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David S. Gunderson University ofManitoba Winnipeg, Canada

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A CHAPMAN &: HALL BOOK

Chapter 9

Identities The business of concrete mathematics is to discover the equations which express the mathematical laws of the phenomenon under consideration; and these equations are the starting-point of the calculus, which must obtain fmm them certain quantities by means of others. -Comte,

Positive philosophy. Identities are tools of every working mathematician. Standard identities involving simple sums, products, fractions, or exponents are used by every student; it may be amazing to see how many of these can be proved by induction. Sorne of the more useful identities involve binomial coefficients or trigonometry, and many of these also can be proved by mathematical induction. Exercises below are often roughly grouped by category; however, many exercises could easily faH into more than one category. Many of the identities below have "proofs without words", that is, a pictorial representation. The interested reader can find many of these in Roger B. Nelsen's two delightful books [403], [404], available from the MAA. A few specific references to these "picture proofs" are mentioned below. Many identities regarding integers also have combinatorial proofs; in fact, many of these identities form the heart of combinatorics. Riordan's Combinatorial Identities [454] is a rich handbook for those studying identities that come from counting in different ways. Many of the identities in that text are provable by induction, so Riordan's text might be a great place to start for references or other resources.

9.1

Arithmetic progressions

For the first two exercises, let On denote the nth non-negative odd number and En denote the nth even number. So 0 1 = 1,02 = 3,03 = 5, ... , and El = O, E2 =

125

126

Chapter 9. Identities

2, E3 = 4, .... The first two theorems have been attributed to Maurolycus (see, for example, [91], where the reference is [376]). This next exercise has a direct proof, and proving it inductively is not nearly as straightforward as one might think.

9.1. Arithmetic progressions

127

Exercise 38. Prove that for n :::: 1, 1 + 3 + 5 + ... + (2n - 1) = n 2 . Exercise 39. Prove that for n :::: 1,

Exercise 32. Prove that for each positive integer n,

2 + 4 + 6 + ... + 2n = n(n + 1). Exercise 40. Prove that for n :::: 1, Exercise 33. Use the result in Exereise 32 and mathematieal induction to prove that for ea eh positive integer n,

n

+ (n -

2 + 5 + 8 + ... + (3n - 1) =

n(3n 2

Exercise 41. Show that for n :::: 1,

1) = On.

3 + 11 + 19 + ... + (Sn - 5) = 4n 2 Although I haven't seen the following referenced, it's very likely that the identity in the next exercise was also used by Maurolycus. Its proof is nearly identical to that of Exercise 32. Exercise 34. Prove that for each positive integer n, En+l = En

In Section 1.6, the numbers T n are called triangular numbers. Recalling Definition 2.5.6, the summation notation is ¿~=a Xi =

Exercise 43. Prove that for n :::: 1, =

n(2n + 3).

Exercise 44. Prove that for n :::: 1, Xa

+ Xa+l +

... +Xb·

i=

n.

Exercise 42. Prove that for n :::: 1,

5 + 9 + 13 + ... + (4n + 1)

t

-

+ 2.

Exercise 35. Define Tn = 1+2+3+·· ·+n. Use induetion to pro ve thatforn:::: 1,

Exercise 36. Prove that for any 1 ::; m

+ 1) .

< n,

(n-m)(~+m+l).

i=m+l

A special case of Exercise 36 is the following (see [404, p. 85] for a pictorial representation) . Exercise 37. Prove that for every positive integer n,

(2n + 1) + (2n + 3) + (2n + 5) + ... + (4n - 1) = 3n 2 • The next exercise generalizes nearly every exercise in this section so far. ExercÍse 45 (Summing arithmetic progressions). Let a and d be fixed real numbers. Prove that for each n :::: 1, n a + (a + d) + (a + 2d) + ... + (a + (n - l)d) = 2[2a + (n - l)d].

As in Section 1.6, for each n :::: 1, the n-th triangular number is T n = 1+2+·· ·+n. Exercise 46. Prove by induetion that for eaeh n :::: 2,

The solution to the next exercise was given by l"laurolycus, and Bussey [91] commented that the proof given was "a clear case of a complete induction proof."


128

9.2

Sums of finite geometric series and related series

9.3. Power sums, sums of a single power

9.3

129

Power sums, sums of a single power

Exercise 54. Prove that lar n 2:: 1,

Exercise 47. Prove that lar every n 2:: 1,

2

2

2

2

1 +2 +3 +"'+n =

+ 1) (2n + 1) 6

.

Exercise 55. Prove that lar every n 2:: 1,

Exercise 48. Prove that lar eaeh n 2:: 1, 2

n (n

3

1 + 3 + 3 + 3 + ... + 3

71

1

-

(1 + 2 + ... + n)(2n + 1) = 3(1 2 + 22 + ... + n 2).

3n -1

= --o

Exercise 56. Por n 2:: 1,

2

The next exercise generalizes Exercises 47 and 48: Exercise 49 (Summing a geometric series). Let a and and l' i: 1. Prove that lar eaeh integer n 2:: 1, n

2

l'

rn+l -

be real numbers with a

#

O

1

a + ar + ar + ... + ar = a --_--=-lr For the particular geometric series with (7.3) in Section 7.3.

l'

By virtue of Exercise 35, the equality in Exercise 56 can also be stated as

= a = ~, Exercise 49 gives equation

which was the form in which it appeared in the 1990 Canadian Mathematical Olympiad (a solution appears in [367], for example). For integers k 2:: O and n > O, define n

Sk(n) = Exercise 50. Prove that lar eaeh natural number n 2:: 1,

L ik =

1k + 2 k + 3k + ... + n k .

(9.1)

i=l

For any m 2:: O define Sm(O) = O.

Rere is the same problem stated slightly differently (and with its own solution).

