

in Problem 2 3 . The stipulations in Problem 2 3 are satisfied in view of Prob­ lem 2 1 . Taking P (x) = Tn (x) , where Tn i s as defined in Prob lem 22 , we see that Markoff ' s Inequality is the best possible . Indeed, Remark .

n sin (n arc cos x) V l - x2

=

Tn' (x)

and so T� (1)

=

�n ne n ss1n e

2 n .

PROBLEM 2 5 .

Let 0

<

x

<

�/2 . Show that

We have

Sol ution .

2 s in x - s in x = 2 sin x ( 1 - cos

i

i

ix)

. 2 1x . . 1 x s 1n 4 s 1n 2 4

Hence . 1x - S ln . 2 S ln 2 since sin t

<

X

� 2 4 � 2 (4 ) ,

<

t for t > 0 . Thus

. 1 x - S ln . 2 S ln 2

X

<

1 3 8x .

(E . 7)

Replacing x by x/2 , x/4 , . . . , x/2 n- 1 , we find (E . 8) . 1 x - s 1n . 1x 2 s 1n 4 8 2 s in _!_X 2n

-

sin

<

1 X 3 8 (4) ,

_!_ IX n2

<

_2_)

3 !:_( 8 \ 2n- 1

(E . 9)

(E . 1 0)

CHAPTER 2

1 02

(E . lO) by 1 , 2 ,

Mult iplying inequalities (E . 7) , (E . 8) , spect ively, and adding them , we get 2 n s in ....!.x... - s in x 2n

+

Pas s ing to the l imit as n

(•

X s 1n 2n ---

l im n -+ co

X -

2n

X

l im n -+ oo

1 + 1 + + -

1

1

i

4

24

oo , we find

1 + --

42

1 + ••• + + - + -

�1

+ ....!.... + ....!.... + • • • +

�

- o in

4

But

(

.!x 3 1 8

<

42

4n-l

_ �t 1 4n- l

and

f

1 1 - 4

= -1 =

.

X s 1n 2n l im --X

Consequent ly X

-

. S 1n

X

-

=

1.

1 x3 . 6

<

PROBLEM 26 .

Show that + ....!.... > 21il+T - 2 . rn

Sol ution .

lil+T

_

Since

rn

=

1 .;:. rn+T

__

w e have

....!.... rn

>

Therefore

2 1il+T - 2 1il.

rn

_ _

+

<

_1_. 2 1il

·

3

4

--)

1 . 2 2n- 2

2n- 1 , re-

I N EQUAL I T I ES

z /2 - 2 ,

1 12 z /3 >

1

- >

z /4 -

1 13

- >

_!.._ li1

1 03

z /2,

z /3

z ....'i1+T - z iTI.

>

Adding these inequal ities , we obtain the required result .

PROBLEM 2 7 . Let lA b e an irrational number and a b e some rational num­ ber such that 0 < lA - a < 1 . Show that

1

A - a2 A - a2 a + 2a+T < lA < a + 2a+l + 4 (Za + 1) Sol ution .

·

Since a < lA < a + 1 , we have

lA + a < Za +

1,

2a+l < 1 ,

lA + a

lA - a

>

0.

Hence ( lA + a) ( lA - a) < lA Za + 1

_

a'

2 A - a < lA - a , Za + 1

---

To prove the second inequal ity, we first note that x ( l - x) = x - x < 1/4 for any real number with equal ity only for x = 1/2 ; indeed , Since it is possible to assume that lA - a f 1/2 , we have

[1 1 -

C IA - a) ] C IA - a) < 1 / 4 , c iA - a ) <

(Za +

1)

--1--1 4 ( /A - a)

- C IA + a) <

--

4 ( /A - a)

Mult iplying both members of this inequal ity by lA - a (Za + l) ( IA - a) - (A - a2 ) <

i·

>

0 , we get

2

1 04

CHAPTER 2

Thus , final ly, .fA

<

A a2 a + 2"a"+1 -

PROBLEM 2 8 . al bl

a2 b2

'

4 (2a

1

+

1)

·

Let

..

'

+

.

an b n

'

be n fractions with b l.. + +

>

0 for i

1 , 2 , . . . , n . Show that the fraction

an bn

is contained between the l argest and the smallest of these fractions . Let m denote the smal lest and M the largest of the given

Sol ution .

fractions . Then a. < m � � b l.. - M for i m

or

1 , 2 , . . . , n . Summing these inequalities , we find n

I b. i=l

l.

�

n

I i=l

a l.. � M

n

I

i=l

b l..

or n

I

a l..

I

b l..

m � i=l n i= l

� M.

PROBLEM 29. Let a , b , . . . , d be posit ive real numbers and m , n , . . . , p be positive integers . Show that

is contained between the largest and the smal lest of the numbers

I NEQUAL I T I ES

1 05

(where , of course , we assume that the principal value of the root is taken everywhere) . Consider the fractions

Sol ution .

log a -m ,

log d p

log b -n -,

and let s be the smal lest and S be the largest of these fractions . By Problem 28, s < log a m+ +logn b+ + •• • +• p+ log d < S •• or s < log

m+n+ • • • +p lab · · · d < S ;

the claim now fol lows .

Let 0 < a < � < o < • • • < A < rr/ 2 . Show that

PROBLEM 30 .

s in a + sin � + sin o + • • • + s in A tan a < cos a + cos � + cos o + • • • + cos A < tan A . See Problem 2 8 .

Sol ution .

PROBLEM 31 . Show that any finite s um o f fractions of the form l/n 2 , where n is positive integer l arger than 1 , is less than 1 . Clearly , any such sum is less than or equal to

Sol ution .

1 ••• + 2 n which in turn is less than n

1 /2

1

x

dx

1

PROBLEM 32 .

-

1 n

Let n be any positive integer . Show that

CHAPTE R 2

1 06 • • •

+ n1

-

- log n

1.

<

Let [t] denote the largest integer less than or equal to t . But it i s clear that Sol ution .

In and that 0

J (� - [�J) n

<

In

<

dx

log n

-1

2

-

1 3

1 n

-

1/2 . Sketch a figure of the integrand !

PROBLEM 33. Let a and b denote real numbers and l a l the absolute value of a, that is , is the larger of the two numbers a and -a if a i 0 and 0 . Us ing the fact that I a l - l b I � I a + b l � l a + l b l , show that

lal

lol

l

l

l

I Xn I ) .

We have l x + x 1 1 � l x l - l x 1 1 and , replacing x 1 by x 1 + x 2 , this yields l x + x 1 + x 2 1 � l x l - l x1 + x2 1 , that is , l x + x 1 + x 2 1 + l x 1 + x2 1 � l x l . But Sol ution .

Hence

or

It is clear that the general case can be proved in an ent irely s imi lar manner .

PROBLEM 34 .

For n

1 , 2 , 3 , . . . , let

Find the largest term of the sequence . It is cl ear that x999 = x 1000 . Since xn+l = n1 000 + 1 xn ' we see that xn is increasing when n goes from 1 to 999 and that xn is decreasing Sol ution .

INEQUAL I T I ES

107

as n goes from 1000 to

Thus xn is largest for n

999 and n

1000 .

PROBLEM 35 . Show Bernoul l i ' s Inequali t y : I f a 1 , a2 , . . . , an with n � 2 are real numbers larger than - 1 and, moreover , all a j ' s with = 1 , 2 , . . . , n have the same sign , then

The case when a 1 , a2 , . . . , an are positive is trivial . It only remains to show that Sol ution .

when n � 2 and a 1 , a2 , . . . , an are pos it ive but less than 1 . Since

the claim is seen to be val id for n = 2 . We proceed by induct ion . Suppose that k � 2 and assume that

Mult iplying by the positive number ( 1 - ak+l ) , we get

+

• • •

+

+

• • •

+

PROBLEM 36 . For any pos it ive integer n let s 1 , s 2 , . . . , sn be arbitrary real numbers and t 1 , t 2 , . . . , tn be any real numbers such that the sum t1 + t2 + • · • + t n 0 . Show that L = l L =l tk t j l s k - s j I � 0 .

�

When n = 2 , we can use the ident ity

Sol ution .

(t l + t z ) 2 but t + t l 2 When n

�

(t l - t 2 ) 2

0.

3,

w e have

4t l t 2 ;

1 08

CHAPTER 2

Thus

But l A - B l � l A - C l + I C - B l for any three real numbers A, B , and C . Hence the claim fol lows for n = 3 . The general case can be proved in an ent irely s imi lar manner .

PROBLEM 37.

Show that

2n_-_ .!..l. � . . . _ l 2 4 6 2n

Sol uti on .

<

--=1"---_ 12il+T.

It is clear that ...,

( 2n - 1) (2n

+

2 1) < ( 2n) .

Thus 1 2 • 32 • 5 2

• • •

2 ( 2n - 1) (2n

+

2 2 2 1) < 2 • 4 • 6

• • •

( 2n) 2 .

1 09

I N EQUAL ! T I E S

PROBLEM 38 . Let f and g be positive-valued functions defined on a com­ mon interval and suppose that for any x 1 and x 2 of this interval we have

Show that

Sol ution .

What we have to show is the following : If a 1 , b 1 , c 1 , a 2 ,

b 2 , c 2 are pos it ive real numbers with a 1 c 1 - b

i

�

0 and a 2 c 2 - b

;

�

0 , then

However

If and then

T �

0 and the claim follows .

Remark . A pos it ive-valued function f defined on the interval I satis­ fying the property that for any x 1 and x2 of I the square of the value of f at the midpoint of the interval [x 1 , x 2 ] is less than or equal to the product of the values of f at the endpoints x 1 and x 2 of the interval [x 1 ,x 2 ] is called a weakl y log convex functi on on the in terval I . What we have shown therefore is that the sum of two weakly log convex funct ions on an interval is a weakly log convex function on the same interval .

PROBLEM 39 . that

Let a 1 � a 2 �

� b . Demonstrate

n

CHAPTER 2

110

(-n1 :En ak)(-n1 :En bk) -< -n1 :En k=l

Sol uti on .

I

a =

b " k=l � k

k=l

n

I k=l

Let ak ,

Ib

n

I

k=l

bk ,

I

ab

then

Hence

= l 2 A a (a A - aa ) (b A - ba ) .

II

But (a A - aa ) (b A - b a ) � 0 for A = 1 , 2 , . . . ,n and a

1 , 2, . . . ,n.

The inequal ity in Prob lem 39 is due to Chebyshew . This inequal­ ity can be generalized as fol lows : I f 0 � a 1 � a 2 � • • • � an , 0 � b 1 � b 2 � � • • • � en , then Remarks .

I aI b ··· I c < ---I ab · · · c n n n n

.

The conditions 0 � a 1 , 0 � b 1 , . . , 0 � c 1 are essent ial ; for example , if a1 = 1 , a 2 = 3, b = 1 , b = 3 , c = -4, c = - 3 , we have 1 2 1 2 a l + a2 b l + b 2 c l + c 2 2 2 2 As an example of the general ized inequal ity we mention : I f a, b , and c are pos it ive numbers and n is a positive integer , then (a + b + c) n � 3n-1 ( an + bn + cn) .

111

I N EQUAL I T I ES

Let n be a pos itive integer l arger than 1 and a > 0 . Show

PROBLEM 40 . that

1 + a + a2 +-• • • + an � n + 1 .,. ---,3::---....2-n---=1 �· • • + a a + a + a + •--Sol ution .

is valid for n

(E . 1 1)

It is clear that (E . l l) holds for n k , that is ,

1 + a + a2 a + a2 + a3 +

2 . Suppose that (E . l l)

k + 1 � 1(:1 k• 1 + a

• • •

(E . 1 2 )

Since a > 0 , (E . l2) can be written as 1 + a +

• • •

k + 1 (a + a2 + + ak � ](:l

1 + a +

• • •

k + 1 ( a + a2 + + ak + ak+l � 1(:1

• • •

+ ak - 1 )

(E . 1 3)

or • • •

k- 1 k+l + a ) + a

We shal l show that k + 1 2 1(:1 ( a + a +

• • •

k + 2 ( a + a2 + � k-

k+ a 1 ) + ak+l

• • •

(E . 14)

+ ak- 1 + ak )

proving that (E . l l) is also val id for n = k + 1 . Let us assume that (E . l4) does not hold, but instead that 2 k + 1 ](:l ( a + a + <

• • •

k + 2 (a + a 2 + -k-

+ ak- 1 ) + ak+l

• • •

+ ak - 1 + ak ) .

Then we find that 2 (a + a 2 +

• • •

+ ak-1 + ak ) + (k 2 + k) ak (a - 1)

<

0,

which is clearly impossible i f a � 1 . Thus , by induct ion , (E . l l) is estab ­ lished i f a � 1 . I f we set a = 1/b (b > 0) , the expression on the left-hand side of (E . 1 1) becomes f (b)

1 + b + b2 +

· + bn- 1

112

CHAPTER 2

But the first part of the proof shows that f (b) � � n - 1

for n > 1 and b � 1 .

Hence (E . l l ) is now establ ished for all real a > 0 .

PROBLEM 41 .

1 + a + 1 + a +

A

p: -

1 +

. . .

. . .

+ an- 1

' + an

1 + b + 1 + b +

B

<

B , where

+ bn - 1

. . .

. . .

+ bn

It is clear that

Sol ution .

1

Show that if a > b > 0 , then A

1 + a +

an

1 bn + ' a= 1 + bn-1 1 + b +

' + an- 1

that is , p: -

1

1 1 + 1 - + 1 + an an-1

.

--

Whence 1/A > 1/B and so A

PROBLEM 4 2 . 1 + X + X

X

<

1 a= 1

' + 1

. .

a

+

1

1 + 1 + -bn

bn- 1

. . .

+

B.

Let x > 0 and n be a positive integer. Show that

n

+ ••• +

2n X

5

2i1+T"

1

-1 2 For t > 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

<

(x - 1 ) 1/n

O;

l} l/n- 1 <

for

X

X

> 1 , we have

> 1.

a/b and not ing that a/b > 1 , we get - 1

PROBLEM 4 7 .

<

( � - lr /n

or

a l/n

_

b l /n < (a - b} l/n .

Let a ,b , x > 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)

<

0,

g ( + co ) = 0 . Thus f is seen to be an increasing function .

I NEQUAL I T I ES

115

PROBLEM k Let a b and n be a positive integer larger than 1. Show that, for � >

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 <

Ill) In+

C

< x < e , where e is the basis of the natural logarithms , x > 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 +

<

rn

rn

(by the induction hypothesis)

+ 1

+ 1

v'i1+l

+ 1.

Thus , P (n) imp lies P (n + 1) .

PROBLEM 5 3 . > 0 . Prove that

Let x

�

�

0 and l e t m and n b e real numbers such that m � n

2n 1

xm+n 1 xn -

-

Sol ution .

The inequal ity is an equal ity if m

n , so suppose that m > n .

Let f (x)

(m - n)xm+n + (m + n) (xn - xm) + (n - m) ;

=

<

we sha l l show that f(x) is negative if 0 5 x 1 and positive if x > 1 , and this wil l be sufficient to verify the inequal ity . Differentiating, we get (m + n) xn-l g (x) ,

f ' (x)

g (x)

=

(m - n) xm - mxm -n + n.

From this , we see that f ' (x) and g (x) have the same sign for positive values of x. Since g ' (x)

=

m (m - n)xm- n- l (xn - 1) ,

we see that g (x) is strictly decreasing on [ 0 , 1 ] and strictly increasing on [ l ,oo) with a minimum value g ( l) 0 at x 1 . Since g (O) n > 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

<

is posit ive if 0 � x 1 and negative if x implies the two inequal ities

> 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

<

is positive if 0 � x 1 and negative if x real numbers and so the two inequal ities 2m ( l - xm+n) 1 - xn

=""-=-__;:.::.� .._

> 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 ,

<

0

we see that f is strictly convex on ( - oo , O ] . Then

l

< xi $ 1 , it fol l ows �

ai �l. --�l. a . ( 1 + exp y . ) - 1 $ l + exp nL a . y . - 1 = 1 i=l 1 + x i i=l 1 i=l l l n 1T i=l

with equal ity i f and only i f y 1 = y 2 = • • • = yn ' that is , if and only if x l = x2 xn . I f some xi = 0 , the above proof breaks down , but this case is easily handled on its own merits . Again , equal ity holds if and only if x 1 = x 2 = • • • = xn , which in this case means they are al l zero . I f we allow al. = 0 (and assume o 0 = 1) , then the condition for equal ity becomes x l. = constant for al l i for which al. 0 . Finally we note that the inequal ity is reversed if x i � 1 for all i . This is because f" (y) 0 on (O, oo) and thus f is concave on [ O , oo) .

>

PROBLEM 55 .

>

I f a i with i = 1 , 2 , . . . ,n denote real numbers , show that

where (n - 1) S 2 equal ity holds if and only if al l ai are equal . Sol ution .

We assume that a 1 $ a2 $

$ an . Then

I N EQUAL IT I ES

1 21

n-:1"

S2 =

1

n i- l

I I i=2 j = l

(a i - a . ) 2 J

< ___1

�L (i - l) (a n - 1 i=2 i ___

_

a1 ) 2

Taking square roots we obtain a 1. - S . Similarly , s2

::;

1 _ n- 1 _ n - 1 � 1

j l

Cn

. - J ) C an - aj ) 2

�

(a - a ) J � j=l n

�l I

2

from which fol l ows that n

I a. + S j=l J

:>

nan .

It is clear that equal ity holds anywhere if and only if it holds throughout and this is true if and only if al l a i are equal . PROBLEM 5 6 . m + ! m � 2

I n the Solution of Prob lem 42 w e encountered the inequality for m

0

>

and equal ity if and only if m m +

��

m

3

for m

>

1 . Prove that

0

and equality if and only if m = 2 .

>

Sol ution . Since m 0 , the proposed inequal ity i s equivalent with the inequal ity m 3 + 4 - 3m2 � 0 . But m3 + 4 3m2 (m + l) (m - 2) 2 , m > 0 and (m - 2) 2 � 0 . Hence (m + l) (m - 2) 2 � 0 and = 0 if and only if m = 0 .

PROBLEM 5 7 . Assume that a i n n a i+l a. n � 1 :E _ :E i=l a i+l i=l a 1.

( )

>

0 , (i

l , . . . ,n) , with an+l

a 1 . Show :

1 22

CHAPTER 2 Sol ution .

product b 1b 2

• • •

Let b i = a i /a i+l for i = 1 , 2 , . . . ,n and bn+l = 1 . Then the bn+l 1 , and the inequal ity to be proved is equivalent with

n+l 1 n+l Ii=l bi � i=lI bn1. . The inequality of the arithmetic and geometric means (see Problem 2) give

Therefore n+l

I

i=l

b�1 .

PROBLEM 5 8 . Show Abel ' s Inequality : Let {a i , a2 , . . . , an } and { b 1 ,b 2 , � b � 0 and put . . . ,bn } be two sets of real numbers with b 1 � b 2 � n s k = a 1 + a 2 + • • • + ak

for k = 1 , 2 , . . . , n ,

with M and m denot ing , respectively, the largest and the smal lest of the .. sn . Then .

Sol ution .

n

I

i=l

,

Clearly

a 1. b 1.

But

because for

j = 1 , 2 , . . . ,n- l ,

1 23

I N EQUAL I T I E S

for j

1 , 2 , . . . , n- l

PROBLEM 59 . Determine al l polynomials of the form xn + a 1 xn-1 + with all ak = ± 1 and which have al l roots real .

• • •

+ an

Sol ution . Let the roots be x , . . . , x . Since the roots are real , the 1 n numbers x21 , . . . , xn2 are pos it ive , and we know that their sum is

x 2l

+

+ Xn2

• • •

1 ± 2

3 (since - 1 is negative) .

The product

Hence , by x2l = x22 = trivial .

Prob lem 2 ' we have 1 :s; 3/n , that is , n :s; 3 , with equality only if x23 = 1 . In case n 3, al l roots are ± 1 and a 2 = - 1 . The rest is =

PROBLEM

Show that

60 .

( t a1·i ) 2

:s;

i=l

Sol ution .

n

Since

a 1. a . I i + jJ- 1 i=l i= j n

I

n

a. a .

:E :E -;---=1'-,-, ,_]_----.i=l j = l i + J 1

n

-

•

( I a.i ) 2 i=l

1

with (k - 1) ak k and

for k

1 , 2 , . . . ,n

b 1. b . I i + jJ- 1 i=l j = l n

I

n

1 24

CHAPTER 2

=

1 n / I b . x i-1 0 i=l 1

I

1

2

dx

�

0,

the des ired inequal ity fol l ows . Moreover, equal ity holds i f and only i f a 2 = a 3 = • • • = an = 0 .

PROBLEM 6 1 . Let f and g be real-valued functions defined on the set of real numbers . Show that there are numbers x and y such that 0 � x � 1 , 0 � y � 1 , and l xy - f (x) - g (x) I � 1/4 . Sol ution .

terval

[0, 1 ] ,

l f (O)

+

I f l xy - f (x) - g (x) I > 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

<0

f' (t)

for 0

1.

< t < 1,

On

the other hand , if a > 1 , then

for t > 1 ,

> 0

for t

0

0.

and f assumes its smallest value at t

1.

Let a and b be positive real numbers and a and 8 be such that a 1 , 8 1 , and a + 8 = 1 , then aab 8 � aa + 8b . Indeed , take t = a/b and denote 1 - a by 8 ; then (E . l9) gives the desired result . Let A > B > 0 . I f we put t = 1 + 1/A and a = A/B in (E . 20) , we obtain

<

Remarks .

<

1 +­ B or (E . 2 1 ) I f w e put t = 1 - 1/ (A + 1) and a

(l

_

)

A+l 1_ B+ 1 > l A + 1

_

_

1_ B + 1

_

(A + 1) / (B + 1) in (E . 20) , we obtain or

A _ A+ l > (B - B+ l (_ A + 1) B + 1)

or (E . 22) Inequalities (E . 2 1) and (E . 22) show that for x > 0 the function g (x) = (1 + .!X. y increases with increasing x and for x > 0 the function G ( x) = (1 + .!Xy +l decreases with increasing x ; in this connection see Problem 2 of Chapter 1 .

PROBLEM 63.

Let s

F

2kn , where k is an integer , and

A = s in t + sin (t + s) + sin (t + 2s) +

• • •

+ sin [t + (j - l) s ]

B

• • •

+ cos [t + (j - l) s] .

and cos t + cos (t + s) + cos (t + 2s) +

CHAPTER 2

1 26

Show that A =

s in (t +

. . s sm _J. -2- -1 s) ·--=. s s1n 2

B

cos (t +

sm J 2 - 1 s) · --. s. 2

J

2

and

Sol ution .

j

. .s

sm 2

Since 2k --1 s ) cos (t cos ( t + -2

2 s in %· s in (t + ks)

+

2k +-1 s ) -2

we get that A• 2

sin �2 cos (t

- %) - cos (t + %)

+ cos (t +

%)

+ cos (t + 3

- cos (t +

I)

3

- cos (t +

%) s

I)

2 · 3 s) - cos (t + � 2 · 1 s) + cos (t + � cos (t

2j - %) - cos (t + 2- 1 s)

. . s· j --1 S ) • Sln . (t + 2 Sln J2 2 Thus A =

1

. .s sm J 2 j - s) •-s in (t + -2 . s . sm 2

To prove the second formula we note that . s • cos (t + ks) 2 s1n 2

=

2k -1 . (t + 2k-+-1 s ) - sin (t + -s1n 2 2 s) ·

I NEQUAL I T I E S

and·so B•2

1 27

sin �2 sin (t + �) - sin (t �) + sin (t + %) - sin (t + %) + sin (t + 5 %) - sin (t + tJ -

3

3

· -- s ) 22 · 1 s ) - sin (t + _J + sin (t + � 2 sin (t + _J_2--1 s) - sin (t I) 2 cos (t + j ; 1 s ) • sin j % Thus . .s . 1 Slll J _ _ _ cos (t + J 2 s) s1n 2s If x 2kn , then 1 and I kI=p cos kx I -:--/ sin ---. ��· Show that for any real number and any positive integer n we have 21) I k=l� sink kx I - 2>'TI. Let 0 < x < TI and let m be the integer such that 22) m ::; - < m + l. Then 3

2.

-

B

z

· --- .

.

Remark .

#

::;

x

PROBLEM 64 .

r-

<

L

(E .

Sol ution .

l7f X

(E .

1 28

CHAPTER 2

I k=lI sink kx I k=lI I sink kx I + I k=m+lI sink kx I .

(E. 23) (If m = 0 the first summand on the right-hand side vanishes, if m n the second summand on the right-hand side becomes zero.) Since !sin tl ltl we get � kx = � sin kx (E. 24) k I I k=l k=l k By Abel's Inequality (see Problem and the Remark to Problem �3, we see that I k=m+lI sink kx I - J sin1 �� , m +1 l" Evidently sin t (2/�)t for 0 t �/2. This, together with (E.22), gives S J.. n X X m + 1 > ;;X We therefore have I k=m+lI sink kx I _1_ = ;;, �.� l:rrX which together with (E.23) and (E.24) gives the desired estimate (E.21) for x (0,�). But the function f(x) = I k=lI sink kx l is an even function, that is, f(-x) = f(x) ; therefore (E.21) also holds for X c-�,0) . Since (E.21) trivially holds for X = �. (E.21) is valid for all x such that -� x �. Since f is a periodic function with period 2�, (E.21) is seen to hold for any real number x. �

<

L

r>"IT .

<

rnx

L

�

-

5 8)

_

<

_

�

Z > n'

�

�

�

E

E

±

�

�

PROBLEM Show that -z1 ·log(2n + 1) 1 + 1 + 1 + + 1 1 + -z1 • log(2n - 1). For k x k + 1 we have 65 .

3

<

Sol ution .

�

�

S

� <

�

1 29

I N EQUAL I T I ES

1_ 1 _ _ 1_ _ 2k + 1 <- 2x - 1 <- 2k - 1 " _

_

_ _

Integration gives 1 2k+l

.(

dx 2X""=l

k+1

<

<

2i(:"l;

1

thus n dx 1 ••• + 1 -13 + --2n - 1 < f 5 + 1 2x - 1 and n+ 1

dx < 1 + -1 + -1 + J 2X""=l 3 5 1

1 + --2n - 1 "

Therefore

r --dx 1 1 2x - 1 < 1 + -3 + -5 +

n

1

n dx 1 < 1 + + --J 2n - 1 2X""=l 1

or 1 2- 1og ( 2n + 1) <

PROBLEM

1

+ -31 + -51 +

1 1 + --2n - 1 < 1 + 2· 1og (2n - 1) .

Show that

66.

n/.2 . 2 nx 1 s 1n 1 1 2• 1og ( 2n + 1) < 0 sin n dx < 1 + 2• 1og ( 2n Sol ution .

nt S 1ll 2 2 o

=

1) .

Since 1 - cos nt 2

( 1 - cos t) + (cos t - cos 2t) + • • • + [cos (n - 1 ) t - cos nt] 2 and cos

A - cos B

it fol l ows that nt s1n 2 2 0

=

(s1n 2t + 0

A B B A

2 S1n -+2- S1n -2-- , 0

o

3t + • • • + s1n (2n - 1) t ) S 1n t 0 s1n 2 2 2 0

0

0

CHAPTE R 2

1 30

Putting t = 2x, it follows that sin 2 nx = [sin x + sin 3x + •• · + sin(2n - l)x] sin x and so rr/2 s1nsin. 2 nxx dx 1 + 1 + 1 + • • • + 1 J; The desired result now follows by invoking the inequality established in Problem Since log(2n + 1) > log 2n log(2n - 1), we see that -21 ·log 2n rrJ0/ 2 Slsin. n2 nxx dx 1 + -21 · log 2n. This in turn shows that rrJ/ 2 s1nsin. 2 nxx dx 1. 2 lim log n n+ <» 0 3

5

2i1="1"

65 .

>

Remark .

<

<

--

PROBLEM Let f(x) = a1 sin x + a2 sin 2x + •• • + an sin nx satisfy jf(x) 1 for all real x. Show that i a1 + 2a2 + ••• + nan ! n. By the Inequality of Bernstein (see Problem 20) we have that if' (x) n for all real x. Hence it follows that if' (O) n. I

67.

s

s

Sol ution .

I

I s

PROBLEM Suppose that -1 ax2 + bx + c 1 for - 1 1, where a, b, and c are real numbers. Without using (see Problem 24) , show that - 4 2ax + b 4 for - 1 1. The graph of the linear function f' (x) = 2ax + b derived from the function

s

68.

s

s

s

Sol ution .

s

S

S X

S

X

S

A. A .

Markoff's Inequality

1 31

I N EQUAL I T I ES

f(x) = ax2 + c is a straight line ; therefore, as x varies over the interval [-1,1], f'(x) assumes its largest and smallest values at the endpoints of that interval. Hence, it is enough to show that the values of f'(x) at x -1 and x = that is, the numbers f' (-l) = -2a + b and f' (l) 2a + b are neither less than -4 nor larger than 4. To estimate these numbers, we substitute -1, 0, and 1 for x in f(x) ax2 + bx + c and use the given in­ equalities. We get � f(-1) a - b + c � 1, (E. 25) -1 � -f(O) (E. 26) -c � 1, (E. 27) - 1 � f(l) a + b + c � l. The sum of (E. 25) and (E. 26) gives (E. 28) -2 � a - b � 2, the sum of (E.26) and (E. 2 7) gives (E. 29) � a + b � 2, and the sum of (E.28) and (E. 29) gives -4 � (a + b) + (a - b) � or (E. 30) -2 � a � 2. Adding first (E.28) and (E.30), then (E.29) and (E.30) , we obtain -4 � 2a - b � 4, or � -2a b � and -4 � 2a + b � 4, which shows that the extreme values of f' (x) (and hence certainly all is values) in the interval - 1 � x � are situated within the specified range. 2 - has the prop­ Note that the quadratic polynomial f(x) = 2x erties � f(x) � 1 for -1 � � 1 and f' (-1) = -4, f' (l) 4. =

1,

=

=

-1

-2

4

+

-4

4

1

1

Remark .

-1

=

X

CHAPTER 2

1 32

PROBLEM Show that if x is larger than any of the numbers a1 , a2 , . . . , an then x - a1 x - a2 • • • ----an -- >- x - (a1 a2 n ••• an)/n' Let 69 .

1 + ---

1

---

+

+

X

1

---�---------------�-

-

+

+

+

Sol ution .

It is clear that x. > for i 1,2, . . . ,n and that we may use the result in Problem 2; thus xl x2 ••• xn n n xl x 2 xn or 2 n ....!.. � x l x2 n n x l x2 an) and everything is clear. But xl x2 n nx (al a2 0

1

-:1:------:1---------,..1 � + + +

1 + 1 + -

. . .

+

+

. . .

+

+

+

+

+

+

+

__ __ _ __ _ _ _ ....__ .;::

X

-

X

+

. . .

+

+

X

+

PROBLEM Let a, b, r, R, and x be positive, r R, and r2x Show that x � 2/ab. Since R2x aR bR3 , we have that a bR. But �

70 .

Sol ution . X

�

R

�

+

+

by Problem 2. PROBLEM Let a1 , a2 , . . . , an and b 1 , b 2 , , bn be positive num­ bers and suppose that M denotes the largest of the fractions 71 .

. • .

I N EQUAL I T I E S

1 33

al , a2 , . . . , an b l b2 n Show that ann M. al a22 b l Mb22 Mn- lbnn It is clear that al M, a22 M, . . . , ann_ - M· 1 Mb22 Mn-lbnn hence we may use the result in Problem 28. b

+

+

+

+

+

. . .

s;

+

. . .

Sol ution .

b

s;

-

__

s;

<

'

PROBLEM 72. Let xi > 0, pi > 0 for i 1,2, . . . ,n. Letting E j pj stand for the sum p1 p2 pn ' show that (E. 31) Put, for i 1,2, . . . ,n, q. p l pi Pn in Problem 9. Setting p1 p2 = 1 in (E.31), we obtain the farniliar inequality (see Problem 2) +

+

+

Sol ution . 1

=

Remark .

+

+

=

PROBLEM 73. Let j ! (nn!- j) ! for 1, 2, . . . ,n. C.

J

CHAPTER 2

1 34

Show that /S + /S + • • • + � s Vn(2n - 1) . The Binomial Theorem gives 1 + C1 + C2 + •• • + Cn (1 + l) n 2n or Sol ution .

=

=

Our claim now follows by setting ai 1 and b i the Cauchy-Schwarz Inequality (see Problem 14) . =

=

�

for

i

1,

2, . . . ,n in

PROBLEM 74. Let y f(x) be a continuous, strictly increasing function of x for x with f(O) Examining the areas represented by the inte­ grals, we see that ab s Ja f(x) dx + Jb f- 1 (y) dy, where f-1 (y) is the inverse function to f(x) and a and b are nonnegative con­ stants. It is easily seen that there is strict inequality unless b f(a) . This is the Show that, for t 1 and s ts s log t - t + es . Let y f(x) log(x + 1) in Young's Inequality, and put a t - and b s. Letting y f(x) p-1 p in Young's Inequality, we obtain =

� 0,

0

= 0.

0

=

Inequality of Young. �

� 0,

t

Sol ution .

1

Remark .

=

X

'

1,

where p and q are connected by the equation 1/p + 1/q 1.

I NEQUAL IT I E S

1 35

PROBLEM Let p, f, and g be continuous functions on the interval [a,b] and suppose that p is positive-valued and f and g are monotonically increasing on [a,b]. Verify the following inequality, due to Chebyshew: � ! p(x) f(x) dx f � � p(x) g(x) dx f \� p(x) dx H! p(x) f(x) g(x) dx } . Consider the difference b b b b J p(x)f(x)g(x) dx J p(x) dx J p(x)f(x) dx J p(x)g(x) dx. a a a a But b b b b J p(x) J p(y) dy and J p(x)g(x) dx J p(y)g(y) dy a a a a and so bb Jj p(x)p (y)f(x)[g(x) - g(y)]dxdy. aa Exchanging the roles of x and y, we get bb J j p(x)p(y)f(y)[g(y) - g(x)]dx dy. aa Forming the arithmetical mean of the last two expressions for yields bb -z fj p(x)p(y)[f(x) - f(y)][g(x) - g(y)]dx dy. aa Since f and g are monotonically increasing, the expressions in the two square brackets have the same sign; hence the integrand is nonnegative. Thus giving us the desired inequality. As we can easily see, the inequality remains true if both functions f and g are monotonically decreasing on [a,b]. If, however, one of these two functions is increasing, while the other is decreasing, then the inequality reve rses. 75 .

�

Sol ution .

•

D =

dx

=

•

-

=

D =

D =

D

D

=

1

D � 0,

1 36

CHAPTER 2

+ ••• +

=

PROBLEM 76. Let f(x) and the coefficients a1 , a2 ,

an , where n is odd an are real numbers. Letting

show that f(t) 0 for t M and f(t) 0 for t M For sufficiently large positive t we have >

<

<

>

-

.

Sol ution .

I

an

I)

0; here the first inequality holds for all positive t, the second for all t n1 and nthe third for all t > M. the other hand, since n is odd (and so (-t) -t ) ' f(-t) -tn a1 tn-1 a2 tn- z an -tn tn-l ( a1 a2 an ) >

�

On

�

�

+

I

+

I

+

I

I

I

I

+

I

+ ••• +

I + ••• + I

I

I

I

0; here the first inequality holds for all positive t, the second for all t � 1 and the third for all t M. But f(-t) 0 for t > M is equivalent to f(t) 0 for t -M. Let f and M be as in Problem 76. Since f is continuous, f must have at least one real root in the interval (-M,M). Moreover all real roots of f must be in this interval. <

>

<

<

<

Remark .

=

PROBLEM 77. Let f be a continuous function on [a,b] and put f(t) 0 for t [a,b]. For h 0, we define the f, denoted by fh , as follows: i

>

Steklow Function of

I NEQUAL I T I ES

1 37

It is clear that fh (x) lkfF Cx + h) - F(x - h)], where F(x) a-hJ( f(t) dt; moreover, F is a continuous function. Show that b b I fh (x) dx I f(x) dx. a a We assume first that f(x) 0 for all x in [a,b] and consider the function f(z + t) on the rectangle [a,b] [-h,h]. We see that hb bh I J f(z + t) dz dt J I f(z + t) dt dz. a -h -h a Since h t+h I f(z + t) dz J f(x) dx Zhfh (t), -h t-h the integral on the left-hand side of equals =

1

X

I

I

<;,

I

I

Sol ution .

�

x

=

(E . 32)

=

(E . 32)

The integral on the right-hand side of can be put into the form h b+z J I f(x) dx -h a+z and we only need to observe that b+zJ f(x) dx Jb f(x) a a+z in order to obtain the desired result in this special case. Indeed, for z 0, we have equality in If, however, z > 0 (hence f 0), then b+zJ f(x) dx Jb f(x) dx Jb f(x) dx. a a+z a+z (E . 32)

<;,

dx

(E . 33)

(E . 33) .

�

<;,

=

CHAPTER 2

1 38

The case is handled with equal ease. We now drop the assumption f(x) and denote the Steklow function of fl l by fh . We get x+h x+h 2h1 1 xrh f(t) dt I 2h1 � lfCt) 1 dt and (E. 34) But, as shown already, b b J fh (x) dx J f(x) dx, a a which together with (E.34) gives the desired result. z <

0

�

0

�

I

I

�

PROBLEM Let N(a) denote the number of times the positive integer a occurs as a binomial coefficient, (�) . We have N(l) N(2) 1, N(3) N(4) N(S) 2, N(6) 3, etc. Clearly, for a > 1, N(a) Show that N(a) 2 + 2 log2a. Let b be the first b such that (2:) > a. Now 78.

=

=

=

�

= oo,

<

oo

=

Sol ution .

is monotonically increasing in i and in j ; hence for a 11 . Thus (i j+ j ) a 1mp 1es 1. b or b. Again by monotonicity, for each value of i (or j) , 1,

·

J· �

=

0

1

·

·

<

<

has at most one solution. Hence N(a) 2b. Now (2bb ) 2b , so we have a >- (2(bb - - 1 1) -> 2b-l ' i.e. , b log2 a + 1 and N(a) 2 + 2 log2 a. <

)

�

�

�

I NEQUAL I T I ES

1 39

PROBLEM 79 . Let a 1 /b 1 , a2 /b 2 , . . . with b 1 > 0 , b 2 > 0 , ly increasing sequence . Denoting by + a.

J

. . . be a strict ­

and

show that A 1 /B , A2 /B 2 , . . . is a strictly increasing sequence . 1 Sol uti on .

�< a. b l.

an+l b-n+l .

For i

1 , 2 , . . . , n we have

that i s ,

whence

Thus

PROBLEM 80 . Let m, n be positive integers and a 1 , a 2 , . . . , an be pos i ­ tive real numbers . For i 1 , 2 , 3 , . . . , put an+i a i and

Show that

except if all the a i are equal . Sol ution .

The des ired inequal ity is equivalent with

Noting that (see Problem 2)

I

b 1. 1 m = m k=l a i+k

�

l/m a i+k ) (n ' k=l

with equal ity only if al l the a i+k are equal , we have

1 40

n

CHAPTE R 2

( b . ) > n ( m a ) 1/m ( I I am. ) l/m }T 1T i+k 1

}T 1=1 :

1= 1

...!!....

k=l

i=l

unless all the a i are equal . Note that for m = 1 and any set { a 1 , a2 , . . . , an } ' the inequality must be replaced by equal ity .

Let a , a2 , . . . , an be real numbers . Show that 1

PROBLEM 8 1 .

2 1.::. -="-2 . M = --2 n (n - 1)

+ ••• +

_ _

Sol ution . We may assume that a � a � 2 1 � ni=l a2i = 1 . Now

-(

n 2 = n L a. i=l 1 Assume a i+l - ai

>

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 .

<

<

X

>

89 .

�'

1_ __ >

1

Sol u tion .

1

.

�

�

>

�

J

A.

1

�

1 J

1J

1

_

=

<

1

� -

1

X

A

1

1J

X

<

>

I L

X

>

<

J

J

1

F ( l)

I N EQUAL I T I ES

1 45

PROBLEM 90. Show that for n 2, n C�) {(2n - 2)/(n - l)}n-l . i=O Since n (�) = n-1 (�) , 2n - 2 n-1l: (�) , i=l i=O i=l the desired inequality is merely the arithmetic-geometric means inequality (see Problem 2) for the n - 1 numbers (�) , i = 1,2, . . . ,n-l. �

TI

�

1

Sol ution .

TI

lT

1

PROBLEM 91. Let b 1 , . . . , bn be any rearrangement of the positive numbers a1 , . . . , an . Show that a1 /b 1 + + an/bn n. Since the arithmetic mean of n positive numbers is at least as large as their geometric mean (see Problem 2) , we have -n1 i=lnl: baii ( i=l..!!... baii ) l/n But, by the conditions of the 'problem, TT;=l ai = TT;=l b i . Thus nl: a. - n. i=l b i PROBLEM 92. Given that Lni=l b i b with each b i a nonnegative number. Show that n-1I b . b . b 2 j=l J J +l Let bk denote the largest of the numbers b i ' where i 1,2, ... ,n. Then bk ki=ll:- 1 b.1 + bk n-1i=kl: b.1+ 1 �

Sol ution .

- �

2.

II -

>

� T·

Sol ution .

.

CHAPTER 2

1 46

Note that there is equal ity if and only if b k 2 Thus for n � 3 , L n-1 i=l b ib i+l < b /4 .

PROBLEM 9 3 .

b/2, in which case n = 2 .

Let n � 2 and 0 < x 1 < x2 < • • • < xn � 1 . Show that

n n L xk k=l + nx1 x 2 • • • x n By Prob lem 7 , if ak > 0 for i

Sol ution .

l , . . . , n , then

n

n 1 2 ak • L � n a k=l k=l k

L

and thus n

n n ak • L a - 1 ( 1 - a ) � n I ( 1 - ak ) . k k= l k= l k k=l

I

Setting ak = xk ( 1 + xk ) - 1 ' we obtain next

(

n 1 L x� • n k= l Multiplying both sides of this result by the product x 1 x 2 • • • xn and observ­ ing that for n � 2 , X

n -1 • • • X • L xk ' 1 2 n k=l X

we obtain the proposed inequal ity , after rearrangement of terms .

PROBLEM 94 . Show the following inequal ity , due to Kantorovich : Let f be a continuous funct ion on the interval [ 0 , 1 ] such that 0 < m � f (x) � M. Then (m

+

M)

4mM

2

(E . 38)

I N EQUAL I T I ES

1 47

Since {f(x) - m}{f(x) f(x) - M} 0 for 0 x we obtain by integrating f - (m + M) + (mM)/f over [0,1] rol f(x) dx + mM.JrO l f(x)dx m + M. Putting u mMJ{O 1 f(x)dx ' we obtain f(x) dx (m + M)u - u2 (m +4 M) 2 In Problem 94 the unit interval [0,1] can of course be re­ placed by any interval of length 1 and in place of requiring f to be conti­ nuous it would have been sufficient to assume that f be integrable. If in Problem 94 the unit interval [0;1] is replaced by the interval [a,b], then (E.38) changes into (m4mM+ M) 2 (b a) 2 . Sol ution .

$

n J

$

$ 1,

$

=

•

$

$

Remarks .

_

F

PROBLEM 95. Let x > 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 <

-

<

-

__!_ rx'

1 +

X

Sol ution .

--

=

• • •

<

<

CHAPTER 2

1 48

X

=

and note that (E . 39) becomes l + t -1 log -t 1 t -

-

for 0 < I t I < 1 .

1 - 0 1 - t2

--- <

Using the power series expansion , we get

I n=l

l_) 2n (1 - _ 2n + 1 t

�

for 0

0

:S

It I < 1,

which is evidently true . To prove Part (ii) ' we set 3 = ( 1 + t) 3 (1 - t)

X

and observe that (E . 40) becomes for 0 < l t l

3 log 1 +- t --t2 + 3 < - 0 1 t 2t 1 - t4

<

1.

Using power series expansion s , we get 2n _t__ t 4n+2 + 3 t 4n - 3 + 1 2n n=O n=O n=O

I

I

�

I

�

�

o

for I t ! < 1 ,

that i s , 3

I

n=O

3_) 4n+2 l_) 4n _ _ I (1 _ _ 4n + 3 t 4n + 1 t + n=l

(l

�

O,

which is obviously true . Another way of estab lishing the inequal ity ( E . 40) runs as fol. lows : For u 1 we have 0 s (u - 1) 3 (u 3 - 1) . Add1ng 3u5 + 6u 3 + 3u on both 4 2 sides we obtain 3u ( l + 2u 2 + u4 ) :S u6 + 3u + 4u 3 + 3u + 1 . Dividing by the 2 posit ive number u ( 1 + u 2 ) 2 , we get Remarks .

�

Letting

I NEQUAL IT I ES

1 49

f(u) 3 log u, g(u) (u2 - u(ll) (u+2u2) u + 1) we therefore see that f' (u) � g' (u). Since f(l) g(l), this leads to f(u) g(u) for all u � 1. Hence, letting u x1/3 with x 1, we obtain � 1 + x 1/3 for x X - 1 - X + X 1/3 Applying the transformation x 1/x, the proof is complete. +

�

>

<

> 1.