Note: manyauthors (e.g. [230]) define Sk(n) = Ok + 1k + 2k + 3 k + ... + (n - 1)k, which can lead to a great deal of confusion when researching such sums. Identities in Theorem 1.6.1 and in Exercises 54 and 56 say

Exercise 51. Prove that lar every n 2:: 1, 1 + 2·2 + 3.2 2 + ... + n2 n -

1

=

(n - 1)2 n + 1.

Exercise 52. Prove that lar every n 2:: 1, 2

1 + 2 . 3 + 3 . 3 + ... + n3

n-1

=

3 71 (2n - 1) 4

+ 1.

Note: Results in Exercises 50, 51, and 52 can all be found by taking the derivative of the identity in Exercise 49 and, if necessary, shifting exponents by multiplying each side by the appropriate termo Exercise 53. Generalize Exereises 50, 51, and 52 to powers and prove your answer by induetion.

01 an

arbitrary k 2:: 2,

The formula for SI(n) was derived independently, and then proved by induction. The expressions for S2(n) and S3(n) were simply given, and an inductive proof was used to check each. Nichomachus (ca. 100 A.D.) knew of the expression for S3(n). Row were they found? Rere are a few more (each of which also has an inductive proof, for those with the energy): n

S4(n) =

L i=1

i

4

1

=

30 n (n + 1)(2n + 1)(3n 2 + 3n - 1)


130 n

¿

S5(n) =

i

5

1 2 = 12 n (n

+ 1)2(2n2 + 2n -

9.4

1)

i=l

S6(n)

n

7 "i = L i=l

Ss(n) =

S

ü

"i =

=L

9

i=l n

SlO(n) =

¿

i lO

Exercise 62. Por ea eh n 2': 1, pro ve that

n 2 - 4n + 2)

-

I " 6 ¿n i = 90 n(n + 1}(2n + 1)(5n + 15n + 5n i=l n

S9(n)

1

-n 2(n + 1)2(3n4 + 6n 3 24

4

-

15n 3

-

n 2 + 9n - 3)

-n 2(n + 1)2(2n6 + 6n 5 + n 4 20

-

8n + n 2 + 6n - 3) 3

-

33n 2 + 5).

66

Exercise 64. Show by induetion that lor eaeh n 2': 2,

12 +32 +52 +

... +(2n_1)2=

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

Exercise 58. Prove that lor eaeh n 2': 1, 2

2

+ 4 + 6 + .. , +

+ ". +

(n - l)(n

+ 1) =

(n - 1)(n)(2n + 5) . 6

Exercise 65. Prove that lor n 2': 1,

1·2·3 + 2 . 3·4 + 3·4·5 + ". + n(n + l)(n

+ 2) =

1

¡n(n + l)(n + 2)(n + 3).

Exercise 66. Prove that lor eaeh n 2': 1,

~ '(' 1)( , 2)( , 3) n(n + l)(n + 2)(n + 3)(n LJ J + J + J+ = 5

Exercise 57. Show that lor eaeh n 2': 1,

2

1·2 + 2·3 + 3·4 +". + n(n + 1) = _n...:...,(n_+_13'c-'(,-n_+_2....:..) ) .

1·3+ 2 . 4 + 3·5

Each Sm(n) aboye is a polynomial in n of degree m + 1 with O constant termo A general express ion for Sm(n), called "Faulhaber's formula", is given in Section 9.6.3; each is indeed a polynomial with properties shared by those listed aboye. (One development of Faulhaber's formula relies heavily OIl binomial coefficients and induction, so it appears in a later section.)

2

(2n)! 2 . 6 . 10" . (4n - 2) = - , - . n. Exercise 63. Prove that lor n 2': 1,

1

= ~(6nl1 + 33n 9 + 55n S - 66n 6 + 66n 4

i=l

Products and sums of products

n!

=" i6 = -n(n + 1)(2n + 1)(3n4 + 6n 3 - n 2 - 3n + 1) L 42 i=l

S7(n) =

131

Recall that the definition of the factorial functíon ís recursive: 01 = 1 and for n 2': 1, = n· (n - 1)1. The first exercise in this section is not really a sum unless the trivial sum is counted as such.

1

n

9.4. Products and sums of products

+ 4) .

j=l

The next exercise generalizes Exercises 63, 65 and 66.

Exercise 67. Por a fixed k E Z+, show that lor eaeh natural number n 2': 1,

()2

2n

=

2n(n+1)(2n+1) 3

~

.

'('

~J J

+

1)

'"

(' J

k

+ -

1)

=

(k

(k+n)!

+ 1) . (n -

1)!'

Exercise 59. Show by induction that lor each n 2': 1,

1 - 4 + 9 - 16 + . " + ( -1 t+ 1 n 2

=

(-1) n+l (1 + 2 + 3 + ' " + n).

Exercise 68. Prove that lor each n 2': 1, n

¿(2k - 1)(2k + 1)(2k

Exercise 60. Show that lor every n 2': 1,

+ 3) = n(2n 3 +

8n 2 + 7n - 2).

k=l

2

n - (n - 1)

2

+ (n -

2)

2

+'" + (-1)

n-1

2

(1) =

n(n + 1) 2 .

Exercise 69. Prove that lor any n E Z+, n

Note: this equality lollows directly lrom that in Exercise 59 by Theorem 1.6.1 and division by (-1) n+l, however this equality is to be proved by induction without Exereise 59 or Theorem 1,6.1. Exercise 61. Prove that lor each n 2': 1, 13 + 33 + 53

+ .. , + (2n - 1)3 = n 2(2n? - 1).

¿

i 2 2i = 2n+1(n2 - 2n

+ 3)

- 6.

i=l

One might want to compare the next exercise to Exercise 593.

Exercise 70. Prove that lor n 2': O,

O· O!

+ 1 . 1! +

2 . 2! + 3 . 3!

+ ... +

n' n!

= (n + 1)! -

1.


132

9.5

Sums or products of fractions

9.5. Sums or products of fractions

133

Exercise 80. Use Exercise 71 to prove that for n 2': 1,

Exercise 71. Prove that for every n 2': 1, 111 1·2 2·3 3·4

- + - + - + ... +

1 n =--. n(n+1) n+1

then give an inductive proof (that does not rely on Exercise 71).

Exercise 72. Show that for each n 2': 1, 1

n-I

L

~=o

Exercise 81. Prove by induction on n that

1

(n+i)(n+i+ 1) = 2n'

Exercise 73. Prove that for each n 2': 1,

1

1

1

1

-+-+- + ... + n(n+1)(n+2) 1·2·3 2·3·4 3·4·5

=

n(n + 3) . 4(n+1)(n+2)

1

Exercise 74. Prove that for each n 2': 1, 111 +- - 2 3 4

1- -

2n

1 1 1 1·33·55·7

1

1

1 + -1- + ... + -. n +1 n +2 2n

1·44·77·10

1

(3n-2)(3n+1)

n 3n+1'

=--

= --

Exercise 75. Show that for each n 2': 1,

- + - + - + ... +

1

- + - + - - + ... +

+ ... + -1- - -1 2n - 1


n 1 =--. (2n-1)(2n+1) 2n+1

Exercise 83. Prove that for each n E Z+,

t m=l

m(m

m +4 + l)(m + 2)

n(3n + 7) - 2(n + l)(n + 2)"


Related to Exercise 75 is the following: Exercise 76. Prove that for each n E Z+,

~ _1_ = (n - 1) (3n + 2) . ~ i2 - 1 4n(n + 1) 2=2

Exercise 85. Prove that for each n 2: 2,

Exercise 77. Show that for every n 2': 1, 1

1

1

1

+++ ... + n(n+2) 1·3 2·4 3·5

=

n(3n + 5) . 4(n+1)(n+2) Exercise 86. For n 2': 3, suppose that al, a2, ... ,an are positive integers so that all the quotients

Exercise 78. Prove that for n 2': 1, 1

1

1

- + - + - - + ... + 1·55·99·13

1

(4n-3)(4n+1)

n 4n+1

=--. Pn =

Exercise 79. Show that for n 2': 1,

12 22 32 + + + ... 1·33·55·7

+

n2 n(n+1) =. (2n-1)(2n+1) 2(2n+1)

are integers. Show that PI

+ P2 + ... + Pn

:::;

3n - 1.

an-l

+ al


134

9.6

9.6. Identities with binomial coefficients

Identities with binomial coefficients

135

= m (m-1). 8

Recall the definition of binomial coefficients: for integers O ::; k ::; n, "n choose k" is defined by

8-1

Exercise 87. Show by induction that for each n 2 O,

(~) = k!(nn~ k)!' Binomial coefficients are so-called because they are the coefficients in the binomial theorem (see Exercise 104). In other texts this same number is represented by various other notations, including nCk, C'k, and G(n, k). If n < k, or if n < O, then by convention put G) = O. One might note that (~) is defined for other choices of n and k, (for example, when k is rational) but these situations are described only as needed. As it is, the number (~) counts the number of different collections of k objects chosen from a set of n distinguishable elements (see Exercise 419).

Exercise 88. For each n

8

and m be non-negative integer8. Jf O ::;

(m) s

=

8 ::;

~(m-1) m -

8

(

=

and conclude that

S

m - s + 1 ( m ),

8

S

and

S -

1

1) .

m) = m (m ( s s s- 1 Proof:

m! (m - 8)!S!

m (m -1)! m - s . (m - 1 - 8)!S!

~(m-1), m! (m - s)!s!

m-s 8 m! m-s+1 s (m - 8 + l)!(s - 1)! m-8+1 m! 8 (m - (s - l))!(s - 1)!

m-8+1( m ), s-l (m - 1)1 (m - 8)!(8 - 1)! (m - 1)1 (m - 1 - (s - 1))!(8 - 1)! s

mI (m - s)!s!

m s m 8

n

(9.3)

(9.4)

1

I: (i+l) = 2 ~=1

(9.2)

+ 2)

Exercise 89. Prove that for every n 2:: 2,

m - 1, then

Jf 1::; s ::; m,

m)

1, show that

n(n-+ l)(n 6

Certain recursions regarding binomial coefficients are often useful, especially in inductive proofs. Lernma 9.6.1. Let

~

2 n

2

+ l'

1 (i+l) = 2. i=l 2 00

I:

The next exercise asks to prove "Pascal's identity" , one of the most useful identities in combinatorics and discrete mathematics. (It's very likely that the original proof was not inductive per se, but instead based on properties of Pascal's triangle.) There are (at least) two standard simple proofs of the identity, however it may also be proved by induction. (The proof by induction is not nearly so elegant as the counting proof, but Lemma 9.6.1 helps.) In the solution to this exercise, one proof of each kind is presented; see the comments following Exercise 94 for yet another (non-inductive) proof. Exercise 90 (Pascal's identity). Prove that for any fixed r 2 1, and all n 2:: r,

The next exercise has a direct counting proof, and another very simple proof invoking the binomial theorem (see Exercise 104); however, as is the case with many identities involving binomial coefficients, the next exercise can also be solved by induction (where at least one proof uses Pascal's identity). Exercise 91. For any n 2:: 0, prove that

t (7) t=O

n = 2 .