+

0 <

<

PROBLEM Let y x. Show that 2 log xx -- ylog y· We integrate the inequality 2 2t ----= (1 + =-t) -:-2 ::- (t 1) , and we get i r x/y �t 2 r x/y dt 2 , Jl (l + t) J l] that is, 21 log X x + y for y x. This inequality is equivalent with (E.41) . � >

96 .

Sol ution .

_!_

>

>

>

y

PROBLEM 2x 2+ 1

>

�

97.

Let x

0 <

>

0.

<

Show that

--- <

Let f(x) 2x 2+ 1 - log(l + �) for x

Sol uti on . =

>

0.

(E.41)

1 50

CHAPTE R 2

Since

f' (x) x(l + x) 1(2x + 1) > for x > f is an increasing function. From this fact and since limx -r oo f(x) we conclude that f(x) for x > Let 1 for x > O. g(x) log l + ..!:.X) - � Since g' (x) 2x + 1 x(l 1+ x) > 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

0.

Remarks .

<

<

0

95 .

0

96 .

PROBLEM Let F be a positive-valued, continuous, and decreasing function over the interval [ a,b]. Show that J:b x{F(x)} 2 dx ib {F(x)} 2 dx ab b Ja x F(x) dx fa F(x) dx For p(x) {F(x)} 2 , f(x) x, g(x) 1/F(x) , the result in Problem yields the desired inequality. 98.

----" ----- < --,-

=

=

Sol ution .

75

•••

PROBLEM Let Sn 1 + 1/2 + 1/3 + + 1/n. Show that n{(l + n) l/n - 1} Sn n l - (n + 1) -l/n - ---n +1--1 -}. 99.

<

<

{

I N EQUAL I T I ES Sol ution .

But

151

By Problem n n+ 1 > (n + 1) 1/n . + -nn n 2,

+ - =

and so Sn > n{(l + n) l/n - 1}. Again by Problem -1 + -3 + -43 + + _n n_1 > n(n + 1) -1/n . Noting that + n + 1 n, we see that Sn n{l - (n + 1) - l/n - -n +1--1 -} . 2,

2

2

+

i1+"T

=

<

PROBLEM 100. Let x > 0 and y > 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-

<

0

for x

>

1,

>

0

for 0

<

x

<

1.

Hence (x lo � x) / (x2 - 1) � 1/2 . The inequality 0 � (x l og x) / (x 2 - 1) is obvious , because for x > 0, the sign of x log x is the same as the sign of x2 - 1 . This completes the proof of the inequal ities posed . We note the fol­ lowing direct consequence : For m � 1 ,

I N EQUAL I T I E S

1 53

PROBLEM 103. x (2

+

cos x)

Sol ution .

> 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.

Let a and b denote real numbers and r satisfy r

�

0 . Show

that (E . 48) where cr

1 for r � 1 and cr = 2 r- l for r

Sol ution .

Since x

+ f (x)

lxl

r (r

>

>

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

<

1) .

0 , this result is trivial ly true . Consider the function

=

g (t)

(1 + t) r - 1 - tr

=

(0

<

r

<

1) ,

which vanishes at t 0 and decreases as t increases . This yields (E . 48) for a 1 0 . (E . 48) also holds when a 0 . =

=

PROBLEM 108.

<

Let 0

a + b 1 (a - b) 2 < -- 2 a 8

-

b

s

r. van b

a . Show that < _

.!_

(a - b) 2 8 b

The claimed inequal ity is trivially true in case 0 a , then , by Problem 2 ,

Sol ution .

If 0

<

b

<

a +b 2

>

Thus , i f 0

<

0

r-;­ vab . <

b

a , then

a + b - 2 /ab

<

and

0

>

2/ab - (a + b) .

This means that a - b 2/a

( 0

-- <

or

(_ a -

_,_

-) -2 -b": 4a

<

( ra -

<

b

<

a)

a + b - 2/ib

<

(a - b) 2 4b

This completes the proof. Remark .

If 0

<

b

<

a, then we can easi ly see that

(0

<

b < a) .

<

b

a.

1 57

INEQUAL I T I E S

a +b a + v'ib + b > liib ab . 3 2 > The middle term in the foregoing inequality is sometimes referred to as the heronian mean of the pos itive numbers a and b .

Consider any sequence (an ) := l of real numbers . Show that

PROBLEM 109 .

:E a n=l n

� .2_

13

t

(rn /n) 1/2 ,

n=l

where rn

:E a� .

k=n

First we observe that

Sol ution .

1 I n=k n 2 (n + 1) 2

Using this , we have by Cauchy ' s Inequal ity oo a oo a n 2 I n (n � 1 ) I k = 2 I k I n (n +n l) k=l n=k n=l k=l

L a n= l n < -

2

I k

CX)

k=l

� .2_

I 13 k= l

PROBLEM 110.

l

�

1/2 I an2

CX)

n=k

1I

�

1/ 2 1 2 2 n=k n (n + 1) CX)

(rk /k) 1/ 2 .

Show that , for n = 1 , 2 , 3 , . . .

n where limn + oo (1 + ! n) = e .

,

CHAPTER 2

1 58

The inequal ity to be proved is certainly true for n = 1 , 2 . Proceeding by induction , we assume that Sol ution .

holds for n = 1 , 2 , . . . , k and show that the inequal ity is true for n as well . Thus , let

we wish to estab lish that <

(k e+ l ) k+l -

that is , .!. (�) k e e

<

(k + 1) ! k ·1

<

<

k + 1

k +---l ) k+l , e (--2

� (�) k 2 2

•

But + k1 ) k

< 1,

+ kl ) k

>1

becaus e , for k = 2 , 3 , . . . ,

Remarks .

Let x 1 = 1 , x 2

x3 =

•••

= xk + l = 1 + 1/k . By Problem 2 ,

hence 1 ) k+ 1 . (l + m In a s imilar way we can show that by Prob lem 2 . Indeed , let x1 = 1 , x2 = x 3 = • • • = xk+2 = k/ (k + 1) . These (k + 2) numbers have an arithmetic mean of (k + 1) / (k + 2) and a geometric mean of [k/ (k + l) ] (k+ l) / (k+2) . Hence

1 59

I NEQUAL I T I ES

�

k + 2

>

k_'\ (k+l) / (k+2) . (k + 1)

taking reciprocals this becomes

On

1 + k 1+ _1 _

Ck+l) / (k+2) (l + !.1 k)

<

1 _'\ k+2 (1 + _ k + 1)

or

k+ l (1 + !.1 . k)

<

If we consider the elementary inequalities

for k = 1 , 2 , . . . , n - 1 , and mult iply them together , we get nn - l (n - 1)

!

...,.:.:..._ ...: ....,. __ .. ..,... .,..

<

en-1

<

nn (n - 1)

!·

"7""".:..:_ " --,--,--,-

This leads to the approximation

PROBLEM 111 .

Show that +

Sol ution .

__!__ < 21i1 rn

We have 1 +-

rn

0

<

- 1.

<

1

n

+

J -1 dx 1 /X

z iTI - 1 .

PROBLEM 1 1 2 . Let a , b , and x b e real numbers such that 0 x < 1 . Show that

<

a

<

b and

(E . 49) We can write (E . 49) in the form b log (l - xb ) - a log (l - x a) - (b - a) log ( l - x a+b )

Sol ution .

>

0.

Usin g the series representation of log ( l + x) , this inequal ity becomes (b - a) x (a+b) k + axak - bxbk k k= l

I

>

0.

CHAPTER 2

1 60

Hence it wi l l be enough to show that (b - a) tb+a + at a - btb > 0

(0 < t < 1) '

that is , 1 - tb > b tb-a a a 1 - t

---

(0 < t < 1) .

But the last inequal ity is a simple consequence of Cauchy ' s form of the Mean­ Value Theorem of differential calculus : 1 - tb = -be b - l = b 8b-a b b-a --> a t a a a 1 1 t ae

(t

<

e < 1) .

_

PROBLEM 113.

Let 0 < a < 1. Show that

a 1 l � < a a + a l-a < 1 . e Putting

Sol ution .

f (a)

(1

(E . SO)

a a) a 1-a

+

(0 < a < 1) ,

we get log f(a)

log ( l

+

a)

f ' (a) 1 (log a f(a) ( 1 a) 2 _

+

�

a

+

log a ,

2 �) 1 + a

and, setting g (a) = log a + 2 ( 1 - a) / ( 1 + a) , 1 - a 2 g , ( a) - -a1 (-1 + a) . _

We conclude that g ( a) < g ( 1) = 0 ,

f ' (a) < O ,

l im f(a) < f (a) < l im f(a) . a -+ 1 a + O+ But l im f(a) = -2e a + 1-

and

l im f (a) a + O+

1.

I NEQUAL I T I ES

PROBLEM 114. - -12 tan � 4 Sol ution .

s

0 <

Let n

L sin kx

k=l

cos

I s in kx =

. 2 sm

this together with - 1 result .

PROBLEM 115 . 1 . Show that

�l.. a k

s

-12 cot �4•

n

k=l

X

2

+

�) x

cos (n + �) x

0 < � <

Let

---- > TT

1 -

s

2TI. Show that

I - cos (n

k l

1

<

x

Since (see Solution of Problem 63)

n

<

1 61

(1

+

s

1 and

1 for k

0 <

x

<

2TI , gives the desired

1 , 2 , . . . , n and a 1 + a2 + · · · + an

n ak ) > 1 + I ak k=l

(E . 51)

k=l

and -�-- > TT

1

1 +

�l.. a k=l k

Sol ution .

n

k=l

c 1 - �)

>

1 -

n

I a . k= l k

(E . 52)

C learly

Hence

and continuing this process we see that (E. 53) In l ike manner we see that

CHAPTER 2

1 62

Thus , s ince 1 - a 3 is positive , we have

and, general ly,

(E . 54) Next , =

1 - a 2l 1 1 - a1 < 1 - a 1 '

---

---

so that

and therefore , s ince a 1 + a 2 +

•••

+ an < 1 , we have , by (E . 54) ,

1 - -­ ( 1 + an ) < --c:-1-----,,.-+---(a la2 + • • • + an ) '

(E . SS)

Similarly, we find (E . 56) By combining the four inequal ities (E . 52) , (E . 53) , (E . 54) , and (E . SS) , we obtain the inequalities (E . 51) and (E. 52) .

PROBLEM 1 1 6 .

Show that

"' w (t) e - t dt < 1 Hn- 1 = (n -1 1) ! / ' (e l) n n where t is real , n is a positive integer, and w (t)

=

(t - 1) (t

Sol ution .

In = Hn en+l

We put

-

2)

•••

(t - n + 1) .

(E . 57)

1 63

I N EQUAL I T I E S

that Hn-l (e - 1) -n is the best possible estimate. Changing from andtoshow t t - n - 1 yields <

Now,

(n +1 l) ! J t (t + 1) (t + n) e -t dt Kn + e-1 In+l ' where Kn (n +1 1) ., /1 t (t 1) (t + n) e- t dt. (E . 58) Thus (1 - e-1) In+ 1 - In Kn (E . 59) and, in particular, (1 - e-1 ) In+ 1 In so that Jn In (1 - e-l ) n+l is a monotonic increasing sequence. Using (E . S9) , we have •••

0

0

+

•••

>

so that It follows from (E . S8) that Kn 1, n 1,2, . . . , so that we may write lim Jn J l + n=lI Kn (1 - e-l ) n+l . (E . n+co Now for t et [1 - (1 - e-1 )] -t 1 + t (1 - e-1 ) + t(t + 1) (12! - e-1 ) 2 + <

60)

;;>: 0 ,

. . .

'

CHAPTER 2

1 64

or

••• .

-1 2 -t 1 = e -t + te - t (1 - e - 1 ) + t (t + l) e 2 !(1 - e ) +

Integrating over 0

�

t � 1 , a clearly j ustified termwise integration yields - l n+l , L Kn ( l - e ) n=l

and by (E . 60) 1 1 1 + J 1 - / e -t dt - (1 - e - 1 ) J te -t· dt . 0 0

lim J n -r oo n

A simple calculation shows that J1 ---n2 (1 - e- 1 ) 2 and l imn + oo Jn = 1 . This is the asymptotic result Hn- l (e - 1) as n + lows immediately from the monotonicity of Jn . =

PROBLEM 1 1 7 . Show that

1

n sin x . TT x. i=l

1

___

Sol ution .

Y1 + Y2 + n

�

Let 0

<

x 1.

<

11 ,

"" ·

X

1 , 2 , . . . ,n,

i

(sinX x) n . y 1. for i

Let f(t) = (sin t) /t and f(x i )

•• • + Yn

because f is concave on > 0 for 0 < t < 11 and f� �t} = ta� t (

The inequality (E . S 7) fol-

-

i-

•••

f (x 1 ) + f(x2 ) + + f(xn ) --�--------n-----------( 0 , 11) .

cot t

�

1 , 2 , . . . , n . Then

f(x)

The latter can be seen by noting that f (t)

-

i<

0

for 0

<

t

<

11 .

CHAPTER 3 SEQUENCES AND SERI ES

PROBLEM 1 .

Let a0 and a 1 b e given and define , for n

=

2,3, . . . ,

Show that lim an Sol ution .

We have

hence

and thus a 3 - a2

a l - ao /

a l - a2 2

a a (-l) n- 1 l n- l o 2 -

.

.

.

,

an - an- 1

Consequently 1 65

=

CHAPTER 3

1 66

a l - ao 2

and so

+

2a l

ao

----=--.c.. +

an Letting n

3

� oo ,

=

- l + __!._ -

>

_1_)

(- l) n-2 n-2 2

0.

an = ak a a liT rr·m·m

>

a n

Thus

But ak /k ! is fixed and

Let

Show that limn � oo Pn Sol ution .

+

For any real number a , show that

We assume that a k w e have

PROBLEM 3 .

••.

a - ao ( - l) n - 1 l n- l · 3•2

Sol ution .

For n

22

2

we obtain the desired resul t .

PROBLEM 2 . n lim � n � oo n !

(1

=

n3 - 1 --. n3 + 1 2/3 .

Note that

0 and let the integer k be such a

<

k

+

1.

SEQUENCES AND S E R I E S

k3 - 1 = k+T - 1 · k22 k 1 � k -k+1 +

k

But

1 67

+

_E_I I k+T k - 1 = n(n 2 1) and nI I k22 k + 1 n2 3n 1 k=2 k=2 k - k + 1 Hence 1n +oo1m. Pn = 2 n1+1m. oo n2n2 n+ n 1 = 2 +

+

+

3•

+

+

+

3"

PROBLEM Let be a positive number. Taking an arbitrary positive number x0 and forming the sequence xn = .!_2 (xn-1 + xn-lK ) with n 1 ' 2 ,3, . . . show that limn+oo xn More generally, show that if m is a positive integer and Xp = m m- 1 Xp-1 m xm-1 with p 1 ,2,3, . . . p- 1 then limp +oo xp m/K. By induction we see that 4.

K

-

'

IK.

--

K

+ --­

Sol ution .

But and so nlim+oo Hence

<

IKI 1

IK

0.

'

1 68

CHAPTER 3

fi(

lim n -+ oo Xn +

0

IK.

and

This proves the first part of the claim. To prove the second part of the claim, we first note that xmp

>

K. Indeed,

m p

X

By Bernoul l i ' s Inequal ity (see Problem 35 of Chapter 2) we have that ( 1 + a) n > 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

<

K

--

m p- 1 = K . m m X X p- 1 p- 1 - X

0.

Hence (xp) ;=O is a decreasing sequence of positive numbers and therefore con­ verges to some real number A with A satisfying m - 1 A + K m lii=T rnA '

A Thus Am X

p

m iK. Moreover,

K or A >

m fl(

>

K

1· Xmp

Remark . It is interesting to note how the approximat ing sequence for was formed above , namely, we took K = x0 (K/x 0 ) and defined x 1 to be the arithmetic mean of the factors x0 and K/x0 , then we took K = x 1 (K/x 1 ) and defined x2 to. be the arithmetic mean of the factors x 1 and K/x 1 and so forth .

IK

PROBLEM 5 . Show the convergence of the sequence J_ J_ n = l + .rz + 13 +

X

1 - 2 1il. +rn

169

SEQUENCES AND SERI ES

We show that the sequence is decreas ing . Since

Sol ution .

2 ( v'ri'"+1

1 and

v'ri'"+1

_

rn = __....:=---v'ri'"+l + rn

1

<

>

-

lri)

_....:1=--­ 2 v'ri'"+l

'

we have that xn+l xn . To see that the sequence is bounded below , we observe that +

__!__ rn

> 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

<

B

k)

1 . Hence , i f x = A = (3 + IS) / 2 , then

- --------=-1--

n+ 1 2 2 4 (1 + X ) ( 1 + X ) • • • ( 1 + X )

1 and S X = B

as n

+ ""

(l/x) • 1 = 2/ (3 + IS) = (3 - IS) / 2 . I f , on the other hand , we have (3 - IS) /2 , then 1 -

1

_ _ _ _ _ _ _ _ _ _ __

n (1 + X2 } ( 1 + X4 ) • • • ( 1 + X2 )

1 -

__::1:_ + X4 + • • •

_ _ _ _ _

tends to 1 - --11--1 - X2 as n

<

+

+ ""

Sol ution .

!

1

-

(1

- X2 )

and S = ( 1/x) x 2

PROBLEM 7 .

n/2

=

_ _ _ _ _

X

2

X = (3 - IS) /2 .

Show that

For any positive integer n , we consider

cos 2n t dt .

Applying integration by parts twice , we get n/2

f cos 2n t dt 0

=

t cos 2 n t

I

n/2 0

+

2n

n/2

J t cos 2n- l t sin t dt 0

CHAPTER 3

1 72

n(t2 cos2n-l t sin t) 1:/2 -n rrI/02 t2 [ -(2n - l)cos2n-2 t s1n. 2 t cos2n t]dt -2n2 1 2n n(2n - l)I 2n_ 2 , where 1 2n rr.[/2 t2 cos 2n t dt. On the other hand, rr/ 2 cos 2n-l t d(sin t) .[ cos 2n-l t sin t I :1 2 (2n - 1) !2 cos 2n-2 t sin 2 t dt, that is, (2n - l)J 2n_ 2 - (2n - l)J 2n or J 2n 2n - 1 J 2n-2 ' where J2n rr/2J cos 2n t dt. Noting that J0 rr/2, we see that - 1) (2n- 2)- •••4•2 3)•••3• 1 .�2" J2n = (2n2n(2n We may thus conclude that (2n(2n)- 1)!! ! ! where, as usual, (2n)! ! 2•4••·(2n - 2)(2n), 0 " 1; (2n 1)! ! 1•3•••(2n - 1) (2n 1) , ( -1) ! ! 1. Therefore =

+

+

=

+

= --zn--

-

0

=

rr

2'

=

+

=

=

+

1 73

SEQUENCES AND S E R I E S

2) ! ! I 2n-2 1 (2n(2n)!- 1)!!! I 2n (2n(2n -- 3)!! n and so (2n(2n)- 1)!! !! I 2n - (-1)0! !!! I o nI (2k)! ! I - (2k - 2) ! ! I k=l ( (2k - 1)! ! 2k (2k - 3) !! 2k-2 = 2:.4. k=lnI -k12 . This shows that 2 n ] 3 n (2n(2n)!- 1)! !! I 2n = �4 - i kL k12 = i [ rr6 k� l k12 · But (2n)- 1)! ! ! ! I 2n = 0. lim "' ( 2n n Indeed, since (2/rr)t sin t for 0 t rr/2, we have I 2n = !o0rr/2 t2 cos 2n t dt 2 2 L0rr/ 2 sin 2 t cos2n t dt = 1[42 [Ialf/ 2 cos2n t dt - l0rr/ 2 cos2n+2 t dt] 3 = 2!_3 _( (2n(2n)- 1)!! ! ! ((2n2n ++ 1)2)!! !! ) (2n( 2n +- 2)1) !!! ! and so 42 1T

_

-

.

)

-

-+

�

�

�

(.:!:.)

- If

_

8

Remark .

�

rr

For related results see Problems 103 and 104.

PROBLEM Show the 2:.2 = nlim-+ "' 1 3 3 5 ( 2n - 2n1)• 2n(2n + l) . 8.

Formula of Wallis :

•

•

2• 2•4•4 •

• • •

• • •

CH A PTE R 3

1 74

Let m be a positive integer and put Jm = l0rr/2 sinm x dx and J'm l0rr/ 2 cosm x dx. Integration by parts yields Jm Jo{o rr/2 sinm-1 x d(-cos x) -sinm-l x cos x l :/2 + (m - 1) �rr/ 2 sinm-2 x cos2 x dx em - 1) J(onrr 2 CS1n. m-2 ) C 1 - S1n. 2 ) d (m - l)Jm_ 2 - (m - l)Jm . Thus m m- 1 Jm-2 · Jm = -Noting that J0 rr/2 and J 1 we obtain (2n2n(2n - 1) (2n- 2)- 3)• • • • 4·2.. 3·1 ·2rr and 32n+l = (2n2n(2n + 1) (2n- 2)- 1)• • • • 4•2• • 3·1· Exactly the same results obtain for Using the notation introduced in the Solution of Problem 7 concerning the symbol m! ! , we have r rr/2 sinm x dx = J{ rr/2 cosm x dx (m m!- 1)! !! 2 for m even, Jo o (m m!!- 1) for m odd. We now suppose that 0 x rr/ 2; for these x we have S1n. 2n+l S1n. 2n S1n. 2n-l Integration over the interval from 0 to rr/2 yields Sol ution .

=

=

I

X

=

X

X

l,

J� .

=

X <

<

X <

<

X.

Tr

!!

1 75

SEQUENCES AND SERI ES

hence

(2n(2n) 1)! ! ! ! < (2n(2n) 1)! ! ! ! < (2n(2n 2)1) !! !! or ( (2n(2n) 1)! ! ! !) 2 1 < < ( (2n(2n) 1)! ! ! !) 2 2n"1 But the difference of the two outer expressions equals 2n (2n1 1) ( (2n(2n) 1)! ! ! ! ) 2 < -2n1 -2 and hence tends to as n becomes arbitrarily large, we see that 1m. ((2n( 2n) 1)! ! ! ! ) 2 1 - n1+oo +

2 11

-

2rl+l

-

-

2 11

-

11

+

-

0

11 2

"2r1+1·

_

-

PROBLEM Show the 2 Let t We commence by showing that sin-t . ) (cos 7t ) • • • (cos tn) -t lim ( cos n+oo 2 Indeed, sin t 2 (cos % ) (sin %) = 2 2 (cos %) (cos 2\ ) (sin :2 ) 2n (cos %) (cos 2t2 ) • • • (cos 2�) ( sin 2�) and so sint t --"t/2-'-'-n--n sin t/2 But sin t/2n -n + 1 as n + ---"t/2 9.

Formula of Vieta :

11

Sol ution .

2 t

f 0.

=

"" ·

(E . 1)

CHAPTER 3

1 76

From (E.l) we get, for t = �1 2, -�2 = lim (cos i) (cos i) (cos 2n�+ 1) Since cos i = ft and cos � = f� + � cos the desired result is easily obtained . · · ·

·

8,

PROBLEM 10. Verify the 4 = 4 arc tan !5 arc tan 2319" We have 2 1 1 4 arc tan !5 = 2 arc tan + 2 arc tan = 2 arc tan ----1 12 arc tan 12 = arc tan 5 + arc tan 5 5 5 arc tan -121 - + 25-12 arc tan 120119 . 144 Further arc tan ii� + arc tan (- 2�9) 120 1 arc tan 1 119+ 120 1 arc tan 1 119 239 From the expansion arc tan x x3 + x5 • • • + (-1) k-1 ;k-1 + where -1 x 1, we see that 1 -31 + -51 - + (-1) k- 1 1 + Formula of Machin :

�

-

Sol ution .

5

2..

5

5

I2

-

I2

239

Remark .

X -

5

5

3

S

-

�

�

2s

1 77

SEQUENCES AND SERI ES

To calculate accurate to seven decimal places, we would have to consider the first ten million terms of the last series. However, by Machin's Formula, = 16 (t t- 512 t• 515 - 9-· 517 i· 519 - 111 • 5�1 + • • -) - 4 ( 2;9 t- 2319 3 + •• -) with the displayed number of terms being sufficient to obtain 3.1415926 TI

-

TI

+

+

-

TI

PROBLEM 11. Let q be a positive integer. Show that � n(n 1) (n 1 2) • • • (n q•q!1 The following identity is easily established by induction: k 1 ·(n + q) = 1 { 1 (k l) (k + 12)·· ·(k q)} . l)·· n(n � nl Therefore the claim follows immediately. The identity kI n(n + l)•••(n q) = ---1--- k(k l)···(k q 1) 2 n=l can readily be established by induction as well. It is also of interest to note that, if p is an arbitrary real number different from -1, -2, 3 .. . , then 1 n=lI (p n) (: n + 1) = n=lI {p ! n - p � 1 I (p + n)(p n 1+ 1) (p n 2) n=l n=lI i{ (p + n) (� + n + 1) - (p + n 1)\p n 2) } 2(p + l)1 (p 2) ' co

+

+ q)

+

0

Sol ution .

+

q q! -

+

q +

+

+

Remarks .

+

-

+

+

,

+

+

+

+

+

}

+

+

+

+

+

+

IJ+l'

0

0

0

1 78

CHAPTER 3

and in general, for q 1 and q an integer, �

q(p + l) (p + 12)• • •(p + q) ' Letting p we obtain the result in Problem 11. To see that q(p l ) (p 12)···(p + q) is true under the stated conditions, we only need to observe that (p n) (p + n + 1l) •••(p + n + q) (p + n l) (p + n 1 2)•••(p + n + q 1) + 1 + n q + 1) (p + n) (p n + l)•••(p and proceed by induction on q. Finally, we note that the result in Problem 11 is equivalent to 1 - q. I n=l (� : �) = 0,

+

+

+

+

+

+

+

_
PROBLEM 12. Let f \(q _ 1) . n k=l l' . n 2 Show that limn_,..,, Sn 1/4. Since _x_ < 1 < -2 for 1 + x 2 we see that, putting k/n2 , s

�

Sol ution .. + X

X

ll+X -

x

Hence

=

> 0,

+

SEQUENCES AND SERI ES

1 79

nI k nI k. � k=l 2n2 k n 2n k=l But 1 2 nI k n(n +2 1) + -41 as n + 2n k=l 4n the other hand, 12 nI k - nI 2k = lim nI 2 k22 � lim n + co 2n k=l k=l 2n + J n + co k=l 2n (2n k) But nI k2 nI 4k2 = n(n 1) (2n + 1) 2 2 k=l 2n (2n k) k=l 4n 24n4 (see Problem 15 of Chapter 1) and so nI k - nI k � = 1 � lim n + co 2n2 k=l k=l 2n2 k and nI k = 1 lim n + co k=l 2n2 k Hence the desired conclusion. +

< s

<

"' ·

On

+

<

+

+

+

+

0

4'

PROBLEM 13. Show that 1 1 1 1 lim n! 3! l! 2! n + co By Problems 2 and 3 of Chapter 1, the sequence n = (1 + .!n) n n = 1,2,3, ... is increasing and bounded, hence converges to a number e, where 2 e 3. The number e is the base of the natural logarithms. It is also clear that (�) i nn)_.!_nn n + - + - + - +

+ - +

Sol ution . x

<

+

• • •

+

(

<

CHAPTER 3

1 80

1 + 1 + 2\ (1 !)n + ...!.3! . (1 !)n (1 - �n) + + k!1 (1 - 1) (1 - � n) + -

n

• • •

. . .

• • •

Since k < n 2 + _!_2! (1 !)n _!_3! (1 - !)n (1 - �n) + + ...!.k! .( 1 !)n (1 - � n) Letting n + we have n,

X

>

-

• • •

+

-

. . .

.

oo ,

Clearly and so limn + oo yn exists and is equal to e. We also note that (n +1 1) ! + ...,.-(n-- +1-:::-2),.! .-:- + + ...,-(n-+1-,... m) ! .,1 2 + + 1 m-l } < (n } 1) ! 1 + n ! 2 + (n + 2) (n + 2) n + 2· < (n +1 1) ! � Keeping n fixed and letting m + we get n+2 e - Yn <- (n +1 l)! �· Thus 1 0 < e - Yn < n!n because (n + 2)/(n + 1) < 1/n. The foregoing estimate can be used to calculate e. It can also be used to show that e is an irrational number. Indeed, suppose that e is ) rational and e p/q, where p and q are positive integers. Then 0 < q!(e - Yq < 1/q. By our assumption q!e is an integer. Since • • •

�--=-,..-,-

$\

• • •

oo ,

Remarks .

=

1 81

SEQUENCES AN D SERI ES

q!yq q! (1 + 1 + }! + + __!__q! ) is integer, we see that q!(e - yq) is integer. Since q 1, 0 q!(e - yq) 1/q would imply the existence of an integer between 0 and 1 and we have reached a contradiction. The number e 2.7182818 ... • • •

=

an

an

<

�

<

PROBLEM 14. Show that the sequence cn + n1 log n converges. Let dn cn - 1/n. Then cn+l - cn 1 log n and Since, for n x n + 1, we have 1/(n + in+l n+l dx n+l dx , n + l Ln x Jn n it follows that • • •

Sol ution .

il+f -

<

�

<

<

=

1) <

..!_ -

dn n log (l + ..!.n.) . 1/x 1/n, and so <

<

This inequality shows that the sequence e decreases monotonically, whereas the sequence dn increases monotonically. nFurthermore, dn en ; hence d1 0 is a lower bound for the en . In other words, the sequence en converges to a limit c; c is known as and c 0.5772156649 ... The foregoing shows that the harmonic series 1 + ..!.2 . + ..!.3 . + + ..!.n . + diverges. Moreover it is easy to see that the sequence en increases very slowly; for example, c1000000 14.39 . . . <

Euler ' s Constant

Remarks .

• • •

=

=

=

CH APTER 3

1 82

PROBLEM 1 5 . Consider the alternating series 1 - I1 + 1 1 + 1 1 + + C-ll n+l 1 + Examine the following rearrangement of this series: Let the first p positive terms be followed by the first q negative terms, then the next p positive terms be followed by the next q negative terms, etc. Show that the resulting series converges to log(2lp/q). Let Hn = 1 + I1 + + n1 By Problem 14, Hn = 1 + I1 + + 1 log n + c + xn ' where c is Euler's Constant and xn as n Thus -21 + -41 + I1 log m + I1 c + I1 xn and + 1 + .!.3 + log 2 + I1 log k + I1 c + x2k - I1 xk . The foregoing shows that the partial sums of the series in question are of the form log(2lp/q) + zn ' where zn as n 1 - l

� - �

�

. • •

. • .

Sol ution .

··•

�

=

+

0

+ oo .

+

0

+ oo .

• . .

PROBLEM 1 6 . Let a and d be positive numbers and consider the arithmetic progression a, a + d, a + 2d, a + (n 1) d. Let An denote the arithmetic mean and Gn the geometric mean of the foregoing progression. Show that -

where e is the base of the natural logarithms.

S EQUENCES AND S E R I E S Sol ution .

1 83

Setting c a/d, we get n cn n - 1 c + nn - 1 =

���.�.� ln

n n It is clear that nc n 2n- 1 21 as n + To see that yn n n c nn - 1 e1 as n we observe that � { log * + log c � 1 log c � 2 + • • log c � - 1 } tends to {J c0 1 log x dx -1 as n + Note the special case lim n lil!n = e G A

il

+

• • •

---zri

- + --- + -

"' ·

�.� .�

+

+ -

+

+ "'

•

+

+

"' ·

Remark .

..!:. .

PROBLEM 17 . Let [t] denote the greatest integer less than or equal to t. Show that as n + We see that n-1� 11/k ([2j - 2 [l]) dx lim n+ oo k l 1/k+l n-1 ( ! - ! ) lim n+ oo k=l k Y, k 1 "' ·

Sol ution .

x

l:

x

CHAPTER 3

1 84

2 3 .!. + .!.5 - 6 + • · ) = 2 log 2 - 1 by Problem 15, or noting that, for positive integers p and q, ( 1 xp-l q dx 1 - 1 + +1 2q - +1 3q + Jo 1 + x =

(.!.

-

.!.

4

=

•

P

P

p+q

P

PROBLEM 18. Let p 1 , p2 , . . . , pk be positive and a1 > a2 > • · · > ak > Show that + ••• + We have ••• + Sol ution .

But the expression in square brackets ntends to 1 as n + Moreover, n � tends to 1 as n + Indeed, let xn � - 1. Then xn if p1 1 and -1 xn if p1 1. In either case 1 + nxn (1 + xn) n p1 by Bernoulli's Inequality (see Problem 35 of Chapter 2) and so xn + as n + If p1 1, then xn for all n and there is nothing to prove. < 0

<

oo .

<

0 <

oo .

<

> 0

>

=

0

oo .

=

=

0

PROBLEM 19. Let c for the moment be an arbitrary real number and put 2 c n and xn+l + for n 1,2,3, . . . Investigate the convergence of the sequence (xn) :=l" We observe that if the limit of the given sequence exists, say limn + oo xn a, then = 2

Sol ution .

=

2 X

=

0.

SEQUENCES AN D SERIES

Thus

1 85

a = c + a2 or a2 - 2a + c

o.

2'

2

�.

a=1 and so the sequence (xn) :=l can not converge for c 1. (a) We assume now that c 1. Then xn It is easy to verify that and xn xn+l for n 1,2,3, . . . Thus limn -+ «> xn exists and cannot be xlarger n 1 than 1; hence the limit is equal to 1 (b) Suppose that -3 c Then xn c/2 and xn for n = 1,2,3, ... Indeed, the first claim is obvious and we need only to have a closer look at the inequality xn for n = 1,2,3, . . . But the inequality is true for n 1. Assuming that xn for some positive integer n, we see that xn lc l /2, (because lcl/4 1) and xn+l is seen to have the same sign as c/2, that is, xn+l is negative. At any rate the sequence (xn) :=l is bounded. However, the sequence is not monotonic. Noting that ±

0 <

<

<

>

>

$

=

$

< 0

< 0.

�

0.

�.

< 0

=

< 0

$

<

and it can easily be established by induction that x2k+l > x2k-l and x2k+2 x2k for k 1,2,3, . . . In other words, a' = nlim-+ «> 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" <

oo ;

-+

CH APTER 3

1 86

Thus a ' - a"

+

and so ( a ' - a") (a' a" + 2) = 0 . As we shal l see at once , the second fac­ tor cannot vanish for c > - 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

<

(e l /e ) - e

=

0.

We consider the three interval s from

0

to

to oo

a,

from

a

t o S , and from S

The function g is pos itive in the first and third interval and negative in the second. Consequently , if we take the initial value x in the first or third interval , the consecutive values of xk increase with k . On the other hand, if the initial value x comes from the second interval , the numbers xk form a decreasing sequence . By putting x S + o , with o 0 , we have a S+o - S -

>

= (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 <

for x

aX

y.

<

By taking, instead of

aa

'

that is to say, have X

n

>

X

n- 1

X

>

]

< a.

. . .

y,

the root

We also note that Xn

> xl > x ,

a,

> S.

we have , for x

< a;

< a

but since , for x

<

a,

we

we see that Xn tends to a l imit which does not exceed a ; but as this limit is equal to one of the roots of the equation a X x we have that this l imit is equal to a , that is , we can write l imn -+ oo Xn a .

SEQUEN CES AN D S E R I ES

1 89

In the same way we can show that if x is between x

>

x1

>

x2

and so l imn + oo xn

•••

>

a.

=

> X

a,

>

n

8,

If x n

8.

and

8,

we have

we have

8

X

and so l imn + oo xn =

a

The l imit ing case a

l im x = e n + oo n

for x � e

lim x n + oo n

for x

e l/e gives

and >

e.

We come t o the case a < 1 . When considering the function g , we noted that its derivative is con­ stantly negative for a < 1 ; thus the function is decreasing ; it is pos itive for x = 0 and negat ive for x 1 , hence has one real root between 0 and 1 . We denote this root by a . W e consider for the moment the equation =

X

aa

x.

=

This equat ion evidently does not have real roots for a > e l/e ; for 1 < a x < e l/e the two roots a and 8 of the equation a = x are the only roots of the equation X

aa = x . Always staying with the case a We take up the function h (x)

1 and putting a

1/b , w e obtain b

>

1.

X

=

By putting y h ' (x)

<

aa - x . ax , we can write the derivative h ' in the form aY y log 2 a - 1 = b -y y log 2 b - 1 .

The second derivative is h" (x)

l og 3 a aY y (1

+

log log ay)

log 3 b b -y y (1 - l og log by) .

1 90

CHAPTE R 3

<

If we suppose b e, we have 1 - log b b -x 0, and we obtain h"(x) < 0; consequently, the first derivative h' is a decreasing function. But we hqve 2 h' (O) = � b - 1 < 0, and so h' (x) < 0. The function h is decreasing and since h(O) = a 0, h(l) aa - 1 < 0, it becomes zero for the root of the equation ax = x. For the case b e, the equation h"(x) = 0 admits a real root between 0 and this root is equal to logloglogb b For 0 ::; x < x0 we have h"(x) 0; but for x0 < x we have h"(x) < 0. In the first interval the function h' increases when we start from the value -log-b2-b - 1 which is negative for all values of b and attains its maximum for x x0 . After this value the function begins to decrease. If h'(x0) < 0, h' (x) will be negative for all values of x and so h(x)x has the only root which will be at the same time root of the equation a = x. If one has h' (x0) 0, it is not difficult to convince oneself that the x equation h(x) = 0, in addition to the root of the equation a x, admits two others. But, since 1 log b - 1 , we prove that if log b e, the equation h(x) = 0 has two real roots included between 0 and 1 , distinct from In this case a < �ee = 0.065948 . . . >

>

a

>

1;

::;

>

1

a

>

=

e

>

a.

SEQUENCES AN D SERI ES

1 91

The function h clearly has one root a which belongs to the equation ax = x. We subst itute this root into the expression of the first derivative ; we have the relation h ' (a) = a 2 log 2 b - 1 = log 2 a - 1

�

0,

where h ' (a) becomes zero for b ee , a = 1/e, and h ' (a) is pos it ive for b e > e Thus it is seen that for E positive and very small h (a

-

E) <

0,

h (a + E)

>

0.

>

But , because o f the inequal ities h (O) 0 , h ( l) < 0 , we arrive at the result that the equation h (x) = 0 has two other real root s , one between 0 and a and the other between a and 1 . We denote the first root by a 1 and the second by a 2 , and have

Thus we see that , for the case 1/ (e e )

xk+l and aa

s

a

<

1 , we have

a from which

When x < a , we clearly have

x 2n+l

> a.

Since the h (x) and g (x) have the same sign , w e have the inequal ities

and we note that the expressions with even indices increas e , remaining less than a; consequent ly they tend to a limit . In the same way, the expressions with odd indices decrease , remaining larger than a and tending to a l imit . It is clear that these express ions cannot posess a l imit different from the

CHAPTER 3

1 92

root of the function h and so l im x n -+ oo 2n

=

lim x = a. n -+ co 2n+ 1

For x > a the reasoning is the same . The express ions with even index approach the l imit a decreasingly. Finally, we take up the case a < 1/ (e e ) . We consider the four interval s

The s igns of the functions g and h for these interval s are tabulated next : I

II

III

IV

g (x)

+

+

h (x)

+

-

+

-

-

-

The table of signs of the functions g and h shows that the root a is to be found between two consecutive terms of our sequence

We wil l show that the terms with even index are to be found in the same interval . Indeed , if x is in the first interval , x 1 wi l l be in the fourth and conversely. This can be shown in the fol lowing manner : If x < a 1 we have ,

or, in other words ,

and conversely; if x

> a2 , we have

We note that al l terms x 2k with even index are situated in the same interval and all other terms x Zk+l in the other. If x is in the second interval , all terms x 2k are in the same interval , but al l terms with odd index are in the third interval and vice versa . Indeed, if a 1 < x < a, we have

1 93

SEQUENCES AN D S E R I ES

or In the same way, if a < x < a 2 , we have or From the preceding discussion we conclude that if x is in one of the two first intervals I , I I , we have x2n � a 1 , x2n+l � a 2 as n � oo . For the intervals I I I , IV, or x > a , we have x 2n � a 2 , x 2n+ l � a 1 as n + oo . We summarize the results in the fol l owing fashion : The l imit L (x) is determined as fol lows :

Theorem .

1 . I f a > e l/e , then for all real values of x . L (x) 2 . I f l < a :<; e 1/e , then L (x) = a L ( S) s L (x) 3 . I f 1/ ( ee ) 4.

If 0 < then L 1 (x) L 1 (a) L 1 (x)

:<;

< X < a, for for x s , for s < X < + oo . a

:<;

1 , then L (x) = a .

a < 1/ ( ee ) and putt ing L 1 (x) = l imn + oo x 2n and L 2 (x) = l imn + oo x 2n+ l '

a l , L 2 (x) a 2 L 2 (a) = a a 2 ' L2 (x) = a l Here a and S are roots of aX a = a2 , a 2 = a l .

< X < a, for for x a , for a < X < + oo . =

X

a and a < S ; moreover, a l < a < a 2 and a l

Remark . The foregoing Theorem is actually due to Euler (De formulis exponential ibus repl icatis , Acta Academia Scient iarum Imperial is Petropo l i­ tanae , 1777 . )

PROBLEM 2 1 . Show that for any irrational number a , there exist infi­ nitely many rational numbers p/q such that

CHAPTER 3

1 94 Sol ution .

Let n be a pos itive integer . Consider the n + 1 real numbers

0 , a - [a ] , 2 a - [2a ] ,

...

na - [na]

'

(E . 2)

(here [r ] denotes the greatest integer contained in r) and their distribution in the hal f-open intervals

[i,n � n )

with j

=

0 , 1 , 2 , . . . ,n - 1 .

The union of these n intervals is the hal f-open interval [ 0 , 1) and hence contain the n + 1 numbers (E . 2) . It is clear that two of these numbers (E . 2) l ie in the same interval ; this is the "pigeonhole principle" : if there are n + 1 obj ects in n boxes , there must be at least one box containing more than one obj ect . Cal l these numbers with

and

Since the intervals are of length 1/n and are not closed at both ends , we must have that

Write q for the positive integer n 2 - n 1 and p for [n 2 a] - [n 1 a] ; we have that l qa - p i < 1/n with q $ n . Hence , given any pos it ive integer n , there exist integers q and p with n � q > 0 such that l qa

PI < n

1

-

or

Ia

-

E.

q

I

<

_!_.

nq

(E. 3)

The latter relation impl ies the inequality stated in the prob lem, because n � q impl ies that 1 nq

$ z·

1 q

To show that there are infinitely many pairs (p , q) , suppos e , on the contrary, that there are only finitely many , say (p 1 , q 1 ) , . . . , (pm , qm) . We prove that this supposition is false by finding another pair (p , q) satisfy­ ing (E . 3) . We define £ as the minimum of

Since a is irrat ional , £ is positive . Choose n so that 1/n < £, and then by the first part of the proof which led to (E . 3) we can find a rational number p/q so that

SEQUENCES AND S E R I ES

1 95

I a E.q \ < nq < n < E . .!_

_!__

-

By the definition of = 1 , 2 , . . . ,m.

E

it fol lows that p/q is different from p 1. /q1. for i

The "pigeonhole principle" has already been used in Prob lems 35 and 36 in Chapter 1 . Remark .

PROBLEM 2 2 . Let a be an irrational number. Show that the set of all numbers of the form ax + y, with x and y integers , is dense in the real num­ ber system, that is , given any real number r and any E > 0 , there are inte­ gers x and y such that ax + y - r < E holds .

l

Sol ution .

l

From Problem 2 1 we see that the inequal ity

I ax + Y l < E < 1

can be sat isfied by nonzero integers x and y . Hence let CI.X l

+

lsi

y1 = S '

< E•

Let g be the largest integer contained in r/S ; then g � rs < g + L

Thus , putting x = gx 1 and y = gy 1 , we get

l ax + y - r l

PROBLEM 2 3 . Cl. =

=

l agx 1

l

l

l

+ gy 1 - r = ga - r <

lsi

< E.

The decimal fraction

0 . 1 2345678910 1 1 1 2 1 314 . . .

(the pos it ive integers written consecut ively) represents an irrational num­ ber. Show that , for n = 0 , 1 , 2 , 3 , . . . ,

]

is dense in the interval [ 0 , 1 . Sol ution .

]

We obtain lOn a - [ lOn a by shift ing the decimal point in a

CHAPTER 3

1 96

by n places to the right and deleting all digits to the left of the decimal point . Let

be a finite decimal fraction . We choose n such that lOn a - [lOn a ] begins with the digits 8 1 , 8 2 , . . . , 8 k and is followed by r zeros ; then we have 1 ' lOk+r

< -----

PROBLEM 24 . na - [na]

Let

be an irrational number and consider the sequence

for n = 1 , 2 , 3 , . . .

Show that the set of accumulation points (i . e . , the set of al l subsequential l imits) is [0 , 1 ] . Sol ution .

any

E >

0,

By Problem 21 , there are integers q 1 and p such that , for

The point q 1 a - [q 1 a ] i s therefore within a distance o f The sequence of points

E

o f either 0 o r 1 .

continued so long as may be necessary, mark a chain ( in one direction or the other) across the interval [0 , 1 ] such that the distance between consecutive points on the chain is less than E . Hence for any point r of (0 , 1) there is a point kq1 a - [kq1 a ] or na - [na] within a distance E and so the set of points na - [na] for n = 1 , 2 , 3 , . . . is dense in the interval [0 , 1 ] . Let e be the base of the natural logarithms . Since e is an irrational number , the set of accumulation points of the sequence Remarks .

ne - [ne]

with n

=

1 , 2 , 3, . . .

is the interval ( 0 , 1) . However, the sequence n ! e - [n ! e]

with n

=

1,2,3, . . .

has 0 as its only point of accumulation .