136


The next result was also proved by Pascal, probably cirea 1659 (see [91], where the reference is "Consequence XII, Vol. III, p.248 of [424J"). The result has a very simple direct proof using the definition of binomial coefficients, namely n! k!(n-k)! n! (k+l)!(n-k-l)!

n!(k + 1)!(n - k - 1)! n!k!(n - k)!

k+l n-k'

9.6. Identities with binomial coefficients which is Pascal's identity.

137

o

The next wonderful identity is often attributed to Euler, but is also caHed "Vandermonde's convolution". This identity is stated next in terms of a theorem, together with a simple counting proof; an exercise is to find a purely inductive proof. Theorem 9.6.2 (Euler). For any non-negative intcgcrs m and n and p,

Exercise 92 (Pascal). Using Lemma 9.6.1, prove by mathematical induetion that for each n 2: 2 and all k satisfying 1 ::; k ::; n - 1,

G) (k~l)

k+1 n-k'

The next exercise asks to reprove the result from Exercise 92, again using induction, but in a different (more cumbersome) way. Exercise 93 (Pascal). Using Paseal's identity, prove by mathematieal induetion that for each n 2: 2 and all k satisfying 1 ::; k ::; n - 1,

G) (k~l)

k+ 1 n-k

In the following exercise, it might help to imagine an m x n rectangular array of city blocks, m blocks east to west, and n blocks north to south. AH streets are one-way either northbound or eastbound. The goal is to count the number of ways to drive from the southwest comer to the northeast comer.

Proof: Let X be a set with m + n elements, and fix a partition X = L U R, where ILI = m and IRI = n. If a subset of X with p elements intersects L in i elements, then it intersects R in p - i elements. This can occur in (7) (P~i) ways. Note that if i > m, then (7) = O or if p - i > n, then (p~J = O. Summing over aH i finishes the proof. O

Exercise 95. Find a purely inductive proof of Theorem 9.6.2. A special case of Theorem 9.6.2 leads to what is sometimes caHed "Lagrange's identity" (after Joseph Louis Lagrange (1736-1813), one of the most powerful mathematicians of his time, rivalling even Euler). [This result is special to me, as it was first shown to me by Paul Erdos when 1 was in grad school; this beautiful simple result had somehow escaped my attention until then.J Corollary 9.6.3 (Lagrange). For each non-negative integer k,

Exercise 94. Consider the integer lattiee grid [O, mJ x [O, n], points in the plane with integral coordinates (x, y) where O ::; x ::; m and O ::; Y ::; n. Prove that the number of walks from (O, O) to (m, n) on the grid that go up andjor to the right is Proof: By Theorem 9.6.2 with m = n = p = k,

Though not an inductive exercise, one can use the result from Exercise 94 to give a new proof of Pascal's identity. The idea is to separate the paths from (O, O) to (n, m), into two groups, those going through the point (m, n - 1) and those which do noto The number of paths from (O, O) to (m, n) that pass through (m, n - 1) is (by Exercise 94) Cn+;-l). All other paths must go through (m - 1, n) and the number of such paths is (again by Exercise 94) (m~~tn). Putting p = m + n, then p 2: m and the total number of paths is

It is easy to check by merely writing out the definitions for corresponding binomial coefficients that they are symmetric, that is, for each i = O, ... , k,

and so the result follows directly.

o

Chapter 9, Identities

138

Exercise 96. Discuss why any inductive proof of Corollary 9,6,3 might be more complicated than the inductive proof of Theorem 9.6.2, if indeed one can be found.

9,6, Identities with binomial coefficients

139

Another interesting identity follows from equation (9.5):

In the next two exercises, one must first decide on the ranges for m. Exercise 97. Prove that for each n 2: O,

The proof actually follows fairly easily from equation (9.5), if the left side is rewrítten as

t

J=O

Exercise 98. Show that for each n 2: 0,

(~) t( _l)i (~), J

2=0

and j is used in place of n in equation (9.5), giving O for each of the outer summands except for when j = O. The "well-known" identity in the following exercise relates to something called the "Euler characterístic" (see [315] for details). Conventions dictate that 1 = (~) and for i > n, (~) = O.

C-c/)

Exercise 99. Prove that for any 1

~

m

~

n, Exercise 101. Prove that for m 2: O and any n 2: 0,

Note: Here is an interesting consequence of Exercise 99; this consequence can be proved in other ways.

t(-l)i(~) = (~) + t(-l)i(~) ,=0

Exercise 103 (Binomial theorem, simple case). Prove that for each n 2: 1,

2=1

=

Exercise 102. Prove that for every n 2: 0,

(~) - ~H)j(l:J

=(~)_(n~1)

(with m = 1 in Ex. 99)

The following more general version of the binomial theorem is sometimes called "Newton's binomial theorem" named after Isaac Newton (1642-1727) (e.g., in [499]), but is most often referred to as simply "the binomial theorem".

=0,

and so for n 2: 1,

(9.5)

Exercise 104 (Binomial theorem). Give an inductive proof of the binomial theorem: ¡or each n 2: 1,

Exercise 100. Find a proaf by induction of the identity (Jor n 2: 1)

In summation notation, the binomial theorem can be written (9.6)

that daes not use Exercise 99. Yau may, however, use Pascal '5 identity.


140

Observe that (1 - 1)n = 0, and expanding (9.6) (with x = 1 and y =. -1), equation (9.5) is obtained, that is, ¿~=o( _l)n = 0, offering another solutlOn to

G)

Exercise 100. Exercise 105. Show that differentiating the simple binomial theorem (Exereíse 103)

with respeet to x gives

141


9.6.1

Abel identities

The next two identities are attributed to the Norwegian Neils Henrik Abel (18021829) [1], perhaps most famous for showing that fifth degree equations are not, in general, solvable by radicals. Exercise 108 (Abel identity 1). Por any a E JR and ea eh n 2': 1, show that

t (:)x

i

= nx(l

+ xt-

1 ,

(x

~=o

+ y)n =

t

(~)X(x -

+ ka)n-k.

ka)k-1(y

k=O

then give an induetive proof of this equality. The identities called "difference of squares", x2 - y2 = (x - y)(x + y), and "difference of cubes", x 3 - y3 = (x - Y)(X2 + xy + y2), are special cases of a more

Exercise 109 (Abel identity 2). Prove that 10r eaeh n 2': 1,

general equality: Exercise 106. Prove that for every positive integer n 2: 1,

xn_yn=(x_y)

(~

~x n-v y

Vandermonde's convolution (Theorem 9.6.2) is the special case of b = 0, a = m, and e = n in Rothe's formula [468], given in 1793: For n 2': 1, and any a, bE C,

V-l) .

The idea in Exercise 105 of differentiating a known equality is very powerful. For example, by the binomial theorem,

(x -

1t =

t (~)x71.(_1)71.-k; k=O

t_a_(a+bk) e (c+b(n-k))= a+c (a+C+bn). k=O a + bk k e + b( n-k) n-k a + e + bn n If one is energetic, one might be able to prove Rothe's formula by induction as well. Rothe's equality is a special case of yet an even more general equality proved by Hagen [251] in 1891:

~ a (a+bk) e (C+b(n-k)) 2o(P+Qk)a+bk k c+b(n-k) n-k

differentiating with respect to x yields

n(x _1)71.-1 =

t

(~)kxk-1(_1)71.-k.

p( a + c)

+ aqn (a a+c+bn

k=O

If n 2: 2, using x = 1 gives a fairly remarkable identity, 0=

t

(~)(-lt-kk,

(9.7)

k==O

+ e + bn) n

Perhaps p and q are meant to be positive integers, so maybe with even more effort, one can prove Hagen's formula by induction as well. In [324], Knuth shows how the identities of Rothe, Hagen, and Abel, all follow from the binomial theorem and Vandermonde's convolution. [Note: 1 have not se en if these proofs are by induction.]

which can be used as a base case for proving a family of (perhaps surprising) equal-

9.6.2

ities: Exercise 107. Induct on j to show that for every 1 ~ j

t

(~)(-lt-kkj =

< n,

O.

k=O

A similar identity to that in Exercise 107, but for j 2': n, appears when counting surjective functions in Exercise 595. For many more identities involving binomial coefficients, see, e.g., [225] or [454).

Bernoulli numbers

Jakob Bernoulli (1654-1705) (and independently, perhaps earlier, Seki Takakazuj see [558]) studied the following numbers: Definition 9.6.4. For n 2': 0, define the Bernoulli numbers En recursively: set Eo = 1, and for n 2: 1,

B=~~(n+1)B. n

n

+lL j=O

.

J

J

142



143

35,

The first few Bernoulli numbers are Bo = 1, Bl = -~, B 2 = ~, B3 = 0, B4 = B5 = 0, B6 = (For n 2: 1, B 2n +l = O.) Bernoulli numbers do not seem to have a simple description. One property of Bernoulli numbers needed below is nearly immediate from the definition:

i2.

t (n ~ 1)

= O.

Bj

(binomial thm)

(9.8)

J

j=o

For the reader who knows about exponential generating functions, m+l( =:L m:- 1) Sj(n). J

j=o

and by induction, one can check that this definition agrees with the aboye definition (e.g., see [558]). The interested reader might try to prove by induction the Carlitz identity: for any non-negative integers m and n,

o

Subtracting Sm+l(n) gives equation (9.9). The second proof of (9.9) uses a collapsing sum: n

(n

+ l)m+l

=

:L((n - k

+ 1) m+l -

(n _ k)m+l)

k=O

=

9.6.3

Faulhaber's formula for power sums

For integers k 2:

°and n 2: 1, define

For m 2: 0, one can also define Sm(O) = O. As observed following Exercise 56, for m::; 12, each Sm(n) is a polynornial in n of degree m + 1 with constant term (so n divides Sm(n)). In fact, for any m 2: 0, this fact is true, and is provable by strong induction on m. The induction can be based on the following trick:

°

Lemma 9.6.5. For integers m 2:

(n

t f (m :- 1)

=

~t. (m; l) 0, =

f (m:- 1) j=O

n-k)j

k=O

°

+ l)m+l

(n - k)j

J

k=Oj=O

o Sj(n).

(9.9)

J

Proof: There are two proofs, both relying on the binomial theorem (Exercise 104). Here is the first: for n 2: and m 2: 0,

°

Exercise 110. Prove by strong induction on m that each Sm(n) is a polynomial 01 degree m + 1 in n with constant term O. As noted in the solution to Exercise 110, the following is a consequence of Lernma 9.6.5:

n+l

Sm+l(n)

+ (n + l)m+l

=

:L im+l

Corollary 9.6.6. For positive integers m, n,

i=l n

=

:L(k + l)m+l k=O

(replace i

= k + 1)

(9.10)

144


Is there a way to calculate the coefficients of the polynomial 8k(n)? Since the degree of 8k(n) is k + 1, any k + 2 values uniquely determine the polynomial and this is seen easily with a little linear algebra. For example, if one sets 82(n) = an 3 + bn2 + en + d, the values 82(1) = 1, 8 2 (2) = 5, 8 2(3) = 14, 8 2 (4) = 30 give the system

a+b+e+d=l 8a + 4b + 2e + d = 5 27a+9b+3c+d= 14 64a + 16b + 4d + d = 30. which has a unique solution a = ~, b = ~, e = ~, d = O. In general, the aboye technique works since the coefficient matrix used is a Vandermonde matrix and so is invertible (see Exercise 661). However, this technique does not yield any simple formula for the coefficients. See [157] for more related matrix methods. Another way to find a polynomial of degree n - 1 that fits n values is to use the Lagrange interpolation formula: if f : lR -> lR is a function and Xl, ... , X n are distinct, then

is a degree n - 1 polynomial so that for each i, p(Xi) = f(Xi)' The following express ion for Sk(n) is now eponymous with Johann Faulhaber (1580-1635), who published a form in the 1631 edition of Academiae Algebrae [186] (though the form here is due to Bernoulli). See [116, p. 106] and [325] for more references. Theorem 9.6.7 (Faulhaber's formula). For íntegers k .2: O and n .2: 1,