1 97

SEQUENCES AN D S E R I ES

Indeed, by Taylor ' s Theorem, e 1 + 1 + ••• + 1 + e n e = 1 + IT (n + 1) ! 2! n!

with

0 <

en

<

1,

hence n!e �

For n

=

en e n ! n ! n ! + IT + 2! + • • • + 1 + n-;-r·

2 , we have

e e n n + 1

<

n

!

<

1

l,

hence e n e n ! e - [n ! e] < n + 1

<

n

:

1'

Another interesting example o f a sequence with an unusual set o f accu­ mulation points is provided by the following s ituat ion . For every integer p � 2 there are exactly p - 1 numbers of the form 1/k + 1/m for which the sum of the pos itive integers k and m equal s p. For p = 2 , 3 , 4 , . . . , consider these numbers enumerated . Then we obtain the sequence 5 5 5 3 3 4 1 4 4 2• 2 ' 2' 3' • 3' ' 6 ' 6•

It is eas ily checked that this sequence has the accumulat ion points

1 • 21 ' 31 ' 41 '

o,

and no other accumul ation points .

Cons ider the fol lowing enumeration of the proper fractions

PROBLEM 2 5 .

0 1 0 1 2 0 1 2 I' I' 2 ' 2' 2' 3' 3' 3'

Here we have ordered ·the proper fractions into groups according to increas ­ ing denominators and, within each group , according to increasing numerators and we have al so l isted fract ions in their unreduced form. It is c lear that in this enumeration the fraction p/q has index k

=

k (p , q)

=

+ Y,q (q + 1) .

CHAPTER 3

1 98

For each k, we now cover the k-th term in this enumeration of rational numhers between 0 and 1 b y an interval of length 2 - k i n such a way that the k-th term is at the center of its covering interval . The sum of the l engths of these covering intervals is

Show that the point I:Z/2 does not belong to any covering interval however. Sol ution . Suppose that t = 1:2/2 belongs to one of the covering inter­ vals . Then there would be integers p and q (q � p � 0, q � 1) such that

1 ::-,__ -;:: q..,... --=-"" .-- )-I .E.q - t I < 2 2p+ (l/2)1 q (q+l) - ----,-2 (1/2) + l --=­ ' (q+l _!_

<

But for every p and q with q

�

1 , we have , since 1:2 is an irrational number

and therefore 2 I Cp/ q) - 1/2 1 (p/q) + (12/2) �

1 . 2 (1/2) q (q+l)

PROBLEM 2 6 . The harmonic series E �=l 1/n diverges , but if we omit the terms corresponding to the integers whose decimal representations contain a specified digit at least once (the digit 0 , for example) , the result ing se­ ries converges . Verify these claims . By Problem 1 4 , the harmonic series E : l 1/n is seen to be = divergent . In the Remarks to Problem 14 we also noted that the divergence was very s low . Another way of seeing that the harmonic series is divergent is to note that the sequence of partial sums of this series does not form a Cauchy sequence ; indeed, setting Sol ution .

sn

+ ll1 '

1 99

SEQUENCES AND SE R I ES

we have that

Il+1

1

1 + ••• + 1 + 11+2 2n "

Denoting the sum on the right-hand side of the last equation by tn ' it is clear that 1/2 < tn < 1 . In fact , 1log (2 - n + 1) < tn < log 2 because , for any positive integers m and j + 1 1 og --m ---

<

1 + ___1___ + • • • + -,-1 m + 1 J

rn

<

j

with m

j log ---m - 1 -.

< j,

The last inequal ity is a consequence of 1 ----- < k + 1

i

k+l dx

)(

k

<

k

1

for k

1,2,3, . . .

Thus tn tends to log 2 as n + ro . The rest of the claim seems surprising at first s ight ; but for very large n the integers that do not have a particular digit in their decimal representations are quite scarce . Indeed, there are 9 • 10n- 1 integers with n- digit decimal representations , but only 9n of them omit 0, and 9n /9 • 10n-l + 0 as n + 00 • The n-digit integers which do not have any digit 0 can be written as a1 . . . an ( un derstoo d to mean , l on- l a 1 + 10n- 2 a2 + • • • + an ) , where 0 < ak � 9 . Since all the terms are positive , the convergence of l: 1/k over al l k of this form is equivalent to the convergence of s

L n=l

1

For a given n there are 9n terms in the inner sum, and each exceeds 10n-1 , so 9n = 90 . S � L n= l l On-1 To go further , we notice that of the n- digit integers , one-ninth have lead­ ing digit 1 , one-ninth have leading digit 2 , and so on ; so of the 9n n-digit . �ntegers , 9n- 1 are between 10 n- 1 and 2 · 10n- 1 , and so on . These integers th en contribute at most 10 - n+l 9n- l ( l + 1/ 2 + • • • + 1/9 )

(0 . 9) n - 1 ( 1 + 1/2 + • • • + 1/9)

CHAPTER 3

200

to S , and consequentl y S < (1

+i+

•••

+ �)

L ( 0 . 9) n - l n=l

10 ( 1

+i+

•••

+ �)

< 2 8. 3 .

the other hand, the integers between l On-l and 2 l On - l are less than 2 • 10n - l , and so on , so that for n � 2 the n - digit integers contribute at least

On

to S . Hence s >

1

+ -12 +

1 +9

+

t L

n=l

( 0 . 9) n-1 (21

+

•••

+ TO1 )

>

2 0 . 189 .

PROBLEM 27. Prove that every positive rational number is the sum of a finite number of distinct terms of the harmonic series

+ -n1 +

Let A and B be positive integers . Then by the divergence of the series under consideration there is a unique nonnegative integer n 0 such that (taking 1/0 0) Sol ution .

=

If equal ity holds the desi�ed representation is at hand and so we assume that

+

n Then A/B E j o= O 1/j = C/D < l/ (n 0 1) . Take n 1 as the unique posit ive integer such that l/ (n 1 + 1) � C/D < l/n 1 . We again suppose inequality as the probl em is otherwise solved, and put f -

D

1_ +1

_ _

n1

+

E. > 0 . F

201

SEQUEN CES AN D S E R I ES

{ C (n 1 + 1) - D}/D(n 1 + 1) and C (n 1 + 1) But E/F in lowest terms we must have E < C. Then =

-

D < C so that with E/F

and so the unique integer n 2 such that l / (n 2 + 1 ) s E/F < l/n 2 satisfies n 2 > n . In a finite number of steps we must obtain the desired representat ion , 1 since if the equal ity does not occur before , it must occur when the numerator of the reduced fraction has become 1 .

PROBLEM 2 8 . Consider an infinite series whose n-th term i s ± ( 1 /n) , the ± signs being determined according to a pattern that repeats periodically in blocks of eight . There are 2 8 possible patterns . Show that a necessary and sufficient condition for the s eries to be conditional ly convergent is that there be four "+" signs and four " - " signs in the b lock of eight . Sol ution .

un + 0 as n + the fact that 1 n

Let un be the n-th term ±1/n and S n = u 1 + • • • + un . Since oo , (S ) : n = l will converge if and only if (S 8n ) :=l does . Us ing

n+k

1

= n (n k+ k) '

that �n=l ( 1/n 2 ) converges (see Prob lem 7) , and that � n=l ( 1/n) diverges (see Prob lem 26) , one shows that with four "+" signs and four "-" s igns in each block , (S 8n ) := l converges as the term-by-term sum of four convergent sequences while an imbalance of signs makes (S 8n ) := l divergent as the sum of a convergent and a divergent sequence . 00

00

the s eries + ••• be divergent and the s eries

where sn assume only the values - 1 , 1 , be convergent . Show that

202

•••

51 + 5 + + 5n 2 l im inf ---=-------� n n -+ co Sol ution .

0

�

�

•••

CHAPTER 3

s + s + + sn 2 ----= .c= -"-l im sup ....ln n -+ co

Put Sn

0 . Then

I f we had Sn > en for n > m, c > 0 , then

m

n- 1

> L S k (pk - P k+l ) + c L k (pk - Pk+l ) + cnpn k=l k=m+l K + c

n L p k=m+l k

would fol low ; here K is independent of n and so the right-hand side tends to + "' · PROBLEM 30 .

Let pn > 0 , p 1

� p2 � p3 � • • • and the series + •••

in which the factors s 1 , s 2 , s 3 , . . . assume only the values - 1 , 1 , be con­ vergent . Show that +

••• +

• • • and s 1

Note the two extreme cases s 1 Sol uti on .

quence

Let S n

s1

+

s2 +

•• • + sn as in Problem 2 9 . Then the se(E . 4)

has the property that between two terms having opposite signs there is a

SEQUENCES AN D SERI ES

203

vanishing term. We distinguish two cases : 1 . in the sequence (E . 4) infinitely many terms vanish; 2 . disregarding a finite set of terms , al l terms of (E . 4) have the same sign . Let them be pos itive for example. In case 1 , let the index M be so chosen that SM = 0 and for M � m < n such that

II l l II s p k=m+l k k n- 1

k=m+l

L

k= l

[ (S k - Sm )

-

(Sk-1 - Sm } ]pk

(Sk - Sm) (pk - p k+l ) + (Sn - sm) pn

I

l

(E . 5}

< E

holds . Let Sm be the closest term on the left of Sn in (E . 4) which vanishes such that Sm+l ' Sm+2 , . . . , Sn have the same sign . Then it fo llows from in­ equal ity (E . S) that l c sn - Sm} pn I = I snpn I < E . In case 2 , let M be so chosen that inequal ity (E . S} holds for M � m < n and that moreover SM , SM+ l ' SM+2 , . . . are pos itive . Let Sm be their minimum . Since in this case S k - Sm � 0 for k > m, it fol lows from the estimate (E . S} that (Sn - Sm) Pn < E , that is, Snpn < E + Smpn . But m is fixed and Pn tends to 0 ; hence for sufficient­ ly large n we have Snpn < E .

PROBLEM 31 .

Let the terms of the convergent series + • • • + Pn + • • • = s

satisfy the inequalities

0

<

Pn � Pn+l + Pn+2 + Pn +3 + • • • .

Show that any point cr in the half- closed interval 0 ed by an infinite subseries + pt + pt + • • • + 2 3 Sol ution .

+ •••

< cr

� s can be represent­

cr .

Put , for n = 1 , 2 , 3 , . . . , k = 0 , 1 , 2 , . . . , + • • • + pn+k = pn ,k

and

Let pn be the first term such that pn 1 l

l im P k + <» n ' k

< cr ;

p

n

.

then either there exists some

204

�

CHAPTER 3

�

k 1 such that Pn , < a , Pn k +l a, k 1 0 , or we h ave Pn s cr . In the l k1 l' l l s econd case we have , since Pn pn _ 1 � a (for n 1 = 1 these inequalit ies 1 l are to read : P 1 s � a) , P = a , that is , a can be represented by an in­ nl finite subseries . In the first case we further determine the first term pn

�

=

2 such that n 2 > n 1 + k 1 , P < a ; then either there is some k sat+ p 2 n2 n 1 , k1 isfying Pn + Pn k < a , Pn , k + Pn k + 1 a , k 2 � 0 , or we have Pn k l' 1 2' 2 l l 1 2' 2 + Pn s a . In the second cas e , s ince 2

�

�

(n 2 > 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 <

1.

�3 . _ 1_----=-3 �5 . _ 1_----=-5 _

_

(2n

1)

+

_

_

(2n

1)

+

•••

we have used the familiar expansion

1

+ X log y--:-x = 2x ( 1

for

-1

<

(1 x

+

�

00

k=l

(k log

n

I (k

k=l

1 4 5x

+

+

+

Show that

;� : i -

1) '.!.(1 = 2

log 2) .

Let

10

1

2k + - 1) g 2i(":"1

1111

log

r· 3·s·7

It is easy to see that sn

_l_ x 2m 2m 1

1.

Sol ution .

sn

+

x)

PROBLEM 35 .

L

1

x2 3

_

+

••

·)

'

1) ; the latter is a simple consequence of

setting x = l/ (2n log

+

1) ! 1)!

n- l (n log 2 (2n -

(2n

+

l)n

-

n.

lxl

< 1,

SEQUEN CES AN D S E R I E S

209

Using Stirling's Formula (see Problem we see that the sequence (sn) :=l tends to the same limit as the sequence (tn) :=l ' where tn as n But, as n tn 1 log n - 1 + 1 og (2n - 2) n-1 (2n + l) n + - 1 1 og 2 + 1 Indeed, as n (1 + 1 ) n-1 + ve and thus lim (� 2n - 1) n-1 1 Moreover, as n + -n 1- -t ) n + e because + --n -1 1) and passage to the limit as n gives that and (l + 1 ) n-l (l + 1 ) have the common limit e. 34) ,

+ oo ,

+ oo ,

�

2

+ oo ,

2ri'"-=-r

�

+ oo ,

2ri'"-=-r

2

r

re

rz

+ oo

Il'=1

PROBLEM

36 .

Il'=1

Let

Show that (Qn) :=l is monotonely decreasing and find its limit. We have (1 + n ! 1) (n+l) 2 (1 + ­n1) -(n2+n+�) Sol ution .

2.

CHAPT E R 3

21 0

2 1( - (n +1 1) 2 ) (n+l) (1 n1) n+!z + -

where

1 2k k=l (k + l) (n + 1) I

and

s2 = k=3 (-l) k (1 2(k 1- 1) ) n 1 are convergent series for all positive integers n. Note that s 1 is a series of negative terms and s 2 is an alternating series whose terms in absolute value decrease monotonely. Thus the sums of both series are less than their respective first terms. Thus it follows that 1_2 for n 1 ,2, 3, . . . s l + s 2 2 (n 1 1 ) 2 + __ 12 n This proves that �+1 /Qn 1; that is, (Qn) �=l is monotonely decreasing. Since it follows that (Qn n= 1 converges. The limit of the sequence (�)�=l is easily calculated if n! is replaced by /2�n(n/e) n (see Problem 34) . Thus nlim-r oo ili nlim-r oo exp {n 2 log (l + *) n } exp { n+oo 1 im ( l2 + 3n -4n1-2 } 12�/e. \' L

k -

<

JC=T

+

-

<

0

<

)

0 > 0, 'n

00

�

I21T

-

__!__ -

+

• • •)

PROBLEM 37. Let n be a positive integer larger than 1. Show that Sol ution .

The inequality

21 1

SEQUENCES AN D SERI ES

m log k hm log x dx, m 2 k=2 gives e 1-m so that >

L

>

Replacing m by n! we get the desired result. PROBLEM Let an + and bn + as n + Suppose, moreover, that (bn) :=l is strictly decreasing for all sufficiently large n. Show that an = lim an - an+l b n :': n n+oo bn - bn+l provided that the second quotient is convergent or properly divergent. Suppose first that the limit is finite and equal to s; then if £ is given, m can be found so that and if n m; thus, for n m, 38 .

oo ,

0

0

1'

Sol ution . > 0

�

�

1,

1

Changing n to n + n + 2, . . . , n + p - and adding the results, we find (s - £) (bn - bn+p) < an - an+p < (s + £) (bn - bn+p) . Taking the limit as p + we obtain oo,

because by assumption an+p and bn+p as p oo Since bn is positive, we have, for n m or nlim+oo ban s. + 0

�

� -

+ 0

+

CHAPTER 3

21 2

the other hand, if (an - an+ 1)/(bn - bn+ 1 ) properly diverges to we can find m so that (an - an+l)/(bn - bn+l) k, for n � m, however large k may be. By the same argument as before, we get an - an+p k(bn - bn+p), which leads to an kbn , or an /bn � k for n � m. Thus . ban 11m n n oo ,

On

>

>

�

->- oo

PROBLEM Let (xn) :=l and (yn):= l be two sequences of real numbers. Suppose that the second sequence properly diverges to + and that it is strictly increasing for all sufficiently large n. Show that 39 .

oo

provided that the limit on the right-hand side exists (be it finite or infi­ nite). We assume first that the limit is finite, that is, n - Yn-1 r, lim Y n n n-1 where r is a finite real number. Then, for any £ there exists a natural number k such that, for n k, we have and We see therefore that all fractions xk+l - xk ' xk+2 - xk+l ' xn-1 - xn-2 ' xn - xn-1 yk+l - yk yk+l - yk+l Yn-1 - Yn-2 Yn - Yn-1 are situated between r - £/2 and r £/2. By Problem 28 in Chapter 2, the :traction Sol ution . X

- X

+ oo

> 0,

�

+

21 3

SEQUEN CES AN D SERI ES

must also be between r - £/2 and r £/2 and hence YIXnn -- Xykk _ r I 2 We now use the identity +

<

�.

and we get that ).

(noting that yk /yn as n We know already that the second summand on thek; right-hand side of the foregoing inequality is smaller than £/2 for n the first summand (whose numerator is a fixed quantity) also becomes smaller than £/2 for n n' because yn Choosing n' k, we see that, for n n', +

+ oo

0

�

+ oo.

�

�

�

This proves the claim for the considered special case. To finish the proof, we observe that the case of an infinite limit eas­ ily reduces to the case of a finite limit. Take, for example, the case where (x - x _ 1 )/(yn - yn- l) as n that is, the sequence properly diverges ton nThen for all sufficiently large n, xn - xn-l Yn - Yn� l ; hence xn with yn and the sequence (xn) :=l must be strictly increasing for all sufficiently large n. Therefore we can use what we have proved in the begin­ ning of this proof and apply that result to the reciprocal expression, name­ ly Yn /xn : + oo

+ oo ,

+ oo .

+ oo

+ oo

and we conclude that limn xn/yn Let an a as n bn

>

+ oo

PROBLEM 40 .

+

+ oo .

Show that bn a as n +

+

oo ,

where

CHAPTER 3

21 4 Sol ution .

sult in Problem

We set xn

n and apply the re-

39 .

PROBLEM 41. Letoo (pn} �=l be a sequence of positive numbers tending to the limit p as n + and p 0. Show that the sequence of geometric means as n + oo. Let >

Sol uti on .

Then

log p1 log p2 • • • log pn lim log p lim lim log T n n+oo n n+ oo n+oo n (by Problem 40} . Thus T = 0 or lim T lim log p. n +oo Pn n + oo n The results in Problems 40 and 41 have many applications; for example, .!_ 1 ••• 1 (i} n + 0 as n + oo because 1/n + 0 as n + oo ; (ii} l!l ·i·� • • • n � 1 + 1 as n + oo because n/(n-1} + 1 as n + 1 • • • n rn--- + 1 as n + oo because n + 1 as n + oo • (iii} ---' n (iv} (1 *) n + e as n + oo implies nn 1 - + e ---lil! or n lil! + -1 as n + oo. (See Remark to Problem 16.} ---n e +

+

+

...E.

Remark .

+ + + ----�2�----�n

oo ;

12 + 3 13 +

+

+

li1

+

+

SEQUENCES AND S E R I E S

215

PROBLEM 42. Let (tn) n=l be a sequence of positiveoo numbers and suppose that tn+l /tn converges to a positive number t as n + . Show that 41,

Sol ution .

Let p1

n lim n� oo /plp2 Pn nlim+oo Pn or tn l1m. -tn+l- = t. lim t n+ oo n-1 n + oo tn

tn/tn-l ' . . . By Problem

0

. .

PROBLEM 43. Let a Show that a- 1 + 2 a-l + 3a-l + + na-1 1 l lim a n + oo We use the result in Problem 39 and let xn = 1 a- 1 + 2 a-1 + na-1 + (n + 1) a- 1 and Then (n + al) a-1a (n + 1) - n But, as n + oo , (n + 1) aa-1- na ((1 + .!..!.n)) a-1a - l1a (1 + .!.n)1 a - la ( l + .!.n) 1-a + a 1+n n (n + l) n because (1 + .!.n) a - l a d a l a, lim 1. lim dx x=l n+oo n + oo n1 Note that > 0.

0 0 0

Sol ution .

+

o o o

-

Remark .

X

216

CHAPTE R ·3

Let (un) n=100 be a sequence with positive terms and l:Ln-*.m00 -uun+1-n = h, where h Show that n-*limoo Put un_ = P ' . . . Pz ' . . . , _un-1 n-1 we get = un1 PROBLEM 44 .

> 0.

vh.

Sol ution .

,

ll'

thus

= -unu-;n- 1

Now, for the sequence

SEQUENCES AN D S E R I E S

21 7

Pn-1 ' Pn-1 ' . . . , Pn-1 ' we have limn+ co Pn-l h; thus, by Problem 41, ( p1lp22p33 pn-1 ) Y,n (n-1) h. n-1 lim n+ co Consequently (pl1p22p33 pn-1 n-1 ) 1/n 2 vh = lim n "n J /n lim n + co n + co ( '"•"z u •••

•••

PROBLEM 45. Let a. Show that Sol ution .

Put n bn . Then x ;a-

=

nlim+co nbn Thus, by Problem 41, n lb1 (2b2 ) (nbn) lim nln!b 1b 2 lim co n+ n+ co and since -n rnrn + e as n + co (see Part (iv) of Remark to Problem 41), we get x ra.

CHAPTER 3

218

n x ra e lim b n b lb l 2 n n+ oo or x n la1a2 an a e . lim n n+oo X

·

PROBLEM

46 .

Let p 1. Show that >

"' ·

tends to 1/(p - 1) as n + +-p 2) -p is the smallest value of-py x-p on Noting that (n the interval [n+l , n+2 ], (n + 2) the smallest value of y x on [n+2,n+3] , and so forth, we see that 1 1 1 -p (n + 2) p + (n + 3) p + < i:l x dx (p - l) (n + 1) p-1 or 1 p+ 1 p+ < 1 P+ 1 p-1 " (n + l) (n + 2) (n + 1) (p - 1) (n + l) Similarly, since (n-p+ 1) -p is the largest value-pof y x-p on the interval [n+l,n+2], (n + 2) the largest value of y x on [n+2,n+3], and so forth, we obtain 1 + 1 + ••• > 1 p-l " (p - l) (n + l) Thus 1 p-l < (p - l) [ --__l___p + -__l__-p + ••• J (n + l) (n + l) (n + 2 ) 1 < -(n--+ l) P-1-=- + (n +- 1)1 P Multiplying by np-l , we get (n n+ -1} p-1 < (p 1) -n!. [ (n n+ -1} p + (n n+ -2} p + ••• J Sol ution .

=

=

. . .

. . .

=

----

p

_

=

21 9

SEQUENCES AND S E R I E S

.E..___:___!_

< (n n+ -1) p-1 n + 1 " n n+ -1) p-1 + (Setting lim Sn n n -1) p + (-nn n [ (+ 2) p + J and n+oo we see that 1 (p - l)S 1 and so S 1/ (p - 1) . Note that n n-1) p + (n n+ 2) p + · ] n n [ (represents the sum of the areas of approximating rectangles under the curve -p y x over the interval (1,"') ; indeed, < � 1+1/n dx 1+2/n l+2/n n-1 (n-+n-2) p i1 + 1/n ( 1 +dx�)n P < 11+1/n xdx and so { "' dx lim = n n+oo J 1 xp· The same method of proof can be used to evaluate many other limits. For example, dividing the interval [0,1] into n subintervals of equal length and considering-l the sum of areas of approximating rectangles under the curve y = (1 + x) leads to the result that l n +1 2 + + 2n + )[0 1 + x log 2 as n + n +1 1 --Again, dividing the interval [0,1] into n subintervals of equal length and considering the sum of areas of approximating rectangles under the curve y Y, = x shows that + __!__ } { l dx 2 as n + "' · 5

=

_!_

�

s,

•••

n

+

�

Remarks . 5

=

_!_

••

+

=

p' X

p'

=

5

�

__!__

--- +

•••

Ill

+

Jo

IX

=

=

"' ·

CHAPTE R 3

2 20

This latter limit can, of course, also be derived with ease from the result in Problem To compute the limit limn +co Sn ' where 1 n Y4n21 - 1 r4n21 2 y'4n - n2 we note first that 1 n = 1 I f4 -1 (.!.) 2 f4 -1 (�) 2 f4 - (-'n�-) 2 I; n n hence l dx arc Sln 1 1 J: = lim n+ co n V'4""7 In a similar way we can show that limn +co An 1/e, where An Indeed, taking logarithms, we get log �) nlim+ co log An nlim+ co * (log * log � - J[0n 1 log x dx = (x log x - x) 1 01 -1 and the desired result follows. 5.

s

s

+

n

+ ••• +

/

+ ••• +

+

s

.

z X

O

6" 7f

+ ••• +

+

PROBLEM 47 . Verify the following claims : (a) If 0 y 1, then nlim+ co n-1j=lL njy- j j 0. (b) If 0 y 1 , then n+ll·mco nyn i=l� i/1 = 1 -1 y· (c) If x 1, then nlim+ co ....!xn !.... i=lLn <

<

>

<

<

_

£.....

__

--

--

221

SEQUENCES AN D S E R I ES

Let nI j yj I. + __y_ + _2l__ + + (n - l) -yn-1 + !!X_n 1 n j=l n + 1 - n n - 1 n

Sol uti on .

and

2

2

s

3

3

-

•••

2

2

Tn y + + n + + n yn It is clear that n Tn Moreover (n 1) n2l+l as n -+ "' ny and yn+l 1; it follows that n yn tends to as n -+ "' because n yn and (n + 1) y tend to the same limit when n We see, therefore, by Problem that Tn tends to as n -+ oo; thus Sn tends to as n -+ This completes the proof of Part (a) . In order to verify Part (b) , we observe that . n n yn- 1. - n-1 n--n-1 n-1 . J (E .11) n jy j J y y ·. + n=-· J J i= l j =O j =O J =O But L j=O yj 1/(1 - y) for y 1. Passing to the limit as n -+ "' in (E.ll) and using the result in Part (a) , we obtain the desired result. The claim in Part (c) follows at once from the result in Part (b); we only need to set y = 1/x. 2 2 2 y

32y3

0 � s

+

<

0

2

. . .

�

2

2

<

2

-+

1 L..

i

_

1 L..

_

-

<X>

0

PROBLEM lim n+oo

48 .

.!... i.�

"' ·

0

0

40 ,

2

0

<

1 L..

"' ·

1 . L..

<

Show that

2 4 6

Sol ution .

Chapter

2.

The claim follows at once from the result in Problem of 37

PROBLEM For each positive integer n, let f(n) (n!) l/n . Show that is monotonely decreasing. f(n+l)/f(n), for n = 49 .

1 , 2,

.

.

.

CHAPTER 3

222

We show equivalently that Fn f(nf(n)+ 1)/f(nf(n)- 1) 1 for n 2, 3, . . . Consider Sol ution .

<

=

Since the geometric mean is less than the arithmetic mean (see Problem 2 of Chapter 2), [(n l) !] l/(n-1) (n - 1) + (n - 2) n+ + 3 + 2 + 1 - n (n 2- 1) n Therefore (Fn) n (n+ 1) /2 .!.2 (l + .!.n) n/2 2 e!-z 1. Thus n 1 and the desired result follows. It is not difficult to see that 1 f(nf(n)+ 1) -n n+-1 '. consult Problem 34. _

•••

<

< 2"

_

< .!_

<

<

<

F

Remark . <

<

PROBLEM Let 1 an bn e- 12 n and bn - (n.en) n-(n+!-z) . Show that each interval (an ,bn) ' n 1,2,3, . . . , contains the interval (an+l ' bn+l) as a subinterval. Since, for -1 x 1, log l + x 2x ( 1 + 1 x2 + 1 x + • • • ) , setting x (2n + 1) -1 , we get SO.

=

_

=

Sol ution .

� =

=

3

<

S

<

4

1

SEQUENCES AND S E R I E S

l 4 + J. log n -2n 2+-1 [1 + -3(2n--1+-� 1) 2 + 5(2n +_----;1) �

Thus

=

22 3

• • •

_ _ _

n + +) log 1 + ft) 1 + 3(2n 1+ 1) 2 + 5 (2n + 1) 4 + which is larger than 1, but less than 1 + 3 [(2n 1+ 1) 2 + (2n +1 1) 4 + J 1 + 12 n(n1 + 1) and so 1 + 12 n(n1 + 1) e (1 + lt+� n e Now bn = -bn+l- e and so 1n 1 b 12 n e 12 n(n + 1) 1 -bn+l- e 1 e 12(n+l) Thus bn bn+l and 1n 1 + 1) 12(n 12 bn e bn+l e (

=

(

_!_

--�-�

• • •

•••

<

<

<

<

>

<

51. Show that the sequence an - (l + l) n+p n = 1,2,3, . . . is monotonely decreasing if and only if p 1/2. The fact that an is decreasing for p 1/2 is clear from the expansion PROBLEM _

n

'

,

�

Sol ution .

�

CHAPTE R 3

224

log an = 2(n2n ++ 1 [ 1 + 3(2n 1+ 1) 2 + 5(2n 1+ 1) 4 + • • • J ( 1 + n -+ Y,) 1 1 + 3(2n 1+ 1) 2 + 5(2n 1+ 1) 4 + • • • f (see the Solution of Problem 50). This leads to 1 + O(n-3) , log an n + + --2 12 n where O(xn) ' with xn 0, denotes a quantity that divided by xn remains bounded; thus log an+l - log an hence an increases for n larger than a certain subscript n0 if p 1/2.l nIf p 0 this is true already for n 1 as can be seen by expanding (1 + n} with the help of the binomial formula . p)

=

p

�

= !L..:_p_ � 1

>

<

�

�

PROBLEM 52. Show that the sequence n 1,2,3, , is monotonely decreasing if and only if x 1/2. Since an the first factor is seen to be decreasing by Problem 51; the square of the second factor is 1 + n + 1 + -n(n--;---"x-'-+2--:;-1),' The condition 1/2 is therefore sufficient. But log an 2n [-2n -1+-1 + 3(2n 1+ 1) 3 + 5(2n 1+ 1) 5 + ... ] + 2 1_ 2n x+_x + .3!. (_ 2n x+_)x 3 + 1 (2n + x) 5 + • • • f =

...

�

Sol ution .

�

X �

=

5

X

SEQUEN CES AND S E R I ES

225

2n2n+ 1 2n2x+ x + 12 1n2 + O(n 3) . Since log an - log an+l 4x4x-2 2 + O(n-3), the condition is also seen to be necessary. Note that log (1 + �n) log 11 +- 2il+X 2n + x --- +

---

-

--

=

------

Remark .

X X

PROBLEM 53. Show that for any positive integer n, 2n e+ 2 e - (1 + -nl) n 2n e+ 1 The first inequality means

___ __ _

<

< ----.

Sol ution .

and this is a consequence of the following inequality f(x) x + x log(l + �) - (1 + x) log (1 + x) (Note that f' (x) x+_2 - log 11 ++ xx x+ 2 l + � + 1 f(O) 2 The second inequality is equivalent to > 0,

=

__ _

X

2

>

__ _ __

X

-

�

and is a consequence of the result in Problem 52. PROBLEM 54. The number e n-lim+ ro (1 + -nl) n =

0,

0.)

CHAPTER 3

22 6

is contained in the interval (1 + *) n < e < (1 + lnt+l (see Problem 13 of this Chapter and Problem 2 in Chapter 1) . In which quarter of the interval is it contained? The number e is situated in the second quarter of the inter­ val because n 1,2,3, . . . The first inequality follows from Problem 53 because 1 + -4n1 < (1 + -n1) (l + -2n1 ) -1 ' the second inequality is contained in Problem 52. Sol ution .

=

;

PROBLEM 55. Show that the sequence an (1 + .!n) n+l ' n = 1,2,3, . . . is monotonely decreasing if and only if < x 2. We see that an is decreasing for < x 2 because (n + 1) log 11 +- X (2n + 2) k=l� 1 (2n + x) 2k-l 2n + x x2k2k-- 11 1 2k-2 + (2 - x) L _2kx2k-l- 1 1 2k-l" L _ k=l (2n + x) k=l (2n + x) Furthermore log an + 3 (2n +1 x) 2 + x(22n -+ x) + (n -3) , log an log an+l (2n +2x(2 x)(2n- x)+ + 2) + (n-3) , that is, log an - log an+l < for n sufficiently large, if x < or x 2. For x = we have an = 1, n = 1,2,3, . . =

,

$

0

Sol ution .

0

2Il+X X

L

-

3 X

X

=

X

0

0

�

X

_ _

_ _

X

$

0

0

0

.

.

>

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

<

<

X

(1

1.

+

3 1.

PROBLEM 57. Show that lim n-+oo

+

Sol ution .

3

II

+

• • •

'\h

+

We have that 2·4

This leads to conjecturing the relation +

• • •

(n -

1 ) /l+TI

=

3.

'

+

-1

CHAPTER 3

228

for all n

�

1 . Proceeding by induction we verify that

(n + 2) 2 = 1 + (n + l) (n + 3)

1 + (n + 1)

,Jcn + 3) 2 .

This given , we must have

+ •••

1/I + (n - 1) ll+Il.

To get an inequal ity in the other direct ion going, observe that for any > 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

<

<

+ oo

Remarks .

=

=

= 12

�

=

�

>

:

>

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 <

k

n

<

1

+

and

holds ; these inequalities are responsible for the fact that kn tends to rapidly as n oo . We also have that

0

2 34

0 <

K (kn) 2

- 2!:.

=

7f 12

CHAPTER 3

d;::x==;::: :::; == ---;:. rl - k� sin2 X kn2 ;=sin2 -X dx l - ==:; - f:=::;: 1 7[ 2 -1 --;::: ::. k2 sin2 X l f n 0

0

I

7f

2

I

<

From this we get that as n hence K(k) nlim (1 + k1 ) (1 + k2) (1 + kn) . this can be based a method of approximate calculation of the integral K(k) ; for sufficiently large n, + oo '•

•••

+ oo

On

Find the sum of the series 3 X - 1 3 3x + 1 3 32X - 1 3 33x + 3 3 Consider the following succession of identities 4 cos 3 x cos 3x + 3 cos x, 4 cos 3 3x cos 32x 3 cos 3x, 4 3 32X 33X + 3 32X, PROBLEM 5 9 .

3

COS

COS

-z

C OS

�

COS

•••

Sol ution .

=

=

COS

+

=

COS

COS

-1 , the If we multiply the first of these identities by 1, the second by (-3) third by 3-2 , . . . , the n-th by (-3) -n , and denote by Sn the sum of the first n terms of the proposed series, we see that 4Sn 3 X + (-3) n 3n+l X. COS

COS

S EQUENCES AND S E R I E S

Letting n

+

235

oo, we get that Sn

Remarks .

+

(3/4) cos x.

In the fol lowing , let f and g be functions such that

f (x) = af(bx) + cg (x) , where a, b , c are given nonzero constants . Then af (bx)

a 2 f (b 2 x) + acg (bx) ,

a 2 f (b 2 x) = a 3 f (b 3x) + a 2 cg (b 2 x) ,

an f (bnx) + an- 1 cg (bn- 1 x) ; thus

f (x) = an f (bn x) + c [ g (x) + ag (bx) + • • • + an-1 g (bn-1 x) ] .

I f the product an f (bnx) converges to a l imit L as n

+

oo, we get

f (x) = L + c l im [g (x) + ag (bx) + • • • + n + oo and the series g (x) + ag (bx) + a2 g (b 2 x) + • • • is convergent to f (x) - L c I f the product an f (bnx) does not converge as n or indeterminate.

+

oo , the series is divergent

In an ent irely similar manner we obtain the set of equalities f (x) = af (bx) + cg (x) , f (x) + a - 1 cg (b - 1 x) , a- 1 f (b - 1 x) + a- 2 cg (b - 2 x) ,

a 1 -n f (b 1-n x) + a-ncg (b -nx) ;

236

CHAPTER 3

thus

Therefore , if a-n f (b -nx) converges to M as n + oo , we obtain that g (x) + a- 1 g (b - 1 x) + a - 2 g (b - 2x)

M - af(bx) c

+

We now consider a number of applicat ions of these results ; among these applications wil l be the solution to Prob lem 5 9 . Recall the funct ional equa­ tion f (x) = af (bx) + cg (x) , where a, b , and c are given nonzero constants . Case 1 : Consider the identity s in x = 3 s in (x/3) s in x , g (x)

here f (x) L

=

sin 3 (x/3) , a = 3 , b = 1/3 , c = -4 . We have

l im [3n s in (3 -nx) ] = x, n + oo

M

lim [ 3 -n s in (3nx) ] = n + oo

0;

thus x - s in x 4 4

3 sm 3 X

or, more s imply , 1 S 1n3 3 2X • • • . 3 X -1 S1n 3 3X + -S 1n + 3 + 32 .

Case

2:

0

.

4

3 S ln X . 0

Cons ider the identity

cos x = -3 cos (x/3) + 4 cos 3 (x/3) ; here f (x) = cos x, g (x) = cos 3 (x/3) , a - 3 , b = 1/3, c = 4 . The product ( - 3) n cos (3 -nx) increases indefinitely in absolute value as n + oo , hence the series 2 3 X X cos 3 3X - 3 cos 3 2 + 3 cos -3 33

is divergent . But l imn + oo [ (-3) -n cos 3 (3nx) ] to 3x ,

M

0;

hence , on changing x

SEQUENCES AN D S E R I E S

237

1 COS 3 3 2 X - -1 COS 3 3 3X + COS 3 X - 31 COS 3 3X + -33 32

• • •

4

3 COS X .

This , however , is the solution to Problem 5 9 . Case 3 : Consider the identity cot x = 2 cot 2x + tan x ;

here f (x) cot x , g (x) = tan x , a does not converge as n � oo ; but =

=

2, c

2, b

1 . The product 2n cot (2n x)

l im [2 -n cot (2 -nx) ] = 1/x n � oo and so we get

+ tan x + .!.2 tan �2 + _!_2 tan � 2 22

and , in particular, tan

8 7T

+ 21 tan

T6 7T

+ 41 tan

32 7T

1

2

+

Case 4 : Consider the identity arc tan x If

0 <

b

<

=

arc tan (bx) + arc tan ( l - b)2x . 1 + bx

1 , we may conclude that arc tan x equals

( l - b) b 2x + + arc tan arc tan ( l - b)2x + arc tan ( l - b)bx l + b 5x 2 l + b 3x 2 l + bx and so, in particular,

arc tan 31 + arc tan g2 + arc tan 343 +

If b

>

• • •

2n- l 2--,,...-- + + arc tan ----,. 2n2 l + 1

• • •

1 , we find

+ arc tan (b - l)bx arc tan (b - l)x 2 1 + b 3x 2 1 + bx + arc tan

+

• • •

arc cot x .

• • •

CHAPTER 3

2 38

For n = 1 , 2 , 3 , . . . , let 4 12 8 12 2n 12 .

PROBLEM 60 .

sn

=

{ 12

·

·

}

· · ·

Find lim S . n -+ co n Sol ution .

log sn

Since

( � + t + i + • • • + 2ln ) log 2 '

we see that lim log sn n -+ co

=

(l :

�

- l ) log 2

and so Sn tends to 2 as n

-+

log 2

co ,

Show that IS is equal to the rat io S/S ' , where

PROBLEM 6 1 .

and 4 21 +

S' Sol ution .

l , 2,

5,

Denoting by Yn the n-th term of the sequence 1 3 , 34 , 89 , . . . ,

appearing in the first series , we see that any three consecutive terms satisfy the relat ion

denoting by Un the n-th term of the sequence l , 3 , 8 , 21 ,

55,

144 , . . . ,

appearing in the second series , we see that any three consecutive terms sat­ isfy the relation

SEQUEN CES AND S E R I ES

2 39

Hence , us ing the theory of difference equat ions , we get y

n

and c2 an

+

with a and 1/a being the roots of the equat ion = 0.

x 2 - 3x + 1

(E . 1 2)

The constants are determined by the conditions c2 c1 a + a

=

C' C 1' a + a2

2

c2

1'

c 1 a2

1'

2 = 3• C 1' a 2 + 2 '

+

a

=

2,

C' a

hence , noting that

we have c1

= -

c2 a

=

1

rsa

and

1 . rs

-

Denoting by Tn the n-th term of the first series , we see that Tn

(a2n- 1 + 1) 2

'

denoting by T� the n-th term of the second series , we have Tn'

±

n ::..s_ --'nc;;.=a_:_rs a 2n - 1

with the + s ign taken when n is odd and the - sign when n is even . We there­ fore have

240

CHAPIER 3

j

4 a l -:--=-2 + -"""'5=--:----=-2 + • • • S a ___..;;:.__.,..2 + --=3 1) 1) (a + l) (a (a

s

and

2 3a2a + -6ars j -2_1_ ­ -4-a - 1 a - 1 a - 1

S'

-

..· I

By taking for a that root of the equat ion (E . l2) which is larger than unity, we can develop all terms of S ' in convergent series and we find 6 a- 2 + a- 4 + a + a - 8 + . . .

S' a IS

2 (a - 3 + a- 7 + a - 1 1 + a - 1 5 + . . . )

+ 3 ( a-4 + a - 1 0 + a - 16 + a -22 + . . . ) 4 (a-5 + a -1 3 + a- 2 1 + a- 2 9 + . . . )

But

+ ... 1 a

-z -

1 a

4 -

2 + 3 4 + 3 4 - 5 a a a

2 3 4 + 7 + 10 - 13 a a a

1 ' (a + 1 ) 2 a2 ' ( a3 + 1) 2

and so forth . Hence , under the stated condition for a, 1 __.::.1_-=- + a2 + 5 1 2 a4 + • • • (a + 1) 2 (a3 + 1) 2 (a + 1)

_

2 1 2a + -63a- ---a 2 - 1 a4 - 1 a - 1

and so S/S '

Remarks .

-

/5.

A s imil ar representation of IS is given by

1 1 • -11 + -12 + 5 + 13 + • • 1 1 1 1 ··· T - 4 + rr - 29 +

in which the denominators satisfy the relation Dn+ 2 - 3Dn+l + Dn

0.

241

SEQUENCES AN D S E R I E S

Here the prob lem reduces to showing that if T

( 1 + q) f ( q)

=

(1

+

q)

{

and T'

( 1 - q)

( 1 - q) g (q)

1 + � + ••• 1 + -q + � 1 + q 3 1 + qs

-

{

1 1 - -q - � 1 - q3

-

+

�

5

1 - q

-

}

•••

}

•

where q = (3 - 15) /2 , that is q is the smaller of the two roots of x 2 - 3x IS. To do this , it suffices to veri+ 1 = 0 , then T/T' = ( 1 + q) / ( 1 - q) fy that f (q) g (q) . But =

=

f(q)

=

1 - q + q2 - q 3 +

q4

+

q - q4 + q 7 - q 10

+

+ q2 - q 7 + q 12 - q1 7

q 13 q22

+

+ q3 - q 10 + q 17 - q24 + q31 + q4 - q13

q22 - q 31 + q40

+

+ •••

and g ( q)

1 -

+

+

q

+

q 2 + q 3 + q4 + • • •

(q + q4 + q 7 + q 10 + q 13 (q2

+

q 7 + q 12

q 17

+

(q 3 + q 10 + q 17 + q24 +

(q 4 + q 13

+

q22

+

+

q22

+

+

q31

• • •) • • •)

+

+

• • •)

q31 + q40 + • • • )

The method of proof in Prob lem 61 can , of course , be modified to cover other suitable s ituations .

PROBLEM 62 . In the Solution of Prob lem 26 and in the Remarks to Prob­ lem 46 we noted that , for n = 1 , 2 , 3 , . . . ,

CHAPTER 3

242

"i1+T

1

+

11+2

1

+

tends to log 2 as n

n+3

1

+

oo.

1 + • • • + 2n

Use this result to show that

.l 1 - .l2 + .l3 - .l 4 + 5 - .l 6 + • • • = log 2 .

The desired result fol l ows immediately from the Identity of Catalan (see Problem 42 of Chapter 1) . Sol ution .

Remark . Note that the result in Prob lem 62 is contained as a special case in Prob lem 15 .

PROBLEM 63. Let a, b , m, p, and q be pos itive integers . Show that , if m = p 2 q, then the sequence of fractions b'

a

al a pa + mb 1 pa + mb ' 2 = 1 = b1 a + pb b2 a l + pb l '

converges to

1m

Sol ution .

as n

.

a . . , bn n

+ oo .

By definition ,

an+l pan + mbn bn+l = an + pbn

__g__

p - a n + p b n

and

Thus an+l bn+ l = p - 2p -

q

2p - _9_ 2p

-

____3_ .

a b+ p

Hence the sought for l imit wi l l be

=

pan - 1 + mbn-1 an-1 + pbn-1 '

SEQUENCES AN D S E R I E S

X

= p - 2p

243

q

---'---

_g_ •

2p - 2p But, if we develop into a continued fraction, we have = �=p p -p-2 --q p - 2p 2p - * -

/iii

/iii

-

---'q"---

+

J.___ ____:

q

Thus we see that

x

liii.

PROBLEM 64. Let bn k=Or:n nk -1 for n 1. Show that limn+oo bn 2. Note that bn � 2n bn-1 1 for n 2. =

�

Sol ution .

+

�

PROBLEM 65 . Let (pn) :=l be a sequence of positive real numbers. Show that if the series L:n=l Pln converges, then so does the series CX>

Sol ution .

Put � p 1

+ ••• +

pn and q0

0.

We get the estimate

CHAPTER 3

244

m

2 c'ln ! (..E._) n=l 'ln

m $

1 + m n2 I p l n=2 'ln'ln - 1 ( qn

s

co

'ln- 1 )

in terms of T = l: n=l 1 /pn . We observe that s

m- 1 2 = __!_ + I (n + l ) P1 n = l

'ln

- qn- 1 ) = -p1l + n=2mI 'ln-n2 1 m

2 m m 1 5 � s - + 2 I ..E... + I n=2 n= 2 'ln P1 n=2 'ln

'ln

I

By the Cauchy-Schwarz Inequal ity (see Problem 14 in Chapter 2 ) ,

{ I ..E... } { I m

n=l

2

�

and thus Sm

s

PROBLEM such that

m n2 n=2 2 Pn

s

�

}{ I -} m

1 n=2 Pn

5/p + 21S T + T . But this inequality impl ies that 1 m

66 .

Suppose that u0 , u 1 , u 2 , . . . is a sequence of real numbers

un

0, 1, 2,

for n

...

Show that if z::=o un converges then �

0 for all k .

j

1 , 2 , . . . I f z: =O uj converges ,

Clearly 0 s un+l s un for n < 1 . Then 1 so that z: =k+l uj

Sol ution .

we take k

�

j

uk+l s � =

I

j =k+l

u: s � + 1

J

Hence � = � + l and so � + l �+ l

=

L

j =k+l

By induction u .

J

u� .

0 for j

<

( j =k+l I uJ. )

s uk+ l '

0, implying u. = 0 for j

k + }.

J

>

k since

245

SEQUEN CES AN D S E R I E S

PROBLEM 6 7 . Show that lim n ��

£ for all n

�

� .

0.

n

Sol ution . <

Given a sequence (xn ) := l such that xn - xn _ 2 � 0 as n �

For £ > 0 , l et n 0 be sufficiently large so that l xn n 0 . Observe that for n � n 0 ,

n- 2 1

- X

Thus

and s o (xn - xn _ 1 ) /n tends to zero as n � � .

PROBLEM 6 8 . Let (xn ) := l b e a given sequence and let yn xn l + 2xn for n = 2 , 3 , . . . Suppose that the sequence (yn ) :=2 converges . Show that the sequence (xn ) :=l converges . =

Let y = 1 imn � � yn and set x yI 3 . We wish to show that x limn � � xn . For £ > 0 there is an n 0 such that for all n larger than n 0 , I Yn - Y. l < £/2 . Now Sol uti on .

l xn- 1

�

< �

x l - l xn - 1 - x l . l xn - xl < £/4 + � � xn- l - xl ,

2 l xn -

This may be rewritten as to give

- xl

(

I 2- i 4 i=l

By taking m large enough , 2 - (m+ l) l xn- 1 - X- �

+ 2xn - 3x l

<

z•

£

)

+ 2 - (m+ 1) I xn - l - x- �

<

-

which can be iterated

£ + 2 - (m+ 1) x I n- l - x- � . 2

CHAPTER 3

246

Thus for all sufficiently large k, lxk - xl <

£.

PROBLEM 69 . Let 0 < x1 < 1 and xn+l Show that nxn 1 as n Observe that (n + l)xn+l = nxn + xn - (n + l)x� +

1 ,2,3,

+ oo .

• • .

Sol ution .

(E .13) To see that nxn is increasing, we need to show that 1 - (n + l)xn 0. From the graph of y = x(l - x) we note that x2 1/4 and that xn a 1/2 implies xn+l a(l - a). So by induction 1 - __!__n2 _< 1. Furthermore, nxn < (n + l)xn 1 and so nxn is bounded above by 1. Thus nxn converges to a limit L with 0 < nxn < L 1. Now summing (E.l3) from 2 to n we obtain 1 (n + l)xn+l (E .14) If L 1 then { 1 - (n + l)xn (1 - L)/2 for all large n and thus (E.l4) would show that E�=l xn is convergent. However, nxn x1 and so En=l xn x1 E�= l 1/n. But r:= l 1/n is divergent. �

�

�

�

�

�

�

�

#

} �

�

oo

�

PROBLEM 70. We say that a sequence of points in an interval is dense when every subinterval contains at least one point of the sequence. With this agreement in mind, let P 1 , P2 , P3 , . . . be a sequence of distinct points which is dense in the interval (0,1). The points decompose the interval into n parts, and P decomposes one of these into two parts. Let an and bn be the lengths of thesen two intervals. Show that

247

SEQUENCES AND SERI ES

3'

1

Sol ution .

The ident ity to be proved can be written in the form an3

(an + bn ) 3

o.

Consider the partial sum

Each of the positive terms cancels one of the preceding negative terms . Be­ cause the sequence P 1 , P 2 , P 3 , . . . is dense , every negative term eventually gets cancell ed out . I f we choose n sufficiently large , the only terms that are not cancel led out in the above sum are the cubes of the lengths of dis­ j oint interval s , each of which is of length £ at most . The sum of the terms that are not cancel led out is therefore maj orized by £ . It fol lows that the above sum is at most £ in absolute value for n large . Now let n + oo ,

PROBLEM 7 1 . Let N denote the set o f all positive integers . What i s the set of limit points of the set { /a - lb : a , b E N } ? Sol ution . The set contains a l l integral multiples o f each o f its mem­ bers . Since it has arbitrarily small members ;n-;-r In, noting that -

( ;n-;-r

-

In) ( ;n-;-r

+

In) =

1

and so

every real number is a limit point of it . Remark . Generalization of the result in Problem 71 : I f a , a , . . . 1 2 and b 1 , b 2 , . . . are unbounded monotone increasing sequences of real numbers and limn (an+ l - an ) = 0 , then {an - bn : n ,m E N} is dense in the set of real numbers .

+oo

CHAPTER 3

248

Indeed , let r be any real number and E > 0 . Then there exists k so that - an < E for all n � k . Pick m large enough so that bm � ak - r and let be determined (uniquely) by the condition an $ bm + r < an+l " This gives I (an - bm) - r l < c . In Problem 7 1 , an bn In satisfies the assumptions of the foregoing general proposition .

PROBLEM 72 .

For k

�

0 , let S be the set of al l numbers of the form

with arbitrary finite sequence of signs . I f k � 2 , then al l s in S are real . Show : (i) if k 2 , then S is dense in the interval (0 , 2) ; (ii) if k > 2 , then S i s dense in no interval . =

The set T of express ions of the form

Sol ution .

t

= ±

Vk

±

• • •

± 1i<

is precisely the set of zeros of the polynomial Pn (t) for some n where p

1 2,3 J

J •

•

•

(t)

In the case k = 2 we may put t = 2 cos x, whence Pn (t) = 2 cos (2 nx) and the zeros of Pn are 2 cos { (2k + 1) 2 -nrr } , k = 0 , 1 , 2 , . . . , so that T is a dense subset of ( - 2 , 2) and hence S a dense subset of ( 0 , 2) . Let k > 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)

can be treated similarly; it converges to

12.

PROBLEM 74 . Let u 1 , u2 , u3 , . . . , v 1 , v 2 , v 3 , . . . be sequences of pos­ itive terms , the first sequence being bounded above , and in the second sequence limn + oo vn = 1. Show that the limit superiors satisfy lim u v n + oo n n

lim u n + oo n

I t i s clear that as u 1 , u2 , u 3 , . . . are pos itive and bounded above and the sequence v 1 , v 2 , v 3 , . . . has positive terms only and is con­ vergent (hence bounded) , the sequence of positive terms u 1v 1 , u2 v 2 , u3v 3 , is bounded. I f possible , let Sol ution .

lim u v = L ' n + oo n n

>

L = l im un n + oo

CHAPTER 3

250

Take 2£ = L' - L, and let the least upper bound of u , u , u , . . . be Since limn +oo vn = 1, there is a positive integer k1 1, such2 that3 when n k 1 . Thus ! unvn - un l = lun l ! vn - 1 1 < 2� < � , when n k1 . Therefore unvn < un + �£, when n k1 . But since nlim+oo un L, there is a positive integer k such that un L �£, when n k2 . Therefore unvn < L + £, when n k, where2 k is the larger of the integers k 1 and k2 . But since nlim+oo unvn L', we know that unvn L' - £, for an infinite number of values of n. Therefore L' cannot be greater than L. In a similar manner it can be shown that L' is not less than Hence L' L. K.

�

�

K

�

<

+

�

�

>

L.

PROBLEM Given a sequence (an) :=l of nonnegative real numbers such anam for all pairs of positive integers, m and n, show that that an+m oo nra( an) n=l converges. Since an+ 1 ana1 , it follows by induction that anoo an1 for all n, and hence that a�/n a1 for all n. The sequence ( an1/n) n=l is thus bounded. Taking the limit superior, let L = lim n ra-.n We have L a 1 . We shall show that an1/n L for all n. Once this is es tablished, it will follow that the limit inferior lim n �, lim n ra-n L n+oo 75 .

$

Sol ution .

0 $

0 $

$

$

$

�

$

�

251

SEQUENCES AND S E R I ES

which implies the existence of limn an1/n . Let m1 m2 m3 be such that <

<

-+ oo

< •••

Let n be any positive integer, and let the rational number mk/n be given by qk + dk where qk is an integer and � dk 1. We observe that qk/mk = (1/n) - dk/mk ' so that as k -+ qk/� 1/n, since dk/mk 1/mk -+ Now by the given relation 0

oo ,

<

-+

<

0.

Hence and letting k we ob1/nta1.. n L � an1/n , follows that limn -+oo an exists. The sequence (an1/n) oon=l need not be monotonic. As an example let an e-n if n is even, e-n+l if n is odd. oo is an oscillating sequence converging to 1/e. For this sequence (an1/n) n=l -+ oo

Remark .

PROBLEM Let u1 + u2 + u3 + be a given series and put sn u1 + + un and +s crn n n By Problem if sn -+ s as n -+ oo, then crn s - as n -+ oo . Show the following theorem due to Hardy and Landau: If crn -+ cr as n + oo and either n(sn - sn+ l ) or n(sn+l - sn) where is some positive constant, then sn as n + oo, •••

76 .

•••

-+

40 ,

< K

< K,

K

-+ cr

CHAPTER 3

252 Sol ution .

o = 0,

K

For if we put u l = u1 etc., and

=

We may without loss of generality take 1, and

o

----, K

we have sn s n Thus

o

and .!.n cs 1 + sn) on if n(sn - sn+ 1 ) In the other case we put o - u1 etc. Suppose then we are given o + and n(s - s ) 1. However, it is clear that limn +oo sn is not equaln tp oo (or n- oo ) ,n+lbecause if it were we would have limn +oo on + oo (or - oo ). (a) If possible, let nlim+ oo sn ' where is oo, or a finite positive number. Then, if A is any positive num­ ber less than for infinitely many of the values of n, say M , M , . . . But to the arbitrary positive number £, there corresponds a positive1 2integer such that when n Let M be the first of the sequence M1 , M2 , . . . which is greater than and such that an even positive integer. Let 2p be the largest positive integer not greater than Then 2p 4p. Also 0 --K --'

K-.

+ •••

-

<

K

.

--K --'

<

0

-+:

=

=

A

+

A

A,

v,

:?:

MA

:s; MA <

v.

v,

:?:

MA.

253

SEQUENCES AND SERI ES

£

sM+p sM - M ' and each of these is greater than A - p/M. But aM+p MaM + (sM+l +MsM+2+ p + + sM+p) M + p _(A - M M M+ -p aM + _.J?..._ _a +a_1 (A - a) - £ , where a � �· Now A - a A/2 and a = A/A/4 4 + 1' since p/M a A/4. Therefore a (A - a) 2(AA+2 4) ' and aM+p 2(A + 4) - £ , 'i"+'T If A2----+ -,.4")' £ = ...,.47(A.,we have aM+p £ , which is impossible. Thus nlim-;. oo sn (b) We shall now show that nlim+oo sn For, if possible, let nlim+oo sn > ••• >

=

•••

> _

E)

>

=

,;

�

'i"+'T

=

>

>

>

,;

--

0.

� 0.

= >. < 0 .

>

CHAPTER 3

2 54

Take B any positive number less than Then sn < -B, for an infinite number of values of n, say Hl ' H2 , and an < £ , when n Let H be the first of the sequence H1 , H2, . . . such that 1 +H �B and such that between 1 +H !zB and 1 +H �B there shall be at least one positive integer. Let the integer next above H/(1 + �B) be q, and write H/q 1 + b. Then since 1 +H !zB < q < 1 +H �B ' we have �B < b < �B. Also as before qaq + sq+l + + sH H But -

I

I

A.

� v.

> v

=

• • •

1 + 1 + • • • + q +1 1 + l)q . sH + (� Therefore each of these is less than -B + (H - q)/q < -B + b. Also aH < H q + H ( - B + b) < - b (B - b) + £ . But B - b l.2 B and b B Therefore li="2

< ••• <

�

>

a

�

r-+lb

r-+lb

> m·

---

255

SEQUENCES AND S E R I E S

b - b) 2(BB2 4) and - 2(B 4) £ . r-+lb(B If we take 2 £ 4(B B 4) . we have £ , which is impossible. Thus nlim+oo sn 0. But we have seen that nlim+oo sn 0. It follows that nlim+oo sn 0. From Problem we easily see the validity of the following statement: Let the sequence of+arithmetic means cr of the series u u u3 converge to a as n oo , If a positive ninteger n0 exists 1 such2that l un nK ' when where K is a positive number independent of n, then the series u1 u2 u3 converges and its sum is <

0H <

+

+

+

+

=

crH <

-

;::

--

s;

=

Remark .

76

+

+

+

• • •

I

< ­

+

+

• • •

+

cr .

PROBLEM Show that if a, b 0, then {n ,;a ; n lb}n lim n +oo Let f (x) ax . Then f' (x) ax log a and f' (O) log a. Thus f(i-) - f(O) lim n cn ra - 1) log a. lim n +oo n1 n +oo Putting 77.

=

Sol uti on .

liib.

>

CHAPTER 3

256

xn n [nra ; n ib - 1 ] -}[n(n va - 1) n(nlb - ) ] we see that xn + 21 c1og a log b) log as n But n ra n v'b 1 2 n and =

=

+

lib

+

+ oo .

.....!!. X

+

+

In the foregoing reasoning we made use of the fact that l/a lim (1 e; a) a+O this fact is the basis for most applications of the number e. Remark .

+

PROBLEM

78 .

Let a1 , a2 , . . . , � be given constants and put

Evaluate limn +oo Sn . In the identi�y y - yk-1 yk-2k k k-1 we put y k v'(n - a1) - �) and n. Then n (k k-1 n(k k-2 nk Sol ution .

- z

z

z + ••• + z

+

• • • (n

�----------------,

s

v•• •) r--'

+

v••·) r--'

z

+ ••• +

=

1

257

SEQUENCES AND SERIES

(al + + ak) + al a2 + n + ak- Iak + + al an2k-1 ak k1 a l l + ( l + a:) } + + 1 V( n ) •••

•••

•••

{ and so nlim+oo n

•••

S

Let (xn) n=l be a sequence of real numbers such that xn for all n and limn -r oo xn = Without using differentiation, show that (1 + xn) l/m - 1 = 1 lim n +oo Xn where m is a positive integer. We set PROBLEM

79 .

0.

Iii'

Sol ution .

1 - I t ! m� 1 + I t ! for I t I 1 and so l/m 1 ' lim (1 + x ) n oo n -r implying that limn +oo yn Consequently (1 + xn) 1/m - 1 lim Yn lim n -roo (1 + yn) m - 1 n +oo Xn and the latter limit is seen to be equal to 1/m. <

<

<

= 0.

PROBLEM 8 0 .

Let b a >

> 0,

d

> 0,

and

1 +

•••

1 0

CHAPTER 3

258

� b(ba(a ++ d)d) (a(b ++ 2d)2d) ((ab + nd)nd) Show that limn +"' qn Let y x and put u y - x. By Problem 62 of Chapter 2, if y and o are positive integers larger than then (I.) X Y + y�X and Thus =

+

·

0.

>

Sol ution .

>

>

=

0

1,

1

or

1

yX x +x-yu) /y . We take for y and o only those values for which u/o d and yy d. Then (x X+ d) o (x +Xu/o Xy (X +X y) y (X : d) y We therefore have the inequality (X x+ d) o y (X x+ d) /y (E wedging in the proper positive fraction x/y and holding for all integers y and o which exceed the largest of the three numbers (y - x)/d, d/(y - x) , and If in (E.lS) we set successively x a, a + d, a 2d, a + nd, y b, b + d, b + 2d, . . . , b nd, we obtain the system of inequalities ( a ) o b (-a a+-d) 1 /y (�) o a + d (�) 1/y a + 2d a + 2d ' (aa + ndmd) o ba ++ ndnd (aa + ndmd) 1/y with m n + (

< - <

<

>

-----

<

__

<

)0

�

<

<

< - <

•

. 15)

-- 1

1.

+

=

+

a+d

<

�

<

+

<

�

'

<

< ------ <

+

1.

SEQUENCES AND S E R I ES

259

Multiplication of the inequalities of the system gives (a : md)'s < � < (a +a-md) 1/y with m n Letting n + co, we see that both (a +a md) and (a +a-md) 1/y tend to zero; thus limn+ co � +

0

1.

= 0.

=

PROBLEM 81. For n 2,3, . . . , let Compute limn + co Pn . We can easily see that (n - 1) 2(n + 1) 21 -n n-1 n and so limn +co Pn 1/2. Sol ution .

+

=

PROBLEM 82. Find the sum of the series 1 1 7 + 1 - 22 [· - �] r �] . [ · - �] [·��] [· - 412 J + (E.l6} From the Solution of Problem 47 in Chapter 1 we know that a1 - 1 a2 - a3 - 1 + an 1 1 - 1 al al a2 al a2 a3 ala2 an al a2 an Thus, if limn+ co (a1 a2 an) exists and is equal to L, then the series a 1 - 1 + a2 - 1 + a 3 - 1 --1 - r1 · a l a l a 2 a l a 2 a3 We now set 1

__!_

• • •

---

Sol ution .

--- +

1

---

-

+

---

+

•••

---

---

+

. . .

. . .

CHAPTER 3

260

and an 1 n1 for n 2,3, . . . By Problem 81, (a1 a2 an) tends to 1/2 as n + thus L 1/2. Moreover, it is clear that the series (E.l6) tends to (1/L) - 1 1. - 2

oo ;

=

PROBLEM 83. Show that 1 10·981 10·98·9602 1 (each factor in the denominator is equal to the square of the preceding fac­ tor diminished by 2) . The series to be summed is of the form Y1o Yo1Y l YoY1lY2 where y0 10, y 1 98, y3 9602, . . . , and, in general, TO +

• • •

+

+

=

S

- 124

Sol ution .

- + -- + --- + =

=

But from the Solution of Problem 6 we may infer that Yo - i2y� - 4 1Y Y 1Y Y Y1 Y o ol ol2 when y0 10 and yn yn-l2 - 2 for n 1,2 ,3, . . . - + -- + --- + =

=

PROBLEM 84. For n 1,2,3, ... define Find the limit limn + xn . We put Ik=l log j1 1 nk2 fl 00

Sol ution .

+

261

SEQUENCES AND S E R I E S

and note that, for x 2 ·< log(l x) < x. 2' Thus k 4k2 < log � k2) k n n n n or nI k n k2 < n log � k ) n k · k=l n k=l 4n k=l n2 k=l zn But n k n(n 1) and nI k2 n(n 1) (2n 1) 2 6 k=l k=l (see Problem 15 of Chapter 1) . Thus n k = n 1 + -1 and n k2 :<; 4n 3 = -1 + 0 as n + k=l -zn 2n 2 k=l 4n n n This shows that limn Yn 1/2; therefore limn n >

X - X .

+

-z -

+

-z -

2:

<

2:

2:

2:

0,

2

+

+

<

2:

+

+

2:

+

---

-r eo

-r eo

X

"" ·

re.

PROBLEM 85. Let x be positive. The sequence n is monotonical (increasing for x < 1 and decreasing for x 1) and bounded. Moreover, X

>

and it is easy to see nthat limn +co xn = 1. Setting xn xl/2 , show that the two sequences and bn = 2� � - ;n ) converge to the same limit, the sequence (an) decreasing monotonically and the sequence (bn) increasing monotonically.

CHAPTER 3

262

If x and with it xn is larger than then 1,

Sol ution .

and thus 2(xn that is,

1) <

xn-l

-1,

Hence This shows that the sequence (an) decreases monotonically. The same result is found for 0 < x Similar arguments show that (bn) increases monoton­ ically. Moreover, we have (whether the xn are all greater than 1 or less than or equal to s 1.

1)

that is and

2n ( xn) 2n (xn Thus bn an . Since xnbn an and limn -+ oo xn we have lim bn . lim an n-+oo n-+c:o We may define, for positive x, 1 - 1

s

s

= 1,

Remarks .

1) .

SEQUENCES AND SER I ES

263

log x nlim-r oo 2n (x1/2n 1) n-+limoo 2n (1 �) . l/2 x - From this definition of log x it is easy to see that log 1 x 1 we have 1 < log < 1, 2 1 -{f) < log x < 2(v'X - 1) . Moreover, for positive x and y, log (xy) log x log y. Indeed, let z xy and n Y yl/2n ' z z l/2n . l/2 n n n then zn xnyn . Hence log z nlim-roo 2n (zn - 1) n-+limoo 2n (xnyn - 1) =

'I

- l

=

X

X

=

0;

for

(

X -

X

X

-

+

'

'

lim xn nlim-+ oo 2n (yn - 1) n-+limco 2n (xn - 1) n-+oo log y log x. In a similar way we can verify that, for positive x and y, log �y log x - log y and log 1 log x. Noting that log x 0 for x 1, this in turn can be used to show that the function f(x) log x increases monotonically with x. +

•

+

=

x = -

=

>

>

PROBLEM 8 6 . Starting with any real number x, consider the sequence n (E. 17) 1 i1 � X

+

+

X

CHAPTER 3

264

For x 0 this sequence increases monotonically, for x 0 it decreases monotonically; in either case it converges to 0 . We use this sequence (xn) :=O to define the two sequences an and bn limn -+ "' bn We limit ourselves to the case x 0 ; the considerations for x are similar. We have 2n+l x 2nx a . 1 + fl + x� n n To see that (bn) increases monotonically, we note that for all n 1,2, . . . �n 1 + xn2 . Hence, <

�

Sol ution .

>

� 0

------'n"'-- <

'

<

or

2xn (1 + Vl + x�) Vl + x�

------- >

or

�n

that is, bn+l > bn . Clearly V l + z- -'V-e:l _:.. +-.1x_..,- 2- • xy1:;,_+� �

,-----z = -

.. !

n

X

- r---'2 ....: Vc...; Y_ -

SEQUENCES AND SERI ES

265

and therefore + 1 - xy + �· � From (E.lS) we get and y Substitution yields X

X

=

Using induction, one sees that xn + Yn ' n 1 nyn or z

- X

Passage to the limit as n oo gives arc tan arc tan x + arc tan y and so, for any real numbers x and y satisfying xy 1, arc tan x + arc tan y arc tan - xyy It is also easy to see from the way we defined arc tan x that !arc tan xl 2 l x l 2 2, 1 + v1 + bn an . Thus both sequences converge. Since Vl + � 1 as n + oo, the two sequences converge to the same limit. +

z =

X +

=

1

-. -

<

<

X

<

X

+

<

CHAPTER 3

266

We may define, for all real x, arc tan x nlim+oo nxn '

Remarks .

2

where

n x, � We note from this definition: If x 0 , then 2x tan x < < arc � ( �) Vl x 2 For x < 0 , 2x 2x < arc tan x < �( �) Obviously, arc tan 0 0 . We next check the functional equation of the arc tangent. Let x and y be any numbers satisfying xy < We then set z = �xy and write out the numerical sequences (xn) ' (yn) ' and (zn) according to (E.l7) , starting with the numbers x, y, and z. X

1 +

>

----;==::;:

2x

-------

1 +

1 +

+

��----

-

1 +

1 +

=

1.

-

1

= 1 , 2,3,

PROBLEM Let n range over all positive integers (n . . ) and p over all prime numbers (p = . . . ). Show that, for x �n n; p l p This is the Let pk denote the k-th prime number; then l l x lx l l - p pk (p�) (p�) k 87.

2,3,5,7,11,

1 .: lT _....'--7 1 - X

Euler Product Formul a .

Sol ution .

1 --1X

+ + --- + X

---

+ ···

> 1,

.

267

SEQUENCES AND S E R I E S

Multiplying the finite number of series that correspond to primes smaller than a certain positive integer N, the partial product p CN) p :<>N -1 -1 -1 = n=ly· n1 n=lN n1 n=N+l (E .18) y· n k pk where the symbol is to indicate that the summation does not extend over all positive integers, but only those (1 is excluded) , which in their prime factorization contain solely prime numbers :<> N (the first N prime numbers do, of course, have this property) . From (E.l8) we get 0 p (N) n=N+l --k·n but n=l (1/nx) is convergent and so lim �l n=N+l I 2..x l = 0 ' N+oo n\ and we obtain the desired result. lT

X

X

-

X

I

- + X

x' 1

L1

<

L

I

X

00

PROBLEM 88. Let p range over all prime numbers (p = 2,3,5,7,11, . . . ) and let pk denote the k-th prime number. Using the Euler Product Formula (see Problem 87), show that the set of all prime numbers is infinite and that the series diverges.

The relation (E.l6) also holds for x 1 and hence N1 1 pk:<>N l - -pk n=l implying that P 1(N) for N + oo, Thus 11k=l1 1 - pk Sol ution .

lT

2..

1

_ _ _

>

I n'

CHAPTE R 3

268

diverges to + "" · If, however, the set of all primes was finite, this product would have to have a finite value. We can also see that (1 - .!.2) (r - .!.3) (1 - 5 (1 - ...!.pk).. = k=lTr (1 - ...!.pk).. 0 . But log (! - ...!.pk).. n-+limco p1k 1 and so 21 -31 + -51 + ... + -p1k + is divergent. The following is another proof of the fact that the infinite series .!.)

• • •

• • •

- +

Remarks .

diverges. We assume that the series converges and obtain a contradiction. If the series converges, there is integer n such that an

Let Q = p 1p2 pn , and consider the numbers 1 + kQ for k = 1,2, . . . None of these is divisible by any of the primes p1 , p2 , . . . , Pn · Therefore, all the pn+l ' Pn+2 ' . . . Thus for each prime factors of 1 + kQ occur among the primes r 1 we have r 1 < ( co ) t ' k=l 1 kQ - t=l m=n+l Pk since the sum on the right includes among its terms all the terms on the left. But the right-hand side of-t this inequality is dominated by the con­ 00 vergent geometric series Lt=l 2 . Therefore the series L""k--l 1/(1 + kQ) has se bounded partial sums and hence converges. But this is a contradiction becau this series diverges. • • •

<:

2

+

2

2

...!...

SEQUENCES AN D SE R I ES

269

PROBLEM 89. Let a 0 and b 0. Putting A(a,b) = � J:n/2 log(a sin 2 x + b cos 2 x)dx, (E .19) show that A(a,b) = 1 A (-a +2-b) 2 ,ab } . (E. 20) Moreover, verify that A(a,b) 2 log la 2+ lb (E. 21) Put x = t/2 in (E.l9) and breaking the interval of integra­ tion into two halves, we get A(a,b) = � Ian/2 log (a ; b - a ; b cos t)dt n1 ln/2n log(-a +2-b - a 2- b cos t) dt. Replacing t by n - t in the second integral and recombining, we find A(a,b) = .!.n l0n/2 log{ (-a +2-b) 2 - (-a -2-b) 2 cos 2 t } dt .!.2 L0n/2 log{ (a +2 b) 2 sin2 t + ab cos 2 t } dt. This proves the invariance property (E.20) . The arguments on the right-hand side of (E.20) are the squares of the arithmetic and geometric means of a and b. To evaluate the integral, that is, to establish (E.21) , let (E.22) >

2

>

{

Sol ution .

+ -

--

=

where n 0,1,2,3, . . . Define also n = .!.2 e ran + Ibn) ' a

8

n

(E. 23)

CHAPTER 3

2 70

for n 0,1,2,3, . . . The recurrence relations (E.22) imply

It follows from s02/a02 that 0 1 1 and hence, as n b � �} 2 l, a2 1. an an 1 + n Successive applications of (E.20) show that � o

o1

� =

<

+

0

(E. 24) + oo ,

(E. 25)

_E. +

(E.26) For strictly positive a and b, the integrand of (E.l9) is jointly continuous in a, b, x and A(a,b) is continuous in a and b. Since A(l,l) 0 it follows from (E.26) and (E.25) that A(a, b) 2 lim 2 -n log an . Finally, because 2 -n log an is independent of n by (E.24) . A(a,b) 2 log a0 2 log la +2 lb establishing (E.21) . =

=

PROBLEM 90. Let S(k) 1 + -21 + -13 + + 1 and define k to be the least integer k such that S(k) n. For example, k 1 1, k 2 n4, k3 11, k4 31, k5 83, k6 227, k 7 616, ... Find 1n 1m. -kn+l k-n . It is clear that, as n 0 S(kn) - n kln 0; but S(kn) - log kn c (Euler's Constant, see Problem 14) . Hence we conclude k

=

=

=

=

=

=

=

+ oo

+ oo ,

Sol ution .

�

<

+

+

�

=

271

SEQUENCES AND S E R I E S

that n - log kn + c. This implies that n 1 - log kn+l + c as n + and subtracting these two limits gives 1 - log -kn+l k-n + 0 as n + Hence 1n +oo1m. -kn+l k-n = e. "' •

+

"' ·

PROBLEM 91. Let a0 be arbitrary but fixed and define an+l = sin an for n = 0,1,2,3, . . . Show that the sequence (nan2) is convergent and find lim n an2 n+oo Suppose first that 0 a0 �. Then clearly an 0 for all n, and if we let f(x) = x - sin x, then f' (x) = 1 - cos x 0 for 0 x �; since f(O) = 0, it follows that <

Sol ution .

<

>

<

>

<

Hence the sequence (an) is strictly monotonically decreasing and bounded below by 0, so if L = limn+oo an ' then L = sin L, implying that L o. The same can be said for any a0 which is congruent (mod 2�) to some a (0 .�) : 0 and an + 0. Similarly, if a0 is congruent (mod 2�) to some a1 a2 0 and an + 0. If a0 is an integral multiple a (-�,0) , then a1 a2 of �. then a1 = a2 = = 0. Consider then for any a0 k� the limit 2 - S . n2 ' an l lim lim lim = •c 2 2 n+ oo (.£. - :�) n+oo t :� ) x+oo sin which is equal to 1/3 by four applications of L'Hospital's Rule. E

>

>

• • •

E

>

<

<

• • •

.

<

!

X

X

X

X

CHAPTER 3

272

If Yn = an-2 an-1-2 ' then (yn) has 1/3 as its limit, hence the sequence Y-n yl Y2 n Yn 1 as n -+ by Problem Thus 1 - limoo Yn lim n1 n (ak-2 lim � n-+ n-+ao k=l n-+ao nan · It follows that 2 n-+limao n an 3 for any ao k1T, for any ao -

+

+

.... 3

+

. . .

"'

•

40 .

3 -

L

-

"f

kTI .

0

PROBLEM 92. Let (an) and (bn) be sequences of positive numbers such that ann a and bnn -+ b as n -+ "' with a,b "' · Let p and q be nonnegative numbers such that p q = 1. Show that 0 <

-+

+

<

Let ao·' then xnn -+ x if and only if n(xn - 1) -+ log x. Both conditions imply xn -+ 1, so we can assume that xn and hence x� x if and only if n log xn log x. If we define yn 1 if xn 1 and yn = xlogn - xn1 = log xnn - log1 1 if n 1, then n log xn = n(xn l)yn and since yn -+ 1, the equivalence is shown. n follows immediately; from a� -+ a The limiting behavior of (pa ) qb n n and bnn -+ b, we conclude that n(an - 1) log a and n(bn - 1) -+ log b. n - 1) = pn(an - 1) Letting x = pa qb , we have n(x 1) = n(pa qb n n n n n qn(bn - 1) -+ p log a q log b log apbq and we have the desired result. 0 < X <

Sol ution .

=

-+

X

--

> 0

X

-

F

+

-+

+

+

+

=

+

=

-+

SEQUENCES AND SERI ES

273

PROBLEM 93. Show that, for any fixed m 2, the series 1 + 1 + ••• + 1 - + 1 + 1 + ••• + + ---2m 1+ 1 -- + 2m 1+ 2 + ••• + is convergent for exactly one value of x and find the sum of the series for this x. Let 5n (x) = kn�l { (k - l)m1 + 1 + ••• + 1 If the series converges for x and y, then the sequence Sn (x) - Sn (y) = m k=ln k1 converges. But this is only possible if x = y, since 1/k diverges. Hence the series converges for at most one value of x. The sequence An = 1 + -12 + ••• + nm1 - log(nm) is known to converge to Euler's Constant (see Problem 14) . Now An + log(nm) - k=ln km1 - k=lnI m - 1 An + log m + j1 log n - k=lI ft_� r c + log m - c log m. Therefore, when x m - 1, the series converges to log m. �

�

2

iii X

iil+l

m+2

------

Sol ution .

�

�

L -

E

I

which

�

a

PROBLEM 94. Find the maximum value of and the minimum value of for

for all positive integers n. taking logarithms we obtain Sol ution .

On

B

CHAPTER 3

274

max infn { log (1 1+ 1/n) 13min supn { log(l 1+ 1/n) - n} . We now show that the function F(x) = log(l 1+ 1/x) - x is monotonically increasing for x > 0 by showing its derivative is positive: F' (x) x(x + l)[log(l1 + l/x)] 2 - 1 .::.sinh�.= u2:--2 u- - 1 > 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, <:

<

Sol ution .

SEQUENCES AN D S E R I ES

275

Letting Sn represent the n-th partial sum, we have Since r 1, there is a positive integer n0 such that for all n n0 , x I rn +l l < 1. Therefore, for fixed x and n n0 , the sequence of partial sums is constant. For x 0 and n n0 , the sequence is [x] and for x < 0, the sequence is [x] + 1 [x + 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 .

<

<

<

+

<

+

=

+

+

/

+

!

+

97

Remark .

• • • ,

PROBLEM Show that the total number of permutations of n things is [n!e], where [x] denotes the greatest integer in x. The total number P of permutations of n things is the sum of the number of permutations taken n, n - 1, n - 2, . . . , 2, 1, 0 at a time. Hence n n! = n!e - I n! . r=O rr r=n+l r. But n! 1 0 r=n+l r! r=l (n + 1) r n 1 ' whence the result. 98.

Sol ution .

I

p

<

....

I

- <

I

.!_ :S

277

SEQUENCES AND S E R I E S

PROBLEM 99. Let Pn = k=O'En log(pkPn) ' sn = n2 Pn ' and n 'E (-l) k log(Pp�) , Tn = -n1 k=O where p is a positive integer. Show that (E.27) n-+limoo Sn = 2 ' and (E. 28) n-+limoo r2n = 0 . It is clear that r2n-l 0 for all positive integers n. We shall make use of the binomial coefficient identity ..!...

0'n '

.E.

Sol ution .

First we establish (E.27) . By means of (E.29) we have, when n 1, nI log(P�) = nI log(P� - P) nI log(pn) - nI log(Pk) , k=l p k=l p - p k=l p k=l p or Pn - Pn-l = log �l (PPn) n k=lTI (PPkr l �l = log (pn -(pn)p) !! nn (pn) ! By Problem 39, if bn increases steadily to oo then 1n -+imoo Pnn = nlim-+ oo bPnn - bPn-1 n-1 , provided that the second limit exists. Here we choose bn n2 , so that bn - bn-l = 2n - 1 = n(2 - 1/n) . Then we have 2 - 1 1/n log (pn p)(pn)! (pn)! ! 1/n . +

b

_

_

CHAPTER 3

2 78

Since the outside factor tends to 1/2 we shall have finished the proof of (E.27) if we can show that the ratio of factorials tends to eP. To show this we need the fact that 1r -1r. meo r! r = e (see either Remarks to Problem 16 or Remarks to Problem 41, Part (iv)) . In­ deed we have n ( n (pn) ! 1) (pn 1) p p lim lim 1/pn p 1/n n -r oo (pn - p) ! (pn) ! n +oo [ (pn) ! ] r(r - 1) (r - p 1) rlim+oo (r! 1/r) p r _ � lim , r +oo r. 1/r_ ; P . (1 - .!.r) (1 - �)r (1 - r for p 2, and it is clearly also correct when p = 1. In the case of we find by means of (E.29) that nL (-l) k log (pn - P) nL (-l) k log (Pn) - nL (-l) k log (Pk) , p k= l p pk p k=l k=l or n n 'n 'n_ 1 = - 1 - 2(-1) log (pnp ) - k=lL (-l) k log (ppk) , n 1. Since Qn 0 when n is odd we have then nI log ( (2k - l)p) nI p k=l k=l = log k=lTI ekpp- P) epkpr l = log k=lTI j=ltr 2pk2Pk- - - j 1 1 Now to show that T2n tends to 0 as n + oo we should have to show that the n-th root of the product tends to 1. For example, when p = 3 we should have to show that l"7T

p +

• • •

• • •

+

• • •

�

�

�

0

.L..l:.)

+

+ 0

�

_

p

But in general, if a > b c >

� 0,

then

j +

+

SEQUENCES AN D SER I ES

2 79

ak - b l/n (E. 30) nlim j1 k=lTr ak - c fl 1. This is immediate from Problem 42: limn+co anl/n exists and has the same value as limn -+ "' an+l/an provided the latter limit exists. Thus by (E.30) any product of a finite number of such factors has the same property, so we see that r2n tends to zero as n increases indefinitely. "* "'

PROBLEM 100. Evaluate Subtracting term by term the denominator series D from the numerator series N (both unconditionally convergent) we find N - D (l/2)N. Hence N/D = 2/3. Sol ution .

=

PROBLEM 101. Let a1 = 1, an = n(an-l + 1) , and define Find limn-+oo Pn . Sol ution . p

Now Hence

n

We have

(an + l)/n!. {(an + 1)/n!} {(an-l + 1)/(n - 1) !} {(an + 1)/n!} {an/n!} = l/n! .

-

CHAPTER 3

280

P1 1/2! + 1/3! + + 1/n! 1 + 1/1! + 1/2! 1/3! + + 1/n! and limn +oo Pn e. • • •

+

=

• • •

+

=

PROBLEM 102. There are given pn = [en!] + 1 points in space. Each pair of these points is connected by a line, and each line is colored with one of n different colors. Show that there is at least one triangle all of whose sides are of the same color. Define a sequence (bn) inductively by b 1 2 and also by bn+l (n + l)bn 1. We will prove: (i) When the segments connecting a set of b 1 points are colored with n colors, at least one single- color trianglen results; (ii) Pn bn 1 for all n. Statement (i) is clear for n = 1. Suppose it is true for n = k, and let a set of bk+l 1 points be given. Starting at any point in this set, there are bk+l segments joining to the remaining points; since bk+l (k + l)bk ' one of the k + 1 colors (call it "blue") must be used at least bk + 1 times in coloring those segments. Thus we have a subset B consisting of bk + 1 points, each joined to by a blue segment. If any segment joining two points of B is blue, they will form with an all -blue triangle; otherwise the seg­ ments of B are all colored with the k remaining colors, and the induction hypothesis assures us that a monochromatic triangle exists in this case also. Now if an = bn/n! for each n, we see that a1 = 2 and an+l = an + 1/(n 1) !. Thus (an) is the (increasing) sequence of partial sums of the usual series for e. Therefore, for all n, an < e and consequently bn < en!, bn � [en!], and finally bn 1 � pn . Sol ution .

=

+

+

�

+

A

A

+

A

>

A

+

+

PROBLEM 103. Let (s 1) . z; (s) Show that, for n = 2,3,4, . . . , z; (2) z; (2n - 2) + z; (4) z; (2n - 4) + + z; (2n - 2) 1; (2) (n + Y,) z; (2n) . >

• • •

281

SEQUENCES AN D SER I E S

The left-hand side, written out at length, is the limit, as N , of 1 _.12Jf l l _ + (E.31) 2n-4 2n-2 k j=lr k=lI �l _l_k2 ._ j 2n-2 + _l_k4 ._ j j (N a positive integer). Summing the expression within braces, and taking note of the exceptional case k = j, this becomes � ( j 2-2n k 2-2n 1 (E. 32) kl k2 - / + (n - 1) j 2n rl Throughout the present discussion, all sums run from 1 to N, unless other­ wise indicated, and an accent on an inner indicates that the index (in this case, k) does not take on the value of the index of the outer sum (j, here). Ignoring the term on the far right, (E.32) is equal to 2-2n .2-2n k (E. 33) J2 j k k - j 2 j k .2 k2· Inverting the order of summation in the second double sum, and noting that the condition k j is the same as the condition j k, (E.33) may be written as Solution .

-+

""

_

_

+

• • •

] L.

L

+ L L'

L L'

J

-

f

f

(E.34) , 1 l _ _ 2 L. . 2n-2 k k2 .2 , the latter form arising out of an interchange of the dummy indices in the second sum. Combining (E.31) - (E.34) , we find that 1 - + + _2n-21 2 �l � _!_._ 2 2n-2 k j j klk j (E. 35) 1 1 1 _ - (n - 1) j 2n_ + 2 j _ j j 2n-2- t k2 . 2 . Now, 2j k k2 -1 J.2 k 1 k 1 j-1L � + N � - N 1 + 1 2j k=l k=l k=j+l L

J J

L L

- J

_._!_

• • •

I

I

I

'

_

J

L , "j(":"] - L ' k+J

L,

J

L

J

L k+J

282

CHAPTER 3

j-1I _!_ + N-jI _!_ _ N+Ij 1 + 1 "k N+jI "k1 + ..,.1 . + NI- j 1 + -1. J -k 2J _- .]_ _ { 2j N - j 1 + 1 + N - j 1+ 2 + ••• + -N +1-j } . (E.35) , k=l k

k

k=l

k=l

2T

k= j +l

k=l

When we substitute this into

we get

IJ _J 1_ _ J. _J. 2n1_-1 { N - j1 + 1 + =---=+ 2 + ••• + N +1 J N - 1 -= O < - j1 + 1 + :;-;N----j1,_ -.: _+--::c2 + • • •

= en + 1:2) . 2n .

I

.}.

(E. 36)

F inal ly,

N

and so

0 < J? J.2�-1 { N - j1 + 1 + .. . + -N +1 j } < 2 ?J J.2�-2 N - 1 + 1 2 ?J j(N - 1j + 1) : 1 J? {t + N - 1 + 1 } = N ! ?J t < N ! + N) .... 0 N (E.36) (E.37) , �

j

= N

l

j

l (l

Statements

log

and

(n � i)

as

(E. 37)

-+ oo .

taken together , complete the proof.

103,

By success ive appl ications of the result in Prob lem one n } . From Prob lem we know can express r; (2n) as a rational multiple of { that = and so forth . Problem tel ls us hence = nothing, however, about itself. In Prob lem we w i l l circumvent this difficulty. Remarks .

r;(2) n2/6;

PROBLEM

104.

r; (2)

r;(4) n4/90 r;(2) ,

For s > 0 , let s (s )

z:j=O

104

103

(-l) j (2j + 1) - 5 •

7

Show that

283

SEQUENCES AND S E R I ES

� (1) � (2n - 1) + � (3H (2n - 3) + + � (2n - 1) � (1) (n - j=OL (2j +1 1) 2n (n 1,2,3, . . . ) The solution proceeds along the same lines as the solution of Problem 103. In the first place, we have, as in (E.31) - (E.35), j +k �-1+ 1 . 1 2n-l + + 1 2n-l · � 1 fi (2k + 1) (2j + l) jl: kl: -1) � 2k _ = n l: (2j + 1 l) 2n (E.38) + 2 jl: (2j + 1l) 2n-l kL ' (-l) j+k (2k + 1)2k2 +- 1(2j + 1) 2 . Here, all sums run from 0 to N, and the accent has the same significance as in the Solution of Problem 103. Again, 4 k C-l) j+k (2k + 1)2k2 +- 1(2j + 1) 2 kL ' C-l) j+k (k + j 2k+ +l) 1(k j) .+ 1 Lk ' ( - l) j+k _k -l_. + kL ' (-l) j+k .,.---k +__:,:.1_7 (E.39) ,-,---N -__:,:.1-+ -,;-2 + - + �___ - 2j 1+ 1 + N + j,�1 +--,-1 } ' the last step involving reasoning analogous to the corresponding calculations in the Solution of Problem 103. Equations (E.38) and (E.39) now yield l:k C-l) k _ 2k l_ + 1 jl: C-l) j (2j + 1l) 2n-l + kl: c - l) k (2k + 11) 2n-l jl: C - l) j 1 l_-,­2n (n - L __ (E. 40) (2j + 1) + 2 jL (-l) N- j (2j + 1l) 2n-l N - j1 1 ,..,.-N ---'- 1::-+�- 2 + - + N + 1 1 . But 0 < -N-j1_+l _ _N-j1_+2 + + _N+j+l1_ <- _1 N-j+l_' • •

�)

·

00

.