9.8. Thgonometry identities

145

For integers O :::; k :::; n and a prime power q, the Gaussian eoejJicient

is the number of k-dimensional subspaces of the n-dimensional vector space Over the field GF(q). These coefficients are sometimes called the q-analogues of binomial coefficients (and this can be justified by examining limits as q -> 1 below). It is known that = (qn - l)(qn-l _ 1) ... (qn-k+l - 1) k q (qk - 1) (qk-l - 1) ... (q - 1)

[

n]

and from this, one can also directly prove a q-analogue of Pascal's identity

Using the last equality, the statement in the next exercise is (relatively easily) proved by induction on n (as in [94, p. 127]). In Definition 2.5.9, product notation for natural numbers was introduced; the same notation applies to real numbers and polynomials since multiplication is associative: rr~=l Xi = XIX2" . X n · [See Exercise 4 for proof that multiplication of positive integers is associative.] Exercise 112 (q-binomial theorem). Prove that for n .2: 1,

9.8

Trigonometry identities

Recal1 the three main identíties in trigonometry:

No solution is given for the following exercise; solving it uses sorne tricky sums involving binomial coefficients; the reader is recommended to see [230]-however, be warned that their notation is different than that used here. Exercise 111. By induction on k, pro ve Theorem 9.6. 7.

9.7

Gaussian coefficients

This next definition, similar to that of the binomial coefficient, relies upon the versatile notion of a finite-dimensional vector space over a finite field (see Sections 19.4 and 19.5 for definitions). The notation and terminology varies in the literature.

sine o: + ,8)

sin(o:) cos(,8) + cosCo:) sin(,8)

(9.11)

cosCo: + ,8)

cosCo:) cos(,8) - sin(o:) sin(,8)

(9.12)

1.

(9.13)

sin 2 (B)

+ cos 2 (B)

From these, all other standard trigonometric identities follow. For example, a standard trigonometric identity is: tan(a + b) =

tan(a) + tan(b) 1 - tan(a) tan(b)

Rere is one derivation of equation 9.14: tan(a+b)

sin(a+b) cosCa + b)

(9.14)


146

sin(a) cos(b) + cosCa) sin(b) cosCa) cos(b) - sin(a) sin(b)

9.8. Trigonometry identities

147

Exercise 117. Prove that 101' any n ~ 1 and any angle B,

cosCO + ní'r)

+

sin (a) cos(b) cosCa) sin(b) cosCa) cos(b) cosCa) cos(b) 1 sin(a) sin(b) - cosCa) cos(b)

= (_l)n cosCO).

Exercise 118. Prove that lar eaeh n ~ 1, and any angle O that is not a multiple 01 2n, sin(n+lo) sine nO) sine + sin(20) + ... + sin(ne) = : ( /) 2 . sm e 2

+

sin(a) sin(b) cosCa) cos(b) 1 _ sin (a) sin(b) cosCa) cos(b)

Exercise 119. Pro ve that lar n ~ 1, and any angle O that is not a multiple 0/2n,

cos(n+lo) sin(nO) cose+cos(2e)+···+cos(nO)=. s~n(B/2) 2 .

tan(a) + tan(b) 1 - tan(a) tan(b)

Exercise 120. Prove that lar eaeh n ~ 1, and any angle O that is not a multiple 01 n,

Remembering that tan( -b) = - tan(b),

tan(a) - tan(b) tan(a - b) = 1 + tan(a) tan(b)

(9.15)

sin(e)

+ sin(3e) + ... + sin«2n _

1)0)

=

s~~:~;~).

Various other identities are required in the following exercises, and the reader will benefit most by deriving each at least once. Complete solutions using induction are given for aH exercises in this section, though some may have direct proofs.

Exercise 121. Prove that lar eaeh n ~ 1, and any angle e that is not a multiple 01 n, sin(2ne) cos 0+ cos(30) + ... + cos«2n - l)e) = . (O)' 2sm

Exercise 113. Show that lar every positive integer n,

Exercise 122. Fíx some O that is not a multiple 01 n. Let So n 2: 2, recursively define

cos(nn) = (_1)n. Exercise 114. Let x E lR be fixed. Show that lar eaeh n

~

¡ sin(nx) I ::; ni sin(x)¡. The result in the next exercise is named for Abraham De Moivre (1667-1754), a friend of Isaac Newton. (In Newman's anthology The world 01 mathematies, his name is spelled "Demoivre".) De Moivre, as mentioned in [4, p. 155] "emigrated from France to England to escape religious prosecution, and made his living as a coffee-house consultant to students of mathematics." Recall that i is a (complex) number satisfying i 2 = -1. Exercise 115 (De Moivre's Theorem). Prove that lar ea eh n 2: 1, [cosCO)

+ isin(O)¡n

= cos(nB)

+ isin(nO).

In [550] it is explained how some of the these next statements aid in high speed computing. Exercise 116. Prove that lar any n sin(e

~

1 and any angle e,

+ nn) =

(_l)n sin(O).

Sn

1,

=

=

0, S1 = 1, and lar

2 cos( B)Sn-1 - Sn-2'

JIO is not a multiple 01 n, prove that for eaeh n 2: O, Sn

=

sin(ne) sin (e) ,

and lar eaeh n 2: 1, cos(nB)

=

cos(e)sn - Sn-1'

The next exercise uses the same recursion as in Exercise 122, however with different initial values. Exercise 123. Let e be any fixed angle. Define a sequence 01 real numbers recursively by S1 = cosCO), S2 = cos(2e) and lar n > 2, define

Sn

= 2cos(e)Sn_1 - Sn-2'

Prove lar every n 2: 1, that Sn = cos(nO). The following was a contest question in [429]; the solution foHows with help from another exercise given in Chapter 10.