Sol ution .

c

z:

• • •

'

=

J

• • •

---

+

�

• • •

�)

.!_

{

+

• • •

+

}

CHAPTER 3

284

and so the absolute value of the last sum in (E . 40) is 1 1_ < ._ L (2j + l ) 1(N - j +l) N j 2nl +l j (2j +l) j

(n

L

s:

= 2N : 3{ l +

1 + 2N+1 + 2N 1+ 2 }

i+t+

�

-+

1)

O

as N

-+ "'

,

(E . 41)

as with (E . 37) . From (E . 40) and (E . 41) the desired result fol lows at once . Remarks .

In the Remark to Prob lem 1 0 we noted the fact that

1 + 1 � ( 1) = 1 - 3 + S

• • •

TI = arc tan 1 = 4.

Hence , taking n = 1 in Problem 104, we find 1 2 TI 2 2 { � (1) } = -g· = j = O (2j + 1) 2

L

Coupling this with the s imple identity 1 = ( 1 - 2 - s ) l; (s) s j =O (2j + l)

L

(s > 1) ,

the famil iar value (see Problem 7) , 1; (2) PROBLEM 105 .

Sol ution .

-1 = m

=

n

2 /6 , emerges at once .

Show that

Observing that

j � O (m + jn) (: + jn + n) '

we get

1 n 1 = n "' L (m + jn) (m1 + jn + n) n m=L l m mL l j =O = -

1 (k k + n) ' k= l

L

285

SEQUENCES AND SE R I ES

because k = m + jn takes al l values 1 , 2 , 3 , . . . if k takes the values of 0 , 1 , 2 , . . . and m the values 1 , 2 , . . . , n . I n the fol lowing al l summat ions run from 1 t o infinity unless indicated different ly. If we denote the sum in question by S , then we have

I

I

I

1 1 1 + + k >j j k (k+n) ( j +n) k<j j k (k+n) (j +n) k=j j k (k+n) ( j +n)

I

I 1 I

1 I. _ _+ 2 2 k < J k J. 2

=

=

!

2

\

1 + 2 I j k (k + n (j + n) 2 k (k + n) 2 k<j

�l I

_ -

-} � (I

k2 j 2

)

_1_ k=j k 2 j 2

_

l 2 k2

1 k (k + m) (k + n) (k + m + n)

I

-

l k4

)l

+ 2

1 I { kmn (k+m+n)

} + 2 I kmn (k } m + n)

_

1 mn (k+m) (k+n)

}

- 2S

according to the third expression for S . Since

I __!__2 i 6

and

=

k

(see Problem 7 and Remarks to Problem 103) , we obtain

(

4 7f 2 38 - 120 1

-

_

10 1 { log ( l L 1O

\'

whence S

)

_

1

X

I

kmn (k

} m + n)

x) } 3 dx X

=

n-1 ( 1 og X ) 3 dX

-

=

r )0 -

=

l

I

1 xk+m+n- 1 Jo kmn dx (

11 (I xk )3 dx 0

T

( log x) 3 � 1 - X

L j "' ent \'

=

0

t

3

dt

_

1 7 7f4 /360 , as required .

PROBLEM 106 .

Any pos itive integer may be written in the form

n = 2 k (2j + 1) .

x

CHAPTER 3

286

Let an e-k and bn The problemk means that every positive integer n can be writ­ ten uniquely in the form 2 (2j + 1) , where k and j are nonnegative integers Thus k = 0 when n is odd, and k -1is a positive integer when n is even. Hence . a 1 when n id odd, and a e when n is even. Therefore, b 2n = b Zn+l andn the series is equivalentn to + But since b Zn+Z/b2n = a2n+laZn+Z e- 1 the latter series converges. Since the terms approach zero, the grouping of the given series does not affect convergence and the given series converges. Sol ution .

s

• • •

s

PROBLEM 107. Let the entries tnm (1 m n) of an infinite triangular matrix s

s

satisfy the following conditions: (i) The elements of each column tend to zero, i.e. , tnm 0 as n (m fixed); (ii) the sum of the absolute values of all elements of any row do not exceed a constant i.e., for all n = 1,2,3, . . . , -+

-+ oo

K,

If (xn) :=l is a sequence that converges to zero, show that the sequence x'n + + tnn n converges to zero also. (b) Suppose that the coefficients tnm satisfy in addition to conditions (i) and (ii) of part (a) also the condition • • •

X

SEQUEN CES AN D S E R I E S

287

(iii) Tn as n If (y ) is a sequence that converges to a (finite) real number a, show that n :=l y'n as n (a) Let £ 0. Then there exists an integer m such that lxn £/2K for n m; for these n we have, by condition (ii) , lx'n I Itnlx1 tnmxm I 2 Since m is kept fixed, there exists by condition (i) an n0 m such that for tnmxm l £/2. Thus lx� l £. n n0 we have ltn1x1 (b) Since + oo .

+ oo .

I

>

Sol ution .

�

<

<

+ ••• +

+

+ ••• +

�

�-

�

<

<

••• +

we may apply part (a) to the sequence (yn - a) :=l and the proof is complete. The result in Problem 107 is due to 0. Toeplitz and has many applications. For example, setting =1 we see that conditions (i) , (ii) , and (iii) of Problem 107 are satisfied. We can now deduce immediately the result in Problem 40. Consider the result in Problem 39 and for simplicity assume that the sequence (yn) :=l is strictly increasing and suppose that (n 1,2, . . . ; xo 0) tends to a; putting tnm Ym -YnYm-1 we can verify that conditions (i) , (ii), and (iii) of Problem 107 are satisfied. We may then conclude that nI m=l Remarks .

fi'

CHAPTER 3

288

converges to a, thereby having another solution for Problem 39. PROBLEM 108. Let (xn) �=l and (yn) �=l both tend to zero and + I Yn (n = 1,2, . constant) . Show that I

.

� K

. ; K

tends to zero also. Set tnm yn-m+l and apply part (a) of Problem 107. Sol ution .

PROBLEM 109. Let xn a and Yn b as n Show that zn xlyn + x2yn-ln + • • • + xnyl ab as n Assume first that a = 0. If we replace yn by yn/n in Problem 108, we get zn 0 as n The condition imposed on Yn in Problem 108 is not violated by this since the yn 's are bounded, I Yn Turning now to the general case, we note that zn (xl - a)yn + n + (xn - a)yl + yl + n + Yn a. But we have already shown that (xl - a)yn + + (xn - a)yl 0 as n n and, by Problem 40, a Y1 + n + Yn ab as n +

+

+ oo .

+

+ oo .

Sol ution .

+

+ oo .

I

_.::._ ..._ ... _ _ __:.::.._____::.. _...:.:._

_ _ _ ...._ .=; ....;.:.

+

+

+

+ oo

oo .

PROBLEM 110. Let xn a as n Show that x'n l·x0 + (�)x1 2n+ • • • + (�)xn a as n +

+ oo .

+

+ oo .

� K.

289

SEQUEN CES AND S E R I ES

We set

Sol ution .

tnm and apply part (b) of Problem 107. Since (�) nm and (nm)/(2n) 0 as n we see that condition (i) of Problem 107 is satisfied. Since �

<

� oo ,

conditions (ii) and (iii) of Problem 107 are satisfied too and the desired result follows. Let xn a as n and z 0. One can show in analogy to the solution of Problem 110 that the sequences (�) zx1 . . . (� xnzn l·x 0 x'n (1 z) n and nxo + (�) zn-1xl z �) xn x"n (1 z) n tend to a as n �

Remark .

� oo

+

+

>

)

+

+

+

• • •

+

->-

+

(

oo .

PROBLEM 111. Evaluate the convergent infinite series We have

Sol ution .

k=1I 1/kCk 1) c11 2 1/3) k=2I 1 /k Ck 1) c114 1/s) k=3I 1/kCk 1) +

+

+

+

+

+

+

+

CHAPTER 3

290

But

k=mI 1/k(k 1) 1/m. Therefore 4S 1 k=2I [l/(2k - 2) l/(2k - 1)]/k 1 + c 11 2J k=2I 1/k Ck lJ k=2I l/k(2k - lJ 1 + 1 1 2 2 k=2I [l/C2k - lJ l/2k] 3/2 2(1/3 1/4 1/5 - 1/6 3/2 2[log 2 (1 - 1/2)] (see Problem 15) 1/2 2 log 2 and log2-2 . s=1 +

+

+

-

+

+

+

+

-

+

-

+

· · ·)

+

8

+

PROBLEM 112. Show that " n=O" (m +m!n!n + 2) .' = --62 Multiplying each term by (m + n 2) (n + 1) and dividing by (m 1) , the summation over n can be written as a telescoping series, and the double sum becomes m! I (m nn' 1) ! (m (n+ n+ 1) 2)! ! ) L m=O n=O m! 0! I 1 2 = 1[62 . (see Problem Lm=O ii1+T (m 1) ! m=O (m + 1) 00

00

m= O

11

�

�

·

+

Sol ution . +

\'

00

ii1+T

\'

00

(

+

+

-

+

+

PROBLEM 113. Euler's Constant c is by definition

7)

291

SEQUENCES AND S E R I ES

c mlim+oo 1 k=lLm 1 log m f (see Solution to Problem 14) . Derive the following series representations for Euler's Constant c: n 2 c = 1 n=l� �n-1 (2m - n1) (2m) ' (E. 42) +1 m=2 n -1 2 1 c = + n=lL: �n-1 (2m) (2m + n1) (2m + 2) " (E. 43) m=2 Let k+l sn n (-l)k (n 1 ' 2 ' . . . ) . Then, since ( - l)kk+ l 2-�n 1 2 2n-1 2k1 ' k 1 k=l we have the relation (n = 1, 2, . . . ) . (E.44) We start from the definition c mlim+oo �( k=lI f - log m fl , and consider the subsequence in the right member corresponding to m = 2n , (n = 1,2, . . . ) . Then, also, (E. 45) c = nlim+oo (sn - n log 2) . But log 2 = 1 - -21 + -13 - -41 + . . . (-l) n+l n (see Problem 15) and so log 2 = + rn ' (n 1 ' 2 ' . . . ) ' I

00

2

Sol ution .

(J

0'

I

_

k

_

I �

_

=

+

(J

n

.!_

+

CHAPTER 3

292

where

(-l)kk+l Since 0 rn l/2n , limn (nrn) 0 , and (E.45) becomes c = nlim+oo (sn - nan) . (E. 46) The sequence in (E.46) is now converted into a series in the usual way, and we have rn

<

<

+ ""

Using (E.44) , this simplifies to

(E. 4 7)

Equivalently, on replacing the an by their values 1 1 + - nl+l) . c 1 - n=l� n/\� (E. 48) 2 -� By combining consecutive pairs of terms in the parenthesis in (E.48), we get (E. 42) . To show the equivalence of (E.43) and (E.42) , expand the general term in (E.43) into partial fractions. Then n (2m1 - 2m 2+ 1 + 2m 1+ 2) = 1 + n=lI n (an n+l 2n+21 ) Since n=l n/2n+2 1/2, the equation =

2

l:

a>

• • •

a>

- a

=

is equivalent with (E.47) .

+ --

·

SEQUENCES AN D S E R I E S

293

It follows readily that the series (E.48) converges also with the parenthesis removed, in which case the resulting representation has the form c = 1 + t=3I (-l) t [log(t t- 1)/log 2] (E.49) (Here and in the sequel a square bracket will denote the greatest integer function.) For choose any integer k 3 and set a = [log(t - 1)/log 2]. Then we have (-l) t [log(t -t 1)/log 2] Thus as k the series in this inequality will approach zero as a limit. As this series represents the difference between the partial sums of the series in (E.49) and corresponding partial sums in (E.48), the convergence of (E.49) is established. Another representation of the form (E.49) is obtained by modifying (E.47) as follows. Since �oon=l n/2n+l = 1, we have from (E.47) that 1 \j " c - n=lI n (crn+ 1 - n - n+l 2 As before, it may be shown that the parenthesis in this series can be removed. The resulting series representation is c t=lI (-l) t [log t/log t 2] PROBLEM 114. Let Sn 1 + 1/2 + + 1/n. Show that c p q - pq 1, where c is Euler's Constant (see Problem 113) . If we set (p,q) p + q pq' then (p,q) - (p-1 ' q) -p1 - pq - 1q + 1 pq - 1q + 2 - ... - pq1 p1 q(plq) = Remarks .

>

< O.

+ oo ,

=

a

=

• • •

< s

+ s

s

�

=

Sol ution .

< - -

s

s

0.

- s

CHAPTER 3

294

Thus (p,q) is a decreasing function of p and therefore also of q (by symme­ try) . The cases p + q + oo and p = 1, q = 1 are the two sides of the in­ equality in question. oo,

PROBLEM 115. Let a > 0 and b > a + 1. Show that -ba + ba ba ++ 11 + ba ba ++ 11 ba + 22 + ba ba ++ 11 ba ++ 22 ba ++ 33 + a where d b - a - 1. Let q_ 1 0, Q_ 1 = a, and, for n 0,1,2,3, . . . � a a + 1 a + 2 ba ++ nn ' � (a + n + 1) . Then -1 - d� . Letting n 0,1,2, . . . ,s in the last expression and then adding, we get -· ---

-· ---· ---

+

d'

-· ---· ---· ---

Sol ution .

'

iJ'iJ+Tb+Z

�

� =

But

aPs with s A(A B(B ++ l)l) (A(B ++ 2)2) (A(B ++ s)s) and A = a + 1, B = b. But Ps + 0 as s + because B > A > 0 (see Problem 80). Hence q0 + q1 + q2 + = a/d. But this is the desired result. p

• • •

oo

PROBLEM 116. Let s0 1 ' sl 3, sn+l 2Sn2 - 1 for n 1. Show that sn lim n n 2 s0s 1 n-1 12. From sn2 - 1 = (Sn + 1) (Sn - 1) 2Sn-12 (2Sn-12 - 2) 2 2Sn-12 (Sn-12 - 1) it follows that ;::

+ oo

Sol ution .

s

295

SEQUENCES AND SERIES

whence

s

n

n

2 s 0s 1 s

s

n- 1

2So s2 ��

n

2s0 f� - 1

s1 +-

which has as l imit

Vcs2 - l)/2s0

1'

n

12.

1

117. lim L 2 .2·

PROBLEM

Evaluate

n + co

J

n2

n j =l n +

Sol ution .

i

2

n +1

n n +

For a l l pos itive integers n

2 .2 J

dj �

n

2

n j=l n +

I

!c

J

and thus it is cl ear that the l imit as n PROBLEM 1 1 8 .

n

2

+ co

If

= -1- + _1_ + _1_ +

s2m 2 2m 42m 6 2m

find the value of

Sol ution .

I f we let

T2m 2•31 (2m1 ) 2 4•51 (2m) 4 6 · 7 2m =

+

1

n n +

2 .2 - 0 2 . 2 rr/ 2. <

+ 1 ( 1)6 +

is

J

dj

CHAPTER 3

296

then the value of the given expression is the same as that of m=lL T2m' By integrating x/(1 - x2) = x + x3 + x5 + lxl 1, from 0 to r twice and then setting r = l/2m, we find T2m 1 -2m -2+-1 1 og 2m + 1 + -2m -2--1 1og 2m - 1 m 1,2,3, . . . , whence nL T n + log 2n (n!) log 2n en (n!) m=l 2m (2n + l) n+� (2n + l) n+� Using Stirling's Formula (see Problem 34), 1n 1m. nennn.+l:' = �2 we obtain 1 m=l T2m = log (D

• • •

----zm•

----ziD

_

=

<

=

1

=

v L 7T ,

2

+ "'

2

\ L

1 •

e' 71

PROBLEM 119. Verify the following identity, due to Gauss: Is=l s(s-1)/2 lxl 1. Let P0 1 and n 1 - 2s for n 1,2,3, . . . n s=l 1 2s-1 We shall show first that n-1 P 2n n s=OL Pns xs(2n+l) s=lI s(s-1)/2 n Indeed, we readily verify that X

'

<

(E.

SO)

Sol ution . p

A

lT

X

-

X

X

s

(E. 51)

SEQUENCES AN D SE R I ES

297

and multiplying by we find -Pns xs(2n+l) - -n-1p-s xs(2[n-l]+l) + as,n s,n where 2s+l x (s+l) (2n-l) 1 x s,n 1 x2n-l and 1 - 2n-l x2s -Pn-1 s(2n-l) x P 1Now s+l,n = as,n (for s 0,1,2, . . . ,n - 2) and since, further, O,n = 0 and an-l,n = xn(2n-l) ' by summing (E.S2) from s = 0 to s = n - 1 we obtain: An = An-1 + x (n-1) (2n-l) + xn(2n-l) But this may be written An - An-l = Sn - Sn-l ' and by induction p

a

_

p

- 13

(E.52)

_

X

S

13

13

·

This proves (E.Sl) . From (E.Sl) we now readily obtain (E.SO) . The leading term in An (that is s = 0 in the left side of equation (E.Sl)) is P . Since the remaining terms (s = 1, 2, . . . , n = 1) are of order x2n+l andn higher, the power series of the2nfunction Pn (x) must agree with that of Sn (x) at least to terms of or­ der x By induction the function P00 must have power series S00 which proves (E. SO).

CHAPTER 3

298

PROBLEM 120. Show that log (l.l01 0010001 . . . ) = II1 + 2-1 1011 + 31 . 1001 1 + By the result in Problem 119, log(l + X + X + X + n=l ( -l) n log(l - xn) n+l "'L !_pnp f = I =I (-l) n+l ( Ppn) = "'L xp P ' (-l) I n=l I p=l p=l n l p=l p(l + x ) where the operations are justified by the absolute convergence. The desired formula results from placing x = 0.1. Sol ution .

3

6

• • •)

L

�

_ _ _ _

PROBLEM 121. Find the sum of Consider the integral j x arc s1nz . z dz 0 and expand the integrand in a power series in z. Then x arc sin z dz = Jx { 1 0 { z Sol ution .

Inasmuch as the series in the integrand is convergent for z 1, the re­ sulting power series in x is convergent for x 1. Moreover, this series converges for x = 1 and is then just the series whose sum, S, is to be found. By Abel's theorem, = 101 arc sin z z dz. The substitution z = sin t yields the result in terms of a standard improper integral l l

8

<

l l

<

299

SEQUENCES AN D SERI ES

- foTI/2 log sin t dt which is easily evaluated by replacing sin t by 2 sin �t cos �t. One obtains = � log 2. Indeed, let J = S and t 2x. Then J = 2 J0TI/4 log sin 2x dx = log 2 + 2 ( TI/4 log sin x dx + 2 faTI/4 log cos x dx. But, under the substitution x = - u, we obtain TI/4 log cos x dx = f TI/2 log sin u du fo TI/4 and so J = log 2 + 2J or J 2 log 2 . Denoting by xs the roots of x2n 1, we get s cos 2sTn I + i sin 2STIn (s 1, 2, . . . , 2n) . Thus 2n - 1 2n (x - xs ) n-1 (x - xs ) 2n-l (x - xs ) • (x2 - 1) ' s=n+l s=l s=l since n = -1, x2n 1. But x2n-s s with xs being the complex conjugate of xs and so cx2 - n-1s=l ex - xs ) ex - xs ) (x2 n-1s=l (x2 - 2x cos -SnTI + 1) or n-1 (x2 - 2x cos -STI + 1) . 2n-2 + 2n-4 + ••• + n s=l S

S

TI

-

2

J

7f

o

2 7f

2 7f

�

Remarks .

X

TT

TT

TT

X

X

X

1) TT

-

X

1 ) TT

X

lT

CHAPTER 3

300

Letting x = 1, we get n = ns=l- 1 (2 - 2 cos S1Tn = ns=l- 1 s1n2 -S1T2n 2 2(n-l) s 1n 2 2n s J.n2 2n s J.n2 (n-1)1T or (n-2n1)1T = n-l s 1n 2n1T s J.n 22n1T s 1n --2 But this identity leads to the evaluation of directly: (1T/ 2 � log n (n 2 log 1T log sin x dx lim log 2. n Jo n +oo 1T

0

0

lT

)

TT

•••

0

21T

0

•••

4

0

----zn

0

li1

0

--0

J

'IT 2

2

1)

PROBLEM 122. Show that tends to exp(1r2/12) as n + oo. taking logarithms we see that On

Sol ution .

0 5

But, for x log (1 x) = 1 1 1 2 and so log(lx x) dx = I (-l) n-1 2 = I � - 2 I _1_2 = �12 n n=l n n=l (2n) n=l because "'..oon=l n-2 1r2/6 (see Problem X

5

+

1,

- ZX

+

+

3X - • • • +

�

7) .

+ •••

SEQUENCES AND SE R I ES

301

PROBLEM 123. Show that tends to 2 exp ---2----4) as n taking logarithms we see that (11

Sol ution .

+

But

+

oo ,

On

1 log(l x2 ) dx as n Jo [

+

+ oo ,

log - 2J(0 1 (1 - � 1 ) dx = log 2 - 2 2 log 2 - 2 log -2 4 and the desired result follows. In an entirely similar way we can find the limit of 2

+

+ X

2 + 1T

4 1T

1T

2 + ------

=

Remarks .

as n

+

oo ,

Indeed, the given expression can be rewritten as

and tends to - J o 1 log(l x2 ) dx 2 - log 2 - 11/2 as n r

+

+ oo ,

PROBLEM 124 . Let An and Gn denote the arithmetic and geometric mean of the binomial coefficients @ , (�) , @ , . . . , (�) , respectively. Show that

CHAPTE R 3

302

n n lim /A = 2 and lim n n +oo n +oo �n = We have n Sol ution .

re.

A

(by Problem 41) and so limn -r oo n� = 2. Moreover (n!) n+l 2 n (n + 1 - k) n+l-2k (1! 2! 3! n!) k=l n (n + 1 - kt+l-2k ' k=l n + 1 because nL (n + 1 - 2k) 0 . k=l But k+ _1) 1 I (1 - n + 1 log(l - _ lim log Gn lim n n n oo oo n -r k=l n -r -- f0 l (1 - 2x) log(l - x) dx 1 and so limn -r oo n �n = TT

• • •

TT

..!..

J

�)

n

2

re.

PROBLEM 125. Find the limit of n as n + oo , where 2 + 2 2 + 3 2 + + n2 1 n n 3 1 3 n3 + 2 3 n 3 + 3 3 n 3 + n 3" Evidently as n + oo. s

s

. . .

+

Sol ution .

303

SEQUENCES AN D S E R I E S

PROBLEM 126. Let x 0 . Show that -1 = -x + 1 + n=l (x + 1) (x + 2) n! (x + n + 1) We easily see that 1 ' X1 - x+l - (x+l)1(x+2) x(x+l)2 (x+2) ' x+l = x(x+l) and, more generally, that n-1 k! (x + n + 1) x(x + n! (x + n) " -(x + 1) (x 2) k=l But, for x 0 , n! lim (x + n) 0 (x + 1) n+ oo by the result in Problem 80. >

� �

1

•••

00

X

·

Sol ution .

1

1

1

X

1

X

X+T

1

L

•••

+

1)

>

PROBLEM 127. Let Tn = n (Sln. -nt + sin -2tn + . . . sin (n -n t) . Show that lim Tn 1 cos t t n+ oo Dividing the interval [O,t] into n equal parts, we see that lln+· moo nt k=l� s1n. kt = !a0t sin x dx - cos t and so n-1 kt = 1 - cos t lim -k s1n. n +oo k=l n t because sin nt nlim+oo n = O. 1

1)

+

Sol ution .

L

L

11

1

CHAPTE R 3

304

PROBLEM 128. Let n v'(n + 1) (n + 2) (n + n) n n Show that limn +oo Pn 4/e. We have p

• • •

= -=--�>..:..:�-->..:.----'� .:-

Sol ution .

and so, as n + oo, log Pn .!.n{ log (1 + .!.n) + log (1 + �n) + + log ( 1 + !!.n) } + fa 1 log(l + x) dx log 4 - 1. • • •

=

=

PROBLEM 129, Let n n = 2il+l + k=l� ( 2k) 31 - 2k Show that limn +oo n log By the result in Problem 105 in Chapter 1 we have n = n +1 1 + n +1 2 + + --. n +1 n Thus, as n + oo , log 2. Sn .!.n [-1 +1-.!. + _1 +1_� + + _1 1_!!.] + 101 � 1 + x n n n From the result in Problem 129 we can easily see that 1 1 + log 2. 3 k=l ( 2k) - 2k n

s

2.

s

Sol ution . s

--

--

• • •

=

=

+

Remark .

L

2

PROBLEM 130. Show that 1 - 1/4 + 1/7 - 1/10

+ • • •

( 1/3) (TI/ 13 - log 2) .

S EQUENCES AND S E R I E S

305

Noting that 1 ta-l dt (a,b 1 1 -a - a +1 b + a +1 2b - --••• = + a + 3b 1 + tb we see that we merely have to evaluate f l dt · a� But 2 + 213 arc tan � t + C (t + l) = !�log f_E__ 13 � 1 + t3 6 l t2 - t + 1 and the desired result easily follows. Sol ution .

1

--

---

---

>

0) ,

0

PROBLEM 131. Show that, with m a positive integer, 1 1 n=lI n(n + m) = l..m (1 + -2 + -3 + For a -1, b -1, and a b we have l a - b dx. 1 = I n= l (n a) (n + b) b = a fo 1 Thus, taking a = and b m, we obtain the desired result. 1

>

Sol ution .

>

X

+

X X

'f

0

>

> 0.

PROBLEM 132. Let b - 2 a Show that a(a ++ 1) + 3• b(ba(a ++ l)l) (b(a ++ 2)2) + ••• (b - a a(b- l) -(b 1)- a - 2) " a + 2· b(b By the result in Problem 115, if b - 1 a then (E. 53) + a + b(ba(a ++ 1)1) + b(ba(a + l)l) (b(a ++ 2)2) + ••• b b- a- -1 1 " Hence, if b - 2 a then 1 + -a b+-1 + (a b+ (bl) +(a1)+ 2) + (a b(b+ l) +(al)+ (b2)(a+ 2)+ 3) + ••• (E.S4) b 1 b - a - 2· b

=

1)

>

Sol ution .

1

b

+

>

>

0,

>

0,

CHAPTER 3

306

Subtracting from the series in (E . 54) the series in (E . 53) , we get b

1 + 2 . a + 1 + 3 • (a + 1) (a + 2) + • • • = ..--:b'--_;1 b - a- -::_,.2 b (b + 1) b (b + l) (b + 2)

b - 1 b - a - 1

b - 1 (b - a - l) (b - a - 2) and the desired result fol l ows immediately.

PROBLEM 1 3 3 .

Let

l { n n - 1 + -n - 2 + Tn = n -1 + -2 3

_ + l n log (n ! ) .

}

Find l imn ->- oo Tn . Sol uti on .

Let +

an

1 n - log n .

-

We have that an ->- c as n ->- oo , where c is Euler ' s Constant (see Prob l em 1 1 3) . But + ••• + a n n

Tn

and so, by the result in Problem 40, l imn -r oo Tn

PROBLEM 1 34 .

Show that

+ 3· · 7 + · · · + • • • 3 + 3· 5 5 3 5 7 9 1

2

Sol ution .

3

5

z·

1

Since

2 + 3 -1 + -::-:- -::;--____:n (2n :;-...,., 1) +� 3 3 · 5 3 • 5 • 7 + • • + 3 · 5 · 7 • • • :... --

•

the desired result fol lows by letting n ->- "" · PROBLEM 1 35 .

Show that

c.

307

S EQUENCES AND SER I ES

whenever ( a 1 + l) (a2 + 1) • • • (an + 1) Sol ution .

= 1

-

�

oo

as n

�

oo .

By Prob lem 106 of Chapter 1 ,

1 (a 1 + l) ( a2 + 1)

and the desired result fol lows . Remarks . The result in Problem 135 eas ily yields a number of interest­ ing identities . For exampl e , l etting

n - 1, . . . ,

an we obtain 1 = 1 1· 2 + 1 • 22 • 3 + 1 · 2 3· 3 · 4 + • • • + n n-! 1 + --

Putting a 1

=

0 , a2 = 2 , a3

and, putting a1

=

PROBLEM 136 .

1 , a2

=

=

4 , a4

=

6 , . . . , an+l

2n + -:-1-·-::-3-,·5::-·-:7 + l'") "' + • • • ::------,("'"2n---:

2! 3! 1 + ••• = x (x + 1) + x (x + l) (x + 2 ) +

sol ution .

an = 2n - 1 , . . . ,

3, a 3 = 5 , a4 = 7 ,

Show that , for x > 2 ,

x

Since

2n , . . . , we get

x--:--2•

1

CHAPTER 3

308

x = x-:-2 -

1!

1

x(x x(x

2!

2! ( x - 2) x ' 3! (x - 2 ) x (x

2! (x - 2 ) x

+

1)

+

3! 1) (x + 2)

3! (x - 2) x (x

1) '

+

(x - 2) x (x

1)

+

4!

+

1) (x + 2) '

and, in general , x(x

+

+

1) (x

n! 2)

(x - 2) x (x

+

(x - 2) x (x

+

n - 1)

n! l) (x + 2)

(x

+

(x

+

(n + 1) ! l) (x + 2 ) • • (x •

n - 2) +

- 1)

n

we see that .!.. +

x

x (x

+

x-:--2

1

�

2!

+

1)

� ,..-;-;3:..;!(x --"7----, x (x + 1) + 2)

- (x - 2) x (x

+ 1) !

(n l) (x

+

+

+

•••

+

x (x

2) • • • (x

+

+

l ) (x

+

n! 2) • • • (x

n - 1) ·

But , for x > 2 , x (x

+

2 • 3 • 4 • • • (n + 1) + 2) • • • (x + n

1) (x

- 1)

+

as n

0

+ "'

by the result in Problem 80 and so our c laim fol l ows .

PROBLEM 137.

Show that

1 1·2·3

+ + -5•6• 7

-- +

1 -3•4• 5

Sol ution .

-!, 2.

Using the identity

10 1 tn- 1 ( 1

<

and observing that , for J t J ( 1 - t) 2 (t 2

1 log 2 - z·

1

+

t4

+

t6

+

n (n

+

1 l) (n

2)

+

1, • • •)

( 1 - t) 2

- t2

+

1 (- 1) 1 - t2

2t - 2

+

t+T

2

+

n)

309

SEQUENCES AND SER I ES

we see that 1·2·31 3 · 4·1 5 5·6·71 = 1 . �. 3 � Ial (- t 2 2t - 2 t : 1) dt log 2 1 -- + -- + -- +

+

+

+

-

2.

138. Show that v'�� = 2. 1n -+. co \ n 2n We have, for n -+ co , PROBLEM liD

vn:1 --- + I2Il=1 + • • • +

Sol ution .

In a completely similar manner we can verify that n lim n-+ co k=l kn - c 2. Remark .

l:

li
139. Show that n (k2n - a) l/3 - 3 lim L n -+ co k=l kn We have, for n -+ co , n (k2n - a) 3/2 = n � 2 k=l kn n k=l k n PROBLEM

2"

Sol ution .

.!_

L

L

�

(�)

140. Show that n-1L n2 = lim n -+ co k=l (n2 k2 ) 31 2 We have, for n -+ co , PROBLEM

+

Sol ution .

_!_ rz

_

�3)1/3 � f l x2/3x dx = 2 0 n �

�.

CHAPTER 3

31 0

PROBLEM 141. Show that lim £: -6-;.,::1== n k= 1 V2a2kn::;:: -==1 = We have, for n Ik=l V2a2kn1 - 1 n k=lI V2a2k/n1 - l/n2 -+}f0 1 _-d2_xa_2_x = a rz

a·

-+ 00

-+ "" ,

Sol ution .

=

l.

_

rz

PROBLEM 142. Let x 0 and put u1 x and un 1 e log un for n 1 , 2, 3, . . . Show that f

-

Sol ution .

Since the law of formation of the sequence is

we can write successively eun-1 - 1

31 1

S EQUENCES AND S E R I E S

The last term of this expansion , that i s ,

is the remainder of the series when w e stop with the n-th term. W e shall show that l imn + co un 0 . The inequalities =

X >

e x ---1 log --X

>

-

0,

X > 0;

X <

e x ---1 log --X -

< 0,

X < 0

imply that the sequence un is steadi ly decreasing in the first cas e , un > 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

<

eu -1 log u

for u

< 0.

-

and -

PROBLEM 143.

Show that

1 1 4 4 · 16 4 • 1 6 • 256 5. 1-7 + 5 • 1 7 • 2 5 7 + -=5-.""""1=4= 5+ . z=-=5=-=70"""·-,6"'5-=5-=3=7 7 ....,

+ •••

Evident ly it is enough to show that

Sol ution .

1 4 4 • 16 4 • 16 • 256 4 = 1 7 + 17 • 25 7 + 1 7 • 257• 65537 + -----

But the latter series is of the form 1 + 1 Yo YoY l

----

where

y

0

4•

17

+

1 + Yo Y l Y z

------

Y1 =

16'

256

Y2

�·

65537

. . .

, and, in general ,

Now from the Solution of Problem 6 we see that 1 Yo

- +

1 + 1 Yo Y l Y o Y l Y z

----

------

+

CHAPTER 3

31 2

Here

4 + 41

Yo

and

Yo2 - 4

In Problem 83 a series was summed by the same method as the one used in the foregoing Solution . Remark .

PROBLEM 144 . Show that 1 + 1 + 1 + 1 1 = 1 + T 1 · 3 1 · 3 · 7 1 • 3 · 7 · 15 + 1 · 3 · 7 · 1 5 · 31 + • • • Sol ution .

Let

z = ( 1 + XZ) ( 1 + X 2 Z) ( 1 + X 3 Z) • • • ( 1 + Xn Z )

where P 1 , P 2 , . . . , Pn , . . . are functions of x only which are to be determined; changing z to xz , we see that

Hence

and so, by comparison of coefficients , we obtain pl =

p2 y-:--j(• X

pn

X3 (1 - x) (1 - X2 )

n (n+ 1) X 2 ( 1 - x) ( l - X2 )

. . .

'

X6 ( 1 - x) ( 1 - X2 ) (1 - X3 )

p3

( 1 - xn )

'

'

In other words n (n+l -2-) 3 X X X 1 + 1-xz + zn + • • • + ••• + 2 n 2 ( 1 - x) (l - x ) ( 1 - x) ( 1 - x ) • • • ( 1 - x ) --

SEQUENCES AN D SER I ES

31 3

= ( 1 + XZ ) ( 1 + X2 Z ) ( 1 + X3 Z ) • • • ( 1 + Xn Z ) • • •

1/2 and

Putting x

z =

1 , we obtain the desired result .

Note that

Remarks .

3•5 ·9 · 17 2 4 8 16 The result in Prob lem 144 shows that 3•5•9•17 2 4 8 16 is between 2 and 12/5 . Indeed , 1 1 1 1 • 3 + 1 · 3·7 + 1 · 3· 7·15 <

1 3C l

+

1 /6

+

1/6 2

+

+

•••

· · · ) = 31 1 - 1 1/6

= s·

2

From the Formula of Wal lis (see Prob lem 6) we know that the product of al l even numbers divided by the product of al l odd numbers gives the value ln/2 . I f in the quotient of the product of al l even numbers divided by the product of all odd numbers we omit all even numbers which are powers of 2 and al l odd numbers which are powers of 2 augmented by unity , then the re­ sulting quotient wil l be between 2 ( h/2) and ( 12/5) ( in /2) .

PROBLEM 145 .

-+ oo

Let k be a pos itive integer. Show that

- 1) = l im 1 · 2· 5• 4 •· 6" (2kn2kn- 1) 1 • 3 2• 5• 4 •• 6• • . .( 2n • 2n 3 n Sol ution .

lk

·

We have

n.:: 1 • 3 • 5 . . • ( 2n - 1) 2 · 4_·.:.. 6 · · ·.,...2k:. :-;; .=. -"-;1)7 -1o-• -:(2kn • 3::-'•:.,5,...: . . 2 · 4 · 6 . . • 2n

(2n) ! / kn [ (kn) ! ] 2 · ( 2kn) ! 2n 2 (n ! ) 2

12nm mm e - m approximately (see Prob lem 34) and the But , for large m, m! des ired result fo l lows . =

PROBLEM 146 .

-j

I 2m -

Find

dx ' 2 (x + a2 ) m

CHAPTER 3

31 4

oo

m being a posit ive integer. In part icular, show that

la

dx 2 (x + a2 ) m

(2m - 3) (2m - 5) (2m - 2) (2m - 4)

1 1 1T 2 a2m- l z·

Then use this identity together with the Formula of Wal l is (see Prob lem 8) to verify that

Let

Sol uti on .

X

p 2m =

· (x 2 + a 2 ) m-1

Since P 2m

----'----=-

(2m _ 2) a 2

I 2 (m- l) =

2 1 + � 2m - 2 a 2 I 2 (m- l ) ' -

p 2 (m-

l) + � _..!.._ I (2m _ 4) a2 2m - 4 a2 2 (m- 2)

and so forth , we see that I 2m

=

p 2 (m- 1) 1 _ p 2m 2m - 3 2m - 2 a2 + (2m - 2) (2m - 4) a4

_ _

p 2) + ( 2m -(2m2) -( 2m3) -(2m4) ( 2m5) - 6) 2 (ma6 -

+

oo lo

• • •

+ (2m -( 2m2) -( 2m3) -(2m4) -( 2m5) - 6) 1

� 2 arc tan a .

Hence

Putting a

(x 2 =

dx +

a2 ) m

1 and x

( 2m ---....,.3),.P. 2m - 5) - 1 1 1T (� *"'2 a 2m- l z · 4) "* ---;:;(2m - 2)..;; (2m - """" z/ liii, we get

(2m - _;3) (2m=---- 5) - 1 1T 2m _-4:;.,)c- -�2 2 · 2�)�(�

(� � 2 m=---

Since

SEQUENCES AN D S E R I ES

31 5

2 2 m z lim (1 z /m) e m+ oo we therefore obtain =

+

rm

(2m(2m -- 3)2) m. By the Formula of Wallis (see Problem (2m(2m -- 3)2) and f�C2m - 1) become infinite in a ratio of equality. Therefore _·.,.3 ·.... 5:-.., -- --7:((2m2,.m_.. --__,.32)f-) m+oo 11m f 2 m+limoo .,.12•4•6 �C2m - 1) and so e-z2 dz 8) ,

1T rm = 2

�

rm

.

Iii = z ·

PROBLEM 147. Show that sn _1_ ___:ol__ • • • -V---;::2 1=:;;:: = + 2 as n + n 2n l(;i � Since 1 � dx Vn2 1 - 1 � Vn2 we see that sn V 2 1 1 2n Vdx2 sn - 1 n n 2n Vr! But the integral dx f 2 dx 1 2n � n x 11 t/n +

=

+

+

_

+

Sol ution .

k

�

+

k -

+

�

�

+

0

1

�

0

+ X

=

0

+

+ k

'

--

"' ·

CHAPTER 3

31 6

l ies between

j02 dt = 2

and

and so s n + 2 as n +

PROBLEM 148 . terms . Show that

(1

oo .

+

r �r� n J

o

2

dt

Let �:= l an be a divergent series of pos it ive decreasing as n +

oo .

Call the numerator and denominator N and D . Clearly N � D , • • • + a 2n- l :<> D . Also

Sol ution .

2 D � D + N - a1

n L a + k=2 k

oo

as n +

oo .

Hence IN - Dl D

PROBLEM 149 . 0

<

an

<

Let a 1 , a 2 , a 3 , . . . be any numbers such that 1 > -

1,

4

(n

1,2,3

• .

.

.

)

.

Show that l imn + oo an = 1/2 . Sol ution .

For any real number x we have x ( l - x)

:<>

1/4 . Hence

that is , the a ' s are monotone increasing . Since they are bounded they ap­ proach a l imit L which must satisfy (1 - L) L � 1/4 . Hence L = 1 / 2 . PROBLEM 150 .

Find l imn + oo n s in (211en ! ) .

SEQUENCES AN D S E R I E S

31 7

Using the famil iar series for e and s in x and the periodicity of the latter, we have Solution .

where

L 1/ (n + 1) (n + 2)

Rn

• • •

k= l

(n + k) .

But 1/ (n + 1)

< R

n

<

k l: 1/ (n + l) k= l

1/n.

Therefore l im R n -r oo n

and

0

l im nR n + oo n

1

and so l im n sin (211en ! ) n + oo

l im n s in (27!Rn ) n + oo l im 2 11nRn

PROBLEM 151 . 71

=

Show that

"' (n ! ) 2 2n+ l (2n + 1) ! ·

Eo

Sol ution .

Using the fact that

�1112 cos 2n+ l x dx we obtain (n ! ) 2 2 n+ l (2n + 1) !

2

(n ! ) 2 2 2n (2n + 1) ! '

fa 1112 2 -n cos 2n+ l x dx

11/2 2 J( ( cos x) (Y, cos 2 x) n dx , o

31 8

CHAPTER 3

and 2 n=O

(n ! ) 2 2n+l = 2 (2n + l) ! =

171 /2 (cos x) { oo2 (� cos 2 x) n f dx n=O 0

( 71/2 (cos x) ( l Jo

� cos 2 x) - 1 dx

71 .

PROBLEM 152. Evaluate 2 e - x dx (x 2 + 2 ·

�)

Sol ution .

f

Applying integration by parts twice, we get

2 2 e -X dx = e -X 2x � (x 2 + 2 x + �

�)

2 J e -X dx . �

-X 2 xe = -2 - + 2 X +

-X 2

- �� x

-X2

2

2 e -X __ + 2 e -X dx dx = ---_,;;----- + _ x 2x(x +

�)

!

The value of the definite integral corresponding to the last term is known to us from Problem 146 , and thus liT.

CHAPTER 4 REAL FUNCT IONS

PROBLEM 1 . The Mean Val ue Theorem states : I f f i s a continuous function in a cl osed interval [a,b] and is differentiable in the open interval (a,b) , then there exists a point c E (a,b) such that f (b) - f(a) b - a

f ' (c) .

We assume the Mean Value Theorem as known and turn to the prob lem of ob­ taining a mean value theorem for higher- order differences . Suppose that f is defined in [a,b] and for x and x + h in [a,b] we define �h f (x) If x

+

=

f(x + h) - f (x) .

Zh is also in [a,b] , then we set

� � f (x)

=

�h { I\ f (x) }

=

f(x

+

Zh) - Z f (x

I f n is a positive integer so that x tively by means of the formula

+

nh

E

+

h) + f (x) .

[a,b] , we define �� f (x) induc­

n- 1 n �h f (x) = �h { �h f(x) } . Prove the fol lowing Mean Val ue Theorem for Hi gher-Order Di fferences : I f a function f is continuous in a closed interval [a,b] and is n times dif­ ferentiable in the open interval (a,b) and if with x E [a,b] we have x + nh E [a,b] for h � 0 , then there exists some e so that 0 < e < 1 and

31 9

320

CHAPTER 4

where f (n) denotes the n-th order derivative of f . For the sake of argument w e shal l suppose that h > 0 and we choose to denote by P (n) the statement of our claim and to proceed by induc­ tion . Clearly, the statement P ( l) is a mere reformulation of the Mean Value Theorem and hence is true . Assume now that P (n - 1) is true for n > 1 and set Sol ution .

6 f(� g (x) = h h The function g is defined and continuous in [a,b - h ] and is n - 1 times dif­ ferentiable in (a,b - h) . I f x + nh E [a, b ] , then x + (n - l)h E [a,b] . Con­ sequent ly, we may apply P (n - 1) to g and find some 8 1 so that 1 (n- 1) (x Anh g (x) = g

+

{ n - 1 } 8 1h) hn-1 ,

o < 81 < 1 .

(E . l)

Now g (n- 1) (x

+

{n - 1 } 8 1 h)

Applying the Mean Value Theorem to the right-hand side express ion , we get 0 <

82

<

1.

I f we set (n - 1) 8 1 + 8 2 8 = -----n then it is clear that 0

< 8 < ' 1 . Further, since (E . 2)

it fol lows from (E . l) and (E. 2) that (E . 3) I f h < 0 , a s imilar argument wil l lead to the same conclusion (E. 3) . Thus we have shown that P (n - 1) impl ies P (n) and the rest fol lows by the principle of mathemat ical induction . This completes the proof of the Mean Value Theorem for Higher-Order Differences .

REAL FUNCTI O NS

321

Since �h x X + h is continuous at x, n � f (x) f (n) (x) = l im h n' �h x + 0 (�h x)

X

Remarks .

h, from formula (E . 3) we get , if

(E . 4)

this j ustifies the otherwise surprising notation

for higher-order derivatives . It does not , however, afford a means of defin­ ing the n-th order derivative ; the l imit on the right-hand s ide of (E . 4) may wel l exist even if f (n) does not exist . For example, let 1 f (x) = X 3 sin X

f(O) = 0,

for x

#

0.

Here f" (0) does not exist since f' (O

+ �h x) �h X

f' (0)

3�h x• sin

1 -- cos � X

h

y;---x_ h ;

1

but �� f (O) c � h x) 2

1 1 8� h x • s in --2 �h X - 2 �h x • sin -�h X

tends to 0 as �h x + 0 . , By Probl em 1 , if f is k times differentiable in (a,b) and continuous in [a,b] , then there exists a t , a < t < b such that (k) �� f (t) = �� f (a) ,

where h

(b - a) /k and

k � h f(a)

k

I j =O

( - 1 ) k- J'

(Jk. ) f(a

+

j h) .

PROBLEM 2. Let c be a real number such that n c is an integer for every positive integer n . Show that c is a nonnegative integer . Sol ution .

The case n

2 shows that c is nonnegat ive . I f the Mean Value

CHAPTER 4

3 22

Theorem (see Problem 1) is appl ied to xc in the interval [u,u + 1 ] there is a t with u < t < u + 1 such that ct c- 1 For any pos itive integer u the right -hand side is a pos itive integer . Now , in case 0 < c < 1 , u could be taken large enough so uc- l < 1/c and so ct c-l < 1. Thus the Mean Value Theorem el iminates all c with 0 < c < 1 . By the Mean Value Theorem for Higher-Order Differences (see Problem 1) if f is a k times differentiable function in [a,b ] , then there exists a t , a < t < b , such that �hk f (k) (t) = �hk f(a) , where h

=

(b - a) /k and

�hk f(a)

k

( - l) k-j (�) f(a + j h) .

I j =O

J

+

We apply this fact to the interval [u,u k ] , where k is the unique integer such that k - 1 � c < k , and obtain that there is a t with u < t < u + k such that c ( c - 1) ( c - 2)

• • •

(c

-

k + 1) t c-k = �hk f (uL

(E . 5)

with u a pos it ive integer. The right-hand side of (E. S) is an integer , and by taking u sufficiently large t c-k becomes sufficient ly small so that the left-hand s ide of (E . S) , though nonnegat ive , is less than 1 . Hence c ( c - 1) (c and so c

-

2)

• • •

(c - k + 1) = 0

k - 1.

PROBLEM 3 . Let f b e a nonzero differentiable funct ion in [a,b] and f (a) = f (b) = 0 . Show that there is a point t in the interval such that

I f ' C t) I > 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)

<

a

t

<

s

M (x

s

M (b - x)

a)

X < S < b.

x,

x

s

a +--b --2 ,

a +--b for --2

s

x

for a

s

s

b,

The function M (x - a) for a s x s (a + b) /2 and M (b - x) for (a + b} / 2 s x s b is not differentiable at x (a + b) /2 . Hence we can not have that f (x} M (x - a) for a s x s (a + b ) / 2 or f (x) M (b - x} for (a + b} /2 s x s b s imultaneous ly. Thus , sett ing m (a + b) / 2 , =

=

=

L

b

or

M >

f (x) dx

4 (b - a} 2

<

M

(

im ( x - a) dx + M ib (b - x) dx

b

}a

M (b - a} 4

2

f (x) dx .

PROBLEM 4 . Let f have a continuous derivative in [a,b] . Are there two points x 1 , x2 with x 1 < x2 for every point t with x 1 , x 2 , and t in [a,b] such that f(x2 ) - f(x 1 } x2 - xl Sol ution .

f (x)

f ' (t} ?

The answer is no in general . Take , for exampl e , t

0.

PROBLEM 5 . Let n b e a pos �t �ve integer . Show that f (n} (x) = 0 , where f(x) (x2 - l) n , has exactly n distinct real root s between - 1 and 1 . Sol ution . Rol le ' s Theorem states : I f f is a cont inuous funct ion in a closed interval [ a , b ] and is different iable in the open interval (a,b} , and if f(a} = f(b) , then there is a point t in (a ,b) such that f ' (t) = 0 . We note that f (x) = (x 2 - l} n (x - l} n (x + l) n and its first n - 1 =

324

CHAPTER 4

consecutive derivatives vanish at x = ± 1 . Thus , by Rol l e ' s Theorem f ' has at least one real root between - 1 and 1 ; again by Rol l e ' s Theorem, f" has at least two distinct real roots between - 1 and 1 , etc . ; f (n- l ) has at least n - 1 distinct real roots between -1 and 1 . Final ly, applying Rolle ' s Theorem to f (n- l ) we obtain the desired result (noting that f (n) is a polynomial of degree n) .

X �

Let f be as in Problem 5 . Show that f (n) (x) is positive for

PROBLEM 6 . 1.

By a theorem o f Leibniz (eas ily verified by induction) , if g and h have derivatives of order n at a point x , then so does f = gh and Sol ution .

n

I

f (n) (x) Taking g (x)

j =O (x - l ) n and h (x)

PROBLEM 7 . e

=

(x + l ) n , we get the desired resul t .

Show the fol lowing theorem of Hermite : The number e , where

l im (1 + n -+ "'

*t,

satisfies no rel ation of the form +

• • •

+ crnern

=

0

(E . 6)

having integral coefficients not al l zero . (Stated otherwis e , e is a tran­ scendental number . ) Sol ution .

Suppose that e is a root of the polynomial +

• • •

+ CrnXrn

with integral coefficients c0 , c 1 , . . . , ern which are not al l zero ; we shal l show that this assumption leads to a contradiction . Let f be a po lynomial of degree n . Then f (n+l ) 0 and repeated inte­ gration by parts yields

REAL FUNCT IONS

ro

b

JC

325

f (x) e- x dx

We put F (x)

f (x) + f ' (x) + • • • + f (n) (x) .

Then F (b) + eb

J'[o

b

f (x) e-x dx .

(E . 7)

In (E . 7) we let b = 0 , 1 , 2 , . . . ,m, successively, and multiply the resulting equations by c 0 , c 1 , c 2 , . . . , em , respectively; final ly, we add up the resulting relat ions and obtain , by (E . 6) , + ••• + m

. + c . eJ J j =l

I

la0

b

(E . 8) f (x) e -x dx ,

which , according to our assumption , must hold for every polynomial f (with integral coefficients) . We shal l show that there is in fact an f for which (E . 8) is fals e ; this wil l , of course , prove our claim . To this end we put f (x) = (p _1 l) ! xp-1 (x - 1) p (x - 2) p

where p is an odd prime larger than both m and i c 0 1 . The derivatives of or­ der p and l arger of this polynomial have integral coefficients which are divis ible by p ; this is an immediate consequence of the fact that the prod­ uct of p consecutive integers is divisible by p ! (see Remark to Prob lem 22 in Chapter 1) . Since the polynomial f and its first p - 1 derivatives vanish for x = 1 , 2 , . . . ,m, we have F ( l) , F (2 ) , . . . , F (m) are integral multiples of p. But the situation is different for F (O) . For x = 0 , besides f (x) only the first p - 2 derivatives vanish and so F (O)

f (p- l) (0) + f (p) (0) + • • •

holds . Al l summands beginning with the second summand are integral multiples of p ; but , s ince

F (O) is not divisible by p. Since p is a prime number both larger than m

CHAPTER 4

326

and lc0 1 , it follows in particular that c0 is not divisible by p. We there­ fore see that the first sum in (E.S) , namely, is not divisible by p and thus can be equal to zero. We now turn our attention to the second sum in (E.S). the interval [O,m] we obviously have m(pmp+p-1 - 1)! . Thus j mCpmp+p-l - 1) .r ) 0 e -X dx m(pmp+p-1 - 1) ! and, on setting not

On

r

<

we obtain j f(x) e -x dx C em mmp+p-1 C em (mm+l ) p-1 j e c la . (p l) ! (p I I j=l� J 0 By Problem in Chapter the last factor tends to zero as p "' · ·Thus, by taking p large enough, the absolute value of the second sum in (E.8) can be made as small as we please. In other words, by choosing p suitably large, the total sum on the right-hand side of (E.8) cannot be equal to zero. We have thus arrived at a contradiction. <

L

2

3,

_

_

l) ! . +

#

PROBLEM 8. Give an example of a function f discontinuous at all x 0, but differentiable at x = 0. Let f(x) = x for rational x, including x 0, X X for irrational x. Actually, f' (O) = 1. Sol uti on .

2

+

PROBLEM Let f be a differentiable function in the interval [a,b] 9.

REAL FUN CT IONS

327

and suppose that f ' ( a) < y 0 < f ' (b) . Show that there is a point x0 in the open interval (a,b) such that f ' (x0 ) y0 .

=

We first consider the special case where f ' (a) < 0 , f ' (b) and show that there is an x in (a,b) such that f ' (x) = 0 . We note that since f i s different iable it must be cont inuous . I t accord­ ingly attains its smal lest value on [a, b ] . S ince f ' (a) < 0 there are points x1 E (a,b) with f (x 1 ) < f (a) . S ince f ' (b) > 0 there are points x 2 E (a,b) with f(x2 ) < f (b) . Thus the least value of f in [a,b] is attained at an x E ( a , b) . But then f ' (x) = 0 . Now suppos e f i s different iable and only that f ' (a) < y0 < f' (b) . We y0 . show that there is an x0 E (a,b) such that f ' (x0 ) f (t) - y 0 t . Then Cons ider the auxiliary function g (t) Sol ution .

=

g ' (a)

f ' (a) - y 0

=

<

0

g ' (b)

and

f ' (b) - y > 0 . 0

S ince g sat is fies the conditions of the special case , there is an x0 E (a,b) for which g ' (x0 ) 0. Now f ' (x0 ) g ' (x0 ) + y0 = y0 , estab l i shing the des ired result . =

=

PROBLEM 1 0 . Give an example of a function which i s differentiable everywhere but whose derivative fails to be cont inuous at some point . Sol ution .

f (x)

=

X

Consider the function

1 2 s in X

0

for X "f 0 , for x

Here f' (x)

0.

1 - cos 1 2x sin X X

for X "f 0 ,

0

for x

0.

Since l im cos X1

X + oo

does not exist , f ' i s not continuous at x

0.

CHAPTER 4

328

PROBLEM 1 1 . Let [a,b] be a given interval . A any finite set of points x0 , x 1 , . . . , xn such that

We write , for 1

�

k

�

partition

P

of

[a,b] is

n,

and denote by

! P I is called the mesh of the parti tion P . A part it ion P * o f [a,b] is cal led a refinement of a partition P o f [a,b] if every point of P is a point of P* . Given two part it ions P 1 and P 2 of [a,b ] , w e cal l P * = P 1 u P 2 their common refinemen t . Let f be a bounded real-valued funct ion on a bounded closed interval [a,b] . Corresponding to each partition P of [a,b] we put

and define the tively, by

upper

U (p , f)

and the

l ower Darboux sums of f rel ative to

and

P , respec­

L (P, f)

finally we put

/b f (x) dx a

b

i

f(x) dx

inf U (P , f) ,

(E . 9)

sup L ( P , f) ,

(E , 1 0)

where the infimum and the supremum are taken over al l part itions P of [ a ,b ] . The left-hand members of (E . 9) and (E. lO) are cal led the upper and the lower Riemann integrals of f over [a,b ] , respect ively . I f the upper and lower Riemann integrals are equal , we say that f is Riemann integrabl e on [a,b] and we denote the common value of (E . 9) and (E . l O) by the symbol

Iab f (x) dx .

329

REAL FUNCT IONS

To see that the upper and lower Riemann integrals exist for every bound­ ed function f on a closed bounded interval [a,b], we observe that the numbers L(P,f) and U(P,f) form a bounded set. Indeed, since f is bounded, there are two numbers, m and M, such that m f(x) M for a x b; hence, for any partition P, we have m(b - a) L(P,f) U(P, f) M(b - a) . Evidently, Lb f(x) dx /a b f(x) dx and if P* is a refinement of P, then L(P,f) L(P*,f) and U(P*,f) U(P,f). Show that f is Riemann integrable on [a,b] if and only if for any there is a partition P such that (E .11) U(P,f) - L(P,f) For any P we have L (P, f) ib f(x) dx /a b f(x) dx U(P,f). Thus (E.ll) implies fa b f(x) dx - Jab f(x) dx Consequently, if (E.ll) holds for any £ then -b fa f(x) dx Jab f(x) dx, implying that f is Riemann integrable on [a,b]. Conversely, suppose that f is Riemann integrable on [a,b], and let be given. Then there are partitions P 1 and P2 of [a,b] such that (E .12) f(x) dx and (E .13) �b f(x) dx - L(P1 ,f) ��

�

�

�

�

�

�

�

�

�

£

> 0

< £.

Sol ution .

�

�

�

0 �

< £.

> 0,

=

r

£

> 0

£ < 2

<

CHAPTER 4

330

Let P be the common refinement of P 1 and P 2 . Then , by (E . l 2) and (E . l 3) , we get U (P , f)

�

U (P 2 , f) <

J(ab f (x) dx

+

thus (E . l l) holds for the partition P

�< P1

L (P 1 , f) u

+ E �

L ( P , f)

+ E;

P2 .

PROBLEM 1 2 , Let f be a bounded real-valued function defined on an in­ terval J ; we define w (f;J) = sup{ f (x) : x E J } - inf{ f(x) : x E J }

and cal l w (f ; J) the oscillation of f over J . Let a and w be arbitrary positive numbers and f be a bounded function on an interval [a,b] of finite length . Show that f is Riemann integrable on [a , b ] if and only if there is a mode of division of [ a , b ] into subintervals such that the sum of the lengths of the subintervals in which the osci llation of f is greater than or equal to w is less than cr . Let P = { a [a , b ] and consider the sum Sol ution .

x0 < . x1 < x2 < • • • < xn

b } be a partit ion of

Z (P , f)

(that is , wk is the oscillation of f in [xk-l ' xk ]) . We let Q

where

M - m,

M = sup{ f (x) : x E [a,b ] } ,

m = inf{ f(x) : x E [ a ,b] } ,

and denote the length of the interval [a,b] by L . W e now derive bounds for Z (P , f) ; incidental ly , Z (P , f) equals U (P , f) - L ( P , f) . Let o be the sum of the lengths of the subintervals obtained by the partition P in which the oscillation of f is greater than or equal to w . Then

331

REAL FUN CT I ONS

Z (P , f) �

ow.

But i n these subintervals the oscillation o f f is l e s s than o r equal to 0 and in the remaining subinterval s (the sum of whose lengths is L o) the oscillation of f is less than w . Thus -

Z (P , f) S ince L

s oO +

o s

-

ow s

(L -

o) w .

L , w e see that s oO +

Z (P , f)

LO .

I f f is Riemann integrable on [a, b ] , then by Problem 1 1 there exists a partition P such that for any preassigned posit ive numbers a and w we have

<

Z (P , f)

wa .

But then o w < wa , that is , o < a . Converse ly, i f there exists a part ition P for which w

=2

E:

and

L

Then Z (P , f)

s oO +

PROBLEM 1 3 . l im

P + oo

la

0 LO

=

<

20"

o

<

a,

we choose

E:

E / 2 + E/ 2

E.

Let f b e Riemann integrable on [a,b] . Show that

b

. px f(x) Sln cos px dx = 0 . a

Sol ution .

For any bounded interval [ a , S ] we have

I J( S sin px dx l = I eo s pa � cos pS I

�-

s

Let \ f(x) \ s A on [a,b] and be the usual arbitrary positive number. There is a partition of [ a ,b] , say such that U (P , f) - L ( P , f) < E/2 . Thus

\J(b f(x) sin px dx l n

s

I

k=l

\ fx

s in px dx +

xk k-1

\ f(x) ­

3 32

CHAPTE R 4 <

2nA p

[U (P , f) - L ( P , f) ]

+

<

2nA p

---

+

£ 2

-

<

£

when p > (4nA) /£ . Hence �b f (x) s in px dx tends to 0 as p + oo . In the same b way, J[a f (x) cos px dx tends to 0 as p + oo.

PROBLEM 14 . Every rati onal number x can be written in the form x p/q, where q > 0, and p and q are integers without any common divisors . When x = 0 , we take q 1 . Consider the function on the interval [ 0 , 1 ] defined by =

f (x)

=

0

for x irrat ional ,

1 q

for x

£

q·

Then f is continuous at every irrational point o f (0 , 1) and discontinuous at every rational point of ( 0 , 1) . Indeed , let x0 be any point of (0 , 1 ) . Given £ > 0 , there is only a fi­ nite number of posit ive integers q that are not larger than 1/£ and this means that in (0 , 1) there are only finitely many rational points p/q for which f (p/q) 1/q � £ . Thus one may construct around the point x 0 a neigh­ borhood (x0 a , x0 + a) with a > 0 such that in this neighborhood there is no point x for which f (x) � £ (except poss ibly the point x 0 itself) . Thus , if 0 < l x - x0 J < a , then for both rat ional and irrat ional x we have j f(x) I < £ . Letting , for h > 0 , =

lim f(x0 h + oo

+

h)

and

l im f (x0 - h) , h + oo

w e get

for every point x0 • I f x 0 is irrational , then f (x 0 ) = 0 , that is , f is con­ t inuous at xo ; if xo is rational , then f (xo ) ' 0 , that i s , f is discont inuous at x0 . Show that f is Riemann integrable on [0 , 1 ] . Sol ution . Let [ 0 , 1 ] be partitioned into subintervals of leng ch �x � A k by a partition P with j P j A . We pick an arbitrary posit ive integer I . The sub intervals fal l into two classes : 1 . To the first class belong those intervals that contain the points p/q with q at most equal to I ; s ince there are only finitely many such points =

REAL FUNCT IONS

333

p/q, say k = k 1 , there are only at most 2k 1 such interval s ; their total length does not exceed 2k 1 A . 2 . To the second class belong those intervals that do not contain the points mentioned in class 1 ; over these interval s the osci l l ation � - � of the function f is clearly smal l er than 1 / 1 . We now observe that U (P , f) - L (P , f) Taking I > 2 / E and then A < E/ (4k 1 ) , we get n L (M - mk } fi xk < £ . k=l k Thus f is Riemann integrable on [0 , 1 ] by Problem 1 1 . Remarks . Noting that there i s only a finite number o f points at which the functional value of f exceeds an assigned pos itive real number, it is cl ear that � l f(x) dx = 0 . We can also see that if

g (y)

1

for 0 < y ::; 1 ,

0

for y

0,

then the function h (x) = g [f(x) ] for x E [0 , 1 ] is not Riemann integrable over [0 , 1 ] , because h (x) = 1 for ra­ tional x and h (x) = 0 for irrational x; however , the upper Riemann integral of h over [0 , 1 ] equal s 1 and the lower Riemann integral of h over [ 0 , 1 ] is equal to 0 .

PROBLEM 1 5 . Let f b e Riemann integrable on [a,b] and g b e a pos it ive­ valued, bounded , and nonincreasing function in [a ,b] . We denote by g ( a+) and g (b-} the l imits lim g ( a + h) h+O

and

l im g (b - h) , h+O

respectively, where h > 0 . Show that

J(b f (x) g (x) dx = g (a+) J(t f (x) dx

for some t

E

(a,b} .

CHAPTER 4

334

I f g is positive-valued and nondecreasing, the corresponding formula is

i

b

f (x) g (x) dx = g (b-)

f

b

f (x) dx ,

where a < t < b . I f g i s any monotonic function , there i s a number t b etween a and b such that b

} (a <

f(x) g (x) dx

g ( a+)

it f (x) dx + g (b-) ib f (x) dx .

Let £ be a pos itive number less than g (a+) - g (b-) . Then there is a point x 1 such that Sol ution .

g (a+) - g (x)

<

£

Similarly there are points x 2 , x 3 ,

such that

g (xk- 1 +) - g (x) < £

so long as g (xk _ 1 +) - g (b-) > £ . Otherwise we take xn = b . The point b is thus reached in a finite number of steps , since the variation of g in each interval (xk- l ' xk ) is at l east £ . Let h (x) g (xk +) in each interval xk s x s xk+l ' Then s

0

h (x) - g (x) < £

except possibly at the points a = x , x 1 , x2 , 0

�

b

h (x) f (x) dx

n-1

L

k=O

g (xk+)

f

xk + 1

xk

.

•

.

, b , and

f(x) dx .

x Let F (x) = J(a f(s) ds ; then , if m and M are the lower and upper bounds of F , it follows from Abel ' s Inequality (see Problem 58 in Chapter 2) that m g (a+) But

s

i

b

h (x) f(x) dx

s M

g (a+) .

335

REAL FUNCTIONS

liab h(x) f(x) dx - Jab g(x) f(x) dxl Jab lf(x) dx, which tends to zero with Hence, making £ 0, it follows that m g(a+) ib g(x) f(x) dx M g(a+) . Since F is continuous, it takes every value between m and M, and so, at every x = t, say, the value b g(x) f(x) dx. ; � g( +) The rest follows easily. :>

+

£.

:>

I

£

:>

PROBLEM 16. Let E , E , ... , E be n intervals which are situated in the unit interval [0,1].1 If2each pointn of [0,1] belongs to at least q of these intervals E j with j = 1, ... , n, show that at least one of these in­ tervals must have length q/n. For x [0,1], define f.(x) 1 if x E . and f. (x) 0 if x E.; then let f(x) j=n l f . (x) . Evidently, f(x) q for every x in the interval [0,1] and so q Jn{0 1 f(x) dx Jc{0 1 j=lI f. (x) dx = j=lI fo 1 f . (x) dx where IE. denotes the length of the interval E .. It is clear that not every summand in the last sum can be less than q/n, for if it were then we would have q n(q/n) . �

�

E

Sol ution .

E

J

J

J

J

�

J

�

=

:>

J

I

J

J

J

<

PROBLEM 17. Let f be a continuous function in the interval [0,1] and suppose that !01 f(x) dx 0, f0l x f(x) dx 0,

336

CHAPTER 4

xn-1 f(x) dx 0, J( l xn f(x) dx 1. Show that if(x) 2n (n + 1) on some part of [0,1]. The conditions imply Jf:l (x - �) n f(x) dx 1. Suppose l f(x) < 2n (n + 1) everywhere on [0,1]. Then 1 J( l (x - �) n f(x) dx < 2n (n + 1) J( l l x - � � n dx 1. But this is a contradiction. =

I �

Sol ution .

=

I

=

PROBLEM 18. Consider a polynomial f(x) with real coefficients having the property f[g(x)] g[f(x)] for every polynomial g(x) with real coefficients. Show that f(x) x. Let g(x) x + h so that f(x + h) f(x) + h and f(x + h)h - f(x) 1 for all h 0. Therefore f' (x) 1 and f(x) x + c. Letting g(x) be the zero polynomial shows that it is necessary that c 0 and thus f(x) x. It is easily seen that f(x) x is sufficient. =

Sol ution .

=

F

=

=

=

=

=

PROBLEM 19. Show that there exists1 a real-valued, everywhere differ­ entiable function on the real line R such that is monotone on no sub­ interval of R1 and is bounded. In other words, a function may be every­ where oscillating and still have a finite derivative at every point. We divide the proof into seven steps. I. Let r and s be real numbers. 2 2 (i) If r s 0, then (r - s)/(r - s ) < 2/r. (ii) If r 1 and s 1, then (r + s - 2)/(r2 + s2 - 2) < 2/s. H

H

H'

Sol ution .

> >

>

>

REAL FUN CT IONS

337

Assertion (i) is obvious. Inequality (ii) is equivalent to (r - s) 2 + (r - l) (s - 1) + r2 r + 3s But this too is obvious when r 1 and s 1. II. Let v(x) (1 + ) -� for x R1 . Then b v(x) dx min{v (a),v(b)} � b�a whenever a and b are distinct real numbers. We may suppose that a b. In case a, we have by step 1 that 2 ((1 + b) -- (1 + a) min{v(a),v(b)}. Since v(-x) = v(x) , the case b needs no special consideration. We there­ fore suppose that a b. Then step I yields 1 ( b v(x) dx 2 (1 + b) ++ (1 - a) - - 2)2 min{v(a) ,v (b)} . Ja III. If v is as in step II and w is any function of the form w(x) j=ln c. v[:\. (x where c1 , . . . , en and t-1 , ... , '-n are positive real numbers and a.1 , . . . , an are any real numbers, then b b � a i w(x) dx min{w(a) ,w(b) }. This follows directly from step II and the fact that 1 lb v[t-(x - a)] dx = :\(b a) 1 :\(a a) 1 '-(b-a) v(t) dt. :\(a-a) a IV. Let (wn) �=l be any sequence of functions as in step III. For x R1 and each n define Wn (x) = }ro x wn (t) dt. Proof.

>

+

>

5.

>

E

X

< 4

Proof.

v'l+b �

<

0

II+a) <

�

4

4

0

< 0 <

( v'l+b

�

l:

J

1["-=-a

< 4

J

< 4

Proof.

�

_

_

_

E

CHAPTE R 4

338

s < oo for some a E R1 . Then the series F (x) n=lI Wn (x) converges uniformly on every bounded subset of R1 , the function F is differentiable at a, and F' (a) = s. In particular, if n=lI wn (t) f(t ) < for all t E R1 , then F is differentiable everywhere on R1 and F' f. Let b E R1 satisfy b Ja!. Then, using step III, -b x b implies 00

$

�

Proof.

$

Thus, uniform convergence on [-b,b] follows from the M-Test of Weierstrass. To show that F' (a) = s, let £ 0 be given. Choose n0 such that 10 n=nI +1 wn (a) < £. 0 Since each wn is continuous at a, there exists some 0 such that >

8

>

whenever 0 l h l and 1 n n0 . Therefore, using step III again, 0 < lh < implies that + h � - F(a) _ s F(a I l _l hl faa+h wn (t) dt - wn (a) I + £ <

8

< - +

2

< 8

$

$

l

339

REAL FUNCT I ON S

V. Let I 1 , In be disjoint open intervals, let a.J be the midpoint of I j , and let £ and y1 , . . . , yn be positive real numbers. Then there is a function w as in step III such that for each (i) w(aJ.) yj ' (ii) w(x) yj + £ if I j ' (iii) w(x) £ if x i I l In Choose c.J y.J £/2 and define v.J (x) to be v.J (x) = c.J v[A.J (x - a.)], J where A.J is chosen so large that v.J (x) £/2n if x i I.J (one needs only to check this inequality at an endpoint of I.)J . Take >

X

<

E

u

<

u

+

Proof.

<

Since the I.'sJ are disjoint and because v.J takes its largest value c.J at aJ. properties (i), (ii), and (iii) are clear. VI. Le� {aj }j=l and {Sj }j=l be disjoint countable subsets of R1 . 1Then there exists a real-valued, everywhere differentiable function F on R sat­ isfying F' (a.)J = 1, F' (S.)J 1 for all j , and F' (x) 1 for all x. We obtain F as in step IV by first constructing F' f n=l wn ' or, more precisely, the partial sums fn k=ln wk' in such a way that fn (a.)J 1 .!.n n) ' 0

<

Proof.

L

L

>

-

wn (SJ.) 2n·21 n

< ---

(1

�

�

( j n) . 1

�

�

<

�

CHAPTER 4

340

Suppose that this were done we would have (a.) = n1+ooim fn ) = 1, (x) = nlim+oo fn (x) 1, and, picking n j, + k=nL wk 1 - 1 + k=n"L � 2k•2k J n- 1 1 - n + _!_2n . 1 = 1 - 2n 1 and thus we would have the desired We proceed inductively. First choose an open interval I with midpoint a1 such that 81 I. Then apply step with £ y 1 = 1/4 to obtain f1 = w1 that satisfies (A1 ) , (B 1) , and (C1 ). Suppose that n 1 and that fn-l and wn-l have been chosen which sat­ isfy (An _ 1 ) , (Bn_ 1 ), and (Cn_ 1). Select disjoint open intervals I 1 , . . . , In such that, for each j {l, . . . ,n}, aj is the midpoint of I j , I j being dis­ joint from {81 , . . . , 8n } and fn- 1 (x) fn- 1 (a.) + o for x I j , where n(n 1 1) _2n•21_n We now apply step with 1 and y. 1 - n - fn- 1 (a.) when 1 n, to obtain wn . It is clear that (Cn) is satisfied. Also F'

(a .

J

J

0 < F'

:<;

>

F • co . ) = f �

<

"

co ) � J-

1:.

(o ) <

_!_ <

n

� J·

F.

i

=

V

>

E

<

J

E

> 0.

+

V

1:.

J

:<;

J

�

when 1 j n, and so (An) is satisfied. To check (Bn), observe that if x I j ' then fn (x) fn_ 1 (x) wn (x) fn- 1 (a.) + o + y. + £ 1- 1 :<;

:<;

E

+

<

n+l ;

J

J

341

REAL FUNCTI ONS

while if x i unj=l I j , then VII. Let {aj }j=l and {Bj }j=l be dis­ joint subsets of R . Apply step VI to obtain everywhere differentiable func­ tions F and on R1 such that F' (a.) (B.) = 1, (a.) 1, F' (x) (x) for all j and x. Now put =F Then (a.) (B.) -1 (x) 1 for all j and x. Since {aj }j=l and {Bj }j=l are both dense in R1 , cannot be monotone on an interval. Proof of the Claim in Problem 1 9 : 1

G

G'

J

:-;;

0 <

H

H'

-

J

G'

J

0

1,

<

G'

<

J

:-;;

1

G.

> 0,

H'

< 0,

J

<

H'

<

H

PROBLEM A set of real numbers is said to be if for every x in there is an open interval I (x - o,x with o such that I is contained in I is called a x. Show that any open set of real numbers is the union of a countable col­ lection of disjoint open intervals. This representation is unique up to the order in which the terms appear in the union. If is an open set of real numbers, we say that the real numbers x and y are "equivalent" and write x y if and only if there is an open interval (a,b) contained in such that x and y are points of (a,b) . The relation is clearly an equivalence relation on and the resulting equivalence classes are disjoint open intervals whose union is The fact that there can be only countably many follows since each must contain a dis­ tinct rational number. Finally, if there were two distinct representations of in terms of countable unions of disjoint open intervals, then there would exist a point x in which would belong to a component interval I in the first represen­ tation and to a component interval I in the second representation. But 20.

G

G

G;

Sol ution .

open

=

> 0

+ o)

neighborhood of

G

�

G

G

�

G.

G

G

J

1

CHAPTER 4

342

then one of these component intervals, e.g., J, would extend beyond the other; it would follow from this that one of the endpoints of I belongs to J , which is impossible, inasmuch as the endpoints of I do not belong to G.

PROBLEM 21. real number x is called a F (of real numbers) if every neighborhood of x contains a point y x such that y be­ longs to F; F is said to be a if every limit point of F is a point of F. Let f be a real-valued function defined on the real line R1 . Let J be 1 any bounded open interval in R . We define1 w(f;J), that is, the oscillation of f over J, as in Problem 12; for a R , we put w(f;a) = inf w(f;J) , where the infimum is taken over all bounded intervals J containing the point a. One can easily see that if f is1 continuous at a R1 , then w(f;a) and if f is not continuous at a R , then w(f;a) Suppose that r and let Er be the set of all a R1 such that w(f;a) 1/r. Show that Er is a closed set. Let x be any limit point of Er . We have to show that x be­ longs to Er ' that is, we must verify that w(f;x) 1/r. To do this, it is enough to show that if J is a bounded open interval containing x, then w(f;J) 1/r, since w(f;x) is the infimum of such w(f;J) . Since the open interval the limit point x of Er , J must contain a point y of Er But then J containsw(f;y) 1/r and we have what we set out to do. w(f;J) Let denote the set of all points x at which f is discontin­ uous. Then, letting E l/n = {a R1 : w(f;a) 1/n}, n=l E l/n " Indeed, if x then w(f;x) For some positive integer n we must have w(f;x) 1/n. This shows that n=l E l/n" A

limi t point of a set

f

closed set

E

E

E

> 0.

E

> 0

�

Sol ution .

�

�

�

�

D

Remark .

D

�

D

E

U

E

c

U

D,

0

> 0.

�

343

REAL FUNCT IONS

Conversely, if x n=l E l/n ' then w(f;x) and so x Hence n=l E l/n and is seen to be a countable union of closed sets. E

U

E

> 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 <

A

a

+

A

<

and we can find closed intervals J1 , . . . , Jn so that for j 1, . . . , n, I. J. and j=lnl: JJ. J b - a At each point t0 of J 1 we have w(t0) 1/K and there is some o(t0) such that, on the interval [t0 - o(t0),t0 + o(t0)], the supremum of f(t) minus the infimum of f(t) is less than By the compactness of J1 , finitely many J

c

J

J

E.

>

<

E.

< E,

REAL FUN CT I ONS

349

such intervals cover J1 . We can use the endpoints of these intervals to form similarly, we form partitions P2 , , Pn of J2 , a partition P1 of J1 , and . . . , J , thus obtaining a partition P of [a,b] which is the common refinement of pl ' n . . . n · Calculating U(f,P) - L(f,P) , we have U(f,P) - L(f,P) = L �) (tk - tk-1 ) - �) (tk - tk-1 ) f � L (Mk - � )(tk - tk-1 ) + P ) n •

.

.

' p

(� -

p

••• +

l

L (�

� k=lnL !Jk l + £2M � £(2M + b - a), where M = sup{if(t) t [a,b]}. It follows that f is Riemann integrable over [a,b]. I:

E

PROBLEM 28. Using the result in Problem 27, show that there exists a function which is differentiable at every point of an interval and whose derivative is a bounded function, but this derivative is not Riemann inte­ grable. The following construction is due to V. Volterra. We remove from the middle of the closed interval [0,1] the open interval of length 1/4. From the middle of the two remaining closed intervals we remove open intervals ofn-llength 1/16. In the n-th step we remove from the middle of each of the 2 closed-nintervals which remain after the (n - 1)-th step an open interval of length 4 . Continuing this process indefinitely we remove from the inter­ val [0,1] a system of open intervals of total length 1 The points which remain form a closed set E; E is clearly not of measure zero. Let dn denote the length of each of the closed intervals which remain after the n-th step. From the construction it is clear that dn+ 1 dn/2; hence dn 0 as n This implies that no subinterval of [0,1], no matter how small, can be contained in the set E. Sol ution .

z·

<

+

+ oo .

CHAPTER 4

350

Now we define the function f. At the points x of the set E we put f(x) 0. If (a,S) is one of the removed open intervals we define f(x) (x S1n. -1 a immediately to the right of a, and f(x) (x - S) 2 sin S -1 immediately to the left of S, until we reach the maximum points a, nearest the middle of (a,S); in the interval we set f(x) equal to this maximum value. We have thus defined the function f on the entire interval [0,1] and it is a continuous function. It is clear that f is differentiable in each of the removed intervals (a,S) , even at the points where it has derivative zero. For near e­ nough to a (x a) we have f' (x) 2 (x - a) s . 1 - c s 1 if x a the first term of the right-hand side converges to 0 while the sec­ ond term oscillates between the values + 1 and -1. The situation is analogous in a left-hand neighborhood of S. f' (x) exists at the points a, S and even at every point x0 E, and we have (E .14) To show this, we assume first that x x0 . If x E, then f(x) - f(x0) o - 0 o , and if x is contained in some of the removed intervals (a,S) then (x - a) 2 (x l f(x) thus in each case f(x f (x I x) xo o) I $ - xo We get the same result for x < x0 . Letting x x0 this implies equation (E.l4) . =

--­

=

-----­

X

X

8

(a , 8)

a, 8

X

>

1TI x--:-c;:

+

0

x--:-c;:;

E

E

>

=

=

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 ,

<

L

�

>

E

>

J l

J

!

<

f

4 7f

> 5

<

-1

±1,

1

J

< E;

355

REAL FUNCT I ON S

m-1I �P P I � = P n=l n m-1 n=l m-1· (To see this we make use of the inequality ! sin tl I t I and the result in Problem of Chapter 3, that is, Abel's Inequality.) (b) All terms of the series for Q(h') with index n > 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 <

<

58

0,

a

4

'Ill

4

0.

,

�

"

= /:2

Hence therem are values h' , as small as we please in absolute value, for which are values h' , arbitrarily small in absolute value, for Q(h') > 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 <

Remarks .

PROBLEM 33. A function f defined on [a,b] is said to have the provided the closed interval from f(x) to f(y) is contained in the image under f of the closed interval from x to y for each inter­

mediate val ue property

356

CHAPTER 4

x and y in [a,b]. One of the more important properties of continuous functions is that they possess the intermediate value property. From the result in Problem we see that if f' exists on [a,b], then f' has the intermediate value prop­ erty. If f and g are continuous on [a,b], then f + g has the intermediate value property since f g is continuous. If f' and g' exist on [a,b], then f' g' possesses the intermediate value property because f' + g' is the derivative of f g. Give an example of two functions f and g that have the intermediate value property but f g does not. Let F (t ) t2 S l.n 1 if t if t and G(t) t 2 cos t if t if t Then F (t) 2t sin t cos t if t if t and G' (t) 2t cos + S l.n if t if t If f(t) [F' (t)] 2 and g(t) [G' (t)] 2 then f and g have the intermediate value property because F' and G' do. But (f + g) (t) if t if t Hence f g does not possess the intermediate value property on any interval containing 9

+

+

+

+

Sol uti on .

=

.

t

.,. 0

0

0

!

.,. 0

0

0.

1

I

.,. 0

1

0

0

t

1

.

t

.,. 0

1

0

0.

=

=

.,. 0 ,

0

+

0.

0.

REAL FUNCTI ONS

357

E

Let, for x [0,1], h(x) x if x is rational, 1 if x is irrational. This function is seen to be a one-to-one mapping between the interval [0,1] and itself, and hence satisfies the intermediate value property. the other hand, the function is clearly discontinuous, except at x 1/2. Remark .

=

-

X

On

=

PROBLEM 34. Let f be a strictly increasing function with a continuous derivative on a compact interval [a,b]. Integration by parts gives lab f(x) dx bf(b) - af(a) - lab x f' (x) dx. (E.lS) Let y f(x), f-1 (y) ; then (E.lS) can be written (E .16) ib f(x) dx bf(b) - af(a) - lf(a)f(b) f- 1 (y) dy. It is easy to see that (E.l6) remains valid in case f is only assumed to be strictly increasing and continuous on [a,b]. Indeed, if f is strictly increasing and continuous on [a,b], then it admits an inverse function f-l of the same type. Moreover, we may assume without loss of generality that the graph of f is situated in the first quadrant of the x,y-plane. But (E.l6) permits a simple geometrical interpretation in terms of areas of regions represented by the integrals and the quantities bf(b) and af(a) viewed as areas of rectangles. The desired extension is therefore immediate. Young's Inequality (see Problem 74 of Chapter 2) says that when f is s strictly increasing continuous function with f(O) 0, f(a) b, and a 0, b 0, then (E .17) Give a proof of (E.l7) based on (E.l6) . For 0 r a it is obvious that (a - r)f(r) ia f(x) dx. =

X =

=

>

Sol ution .

<

<

<

1

>

CHAPTER 4

358

Writing this as af(r) - loa f(x) dx < rf(r) - lor f(x) dx, we apply (E.l6) to the integral on the right. We get af(r) - }of a f(x) dx < J{o f C r) f- (y) dy. If < b < f(a), we can take r f (b), and (E.l7) follows. 1

-1

0

PROBLEM sequence (fn) :=l of functions defined on an interval [a,b] is said to be if there exists a constant such that I fn (x) I < for all x [a,b] and all n. Prove the following theorem, due to Osgood: If(f ) := l is a uniformly bounded sequence of continuous functions on a closed nbounded interval [a,b] converging (pointwise) to a continuous function f on [a,b], then b b b lim ( fn (x) dx = }( lim fn (x) dx = { f(x) dx. }a n +oo }a a n+oo To prove the theorem, we first observe that the general case can 00be reduced to the special case in which it is assumed that all functions {fn }n= are nonnegative and f Indeed, since I fn (x) I < for all x [a,b] and all n, we have that I f(x) I for all x [a,b] and so I fn (x) - f(x) I < for all x [a,b] and all n. Since fn (x) -+ f(x) , we see that Ifn (x) - f(x) I -+ since fn and f are continuous functions on [a,b], so are the functions I fn - fl . If therefore the theorem holds under the assumptions made in the special case, then the integral of I fn - fl over [a,b] tends to zero as n -+ oo and, using the estimate b b Ifa fn (x) dx fa f(x) dx l 35 .

A

G

uniformly bounded

E

G

Sol ution . 1

=

0.

E

G

:>

G

2G

E

E

0;

-

> 0

REAL FUNCTIONS

359

we obtain b f (x) dx { b f(x) dx. lim 1 n-+oo a n J�a It therefore remains to show that for a uniformly bounded sequence of continuous, nonnegative functions fn which converge to zero along the inter­ val [a,b] we have b lim n-+oo fa To begin with we also impose the further restriction that for all n for all x [a,b], that is, we assume that the sequence is monotonically decreasing. Then, by a well-known theorem of Dini, the convergence is uniform and termwise inte­ gration is permitted. Without Dini's theorem we may reason as follows: The integrals form a monotonically decreasing sequence of nonnegative numbers and converge therefore to a limit 1 0. Dividing the interval [a,b] into two equal parts, the same holds for0both subintervals, and for at least one of them the corresponding limit 1 1 satisfies 1 1 1 0/2. Continued bisection of intervals generates an infinite sequence of nested intervals which have a single point x* in common; for all these intervals the corresponding lim­ its satisfy =

E

�

�

the other hand, let be an arbitrarily small positive number; since fn (x) -+ 0, there is an integer n such that

On

�

moreover, by the continuity of fn ' there exists a neighborhood x* o < x < x* o of x* such that for all points x of this neighborhood -

+

holds. Then for all points x of the neighborhood I fn Cx) I < �.

CHAPTE R 4

360

However, for sufficiently large values of k the intervals we considered in the bisection process further above must be contained in the neighborhood (x* - o,x* + o); thus for these intervals b ---a . I k E ---k 2 Hence Io I E b - a 2 k ---k2 --' that is I 0 E (b - a) . Since E may be chosen arbitrarily small, while I 0 is a fixed nonnegative number, it follows that 1 0 0. Thus the considered spe­ cial case is settled. We now suppose that (fn) :=l is a uniformly bounded sequence of contin­ uous, nonnegative functions which tend to zero along the interval [a,b]. We denote by �(x) (where m n) the largest of the values fm (x) , fm+ 1 (x) , n. . . , m fn (x); the function �m isnclearly continuous on [a,b]. The integral of fm over [a,b] we denote by Im . Holding m fixed and letting n tend to infinity, the numbers I� form an increasing sequence which is certainly bounded, since if the fk (x) nare below a bound G, then the �(x) must also be below the bound G and thus Im < (b - a)G. Therefore the sequence in question tends to some limit Jm . Since fm (x) �(x) m for all x [a,b], it is evident that Jm . If we therefore succeed in showing that Jm 0 as m then we may conclude that <

�

<

�

<

=

�

E

�

+

+ oo ,

and the proof would then be finished. Now, let a be an arbitrarily small positive number and let n n be such that In1 J1 - a/2, for example, the nsmallest such number, then let1 n = n2 a number larger than n 1 such that 1 2 J2 - a/4nand so forth,m in general, let n = nm a number larger than nm- such that Im Jm - a/2 . Finally, =

>

1

>

>

36 1

REAL FUNCT I ONS

denote by gm (x) the smallest of the values . . . fnmm (x) . Evidently, gm is a continuous function on [a,b]. We assert that '

=

holds. For m 1 the assertion true on the basis of the suppositions made nf l . It isis enough therefore to infer m from m - 1. 1 Let us suppose then that b g (x) dx J - cr j l - � l ( }aa m- 1 m- 1 l 2m- f is true. Since the function gm at the point x equals the smaller of the two n values gm_ 1 (x) and fmm (x), that is, gm (x) equals the sum of these two values minus the value of the larger amongst the two, while this larger value be­ cause of gm-l (x) fmn�� l (x) nm-l (x) and fnm (x) , that is ' does not exceed the larger of the values f m-1 m nf m (x) , we have m-l gm (x) gm-1 (x) + fnmm (x) - fnm-1m (x) ; the corresponding inequality is also valid for the integrals of these func­ tions and so, since Inm-1m we obtain the desired inequality lab gm (x) dx Jm-1 - cr � l - 2ml-l l + Jm Jm cr � - 2� � Jm >

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

>

s

<

s

s

LEMMA .

± co

•

s

s

D+ < D .

+

364

CHAPTER 4

In fact, applying (ii) to the function -f(-x), it follows that we have al­ most everywhere and combining this with (i) and (ii) we obtain 0 almost everywhere on [a,b]; hence the equality signs must hold, which was to be proved. Assertion (i) means that the set E of points x for which is of measure zero. This set E is included in00 the set EC of points x for which we have C, where C denotes a number chosen as large as we please. But the relation C implies the existence of some t x such that f(t)t -- f(x) that is, g(t) g(x), where g(x) f(x) - Cx. Hence the set EC is included in the set E attached to the function g by the and so it can be covered by a sequence of disjoint open intervals (ak ,bk) for which that is, hence D+ � D � D-

�

�

D+ � D+ � oo

D+

D+

=

oo ,

>

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+

>

D <

REAL FUNCTIONS

365

D

Consider first the condition < c. Applying the to the function g(x) f(-x) ex on the interval [-b,-a], we obtain, for reasons similar to those just used, that the set E cC can be covered by a sequence of disjoint open intervals (ak ,bk) ' such that (E. 21) Next we consider, inside each of the intervals (ak ,bk ) ' the points where C; applying the to the function g(x) f(x) - Cx on the interval (ak ,bk) ' we see that these points can be covered by a sequence of disjoint intervals (akm ' bkm) such that +

=

>

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) .

�

<

<

+

�b

f(t) dt J:c f(t) dt J(b f(t) dt. (c) The upper and lower integrals are unchanged if f is replaced by a new function which differs from f at one point only. Show that if f is a bounded function on [a,b] and has a right-hand lim­ it at each point of [a,b), then f is Riemann integrable over [a,b]. For any o define h0 (x) J:x f(t) dt - �x f(t) dt - o(x - a) for a x b for x a. It is enough to show that h0 (b) for all o since this will imply Sol ution .