148

Exercise 124. Let x be any real number. Prove that for any n 2::: 1,

cos 2n(x)

+ sin 2n (x)

2::: 2

9.9. Miscellaneous identities

Exercise 130. For any x that is not a multiple of 7f, prove that for each n 2::: 1,

L1 '

~tan (~) + 212 tan ( ; ) + ... + 2~ tan (~) = 2~ cot(~) -

Exercise 125. Prove that for n 2::: 0,

cot- 1 (3)

=

sin((2n + 1)t/2) 2sin(t/2) .

1 = -

2

cos(jt)

.

sinn(x) = L

Exercise 127. Using the notation from Exercise 126 put = N

+1L

Dn(t).

tan- l (l).

+ br sin(rx)) .

(a r cos(rx)

Exercise 133. Let x and o: be real numbers so that x for every n 2::: 1,

xn

9.9

+~ =

2cos(0:). Pmve that

+ -xn1 = 2cos (na).

Miscellaneous identities Xl, ... ,

x n,

Exercise 135. Por x, bE IR.+ with b :/1, prove that for all n E Z+,

KN(t) is called the Fejér kernel (see, e.g., [154, p.64 ¡J. Note that when n sum in Dn(t) is empty, so Do(t) = 1/2. Prove that for N 2::: 0, sin 2 ((N + 1)t/2) KN(t) = 2(N + 1) sin2(t/2)'

=

0, the

Exercise 128. Pmve that for each n 2::: 1 and any real number x that is not a multiple of 27f,

+ 2sin(2x) + 3sin(3x) + ... + nsin(nx) (n + 1) sin(nx) -

nsin((n 4sin 2(x/2)

+ l)x)

+ 2cos(2x) + 3cos(3x) + ... + ncos(nx) (n + 1) cos(nx) - ncos((n + l)x) 4sin2 (x/2)

Exercise 136. For any real numbers al, a2, ... , an and bl, b2, ... , bn ,

n L(ai +bd

=

n Lai

+

i=l

i=l

n Lbj

.

j=l

Exercise 137 (Telescoping sum). Pro ve that if al, a2, a3, ... , are real numbers, then for each positive integer n 2::: 1, n

Exercise 129. Pmve that for each n 2::: 1 and any real number x that is not a multiple of 27f,

cos(x)

n

Exercise 134. Prove that for any n 2::: 1 and non-negative real numbers if Xl + ... + x n = O, then Xl = ... = X n = O.

N

n=O

sinx

1) _

r=O

)=1

is called the Dirichlet kernel, arising in the theory of convex functions and Fourier series (see, e.g., [154], p.64). The next exercise regards the average of Dirichlet kernels, and was developed by Lipot Fejér, perhaps appearing first in R. Courant's 1937 book Differential and Integral Calculus, 2nd. ed., [although I have not yet been able to verify this]. It is also interesting to note that among Fejér's students was Paul Erdos, and that Fejér's advisor was Schwarz (as in "Cauchy-Schwarz inequality").

KN(t)

(n:

n

n

+L

1

(~) + ... + tan- l

Exercise 132. Use mathematical induction to show that for n 2::: 2, there exist constants ao, al, ... , an , and bo, bl , ... ,bn so that

The expression

Dn(t)

+ cot- 1 (5) + ... + cot- 1 (2n + 1) = tan- I (2) + tan- l

Exercise 126. Prove that for n 2::: 1, and any angle t that is not a multiple of 27f, 1

cot(x).

Exercise 131. Prove that for each n 2::: 1,

sin(2 n+lo:) cos(o:) cos(20:) cos(40:)··· cos(2no:) = 2n+1 sin(o:)'

2" + cos(t) + cos(2t) + ... + cos(nt)

149

L(ai - ai+¡)

=

al - an+l.

i=l

Exercise 138. Pmve that for any n 2::: 1, n

- 1

L(3i - j j=1

+ 2)

= n(n 2

+ n + 2).


150

In [366], José Nieto, Unversidad del Zulia, Venezuela, posed the problem of showing that there are infinitely many a > b > e > d > 1 with a!d! = bId. The equality in the next exercise exhibits such families of four numbers. The equality is trivial to verify, but with a little more work, one can prove it inductively as well.

9.9. Miscellaneous identities Exercise 145. Show that for each non-negative integer n,

1 + n(n + l)(n + 2)(n + 3) is a perfect square. Hint: Work out the first five or six cases, conjecture an inequality, and prove it by induction.

Exercise 139. Prove by induction that for every n 2: 3,

(n 2 + n)!(n - 1)! = (n 2 + n - l)!(n + 1)!.

This next exercise contains a surprisingly elegant resulto Recall that the notation [nJ means that S is a set of positive integers all at most n, and that TIsEs s is the product of all elements in S. (See the example given after the exercise.)

S

Here is an elegant result that seems related to Exercise 139:

Lemma 9.9.1. Show that any rational number can be expressed as a quotient, where each of the numerator and denominator is a product of factoríals of príme numbers.

S(k + 1).

.

k(3k + 1) 2

+ 3k + 2 6k + 4

(by indo hyp.)

3k 2 + k =--2-+-2-

+ 1).

By the principIe of mathematical induction, for aH n 2: 1, the statement Sen) is true. O Exercise 42: For n 2: 1, let Sen) be the statement

3k 2 +7k+4 2 (k + 1)(3k + 4) 2

which agrees with the right side of S(k

431

a bit first so that it is easier to see what

+ 1),

+ (3k + 2) =

27.1. Solutions: Arithmetic progressions

BASE STEP (n = 1): S(l) says 3·1 - 1 = 1 + 12 , which is true.