=

=

+

>

0,

<

0

=

< 0

�

0,

�

367

REAL FUNCT I ON S

f(t) dt $ J:b f(t) dt. Suppose this is not true. Then there exists a 0 such that ha (b) 0 . Let x inf{t [a,b]: ha (t) 0 } . We shall show that the assumptions ha (x) 0 and ha (x) $ 0 both lead to con­ tradictions. If h (x) 0 , let M sup{lf(t) I : a $ t $ b} and choose 0 < y < x - a and 2yM
>

E

=

>

>

=

>

=

�

=

�

>

>

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. �

=

=

=

=

+

<

=

C

Cj =

REAL FUNCT I ON S

371

The first of these equations is an identity, the second is and defines c0 , the third defines c 1 , and so on. The two polygonal functions g and g0 coincide at all basic points and are therefore identical. This shows that g admits approximation by polynomials. PROBLEM 40. Prove the following Let [a,b ] be a closed bounded interval and f a continuous function on [ a,b ] . Then, for any > 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. <

I

�

=

+

+ oo ,

C

PROBLEM 42. Let f be a continuous function on a closed bounded inter­ val [a,b] and suppose that b (a xn f(x) 0 for n 0,1,2, . . . Ja Show that f 0. By Problem 41, f admits uniform approximation by a polynomial P; thus, for all x [a,b], f(x) P(x) Eh(x), where is an arbitrary positive real number and l h (x) 1 on [a,b]. We see therefore that Sol ution .

E

=

+

e

But

I <

� f(x) P(x) dx 0

by hypothesis. Noting that F

we see that we have reached a contradiction, for, if f 0, then would be bounded below by f2 (x) dx lf(x) l dx e

REAL FUN CT I ONS

373

If f # 0, then lf(x1) 1 = d > 0 for some x1 (a,b) and there is some o > 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,

<: 1

PROBLEM Let f be a continuous function on a closed bounded inter­ val [a,b]. Show that there exists a monotone increasing sequence of polyno­ mials (pn) :=l converging uniformly to f on [a,b]. Let (en) :=l be any sequence of positive numbers such that �� e = 1. By the Weierstrass approximation theorem (see Problem 40) we can=l findn polynomials p1 , p2 , . . . , pn ' . . . such that for all x [a,b] 44 .

Sol ution .

E

Letting f0 (x) = f(x) 1, and +

we have +

for n

<: 1 ,

But fn f uniformly on [a,b]. (Clearly, a similar result holds for the mo­ notone decreasing case.)

375

REAL FUNCT I ONS

PROBLEM 45. Let f be a continuous finction on a closed bounded inter­ val [a,b]. Show that f admits uniform approximation by a polynomial with ra­ tional coefficients. Let E > 0 be given. By Problem 40 there exists a polynomial P such that sup{lf(t) - P(t) l : t [a,b]} < E/2, where (t) ao + a1 + • • • + antn and a0 , a1 , . . . , an are real numbers. Let c max{lal, l b l}; for each j, 0 j n, choose a rational number bj such that l b . - a . l < 2(n +E l)cJ. and let Sol ution .

E

p

=

�

�

J

J

Then

I P Ct) - Q(t) I j I=0 (b.J - a.)tJ j \ < E/2 and so sup{ IP Ct) - Q(t) I : t [a,b]} < E/2. Hence sup{ lf Ct) - Q(t) I : t [a,b]} < E. I

E

E

PROBLEM 46. Let the function h(x) be continuous on the interval 0 x 1, and let f(x) be an arbitrary continuous function on 0 x 1. Show that a necessary and sufficient condition that every such f(x) can be uniformly approximated on [0,1] as closely as desired by a polynomial in h(x) is that h(x) be strictly monotonic in [0,1]. We say that h is strictly monotonic on [0,1] provided that, for x1 and x2 in [0,1], the inequality x1 < x2 always implies h(x1 ) < h(x2), or always implies h(x1 ) h(x2 ). �

�

Sol ution .

>

�

�

CHAPTER 4

376

If h(x) is continuous and strictly monotonic on [0,1], the transforma­ -1 tion z h(x), x h (z) , sets up a one-to-one continuous correspondence between the points of 0 � x � 1 and the points of some finite interval a � z � b, with the endpoints of the one interval corresponding to the endpoints -1 of the other interval. The given function f(x) = f[h (z)] is continuous on the latter interval, so by the Weierstrass approximation theorem (see Problem 40) it can be uniformly approximated on a � z � b by a polynomial in z. If £ > 0 is given, there exists anzn ao + a 1 z such that l f[h- 1 (z)] - k=OnI ak zk l < £, a � z � b, and this is equivalent to l f(x) - k=OI ak [h(x)]k l < £ ' 0 � � 1; here we define [h(x)] 0 even if h(x0) = 0 for some x0 in [0,1]. Conversely, suppose h(x) is not strictly monotonic on [0,1], and hence we have for two numbers x1 and x2 on [0,1] the relations x1 x2 and h(x1) = h(x2 ). If an arbitrary function f(x) can be uniformly approximated within in h(x) , we so approximate the function f(x) x, where £2 > 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.

=

�

<

Sol ution .

+

+

+

< e J. <

1!'

�

+

(8

•

J

'

+

.2

2!' 1

<

. k- 1

1

• • •

+

J

CHAPTER 4

378

cients equals (j)The system of equations has therefore a solution and this solution gives f (x) as a linear combination of f(x + f(x + 2), Now if x t x + k, then rf(t) limoo (f) rtrf(t) limoo (f) r lim trf(t) lim x oo x-+ x-+ x-+ oo x-+ and similarly, (k) (t) r lim x f x-+ oo Hence xrf (j) (x) is a linear combination of terms of the form or where x t x + k. Since these all go to as x -+ .,, the result follows. 1.

1) ,

<

<

1•0

0,

0.

<

<

0

PROBLEM 48. Let f be Riemann integrable over every bounded interval and (E. 24) f(x + y) = f(x) + f(y) for any real numbers x and y. Show that f(x) = ex, where c = f(l) . Integrating f(u + y) = f(u) + f(y) with respect to u over the interval [O,x], we easily see that (E. 25) yf(x) = Jor x+y f(u) du - Jrox f(u) du - J(oY f(u) du holds. Since the right-hand side of (E.25) is invariant under the -linterchange of x and y, it follows that xf(y) yf(x) . Thus, for x f 0 , f(x)x = c, a constant; hence f(x) = ex. Since (E.24) implies f(O) f(x) = ex also holds for x = Taking x in f(x) = ex, we obtain c = f(l) . If f satisfies (E.24) , then f being continuous at a single point implies that f is continuous everywhere. Indeed, l f Cx + h) - f(x) I = lf Ch) = l f Cy + h) - f Cy) l . Suppose that f satisfies (E.24) but f(x) is not of the form ex, where Sol ution .

=

0,

0.

=

1

Remarks .

I

379

REAL FUN CTIONS

c is a constant. Then the graph of f is dense in the plane. Indeed, let c = f(l) and choose x so that f(x) ex. The graph of f contains all points of the form [u + vx,uc + vf(x)], u, v rational. Let A be the matrix t

Then A is non-singular, hence a homeomorphism of the plane onto itself. In particular, A preserves dense sets; one dense set is {(u,v) : u and v ration­ al} and A maps this set [via multiplication, XA, where X = (u,v)], onto the subset of the graph of f mentioned above. Thus the graph of f is dense. PROBLEM Show that m-l mj m-1 j=OL: -J.! � e for all positive integers m. By Taylor's theorem m-1I mj = em em m e-t tm-1 dt. j=O J (m - 1)! la Hence it will suffice to establish that (E. 26) �m e-t tm-l dt � (m - 1) ! 1 - �) for all positive integers m. This will be done by induction. The relation (E.26) holds, with equality, if m = 1. Suppose that the relation (E.26) holds for m = k. Integrating the left member by parts and multiplying both members of the last mentioned inequality by m shows that 49 .

Sol ution . -=--t •

(

But Therefore

max k�t�k+l

e-k kk .

380

CHAPTER 4

k+l e-t tk dt = k+l e-t tk dt + lk e-t tk dt L !a kk e-k + Lk e-t tk dt k! 1 - .!.e) . $

<':

1:

$

(

PROBLEM 50. Show that the following relation is valid for all real n en-1

Let I(t) = laOy t [x] dx; if m is a positive integer and m < t < m + 1, then k ., + tm (t ,- m) I (t) = m-1 L !. k=O k. m. Direct calculation yields I' (t) - I(t) tm-1 (t - m) (m +. 1 - t) , which is positive on (m,m + 1) . If G(t) = log I(t) , we have, therefore, G' (t) > 1 on (m,m + 1) , and G is continuous at m, whencem-1by the mean-value theorem, G(t) G(m) + (t - m). By Problem I(m) e , so G(m) m - 1. Thus G(t) > t - 1, and therefore I(t) > et-1 on (m,m + 1) . Sol ution .

r=11 ;

LXJ

-roy-

49,

>

<>:

<>:

PROBLEM 51. Let f be a continuous function on the interval [a,b] and xsuch1 , xthat 2 , . . . , xn be points of [a,b]. Show that there is a point t in [a,b] Sol ution .

By Problem 28 of Chapter 2,

381

REAL FUNCT I ONS

The rest follows by the intermediate value property (see Problem

33) .

PROBLEM 52. Give a proof of the relation

(E. 27) Our proof will be based on the result in Problem 13: If f is Riemann integrable in the interval [a,b], then b f(x) sin px dx 0. lim J (E. 28) p a First we note that if x 2k1r 0, 1 , 2 , . . ) , then (E.29) !2 + cos x + cos 2x + + cos nx sin(n2 sm. + �)x Indeed, since S1n. -2n -2+-l S1ll + S1n S1n ) + (S1n. 5 S1ll ) + + S 1n. 2n 2+ 1 S1n. -2n -2- -1) , we get S1n. -2n -2+-1 + COS + COS 2X + + cos nx) 2 s1n by using the identity sin - sin B 2 s1n -2--B - cos ---+2--B -. Integrating both sides of the identity (E.29) , we obtain +x/2�)x dx � (n 0,1,2 , . . . ). (� sin(n (E. 30) 2 sin 0 Let (E. 31) g(x) x1 2 sin1 x/2 2 sin2x sinx/2x/2- x , 0 Sol ution .

4- oo

-f

(k =

.

.:...::.::. ::. c;._ ..>.; -'�

=

X

2

X

.

2

( • 2 3

X

(

• • •

X

= (12

=

=

-

-

X

---

2

2

-

X

X

2

3

X

• • •

=

.

-

-

X

A

Jn

.

X

.

A

.

A

-

2>

<

X

$

�.

2

X

CHAPTE R 4

382

�

The function g is continuous for 0 x By applying twice L'Hospital's rule to (E.31) , we see that limx+O g(x) 0. Consequently, if we put g(O) 0, g will be continuous for 0 x n. Therefore g certainly satisfies the condition of the proposition in Problem 13; then the relation (E.28), with p n Y,, gives nlim+"' }or n (� - 2 si� x/2) sin(n !z) x dx 0. Taking into account (E.30) , we see that 1n+1m. co io0n sin(n + !z) x d or, making the substitution u (n + !z) x, lo (n+Y,)n Sln. u du (E. 32) lim u n+ co If we can show that {"' sinu u du Jo is convergent, then (E.32) yields the relation (E.27) and we are done. But from the result in Problem 15 we get, for 0 a t b, {b sinU--u du 1 lt Sln. U du 1 lb sin U du. laa --a t Since sin u dul 2 for any and we therefore get (E. 33) hence the quantity on the left-hand side of (E.33) is less than £ when b a provided 4/£. Another interesting method for the evaluation of the integral I Jo{ "' sinx x dx depends on the partial fraction decomposition of the function 1/(sin t), that is, on the relation <

�

=

=

�

n.

=

+

+

X

X

_ -

2' rr

=

<

= a

�

I Jfa S

+ b

a

�

�

S,

>

� A,

A >

Remarks . =

383

REAL FUNCT I ONS

(-l) n (t -1-nn + _1 t +_nn) ' where t is arbitrary, but not a multiple of n. To verify this relation, we consider the function f(x) = cos ax, where a is not an integer and -n x n. Since .!2 ao = n )[0n cos ax dx sinanan and, for n > 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

=

<

+ +

0.

95

53

Remarks .

49

53.

l:

53)

+

Since (mm/m!)e-m is a decreasing sequence (m 1,2, . . . ) , f(m) is increasing, and since f(m) for m = 3, we get m-1L �x em-1 x=O ! for m = 3,4, . . . But this gives us the desired result, since the inequality is obviously true for m = 1 and m = 2. We also note that en = n-1 k!k n!n n ' 1/3 < en < 1/2, k=O may be rewritten in the form 2

0

2

2

'\ !...

!!._ + !!._ 6

CHAPTER 4

388

nn! � 2 I nk!k - en } = 2 (1 - en), l k=O from which it is evident that (n!/nn) (2 � O nk/k! - en) 4/3 as n + be­ cause en + 1/3 as n + (see Solution of Problem 53) . �

oo

oo

+

=

PROBLEM 54. Suppose that f is Riemann integrable over [a,b], m f(x) M for all x in [a,b], g is continuous on [m,M], and h(x) g[f(x)]. Show that h is Riemann integrable over [a,b]. Let £ 0 be given. By the uniform continuity of g on [m,M] we can find some o 1 0 such that i g(s) - g(t) I £ if is - tl o 1 and s, t [m,M]. Let = min{o 1 ,£}. Corresponding to o2 , choose a partition P {x0 , x1 , . . . , xn } of [a,b] such that U(P,f) - L(P,f) 2 which is possible by Problem 11. As usual, let s

=

s

>

Sol ution .

>

<

E

<

<

o

0 '

and let and M be the analogous numbers for the function h. Divide the numbers 1,2, . . . ,nk into two classes: k A if Mk - mk o and k B if Mk - mk o. For k A, our choice shows that Mk - mk £. For k 2 B, 2K, where K sup{ig(t) I : m t s M . But U(P,f} - L(P,f) o and so, letting �xk = xk - � -l ' �

E

�

E

E

<

Mf - � s

=

E

<

s

s

It follows that U(P,h) - L(P,h) £(b - a) + 2Ko £(b - a + 2K). Since £ 0 was arbitrary, the result in Problem 11 implies that h is Riemann integrable on [a,b]. s

<

>

REAL FUNCT I ON S

389

Letting g(t) t2 2, we see that if f is Riemann integrable on [a,b], then so is the function f . In view of the identity 4uv = (u + v) 2 - (u - v) 2 , we can easily see that if u and v are Riemann integrable functions on [a,b], then so is the product uv . Letting g(t) It!, we see that if f is Riemann integrable on [a,b], then so is the function If!. Moreover, choosing c as either or to make C fab f(x) dx it follows that l fab f(x) dx l = c fab f(x) dx = fab c f(x) dx Jab jf(x) I dx because c f(x) if(x) I for all x [a,b]. =

Remarks .

=

1

-1

�

0,

�

E

�

PROBLEM Let f and g be Riemann integrable functions on [a,b]. Show the following Cauchy-Schwarz Inequality for integrals: 55 .

For any real number t, (tf(x) g(x)) 2 dx

Sol ution .

+

�

0,

that is, t2 Jab {f(x)} 2 dx + 2t fab f(x) g(x) dx lab {g(x)} 2 dx which directly implies the desired inequality. +

�

0,

PROBLEM Let f and g be two Riemann integrable functions on [a,b]. Let u f(x) v g(x) 56.

�

� U,

�

� V

CHAPTER 4

390

for all x [a,b], where u, U, v, V are fixed real constants. Show that l b = a fab f(x) g(x) dx (b -1 a) 2 ab f(x) dx·Jab g(x) dx l (E . 37) � (U - u) (V - v) . By making the substitution x (t - a)/(b - a) the problem is reduced to the special case a 0, b In that case we write F fol f(x) dx, G !o 1 g(x) dx, and D(f,g) --J(0o 1 f(x) g(x) dx - FG. Then (E . 37) reads (E . 38) \ D(f,g) 1 u) (V - v) . Note that ( E . 39) D(f,f) �l (f(x)) 2 dx - 1 fol f(x) dx } 2 0 holds by the Cauchy-Schwarz Inequality (see Problem On the other hand, D(f,f) (U - F) (F - u) - JC0 1 {U - f(x)}{f(x) - u} dx, which implies that (E . 40) D(f,f) (U - F) (F - u) . One can easily verify that D(f,f) Ial {f(x) - F}{g(x) - G} dx. Using the Cauchy-Schwarz Inequality, we get (D(f,g)) 2 J( l {f(x) - F} 2 dx·fol {g(x) - G} 2 dx D(f,f)D(g,g) . According to (E . 39) and (E . 40) , we infer that E

r J

�

Sol ution .

=

=

=

1.

=

=

I � 4 cu -

=

�

55) .

=

�

�

�

=

391

REAL FUNCT I ON S

(D(f,g)) 2 � (U - F) (F - u) (V - v) . (E. 41) Since F) (F - u) � (U - u) 2 ' 4(V - - v) � (V - v) 2 ' we conclude that (E. 1) implies (E. 38) . Taking f(x) -1 for 0 � x � 1/2 1 for / � 1 and putting g(x) f(x) , we see that the constant in (E.38) is the best possible. G) (G -

4 (U

-

G) (G

4

Remark .

1 2 < x

=

l/4

PROBLEM Let f be a function on an interval whose length is not less than 2 and suppose that I f(x) I � l and If" (x) I � 1 for all x of this interval. Show that (E. 2) l f' (x) I � 2 for all x of this interval, where the constant 2 is the best possible. Without loss of generality we can suppose that 0 � x � 2. ' Then, by Taylor s theorem, for 0 � t 1 � x � 2, f(x) f(O) f(2) - f(x) and, therefore, f(2) - f(O) that is, 2lf' (x) I � 1 + + i1 x2 + i1 (2 - 2 - x(2 - x) � or 57.

4

Sol ution .

l

x)

4

4,

CHAPTE R 4

392

If' (x) I 2. The function f defined by f(x) 1 x2 - 1 shows that the sign of equality can actually hold in (E.42). s.

2

=

PROBLEM In this problem we consider one of the four proofs which Gauss gave of then Fundamental Theorem of Algebra: If f(x) x a1xn-1 + • • • + an (n 0) , where a1 , . . . , an are real or complex numbers, then f has at least onek realk or complex root. Put x r(cos e i sin e) ; then x r (cos ke i sin ke) , hence f(x) P + iQ, where P rn cos ne Q rn sin ne and all other terms of P and Q contain only smaller powers of r and terms not containing r will be constant. Theorem of Algebra will be proved if we can show that 2P + QThe2 isFundamental zero for a certain pair of values r and e. We introduce the func­ tion u arc tan p Then au (aP;ae)Qp2 - P(aQ/ae) as arau (aP/ar)Qp2 +- QP(aQ/ar) 2 Q2 . hence 58.

=

+

>

+

=

=

+

· · · ,

+

=

+

=

"

Q

_

+

Here H(r,e) is a continuous function of its variables whose exact form is of no interest to us. Finally, we need the double integrals and R is a positive constant whose value we shall determine later on.

REAL FUN CTI ONS

393

would not become zero anywhere, then the inte­ If the function grand would be continuous and this would, of course, imply that But we shall see that when R becomes large. This means, however, that the function must become zero at some point in the interior of the circle x + y R , proving the Fundamental Theorem of Algebra. Show that the equality cannot be fulfilled. We compute the interior integral in and obtain P 2 + Q2

2

P 2 + Q2 2

2

11

11 # 12

11

=

=

12 .

12

11

Sol ution .

since clearly is a function dependent on 6 and having period From this it follows that We next consider the integral Here au; ar

2n .

11

=

12 .

0.

For what follows it is important to consider the highest power of r in the numerator and the denominator of Since n as - n r sin n6 n rn cos n6 we get -nrn But auj a 6 .

aP

aQ a6

+ •••

=

+ •••

2

+ •••

+ •••

and so, finally, - n nr n r Since the remaining terms in the numerator and the denominator are made up of smaller powers of r whose coefficients are bounded functions of 6, we have not only that ll·m as = - n, r+ao 2

au a6

2

au

+ ••• + •••

CHAPTER 4

394

but even that this convergence to -n is uniform in 8 , that is, given any there is a real number M M(£) > 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

<

x sat­ (E. 43) < oo

CHAPTER 4

396

and f(x) I (E . 44) for all x and y of the number line (- "') .oo Show that if f is continuous on ( - , ) , then f(x) eiAx - oo < x < oo , where A is real. Suppose first that f is everywhere differentiable. Then dif­ ferentiation of (E . 43) with respect to x gives f'(x + y) f' (x) f(y). Putting x and f' (O) iA, where A is a complex number, we get f' (y) f(y) and integration gives (by changing back to the variable x) f(x) e iAx Condition (E . 44) tells us that f(x) for all real x, hence A necessarily has to be real. Thus, under the additional assumption that f is differentiable, the claim is established. To complete the solution, we show that any continuous solution of the functional equation f(x + y) f(x) f(y) is automatically differentiable. We put FE (a) Jraa+E f(x) dx. For fixed E this function is differentiable with respect to a. We have FE (a) f(y) ia+E f(x) f(y) dx ia+E f(x + y) dx i a+y+E f(x) dx FE (a + y) . a+y Assuming that FE (a) we obtain f(y) l

= 1

"'

,

"'

=

Sol uti on .

=

=

0

=

L\

=

I

l

=

=

=

=

=

=

=

f 0,

1

397

REAL FUNCTI ONS

Here F£ (a + y) is a differentiable function with respect to y, hence f(y) is seen to be differentiable. It only remains to verify that there exist an a and an £ such that F £ (a) 0. Keeping a fixed and differentiating F (a) with respect to £, we obtain = �d£ faa+£ f(x) dx f(a + £) . If for some a we had F£ (a) - 0 for all £, then -d£ F£ (a) 0 would follow and thus f(a + £) 0. This (trivial) case we can exclude. Then, however, necessarily F£ (a) 0 must hold and thus there has to be some £ such that F £ (a) 0 and the proof is complete. �

a

-

=

F

�

=

PROBLEM real-valued function f defined on a set E of real numhers is said to be E if, given any £ > 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 <:

< a,

CX>

£.

CX>

CX>

-CX>

=

£

>

<

£.

<=

) I <= £ .

PROBLEM 62. Let E be a set of real numbers and be a subset of E. We E if the closure of is E. say that is Prove the following proposition, called the Let E be a set of real numbers and be a subset of E which is dense in E. Moreover, let f be a real-valued function defined and uniformly continuous on Then there exists a unique continuous function g on E such that g(x) = f(x) for every x in In addition, if E is bounded, then g is uniformly continuous on E. First, we observe that if there is at all a continuous func­ tion g on E, with the properties stated, then, for every x in E and for every A

A

dense in

A

Principle of Extension by

Continui t y :

A

A.

A.

Sol ution .

399

REAL FUNCT I ON S

x in and for every sequence (xn) :=l of elements of A such that limn + ro xn = x (which exists because x is a limit point of A) , we must have g(x) = n-+limco g(xn) = nlim+oo f(xn) since xn A. Thus, g(x) must be the unique limit of the convergent sequence {f(xn)}�=l and, consequently, only one continuous function g can exist with the required property. We have also found out how we should define the value of g at every point x In"'fact, if x it is a limit point of A, hence there exists a sequence (xn) n-_1 of elements of A such that limn + ro xn = x; thus (xn n= is a Cauchy sequence because any convergent sequence is a Cauchy sequence. But f is uniformly continuous on A and so, by Problem {f(xn)}:=l is a Cauchy sequence. However, any Cauchy sequence of real numbers is a convergent sequence, and so there exists a real number y = nlim+ ro f(xn). We define g(x) = y. Now, we justify this definition by showing that if (zn) := l is another Cauchy sequence in A such that limn + ro zn x, then we have limn + ro f(zn) = limn + ro f(xn). This, however, follows from the uniform continuity of f on A. Given any 0, there exists a 0 such that if x,z A and J x - z < then f(x) - f(z) < Associated with this number o, there exists an index n such that both x - x < and zn - x J < for every index n n0 ; 0hence, for any indexn n n0 , l xn - zn I lxn - x l lx - zn l < o. Thus, for n no , E

E

E

E E,

E.

)

1

00

57,

=

£

o

J

>

J

o

£.

J

�

s

J �

J

E

>

o/2

l

o/2

+

�

which shows that {f(x )} and {f(zn)}:= l are Cauchy sequences converging to the same limit. n �=l Finally, we must show that g is continuous on Since every point x of is in some bounded subset T of the set it is enough to prove that if T is any bounded subset of then g is uniformly continuous on T. This implies, indeed, that g is continuous on T, and, therefore, at the arbitrary point x of At the same time, since we have not excluded that T = whenever is E

E,

E.

E,

E.

E

E

CHAPTER 4

400

assumed to be bounded, this will also establish that g is uniformly contin­ uous on when is bounded. We therefore assume that is a bounded subset of By Problem 61, it is sufficient to show that if (xn) �=l is any Cauchy sequence of elements of then {g(xn)}�=l is also a Cauchy sequence. Given any £ 0, since f is uniformly continuous on there exists 0 such that x',z' l x' - z' I implies l fCx') - f(z') l � Since (xn) �=l is a Cauchy sequence, associated with this there exists an index n0 such that if m,n n0 then E

E

T

E.

T,

>

o

A,

E

>

A,

<

<

o,

.

o,

�

Consider any two indices m,n n0 . Since m klim-+ co x'm, k ' n klim-+oo x'n, k ' (x'm, k 'x'n, k there exists a sufficiently large index k0 (dependent on m,n) such that, for k k0 , l xm - x'm,k I 3 and also l xn - x'n, k l therefore, if m,n n0 , and taking any index k k0 , we have lx'm,k - x'n, k l lx'm, k - xm + l xm - xn I + l xn - x'n,k + + Hence, by the uniform continuity of f, l fCx'm, k) - f(x'n, k) I < £ provided that m,n n0 and k k0 . the other hand, considering any two natural numbers m,n n0 , since g(xm) = k-+limoo f(x'm, k) and g(xn) = klim-+oo f(x'n,k) , given the number £ 0, there is a sufficiently large index k 1 , which we may take such that k1 k0 , with the property: if k k 1 , then l g Cxm) - f(x'm, J I £/3 and l gCxn) - f(x'm, J ) £/3. Combining these inequalities, if m,n no and using k k1 , it follows from g(xm) - g(xn) {g(xm) - f(x'm, k)} + {f(x'm, k) - f(x'n, k)} + {f(x'n, k) - g(xn) } �

X

E

X

A) ,

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401

REAL FUN CTIONS

that

E E E 3 + 3 + 3

E,

I g(xm) - g(xn) < This shows that {g(xn)}:=l is a Cauchy sequence. This completes the proof. The result in Problem 62 shows that if f is a uniformly con­ tinuous function on the set of all rational points of an interval I, then there exists one and only one continuous extension of f, namely the function g, which is continuous on all of I. As a simple application of the result in Problem 62 we may consider the definition of the exponential function with base a that is, g(x) ax < x < oo. Here we first consider the functions fn (r) ar with rational r and the domain of definition of fn restricted to the interval [-n,n]. Applying the result in Problem 62, we obtain gn , namely the continuous extension of fn . Fi­ nally, we define g on the whole real line as follows: if x is any real num­ ber and n is a positive integer such that l xl n, we put g(x) gn (x); this definition is independent of the integer n. I

=

Remarks .

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0,

=

=

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=

PROBLEM 63. If, for a (E. 45) f(x) cos ax, resp. f(x) cosh ax, then for any real numbers x and y we have the functional equation (E. 46) f(y x) f(y - x) 2f(x) f(y). This follows immediately from the familiar formulas cos(y x) cos x cos y sin x sin y, cosh(y x) cosh x cosh y sinh x sinh y. Show that the only nonzero continuous functions on the real line R1 that satisfy the functional equation (E.46) are the trigonometric and the hyperbolic cosines (E.45) . Let some function f satisfy the requirements of the claim. We let x and choose y such that f(y) f then >

0,

=

+

+

±

±

=

+

±

Sol ution . =

0

0;

402

CHAPTER 4

= 1.

f(O) (E. 4 7) For y we get f(-x) f(x) , (E. 48) that is, f is an even function. Since the continuous function f is positive for x = there exists a positive number c such that f(x) is positive on the entire interval [O,c]. The following considerations follow two different lines of reasoning depend­ ing on the two cases: Case f(c) Case 2. f(c) We consider Case first. Since f(c) there is a e with e rr/ 2 such that (E. 49) f(c) = cos e. We now write (E.46) in the form f(y x) 2f(x)f(y) - f(y - x) and put, successively, X = y c, X = y 2c, y 3c, etc. Using (E.47) and (E.49) , we obtain f(2c) 2 cos 2 e cos 2e, f(3c) 2 cos e cos 2e cos e cos 3e, f( 4c) 2 cos e cos 3e cos 2e cos 4e, etc. By induction, we get for any positive integer m, f(mc) = cos me. (E. If in (E.46) we put x y c/2, we get, using (E. 4 7) and (E.49) , [f(c/2) ] 2 f(O) +2 f(c) + cos2 e {cos(e/2)} 2 ; since f(x) is positive between and c and cos x is positive between and e, we may take the root and obtain f(c/2) = cos( /2). 0

=

0,

1.

$ >

1 1. 1

0

<

:s; 1 ,

0 :s;

<

+

=

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C,

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50)

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1

0

0

REAL FUNCT I ONS

403

In the same fashion we get, putting x y (1/22 )c in (E.46) , f(c/2 2) cos(e/22 ) , etc. By induction, for n 1,2,3, . . . , we get =

=

(E. 51) By repeating the process that got us from (E.49) to (E.SO) , we obtain from (E.Sl) Thus, if x is of the form m/2n , we have f(cx) cos ex. (E.S2) But every positive real number can be represented as a p-adic fraction with p 2. Since f and the trigonometric cosine function are continuous, we see that (E.S2) holds for any x That (E.S2) also holds for x and x < can be seen from (E.47) and (E.48) , respectively. If in (E.S2) we replace x by x/c and put 6/c a, we obtain f(x) cos ax. In Case 2 we have f(c) 1. Then there exists a e such that f(c) cosh By repeating the reasoning used in Case 1 and applying the corresponding formulas for the hyperbolic cosine we get, for a f(x) cosh ax. For a we get, of course, in both cases f(x) 1 for all x R1 . This completes the proof. =

=

> 0.

0

=

0,

=

=

>

=

e.

> 0,

=

=

E

0,

PROBLEM 64. Show that any continuous mapping g of the real line R1 into itself such that for any x, y R1 is of the form g(x) ex + a, where c and a are constants. =

E

404

CHAPTE R 4

Indeed, for y = we obtain from the functional equation g(x) + g(O) g(x) + a where a = g(O) ; thus g(x) ; g(y) = g (x ; Y) g(x + y) + a that is, g(x + y) = g(x) + g(y) a, and, with f(x) g(x) - a, f(x y) = f(x) + f(y) for any x, y R1 . But g is continuous; hence f is continuous and so, by the result in Problem f is of the form f(x) = ex with c = f(l) . Thus g is of the form g(x) = ex + a, where c and a are constants. 0,

Sol ution .

2

2

2

E

+

48 ,

real-valued PROBLEM Let f be an everywhere finite, nondecreasing, 1 function whose domain of definition is the real line R . In addition, let f be right continuous at every point; that is, lim-+ y+f(x) = f(y) . Suppose a half-open interval (a,b] of finite length is covered by open inter­ vals (ai ,bi) ; that is, (a,b] i=lL., (ai ,bi ). Show that f(b) - f(a) i=lL., [f(b i) - f(ai) ]. By right continuity of f, given there exists o such that a o < b and f(a + o) < f(a) + By the Heine-Borel theorem, the closed interval [a + o,b] is covered by a finite set (a.1 1 ,b.1 1 ), . . . , (aim ,bim) of the given intervals (ai ,bi ). Clearly, for this finite covering 65 .

X

c

s

E

Sol ution . +

E.

> 0,

> 0

REAL FUNCT I ON S

a.l.l a + <

405 .S ,

and the intervals may be so numbered that for each k (k 1,2, . . . ,m) , Since f is nondecreasing, we have Li=l [ f(b.)l. - f(a.)]l. k=lmI [f(b.1k) - f(a.1k) ] f(b.1m) - f(a.1 1) + m-1 k=lL [f(b.l.k) f(a.1k+l)] ;:: f(b) - f(a + f(b) - f(a) - E for every E 0. Let A and B be given real numbers and suppose that A < B + for any 0. Does it follow that A � B? Yes, indeed! Note that A B im­ plies A ;:E: B + E for some 0. <=

.S) >

>

Remark .

£

>

>

£ >

PROBLEM 66. Let f be a continuous function on the interval [ 0,1]. Show that n n . � z ) f lim ; :E c-1) k n ( k n n k=O We note that f is uniformly continuous on [ 0,1]. Hence, for any given E 0, there exists a positive integer N(E) such that (n N; k = O,l, . . . ,n-1) . lfCk/n) - f({k + 1}/n) l For n N, we have o

+ oo

Sol ution . >

< £,

>

=E

2'

>

CHAPTER 4

406

from which the desired result follows. alternate way of solving Problem 62 is as follows: Given any positive number we can find a polynomial function P such that lf(x) - P(x) for all x in [0,1]. This is the Weierstrass Approximation Theorem (see Problem 40) . If n is greater than the degree of P, then by the theory of finite differences (or otherwise) Remarks .

I �

An

£,

£

Thus, if n is sufficiently large, 2 -n k=OnL (� Since is arbitrary, the assertion of the problem is established. )£

�

£.

£

PROBLEM Let the function F(t) have a continuous derivative on [0,1] and put s 1 {t: F'(t) = 0}, s2 = {F(t) : t s1 }. Show by an example that the set s 2 may be uncountable. Let g(x) 0 on the Cantor set (see Problem 23) and g(x) (x - a) (b - x) in each interval (a,b) forming the complement of the Cantor set. Let F(t) = fat g(x) dx; Then F' = g is continuous, and s 1 is the uncountable Cantor set. Since {x: g(x) 0} is everywhere dense, F is strictly increasing, and hence one-to­ one; therefore s2 is also uncountable. 67.

E

Sol ution .

=

>

PROBLEM The function given by f(x) = 0, if x is irrational or x = 0, and 1/q, if x = p/q, where q 0, and p and q are integers without com­ mon divisors, was considered in Problem 14. We saw in Problem 14 that this 68 .

>

407

REAL FUNCTIONS

function f is not continuous on the set of nonzero rationals of the unit interval, the only possible points of differentiability are the irrationals of the unit interval and zero because f is continuous at these points of the unit interval [0,1]. Show that the function f is not differentiable at its points of contin­ uity on the interval [0,1]. If x = 0 and h 0, [f(x + h) - f(x)]/h = f(h)/h. Let (h.) be a sequence of irrationals having zero as limit. Then lim f(hi)/hi 0. Now let hi = 1/i, i = 1,2, ... Then f(hi)/hi = (1/i)/(1/i) = i+co 1. Therefore li� + 0 f(h) /h does not exist and f is nondifferentiable at x = 0. If x is an irrational number 0 x 1 and if h 0, [f(x +h) - f(x) ]/h = [f(x +h) ]/h. If (h.) is a sequence of real nwnbers having zero as limit such that x + h. is irrational for each i, then lim. [f(x +h.) ]/h. = 0. Let the decimal representation of x be O.a a . . . a ... Choose h = O.a1 a2 . . . a.-x. Since x 0, a. 0 for some i. Let N1 be2-i then least integeri such that 0. Then f(x + hi ) = f(O.a1 a2 . . . ai) 10 for all i N, and i h i l 10 -i Hence, f(x +--hi-) l -> l for all i N. l --,h. Therefore li� + 0 [f(x + h) ]/h does not exist and f is not differentiable. F

Sol ution .

1

<

<

F

1

1 4- CO

1

F

� F 1

1

F

�

1

�

1

�

�

1

PROBLEM Find an explicit function f on [0,1] to the real numbers, such that f' is defined on [0,1] and discontinuous on the rationals in [0,1]. Let f(x) = x2 sin(l/x) ; then (see Problem 10) f is differen­ tiable everywhere, and f' is discontinuous only at x = 0. Let r1 , r2 , . . . be an enwneration of the rationals in [0,1]; define g(x) n=lL 2 -n f(x - rn) . Then g' (x) = L�=l 2 -n f' (x - rn) ' since both series are uniformly convergent. Then g' exists and is discontinuous at each rational in [0,1]. PROBLEM 70. Let bn+l = J0 1 min(x,an) dx and an+l 69 .

Sol ution .

CHAPTER 4

408

Show that the sequences (an) and (bn) both converge and find their limits. For any a0 , b0 , it is easy to see that an and bn both lie between 0 and 1 for all n 2. The recurrence formulas then become (for n 2) Sol ution .

�

�

If we assume that limn +oo an a, limn +oo bn b, they must satisfy a = (1 + b 2)/2, b = a - a /2, from which we get a + b - 1 = (a + b - l) (b - a + 1)/2. Since the factor (b a + 1)2 1, we have a + b = 1, and this yields a = 2 - 12, b = 12 - 1. To show that an + a, bn b as n oo , we write an a + tn , bn b + sn . The recurrence formulas become, after an easy reduction, (b - tn/2) tn . Now sn I = lbn bl $ max(b,l-b) a, and tn I = an - al $ max(a,l-a) = a. Hence lbn + sn/21 $ b + a/2 = 1/12, b - tn/2 1 $ b + a/2 = 1/12. Therefore t $ s l/12 and s 1 1 $ tn l/12. This shows that tn 0, sn 0 as n n+oo,1 1 and nthe proof isn+complete. 2

�

I

I

+

I

I

I

+

l

I

+

I

=

+

+

PROBLEM 71. Let an+ 1 [0 1 min(x,bn ,cn) dx, bn+1 = 0 l med(x,cn ,an) dx, and cn+ 1 = J01 max(x,an ,bn) dx, where med(a,b,c) = b if a $ b $ c. Show that the proposed sequences (an) ' (bn) ' and (en) are convergent, with limits 3/8, 1/2, 5/8, respectively. Evidently min(x,bn ,cn) $ x $ max(x,an ,bn) so that, if x med(x,an ,cn) ' we have =

Sol ution .

r )

409

REAL FUN CT I ONS

(E. 3) Now, if med(x,an ,cn) = an ' we have either x an or en an , so that an max(x,an ,bn) implies (E.S3) in this case also. similar argument holds for med(x,a ,c ) = e , so that (E.S3) is true in all cases. By integration there resultsn nan+l nbn+l cn+l ' n = 1,2,3, . . Now we have 5

�

�

�

A

�

�

.

and similarly cn+l 1/2. Using this �

s·

5

�

Dually, bn+2 3/8. Since 3/8 bn+2 cn+2 , �

�

and similarly cn+3 < 1. It is now assumed that n is so large that bn an+l =i x dx + h l bn dx 2bn 2- bn n an cn bn+l = lo an dx + i x dx + 11 cn dx n n bn 2+1 l b n cn+l = !c bn dx + h x dx = --2-. n Thus 0

0

< an

�

bn en < 1. �

an2 - cn2 - 2cn 2

CHAPTER 4

41 0

1 (2bn - 1) ( -2bn2 + 2bn + 1) bn (2bn - 1) 16+ 5(2bn - 1) Since 0 < -2bn2 2bn + 1 whenever 0 < bn < 1, either or .!.2 > bn+2 > bn· It follows that limn -r oo b 2n limn -r oo b 2n+l limn+oo bn 1/2. Then 2 b 2b n n lim an n+limoo an+l n+oo lim 2 n+oo 2 -+-1 5 b n lim cn nlim+oo cn+l nlim+oo 2 n+oo Note that med(a,b,c) min{max(a,b) ,max(b,c) ,max(c,a)}. = 2

+

8

3

+

=

8'

3

= s·

Remark .

=

72. Let f be a polynomial of degree n such that {Jo0 1 xk f(x) dx 0 for k 1,2, ... ,n. Show that (n + 1) 2 �fol f(x) dx f 2 Set f(x) anxn + an-lxn-1 + + ao . ( . 5 ) l {f(x)} 2 dx ao fo l f(x) dx. L Also l k f(x) dx an + -an- 1 + ao n+k+1 n+k !a PROBLEM

=

Sol uti on .

• • •

=

X

+ k+l;

(E .

54)

(E .

55)

(E .

56)

By E 4 ,

REAL FUN CTIONS

41 1

--n+--ak=-n-.- +-;-1 + -na-n-1+-k + k 1, 2, . . . ,n. Now -n-+--akn---:- +-;-1 + -a-nn-1+-k + (k + 1) (k + 2)Q(k)• • • (n k + 1) ' where Q is a polynomial of degree at most n. Since Q(k) 0, k 1, 2, . . . ,n, Q(k) C(k - l)(k - 2) •••(k - n) with C a constant. Hence -n-+--ak::.:._n-, +--1 -,- + -na-n-1+-k + + k a+o 1 (E. 57) (kC (k+ -1) l)(k (k+ -2) 2)• • •• • •(n (k+ k- n)+ 1) For k 0, 1 a ao (-l) n --. }{ f(x) dx n +n 1 + + n+1 1 o0 Multiply (E.57) by k + 1 and set k -1 to obtain a0 (-l) n (n + l)C or a0 (n + 1) 2 � l f(x) dx. Thus, by (E.S6), we obtain the desired result (E.SS) . =

+

=

-­

·

=

c

=

--

=

=

=

PROBLEM 73. Let a < b be given numbers and let f(t) be defined, con­ tinuous, non-negative and strictly increasing for a t b. By the law of the mean for integrals, for every p 0 there will exist a unique number xp , a xp b, such that �

�

�

>

Find limn +co xp .

�

Let £ be a given number such that 0 < £ < (b - a)/2. Since f is strictly increasing, f(b - E)/f(b - 2£) 1. Hence there exists a pos­ itive integer P such that for p ! f(b - E ) I p £ l f(b - 2 ) f Sol ution .

>

�

> P,

>

CHAPTER 4

41 2

or We have also and therefore -b -1-a fab fp (t) dt b � a J(b-£b fp (b - £) dt >

Hence xp cannot satisfy fP (xp) b � a ib fp (t) dt unless xp b - 2£. It follows that there exists a such that, whenever p P, xp b - 2£. Hence limp+oo xp b. =

>

> >

P

=

Consider the ·function Hn (t,x) 2n (tn - x) 2" If n and x are kept fixed, while t vaires between 0 and the function Hn (t,x) is seen to be a continuous function of t. Hence, for any integrable function f(t) (0 t we can form the integral fn (x) n la l n2f(t)(t -dtx) 2 " Show that for any point of continuity x (0 < x < of the function f we have (E. n+limoo fn (x) f(x). First we note that, for n oo , PROBLEM

74 .

= n" 1

1

+

1,

� 1)

�

= Tr

0

1

+

1)

58)

Sol ution .

+

41 3

REAL FUNCTI ON S

rn(l-x) � [Jo1 n (t,x) dt = .!!.TI J[o + n 2dt(t - x) 2 � 1-nx +z _!_TI !- + + z 2 To prove the relation it therefore will be sufficient to show that the difference rn fn (x) - f(x) J'�ol n (t,x) dt = .!!.TI J'/"o l 1 +f(tn1 (t- -f(x)x) 2 dt tends to zero as n For every there exists a o such that for I t - xl o the inequal­ ity lfCt) - f(x) I holds. Under the assumption that x - o < x + o 1 we can express the difference rn as follows: x+ o f(t1 - f(x) dt rn TIJ'o/"x-o 1 +f(tn1 (t- f(x) �� dt 2 x-o 1 + n (t - x) 2 - x) + Jx( + o 1 f(t)n2 (t- f(x) - x) 2 dt 1

H

1

-r

00

oo

� 1

1

1.

=

(E . 5 8)

=

H

-r oo .

€

<

=

<

0

>

€

0

.!!.

<

+

.!!. TI

1

+

The integral Bn can be estimated as follows: 1 Bn 1 - J:x-ox+o 1 +f(t)n2 (t- -f(x)x) 2 dt <

I

\

In the integral An we have It - xl o; thus �

A(n , where A(o) is independent of n. In a similar fashion we get o)

< -­

hence

<

414

CHAPTER 4

l rn E + n+ and so, for sufficiently large n, rn 2 E . Thus rn as n and we have the desired result. I <

I

C (o)

A(o)

I <

+

+ oo

0

PROBLEM real-valued function f defined on an interval is said to be if f[Ax + - A)y] $ Af(x) + (1 A)f(y) whenever x and y belong to and $ A $ Geometrically, this means that if P, and R are any three points on the graph of f with between P and R, then is on or below chord PR. Prove that if [a,b] is any closed subinterval of the interior of then on [a,b] and there is a constant K so that for any two points f is bounded x, y E [a,b], lf(x) - f(y) I $ Klx - y l . Observe that M max{f(a) ,f(b)} is an upper bound for f on [a,b], since for any point z Aa + - A)b in [a,b] f(z) $ Af(a) + - A)f(b) $ + - A)M M. But f is also bounded below because, writing an arbitrary point in the form (a + b)/2 + t, we have f (-a +2-b) $ 2 f (-a +2-b + t) + 2 f (-a +2-b - t) or f (a ; b - t) ;:: 2f (a ; b) + f (a ; b - t) . Using M as upper bound, -f (-a +2-b - t) ;:: -M, so f(a ; b + t) 2f(a ; b) - M A

75 .

I

convex

(1

-

0

I

1.

Q,

Q

Q

I,

=

Sol ution .

(1

=

AM

(1

1

1

<:

=

m.

(1

=

415

REAL FUNCT I ONS

This shows that M and m are upper and lower bounds of f on [a,b], respective­ ly. We next pick h 0 so that a - h and b + h belong to the interval I, and let m and M be the lower and upper bounds for f on [a - h,b + h]. If x and y are distinct points of [a,b], set z = y + l y h- x i (y x) ' A. = h +l y l y- x- i x l ' Then z [a - h,b + h], y = A.z + (1 - A.) x, and we have f(y) A.f(z) + (1 - A.) f(x) = A.[f(z) - f(x)] + f(x) , f (y) - f (x) A. (M - m) < I Y h- x (M - m) = K I y - x I , where K = (M - m)/h. Since this is true for any x, y [a,b], we conclude that i f(y) - f(x) l Kly - x i as desired. Recalling from Problem 30 the definition of absolute contin­ uity, we see that the choice = £/K meets the requirement for asserting that f is absolutely continuous on [a,b]. To see that f must be continuous on the interior of the interval I, we note that [a,b] is an arbitrary closed sub­ interval in the interior of I. Hence, if f satisfies the conditions in Prob­ lem then f is absolutely continuous on any closed interval [a,b] in the interior of I and continuous on the interior of I. >

E

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I

E

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Remarks .

o

75 ,

PROBLEM Let f be defined and bounded on [0,1], and f(ax) = bf(x) for 0 x 1/a, with a and b numbers larger than Show that f(O+) = f(O) . Clearly f(O) = 0; -nif M isn an upper-nbound for l fl on [0,1], -n then, for 0 x a , jf(x) I = I b f(a x) Mb . 76 .

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1.

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Sol ution .

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I �

PROBLEM point x is called a E of real numbers if every neighborhood of x contains uncountably many points of E. Show that any uncountable set E of real numbers has at least one point of condensation. Suppose by way of contradiction that no point of E is a 77.

Sol ution .

A

point of condensation of a set

CHAPTER 4

41 6

condensat ion point of E . Then for each x in E , there is an open interval I x , containing x, such that I x n E is countab le . Let Jx be an open interval con­ tained in I x , and containing x, but having rational endpoints ; indeed, Jx n E is again countab l e . Moreover , the col l ection of al l such interval s J x is countable and we may enumerate it as follows

Since each point of E is in some Jk , we may conclude that E

U (Jk k=l

n

E) .

The countable union of countab le sets being countable , we have the desired contradiction.

PROBLEM 78 . Show that any uncountab le set of real numbers includes a sequence (xn ) = l of distinct numbers such that � = l xn ± co

:

:

Since the set is uncountab ly infinite , it must have a point of condensation, say a , by Problem 7 7 . For definiteness we assume that a � 0 . I f a > 0 , the neighborhood (a/ 2 , 3a/2) contains uncountab ly many points of the given set and the des ired result follows . If a = 0 , the neighborhood ( - 1 , 1 ) contains uncountably many points of the given set , hence at least one of the intervals ( - 1 , 0) or ( 0 , 1) must contain uncountab ly many points of the given set ; let ( 0 , 1) contain uncountably many points of the given set . Con­ sidering all subinterval s of the interval (0, 1 ) that are of the form Sol ution .

k = 1,2,3,

•

.

.

'

we see that at least one of them , say the interval

�' k 0)

( 0

l

must contain uncountably many points of the given set . P icking a sequence (xn ) = l of the prescribed kind in

:

co

clearly leads to �n=l xn

REAL FUNCT IONS

41 7

PROBLEM 79 . Give an example of a continuous curve in the x , y plane that passes through every point of the unit square [0 , 1 ] x [0 , 1 ] . Sol ution .

Let g be defined on the interval [0 , 2 ] by 5 $ t $ 2, for 0 < t < l and for 3 3

g (t) = 0

_

3t - 1

for

-§- $ t

1

for

f$t

- 3t

+

_

2 $ 3'

<- i' 3

4 for 3 $ t <- � 3'

5

1 We extend the definition o f g t o all o f the real l ine R by setting g (t

+

2) = g (t) .

Note that g is continuous on R 1 and has period 2 . We now define two functions x and y by the fol lowing equations :

I

x (t)

y (t)

n=l

I

n= l

Both series converge uniformly on R 1 ; moreover , both functions x and y are 1 continuous on R . Let z = (x,y) and G denote the image of the interval [0 , 1 ] under z . We will show that G = [0 , 1 ] x [0 , 1 ] . Evident ly, 0 $ x (t) $ 1 and 0 $ y (t) $ 1 for each t , s ince the series -n 1 1 � = l 2 = 1 . Hence , G is a subset of [ 0 , ] x [0 , ] . It remains to show that (a,b) E [0 , 1 ] x [0, 1 ] . For this purpose we express a and b as p-adic fractions with p = 2 , that is , we put

�

a =

a

I 2n n=l -n·

b

b

I 2n n=l -n·

where each an and each bn is e ither 0 or 1 . Let c

=

2

c

I 3n n=l -n·

where c 2n- l an and c 2n = bn ' for n = 1 , 2 , . . . "" Since 2 � n=l 3 -n = 1 , we see that 0 $ c $ 1 . We wil l show that x (c) and y (c) = b .

=

a

CHAPTER 4

41 8

I f we can verify that g ( 3k c)

(E . 59)

holds for k = 1 , 2 , . . . , then we will have that = c 2n- l

and

thi s , however, will give us x (c) = a and y (c) 2

k

c

_n_ + I n=l 3n-k

2

I n=k+l

c � n 3 k

=

b . To prove (E . 59) we put

( an even integer) + dk '

where

Since g has period 2 , k g ( 3 c) = g ( � ) . I f ck + l = 0 , we have

0

� � dk

2

' 3 -n L. n=2

-

1 3'

and hence g ( � ) = 0 ; thus (E . 59) is satisfied in this case . The only other case to consider is ck+l = 1 . But then we get

and so g ( dk ) = 1 . Therefore ( E . 59) holds in all cases and we have obtained the desired result .

PROBLEM 80 .

L:

If

1 nn- xn-1 -nx e n! n= l co

F (x) =

Show that F (x) Find the l imit

(E . 60)

1 when x is between 0 and 1, and that F (x) i 1 when x > 1 .

41 9

REAL FUN CT I ONS

F (l + h) - F (l) h +

as h

0 through pos itive values .

Sol ution . The solution of the prob lem at hand essentially rests on the fact that the function F (x) in (E . 60) is given by

xF (x)

x,

0 < X :'> 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

<

1/e,

CHAPTER 4

420

it suffices to set A = xe -x in order to obtain from (E . 60) the relat ion of (E . 61) , where w (x) is given by (E . 62) . Now , putting in (E . 62) w (x) = xs (x) , we obtain the equation (E . 65) , that is , s (x) is the inverse funct ion of (log x) / (x - 1) , while w- 1 (x) = w (x) ; but s (x)

F (x)

for x ;:: 1

- 2 . Hence

and s ' ( l )

F ( l + h) - F (l) h as h

�

=

1 -2

0 through positive values . The question posed in Problem 80 is due to Ramanuj an .

Remark .

PROBLEM 81 . R (x)

=

I f x is positive , show that

20 42 1 53 31 x . + 1 + (x + 2) 2 + (x + 3) 3 + (x + 4) 4 + (x + 5) 5 +

--

. . .

(E . 6 7)

1 x'

< -

and find approximately the difference when x is great . Hence show that 1 + -1 + -3 + -42 + 53 + 64 + 1001 1002 2 1003 3 1004 4 1005 5 1006 6

--

--

--

is less than 1/1000 by approximately 10 -440 Sol ution . We observe that the function R(x) in (E . 67) is the Laplace trans form of the function F (x) in (E . 60) and that F (x) is given by (E . 61) and ( E . 64) . It is therefore sufficient to express R(x) by s (x) as follows

R (x)

=

fa

00

e - xt F (t) dt

=

.!.f oo e- xt ds . X + X

= .!_

l

Using the substitution

fa 1 e-xt dt + J1oo e -xt s (t) dt

( E . 68)

421

REAL FUNCT I ON S

a = s (t) ,

s - l (a)

t

- cr:-f' _

log a

we therefore get 1 R(x) = X But e ( l - a) l/a

::;;

( 1 - a) l/2 '

e ( l - a) l/a

:2:

(1 - 4a/3) 31 8 ,

0 ::;; a

:2:

0,

3/4 ::;; a ::;; 1

0 ::;; a

::;;

1,

and ::;;

3/4 ,

and thus l r {e ( l - a) l/a } x da -< 2 ::;; X + 8/3 )0 X + 2' 2

(E. 69)

______

Consequent ly -x R(x) = x1 - x (x 2+ e 2 + 8) '

0 < 8 < 2/3.

The l imits 8/3 and 2 in (E . 69) are actual ly attained for x respectively; that is , 8 (oo)

2

and

= 3

oo and x

8 (0) = 0 .

where the function 8 (x) is defined by (E . 70) . From (E . 70) we get , setting x = 1000 , 1 10 -439 . 994608 < _ 1000

_

_

R(lOOO) < 10 - 439 . 994319 '

that is , the factor multiplying l o -440 is s ituated between and

1 . 0 124978 Remark .

1 . 0 1249445 .

The question posed in Problem 81 is due to Ramanuj an .

PROBLEM 82.

Show that the series

=

(E . 70) 0,

CHAPT E R 4

422

X X 2 + X4 + + --1 - X2 1 X4 1 - X8

---

---

-

converges to x/ (1 - x) i f l x l < 1 and to 1/ ( 1 - x) if l x l

X

y-:---x =

X

X2

1

-

1.

We observe that

Sol ution .

---

>

-

X2 . 1 - X2

---

Therefore

1

-

X2

n+l

1

-

X2

n+l '

and by induction X X2 + --X4 + ---2 + 4 1 - X 1 X 1 X8 ---

• • •

X2

+

n

=

n+ 1 1 - X2

-

y-:---x

X

2n+l X - ----n-+-=-l· 1 - X2

The l imit of the second term is 0 for l x l < 1 and 1 for l x l

PROBLEM 83.

•

• •

n- l (1 + X2 )

1 1 - X as n

-+ --

Since ( 1 - i ) ( l + x 2 ) ( 1 + x4)

(1 - X4 ) (1 + X4 )

• • •

and so forth, we see that 1 or Pn (x)

1.

For l x l < 1 , show that

(1 + X) (1 + X 2 ) (1 + X4 ) Sol ution .

>

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 + "'

<

+

---

x3 - � + 1 + X 3 1 + X4

X3 +

J


1

1,

--3

1

• • •

X4 + - --X 1 + X4

x ( l - 2x

+

3x 2 - 4x 3 +

• • •)

X � X ) = ( 1 + X) 2 + i as X + 1 - 0 .

PROBLEM 9 1 .

Show that

n-1 Xn lim L (-l)n x + l-0 n= l 1 + xn

---

Sol ution .

I n=l

( - l) n-1 n

I n=l

( - l) n - 1 n

Let 0

<x

<

=

2

1 log 2 .

1 . The series

converges to log 2 and the factor xn I (1 + xn ) is bounded by 1 and is mono­ tone decreasing as n increases ; hence the series

converges uniformly on (0 , 1) by Abel ' s test for uniform convergence . The de­ sired result is obtained by letting x + 1 -0 termwise in the series (this is of course permissible here) .

PROBLEM 92 .

Let x > 0 . Show that

REAL FUNCTIONS

429

= x-1 · • l ( 2x) • + (l nx) + x) • � 1 + ( n� 2 Let

Sol ution . s

1 +

4 3 + ( 1 + 2x)r ( 1 + 3x) + ...,.(-=-l-+--:2,..-: x""'")"""'(-=-1...;r+�3,-x""'")"""'(-=-1-+--4,----, x"") +

r2

r+2x

Then un r = ' u--n - 1 --- 1 + nx

(E . 72)

where un is the n-th term of S . Proceeding as in the Solution of Probl em 8 7 , w e obtain the differential equat ion dS dr +

.!__::_2:.

xr

S = _!__ xr

because (E . 72) gives L (nx + l ) un = r L un + r. n=2 n=2

But n � 2 (1 + 2x) ( l + 3�) · · · (1 + nx) is the value of dS dr when r = 1 . PROBLEM 93. s

\

n � 2 p (p

Sol ution .

Putting

Show that +

1 n! l) (p + 2) • • • (p + n) = p-:-r

We may write

·

CHAPTER 4

4 30

then S becomes s 1 when r 1. But from s1 we have _un-1un_ = r -p.-n+­n which gives the differential equation dS 1 + n - r 5 dr r(l - r) 1 r(l - r) Therefore S l = p(l r-pr) p-1 { c + f r _(1t_P-_-_t)1_p dt } . But (r _t.;_P__ -_1_ dt p-2I (-ll rp-k-1 p-k-l + (-l) p log (l - r). k=O (p-k-1) (1 - r) Jo (1 - t) P Thus k (1 - r)k+lk + (-l) p (1 - Pr) p-1 log(l - r) + C(l - Pr) p-1 (-l) I 51 = pl p-2 r 1 k=O (p-k-l)r r Multiplying both sides of the last equation by rP , we get C if r Therefore S = 1/(p - 1) p

·

0

·

0

0.

PROBLEM 94. Show that 1 for x l 1.

X

I

>

From the binomial expansion theorem we have 1•3 u3 - 21•3•••(2n-3) - 2•4•6 •4• 6• • • (2n) un for lui < 1. Thus, for ltl < 1, 1•3 t + ••• 1 - (1t - t 2 !2 t + 2 1· 4 t3 + 2·4·6 (E. 73) But (2x)/(l x2 ) 1 for all real numbers x with equality only if x 1 Sol ution .

--

5

__ __

+

�

)

!.<

2

REAL FUNCTI ON S

431

because 0 (1 - x) 2 for all real x. Setting t (2x)/(l + x2 ) with lxl 1 2 in (E. 73) we get the first part of our claim because {1 - (1 - t ) 2} /t x. 2 Finally, for x 0 the expression (2x)/(l + x ) does not change if we replace x by 1/x and so the second part of our claim follows as well. =

$

!<

#

PROBLEM Let a be a parameter with 0 < a 1, and define fa (x) = [a/x] - a[l/x] for 0 < x < 1, where [t] denotes the integer part of t. Show that 1 f (x) dx a log a a and, for m a positive integer and � (m + 1) .,"'."'k=l k-(m+l) ' i0l xm f (x) adx = (am+l - a) �m(m+ + 1 1) We write fa (x) = - (a/x - [a/x]) + a (l/x - [1/x]) . Since, for j 1, 2, 3 , . . a/j (a/x - [a/x]) dx aj( 1/j (1/x - [1/x]) dx fa/(j+l) 1/(j+l) it follows that 95 .

Jr o

$

=

Sol ution .

.

'

=

We set I(m) = (01 xm (a/x - [a/x]) dx. Since fa/a/j(j+l) xm (a/x - [a/x]) dx = am+lf1/l/j(j+l) xm (1/x - [1/x]) dx, we obtain =

Jr

<

=

CHAPTER 4

432

But

m+l )(I(m) - 1/m).

(a - a

1/n (m + l)xm dx (m + 1) 10 1 [1/x]xm dx = n=looL n /1/(n+l) 1 _2m+l1 _ + _3m+l1 _ + r,;(m + 1) +

and

l m1 I(m) = J( [1/x]xm dx. Note that if P is a real polynomial, then by Problem 94 la l P(x) fa (x) dx =. 0 for all (0,1] only if P = 0. It is easy to see that 1 l lim (n/k { (1/x - [1/x]) dx lim { (1/x - [1/x]) dx [n/k]) = I n Jo0 n-r oo } l/n n -r oo k=l = 1 - nlim-roo (1 + 1 + 1 + + n - log n} = 1 where is Euler's Constant (see Solution to Problem 14 in Chapter 3 or Prob­ lem 113 in Chapter 3) . -

-

Remarks .

a E

.!_

=

2

3

.!_

c,

c

PROBLEM Let n be a positive integer larger than 1 . Show that {}aab (x - a) n (b - x) n dx = 2·�3•·±5 ·•�7 •• • 2n + 1 (�) 2 2n+l and, setting 2n 96.

�-

2il+T'

REAL FUNCT IONS

433

verify that n+l l2(n + l)K Since, for k an integer larger than 1, Jsink x dx /sink-l x d(-cos x) sink-l x cos x + J(k - l)sink-2 x cos 2 x dx sink-l x cos x + (k - 1) I (sink-2 x - sink x) dx we see that S l. nk-1 X COS X (k - 1) Isink - 2 X dx. Thus �Tr/2 {1r/2 k Jo0 sin x dx � k 0 sink-2 x dx for k 1 and so {1 Tr/ 2 sin2n+l x dx = 2n 2n - 2 -32 1oTr/ 2 sin x dx K 0O 0 Substituting x a cos2 t + b sin 2 t with t [O,Tr/ 2], we obtain x - a (b - a)sin2 t, b - x (b - a)cos 2 t, and dx 2(b a)sin t cos t. Therefore 2(b - a) 2n+l J{0o Tr/ 2 (s1n. t cos t) 2n+l dt � 4.

Sol ution .

=

= -

+

>

z·

z!1+T2Il="l"

=

=

E

=

Now, by Problem 15, if f and g are continuous function on [a,b] and g is monotonic, then there exists a point t in [a,b] such that

CHAPTER 4

434

ib f(x) g(x) dx g(a) it f(x) dx + g(b) 1b f(x) dx. =

Putting f(x) (x - a) n and g(x) (b - x} n , we obtain {} b (x - a) n (b - x) n dx (b - a) n t (x - a) n dx (b - a) n (t n +a) n+l a 1a But {} b (x - a) n (b - x) n dx K (b - a) 2n+l a where ·.i_·.§_ 2n K z .�3·5·7 Hence t a + b - a n+l 12(n + l}K. Putting f(x) (b - x) n , g(x) (x - a} n , we obtain 1b (x - a) n (b - x) n dx (b - a) n 1b (b - x) n dx and the intermediate value t' b - � n+l lz(n + l}K. The intermediate values to and t' are connected by the relation t + t' a + b. Since a s t s b and a s t' s b, we see immediately that n+l /2(n + l}K s There is equality when n and n 12(n + l}K as n =

=

_

=

=

2

=

------

,

2il+l"

=

r;:-;--....,.-:;­ ...,-

4

=

-----

=

=

=

=

4

=

4

.

0

PROBLEM

97.

Show that

+

1

r=-..,.--..,.,.-=

+

1

+

oo .

1

REAL FUN CTIONS

435 Let

Sol ution .

Then Since

and bn,k

1

an,k+l - an,k bn,k+l - bn,k

k 1+ 1 log (1 + _kl) < _kl or n 1 og (1 + 1) n n and get that n lim log l + � or klim+co ( 1 + �t = exp ( klim+co �) . k +co Letting n + "' in such a way that n/k tends to 1 as k + "' , we get the desired limit with the help of the result in Problem 39 of Chapter 3. <

k+l <

(

k

)

PROBLEM 98.

Let

le

Show that Tn + as n + Let Ip=l log ( 1 + ...E.n2 ).. . n Since x 22 log(l + x) x for x we get, taking x = p/n2 , "' ·

Sol uti on . s

- !_

<

<

> 0

p=lI �n � p=lI �n p=lI log (1 + _E_n2) p=lI �-n -

<

<

< k'

CHAPTE R 4

436

But

n p n(n 1) implying nL � as n + oo 2 p=l p= 1 n2 2n 2 and nl: p2 < n•n2 implying n £_2 < �3 .!. -r o 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 \

<

1,

1 + !2 x 2 + 21 •• 43 X4 + 21 •• 43 •• 56 X6 + • • •

and so X

2n+ l

X

2n+l + 1 2n+3 + 1 • 3 2n+5 + 1 • 3 • 5 2n+7 + • • • 2•4•6 X 2•4 X 2X

CHAPTER 4

438

Hence 12 2n 1 4 x2n+l dx = 1 11 v� 2n 2 -·--the other hand, setting x sin t, lol V1x2n+l 2 dx = 1rr/ 2 sin2n t dt ---

+

+

+

+

On

-

1•3•5 ·----·--2•41•3 2n 1 6 -2•4•6 2n 1 8 +

+

+

3• 52·4·6 • 7 ••••••(2n(2n) 1) +

X

PROBLEM 103. Show that = i-(� - log 2) . 2·51 8•111 14·171 We have, for x i < 1, 6 9 12 15 ) --x l - 3 = 3 1 4 7 10 13 16 - + -- + -- +

I

Sol ut:ion .

(

X

+

Thus

X

X

+

X

-

+

X

X

X

+

X

-

+

X

+

X

X

+

X

X

...

...

3 (-2•1-5 _8•111 _ _14•171_ ••• ) . But 1 {}0 1 1x d\x = t { log x2(x- x1) 2 1 2 13 arc tan � } 1 13 0 = �(� - log 2) . +

+

+

+

+

+

+

PROBLEM 104. Solve the system of ten equations y 2 px qy rz su tv 3 X

+

+

+

+

Z

+

+

U

+ V

+

·

+

439

REAL FUN CT IONS

p

3x

p

+ q 3 y + r 3 z + s 3u + t 3

5 x + q5 y 6x

p

v

+ r5 z + s 5 u + t 5

+ q 6 y + r6 z + s 6 u + t 6

p

7x + q 7y

p

8x

+ r7z

+ q 8 y + r8 z

Sol ution .

+ s 7u

8 + s u

31

v

235

v

674

+ t7

v

1669

+ t 8v

4526

First we consider the general system

l + • • • + x y 2n- l + x2 y2nn n 2

l x l y2nl

a 2n "

We let F (e)

xl 1 - ey l

x + 1 - 2ey + • • • + 2

But

X

n

1 - eyn

xl 1 - eyl

x 1 ( 1 + ey 1

+ e 2 y i + 8 3y3l + • • • ) '

x2 1 - ey 2

x 2 ( 1 + ey2

+ 8 2 Y22 + e 3Y32

X

n 1 - eyn

xn ( 1

+

+ eyn + e 2 yn2 + e 3yn3 +

. . .

· · ·

)'

)

.

CHAPTER 4

440

Thus

+

+

• • •

• • •

+

+

• • •

and s o

Reducing the fractions t o a common denominator, w e find F (8)

Hence

Therefore

+

An e n- 1

441

REAL FUNCT I ONS

Since the quantities a1 , a 2 , . . . , an ' an+l ' . . . , a2n are known , the last e­ quations enable us to first find B 1 , B 2 , . . . , Bn and then A 1 , A2 , . . . , An . Knowing the Ai and B i , we can construct a rational function F (8) and then expand it into partial fractions . Doing so, we get pl p2 p3 + + 1 - q l 8 1 q2 8 1 - q3 8 +

F (8)

-

. . .

P + 1 - n 8· �

It is clear that xl X

=

2

Xn

pl '

yl

ql ;

p2 '

y2

q2 ;

Pn '

Yn

=

�·

The general system is solved. For the given case we have 2 + 8 + 38 2 + 28 3 + 8 4 1 - 8 - 58 2 + 8 3 + 38 4 - 8 5 "

F ( 8)

Expanding into partial fractions , we obtain the following values for the un­ knowns : X =

y z

=

u

=

v

=

5'

3

p

-1,

1 8 + IS 10

q

3 +IS -2

18 10

r

IS

- 8 + IS 2 IS

s

8 - IS ' 2 IS

t

--'

Remark .

'

IS 3 --2 '

=

IS -- 1 2-

'

IS +-1 - 2

Problem 104 and its solution is due to Ramanuj an .

PROBLEM 105 . Let k be a nonnegative integer and 0 b J:a xk dx as l imit of a sum.

<

a

<

b . Evaluate

CHAPTER 4

442 Sol ution .

< aq <

a

Put q aq2

=

n lbli and consider the partition

<

of the interval [a,b ] . We have to sum ak (aq - a) + (aq) k (aq 2 - aq) + ak+l ( q

•••

l) { l + qk+l + q2 (k+l) +

__

(k+l) - 1 qn ak+l ( q - 1) ..::L. .-, qk +--;-1 - 1 .::_ 1 k k1 + q + q

•••

+ ( aqn-1 ) k ( aqn - aqn-1 )

•••

+ q (n- l) (k+l) }

+ 1

But l imn + oo n ib/a = 1 (see Solution of Problem 18 of Chapter 3) and so l im ( qk + qk-1 + n + oo

•••

+ 1)

=

k + 1.

To show that Ji b 1/x dx = log b for b > 1 , we can proceed as _ fol lows : we choose partition x l. (b) 1 /n with i = 0 , 1 , 2 , . . . ,n and consider the sum Remark .

=

n (b) l/n (b) (i-1) /n i=l (b) (i- 1 ) /n

l:

_

n L [ (b) 1/n - 1 ] i=l

n[ (b) l/n - 1 ] .

But l im n[ (b) l/n - 1 ] n + oo

PROBLEM 106 .

log b .

Give an example of a series

L fk (x) 00

k=l

o f functions continuous on a c.l osed bounded interval [a,b] that converges absolutely and uniformly for which the Weierstrass M-test fails . Sol ution .

On the interval [0 , 1 ] let the function fk be defined by

REAL FUN CT I ON S

443 $

_

<

2i<+1

1

Zi(::"l;

1

fk (x)

0

for 0

fk (x)

k

for x

fk (x)

is defined linearly in the intervals

1

and

x

and

1 2k ;

[ A · zk 1- 1 ]

·

[ 2k 1+ 1 '2�J

I t is easy to see that the series L �= l fk (x) satisfies the fol lowing condi­ tions : (a) the series is uniformly convergent , (b) any rearrangement of the series is uniformly convergent (i . e . , the series is uni formly and absolutely convergent) , (c) L � = l Mk diverges , where Mk is the upper bound of [ fk (x) [ as x ranges over the interval [0 , 1 ] . PROBLEM 107. Let b 1 � b 2 � • • • � bn � 0 . Show that a necessary and 00 sufficient condition that the series Ln= 1 bn s in nx should be uniformly con­ vergent throughout any interval is that nbn + 0 as n + oo To see that the condition is necessary, observe that , if x = TI/ (2p) , and n = [Y,p + 1 ] , where [x] means the integral part of x , Sol ution .

bn s in nx + bn+l s in (n + l) x + >

bp (sin nx +

•••

+ s in px)

>

+ bp sin px bp (�p

l) sin �TI ,

since there are at least Y,p - 1 terms in the bracket , in each of which mx �TI . Since the given series is uniformly convergent in an interval includ­ ing the origin , the left-hand side of the above inequal ity tends to zero as p + oo . Hence pbp + 0 . Next we show the sufficiency of the condition . It will be enough to consider the interval [ O , TI] , since each term of the series is an odd func­ tion and has the period 2TI . Consider the sum >

5n , p

bn sin n x +

+ bp s in px,

where now n and p are unconnected . Let Tn

sup{mbm : m

� n},

so that Tn

+ 0.

CHAPTE R 4

444

If x

�

u/n , we apply Abel ' s Inequal ity (see Problem 58 of Chapter 2) . We have ••• +

. nx + I Sln

l

x - cos (r Sl·n rx l = cos (n - Y,) 2 s in Y,x

+

Y,) x

l - __ 1_ sin Y,x <

for all values of n and r , and, s ince (sin 8 ) / 8 is steadily decreasing for 0 < 8 < Y,u , 1/ (sin Y,x) � u/x, and we deduce that n <- ­ b u X

� nbn � Tn

•

I f x � u/p , we have , s ince sin

I f u/p

x

<

<

8

�

8,

u/n , we combine the two arguments . We have

and, applying Abel ' s Inequal ity (see Problem 58 of Chapter 2) to the second part , and the other method to the first part , obtain I sn , p I � kTnx + b k + 1 u/x � Tn [kx +u/ { (k + l ) x} ] .

Taking k = [u/x] , we have I sn , p I � Tn (u + 1) . Hence in any case I sn , p I � ATn ' the solution is complete . and , s ince Tn + 0 as n + oo ,

PROBLEM 108.

Let

f(x)

for

X r

0 , and f (O)

=

0.

Show that al l the derivatives of f have the value 0 at x Sol ution .

f ' (0)

0.

Cl early, f i s continuous at x = 0 . By definition ,

l im f (x)X - 0f (O) x+O -

- 1/x2 e l im --x- · x+O

But l im x+O

2 e - 1/x X

---

1 1 2 x = l im x 2 - l im x + O e l/x x + O - 2 e l/x2 3 X

This shows that f ' (O)

0.

l im x+O

0.

445

REAL FUNCT I ONS

�

To deal with higher derivatives we note that, if x 0, f' (x) = 2 x-3 e-1/x2 , f"(x) = x-6 e -l/x2 (n) (x) is a linear combination of It is easy to see, by induction, that f -l/x2)/xm with 0 < m 3n. Consequently, to see that terms of the form (e f (n) (O) = 0, as well as to see that f (n) is continuous at x = 0, it will be sufficient to show that -1/i = 0 e lim _ x+O m for all positive integers m. But -m - lim - mx-m-1 = m 1 . m x-m+2 ; x _ lim x + O e l/x2 x + O - 2 x-3 e 1/x2 z- x � o e l/x2 after a finite number of steps the exponent in the numerator will be posi­ tive, and then the limit is seen to be 0. 4

$

_ _

X

_ _

PROBLEM 109. Let (aj ) ;=O be a given sequence of real numbers. Show oothat there is an infinitely often differentiable real-valued function f on [O, ) (j) whose support is contained in the interval [0,1] such that f (O) = a.J for j 0,1,2, . . . - By the support of f we mean the closure of the set {x: f(x) f 0}. We first consider two lemmas. 1: There is an infinitely often differentiable function g: [O, oo) 1 + R with g(O) = 1, 0 g 1, and whose support is contained in the interval [O,a] with a 0. Let h: R1 R1 be defined by h(x) exp(-1/x2) when x 0 and h(x) = 0 when x 0. Then define - x) - a/2)-" g(x) = h(a - x)h(a+ h(x Clearly, g is infinitely often differentiable, g(x) = 0 for x � a, g(x) 1 for x a/2, and 0 g(x) 1. This proves the lemma. =

Sol ution .

LEMMA

$

$

>

$

$

>

+

Proof.

$

$

CHAPTER 4

446

2 : For any integer n � 1 and any E > 0 , there is an infinitely often differentiable funct ion f : [O ,oo) - R 1 whose support is contained in the interval [0 , 1 ] such that LEMMA

(i) (ii)

f (k) (O)

f (n) (0) = 1 , I f (k) I

<

0 for k < n ;

for k < n .

E

Take g as in LEMMA 1 with support contained in [O , a] and also a function v as in LEMMA 1 with support contained in [0 , 1 ] . Define Proof.

f(s) = v (s)

fs

t r n 0 J r0

•••

( t3 ( t2 Jn0 Jn0

and choose the number a small enough such that (ii) holds . Obvious ly (i) hol ds . This proves LEMMA 2 . Returning now to the problem, we define

JJ where f . and c . are defined recursively : f . : [O ,oo) R 1 , f . infinitely often J 2 , c0 J J J differentiable and with support contained in [0 , 1 ] , f0 as in arbitrary . Then by 2 we can choose f . in the fol l owing way : J f = L c . f. , j=O

+

LEMMA

LEMMA

(1)

fn(n) (O)

(2)

(n) (O) cn = an - [ cn 1 fn-1 -

(3)

l cn fn(k) I

1,

�

fn(k) (O)

0 for k + ••• +

< n;

c 0 f0(n) (O) ] ;

l/2n for k < n .

By (3) , the series E ;=O c j fj k) converges uniformly for every k . On the other hand E�- O c . f . (O) = 0 . Therefore , f is infinitely often different iable and its support is contained in [0 , 1 ] ; moreover, f (k) (O) = E00. = 0 c . f (k) . (0) . By (1) ,

J- J J

J JJ

k f (k) (O) = L c . f � k) (O) j =O

JJ

= [c 0 f0(k) (O) Remarks .

+ ••• +

(k) (O) ] c k- 1 fk-1

+

ck fk(k) (O)

The result in Problem 109 can readily be extended to the

447

REAL FUN C T I ON S

fol l owing claim: Given a sequence of real numbers (aj ) �= O ' there exists an infinitely often differentiable function f : R 1 + R 1 with support contained in the interval [ - 1 , 1 ] such that f (j ) (O) = a . for j = 0 , 1 , 2 , . . . J By taking a sequence (aj ) j =O such that <X>

l im sup C i an i ) l/n = + a> n+m and constructing f as in the foregoing claim, it fol lows that there exist infinitely often differentiable functions f: R 1 + R1 that cannot be repre­ sented by power series .

PROBLEM 110 .

10 1

X

-X

Show that

:E 1 m= l mm

dx

Sol ution .

Since

x -x = e - x log x ' we see that X

-X

=

1 +

I

n=l

the series is uniformly convergent for 0 the series

<x�

1 as it can be maj orized by

Observing that (-l)n

n! ' ( n + l) n+l

we obtain what we have set out to do . Remarks .

In an ent irely s imilar fashion one can show that

I

m=l

... ( - l) m+ l ...!. mm

CHAPT E R 4

448

Using Simpson ' s rul e , the integrals

i0

l

x

-

X

dx

can be calculated with a high degree of accuracy .

PROBLEM 1 1 1 . Let f b e a continuous funct ion on a closed bounded inter­ val [a,b] and let fh denote its Steklow funct ion (see Problem 77 of Chapter 2) . Show that

f

b I fh (x) - f(x) I dx l im h -+ 0 a

0.

(E . 74)

Let h be sufficient ly small so that the interval [x - h , x + h] is contained in the interval [a, b ] . Then , by the result in Prob lem 15, Sol ution ,

fh (x)

=

f

x+h 1 Zh x-h f(t) dt

=

f(s) .

where s E [x - h , x + h ] . Thus , for x E (a,b) , l i� -+ O fh (x) = f(x) . Since f is continuous on [a,b ] , it is bounded . But l f(x) I s M for al l x E [a,b] im­ pl ies l fh (x) I s M for all x E [a,b] . Hence l fh (x) - f(x) I s 2M for all x E [a,b ] , independent of the parameter h . Using the result in Problem 35 we see that the order of taking the l imit and integration can be interchanged in (E . 74) , completing the proof .

PROBLEM 1 1 2 .

Define a sequence vn = vn (x) recursive ly by

vn+l

=

.!.)

( 2 + n vn - 1 , n

�

1.

It is not hard to show that the sequence v 1 , v 2 , v 3 , . . . converges for at most one real value of x . Find x such that v 1 , v 2 , v 3 , . . . converges . Sol ution .

vn where

1•3

By induction we have •••

( 2n - 1) ( x - sn ) (n - 1) !

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

449

REAL FUN CTI ONS

n! sn+l = s n + 1 • 3 • • • (2n + 1) .

and

I f v 1 , v2 , v 3 , . . . converges , then x - sn + 0 as n + oo , whence n! � = 2n - 1 · L n=l 1 · 3 • • • (2n + 1)

x = l im sn

(See the Remark at the end of the Solution . ) For this x we find that n (n + 1 ) n_ 2n + 1 + (2n + 1) (2n + 3) + • • • '

vn

_ _

and s ince the k-th term here is increasing and tends to 2 -k , it follows that 1 + ••• = 1. 1 + 4 vn tends to 2 Remark .

To see that

n! L n=l 1 • 3 • • • ( 2n + 1) we note that , for m

f

n

/2

0

. 2m+l S ln

X

TI 2- 1

1,2,3, . . . dX

_

-

,

2·4•6 2m 1 • 3 • 5 • • • (2m + 1)

1 • 3 • 5 • • • (2m + 1)

(see the Solution of Problem 96) and so 1 1•2 + 1•2•3 1•3•5•7 + 1 · 3 + 1 • 3•5 =

l ( TI/2

--- ---

. 5 Sln 3 X + S ln --2 22 •

X

. 7 + S ln 3 2

X

+

) /o _ dx

· · ·

2

TI I 2 __::s..::i;:.:: n....;x:.;; _ 2 - dx - 1 . 0 1 + COS X

Setting tan � = t , the l ast integral becomes 4 and we have obtained the desired result .

PROBLEM 1 1 3 . g (t)

Show that if g (2t)

�

g (O) • s i t .

�l

( 1 + t 4 ) - l t dt

g (t) cos t , then

n/ 2

450

CHAPTER 4

I f in g (2t) = g (t) cos t we replace successive ly t by t/2 , n we obtain the fol lowing equalities : t/4 , . . . , t/2 , sol ution .

g (t)

g (t/2) cos (t/2) ,

g (t/2) = g (t/4) cos (t/4) ,

Hence g (t/2n ) cos (t/2) cos (t/4)

•••

cos (t/2n )

g (t) .

But cos (t/2) cos (t/4)

•••

s in t t/2n t s in (t/2n )

co s (t/2n )

= -- --'---'--

(see Solution of Problem 99 in Chapter 1) . Thus n t/2 sin t ---'-­ t s in (t/2n ) As n

+ oo ,

t/2n tends to zero ; g (t/2n ) tends to g (O) and

tends to 1 . Thus g (O) {sin t }/t

PROBLEM 114 .

x n

(i)

+ oo

0 show that

I f a > 0 , xn 1 when a > e e

�

a

�

1 e e ;

< -e n does not tend to a l imit when a e

X

(iii)

Moreover , in ( iii) , x 2n =

g (t) .

-e xn tends to a finite limit when e

(ii)

aA

g (t) .

B,

aB

A.

+

A, x 2n+l

+

B , where A , B sat i s fy

451

REAL FUNCT I ONS

We have

Sol ution . X

X -X n _a = a n n-1 x a n- 1 __

Suppos e a > 1 . Then xn+l > xn if xn > xn - l ' Now x 2 = 1 > 0 = x 1 , so that xn increases with n and must tend either to a limit L or to infinity . Suppose k exists such that ak = k . Clearly k > 0 and k-xn

k

Thus k > xn+l if k > xn . But k > 0 = x 1 . Hence if there exists any root k of ax x it follows that xn + L � k . But L

lim x n + oo n+l

X lim a n n + oo

aL .

Thus if there is any real root of the equation ax root ; if there is no root , xn + oo The equation

tends to the least

ax = x (x > 0 , a > 0) .

Putting X y = aX

we have 1 1 dy y dx - log a - X . -

��

�

�

Thus if a > 1 , < > 0 according as x < = > lo a and x = lo a gives a minimum value e l og a for y . I f a > 1 , then for large x , and for smal l posi­ tive x, y > 1 . Hence ax = x has 2 , 1 , or 0 roots according as e log a < = > 1 ; is always negat ive , and the equation i . e . , as a < = > e l/e . I f a < 1 , ax x has one and only one root . The case a = 1 gives xn 1 and therefore xn + 1 . We now turn to the case a < 1 . Our original equation shows that i f xn > xn- l then xn + l < xn and so xn + Z > xn+l ' We are led to consider separately the odd and even successions x 1 , x 3 , x5 , . . . and x 2 , x4 , x6 , . . . The equat ion

��

CHAPTE R 4

452

x -x a n+l n- 1 shows Since x2n +

(by induction) that these are respectively increasing and decreasing . xn is essentially positive it follows that x 2n- l + B and hence that A � B . We have

A

k I f k is any root of aa x ak -a n-2 a X n

.!.. =

A aa

l im n + co

l im x n + co 2n+2

k we have clearly 0 < k < 1 . Also

=

so that xn < > k according as xn _ < 2 X greatest and least roots of a a x.

>

=

The equation

B aa

and similarly B

X

a a = x (a < 1 , x

>

k . Thus A, B are respectively the

0) .

There is always one root of this equation - the root k of ax = x which we have shown to exist . For convenience we put a = e -b . Taking logarithms twice we have - bx + log b

1 log log x"

Put y

=

1 - log b , bx + log log x

s o that

�= dx

b ­

1 1" x log x

Now

so that x log

� has

a maximum value

�� is

maximum value b - e . I f b � e , only of the equat ion y 0 . Thus

=

� when

A=

log

�=

1, x =

!

Hence

� has

a

always negat ive , and there is one root at B if a � e -e . If b > e consider

��

453

REAL FUNCT IONS

the point x = k , where k is the root of ax already referred to. Then k log a

=

log k,

that is ,

x (and therefore of a

aX

x)

log k ,

- kb

and so dy b 1 dx 1 x log x -

b -

1 . k2b

To decide the s ign o f this expression w e note that the increasing function bx + log x is negative when x = 1/b and b > e . Thus 1/b < k , the only value > 0 . It follows that y of x for which bx + log x 0 . Thus , when x k , k + o and y 0 for x = k - o , o being smal l . But y > 0 for smal l > 0 for x values of x and y 0 when x approaches 1 . Thus there is a real root of the equat ion y 0 in each of the interval s (O, k) , (k, l) , and it fol lows that A > B . Thus x does not tend to a l imit . The properties ( i) , (ii) , and ( ii i) n are therefore proved . From these results and the fact that

<

��

<

=

xn+l

=

X a n

it fo llows at once that Remark .

a B and B

A

aA .

Compare Prob lem 1 1 5 with Problem 20 in Chapter 3 .

PROBLEM 1 1 5 . Let f b e a polynomial without constant term, defined on the interval [ a , x] . We insert , between a and x , the n - 1 geometric means . and denot e by

A

.

a

'

V (x/a) n

1,

the arithmetic mean of the values f(x) .

f(a) , On

.

the other hand, we ins ert , between a and x , the n - 1 arithmetic means ql

=

x - a a + n -,

q2

=

a + 2 • �' n

.. ' .

and denote by B the arithmet ic mean of the values

�- 1

x ---, - a a + (n - 1) · n

CHAPTE R 4

454

f (
f(a)- ' a

a

'Il - l

'

f (x)-· x

Show that , as n + oo , the fract ion B/A tends to a l imit which is ent irely in­ dependent of the polynomial f; this limit is log �a x - a

--.

Suppose first that

Sol ution .

f (x)

=

xm+l ,

where m is a positive integer . Then x - a) m + • • • + ( a + n x --a) m , am + (a + -n n-

B = n +1 1

--

or, expanding and setting

we have B

sl m- 1 (x a) m am + I (n + l ) n a s2 + m(m1 • 2- 1) --�--=am- 2 (x - a) 2 + • • • 2 (n + l ) n sm + m (m 1 -• 2 1). . . m 2 • 1 (x - a) m . (n + l ) nm

By the result in Problem 15 of Chapter 1 , we have (p + l) Sp

=

(n + l) p+l - (p +1 · l2 ) p sp- 1 (p + l ) p (p - 1) s 1·2·3 p-2 - (p + 1) s l - (n + 1) '

hence (p + 1) s p (n + l ) np

(�) P n

_

(p + l) p s 1·2 p-1 + • • • + (p + l ) S 1 + (n + 1) (n + l)np

455

REAL FUN CT I ONS

The numerator of the last fraction is of degree p in n ; thus the limit , as n � oo, of this fraction vanishes , and we have (p + 1) s E l im n � oo (n + l ) np

l im (n n � oo

� l) p =

l,

hence l im n � oo (n

s

1 p+T·

l ) np

+

and - 1) m- 2 (x _ a) 2 + • • • am + 1m· 2 am- 1 (x _ a) + m(m 1•2•3 a

l im B n � oo

+ m +1 1 (x - a) m . ______

Multiplying both sides of this equal ity by (m + l) (x - a) , and adding am+ l on both sides , we see that (m + 1) (x - a) am+l

+

(nl�imoo B )

+ am+l

m +---1 am ( x - a) + (m + l ) m am- 1 (x - a) 2 + • • • --1 1 •2

[a + (x - a) ]m+l

+

(x - a)

m+l

xm+l

hence lim B n � oo On

A

xm+ l - am+l (m + 1) (x - a) ·

the other hand, 1 -- [ m+l + ---n + 1 a xm+l

(�) [(�) � J n -

m+l

]

• • • + Xm+l am+l

m 1

(n + 1)

- 1

As n � oo, the numerator reduces to xm+l - am+l the denominator presents itself in the indeterminate form oo x 0. But we can write the denominator in

CHAPT E R 4

456

the form m+l � ( a) n 1

-

1

and, us ing ! ' Hospital ' s Rule twice , we get

l im n + oo

m+ l @ n

-

1

n+l

1

lim (m + 1) n + oo

l im n + oo

m+l � (�) n log � a n2 a 1

2 log � = (m + 1) log x (�) a n a

Thus xm+l - am+l x' (m + 1 ) log a

l im A n + oo

and, consequently, . B 1 1m n + oo A

=

log �a x - a

This limit is independent of m. I f we take f(x) = K · xm+l , where K is a constant , the fraction B/A is clearly not affected ; the factor K introduced in the numerator and denomin­ ator in B/A cancels out . In the case f(x) = x , the fraction B/A also conx) / (x - a) as n + "" , as a simple calculation shows . verges to the limit (log a Final ly, to see that the cl aim is valid when f is a polynomial without constant term, in other words , when f(x) is a linear combination of x, x 2 , x 3 , . . . , xk , we may invoke the result in Problem 2 8 of Chapter 2 .