This completes the inductive step.

INDUCTIVE STEP (S(k)

->

S(k

+ 1)):

Fix sorne k 2: 1 and suppose that

ConsequentIy, by the principIe of mathematical induction, for aH n 2: 1, the statement Sen) is true. O Note: Observe how much easier the proof was made by cleaning up the expression in advance that needed to be derived at the end of the inductive step. Without this preliminary calculation, one must transform 3k 2 + 7k + 4 into (k + 1)(3(k + 1) + 1), a slightly clumsy calculation.

is true. Yet to be proved is

Exercise 41: For n 2: 1, let Sen) be the statement

Beginning with the left side of S(k

3 + 11

S(n):

BASE STEP (n

=

+ 19 + ... + (8n -

1): S(1) says 3 = 4 . 12


->

S(k

+ 1)):

3 + 11

S(k):

-

5) = 4n

2

-

n.

Let sorne k 2: 1 be fixed, and suppose that

+ 1):

3 + 11

+ 19 + ... + (8k -


5)

1) =

(~i) + (k + 1)2.

+ 1),

k

I:(3i - 1)

I : (3i - 1)

i=l

i=l

(ti) (t i) + (t i) +

+ 3(k + 1) -

1

+k2+3(k+1)-1 (byS(k))

5) = 4k 2

-

k

is true. The next identity to be shown is S(k

~(3i -

+ 1),

k+l

1, which is true.

+ 19 + ... + (8k -

S(k

+ (8(k + 1) -

5) = 4(k

+ 1),

+ ... + (8k - 5) + (8(k + 1) - 5) 4k 2 - k + (8(k + 1) - 5) (by S(k)) 4k 2 - k + 8k + 3 4k 2 + 8k + 4 - k - 1

+ 1)2 -

(k

+ 1).

3 + 11

k'

+ 3k + 2

(k + 1)

+ k' + 2k + 1

k+l ) (

~i +k2 +2k+ 1

k+l ) (

~i

+(k+1)2,

Chapter 27. Solutions: Identities

432

the right-hand side of S(k

+ 1).

27.1. SoIutions: Arithmetic progressions

(2k + 1) + 2 + ... + (2k + 2k - 1) + 2 + (2k + 2k + 1) + 2 (2k + 1) + (2k + 3) + ... + (2k + 2k - 1) + 2k + (4k + 3) 3k 2 + 6k + 3 (by S(k)) 3(k 2 + 2k +1)


Thus, by the principIe of mathematical induction, for aH n 2:: 1, Sen) holds.

O

Exercise 43: For n 2:: 1, let Sen) be the statement

5 BASE STEP (n

+ 9 + 13 + ... + 4n + 1 =

= 1): S(l)

says 5

INDUCTIVE STEP (S(k) ~ S(k

S(k):

=

n(2n + 3).

which is the right side of S(k

1(1 ·2 + 3), which is true.

+ 1)):

+ 1):

+ 1) + 4(k + 1) + 1 =

Beginning with the left-hand side of S(k 5 + 9 + ...

S(n): (k

+ 1)(2(k + 1) + 3).

+ 1),

S(k):

O

a + (a

+ d) + (a + 2d) + ... + (a + kd)

.!.[k(2a) 2

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

3)

+ (2n + 2n -

+ 3) + (2k + 5) + ... + (2k + 2k -

+ (k

- l)d]

3) + (2k + 2k - 1) = 3k

+ k 2d _

=

k+1 --[2a + kd] 2

+ (a + kd) kd]

- l)d) + (a

+ kd)

(by S(k))

+ 2a + 2kd 2

1

1) = 3n 2.

"2 [k(2a) + 2a + k 2d + kd] 1

"2 [(k + 1)2a + (k + l)kd] k+1 -2-[2a + kdJ,

2

precisely the right-hand side of S(k is true. To be shown is

S(k+1):

k 2

= -[2a

+ 1),

+ d) + (a + 2d) + ... + (a + (k 2[2a + (k - l)d]

1) = 3n 2.

INDUCTION STEP: Fix sorne k 2:: 1, and assume that

(2k + 1) + (2k

+ 1):

k

BASE STEP (n = 1): The statement S(l) says 2· 1 + 1 = 4· 1 - 3, which is true.

S(k):

l)d], which holds.

+ d) + (a + 2d) + ... + (a + (k - l)d)

a + (a

+ 1) + (2n + 3) + (2n + 5) + ... + (4n -

(2n + 1) + (2n

a + (a

foHows, starting with the 1eft side of S(k

It is easier to see what is going on Sen) is rewritten as

S(n):

+ (1 -

n l)d) = 2[2a + (n - l)d].

completing the inductive step.

Exercise 44: For n 2:: 1, let Sen) be the statement

(2n

+ ... + (a + (n -

holds. To show that

S(k

Hence, by the principIe of mathematical induction, for aH n 2:: 1, Sen) holds.

S(n):

o

INDUCTION STEP: Let k 2:: 1 be fixed and suppose that

k(2k + 3) + 4(k + 1) + 1 (by S(k)) 2k 2 + 3k + 4k + 5 2k 2 + 7k + 5 (k + 1)(2k + 5),

which agrees with the right-hand side of S(k

a + (a + d) + (a + 2d)

BASE STEP: The case n = 1 says a = ~[2a

+ 1),

+ 4k + 1 + 4(k + 1) + 1

the inductive step.

Exercise 45: Let a and d be fixed real numbers. For any natural number n 2:: 1, let S (n) be the statement

5+9+13+···+(4k+1)=k(2k+3)

5 + 9 + 13 + ... + (4k

+ 1), eompleting

Henee, by mathematical induetion, for all n 2:: 1, Sen) is true.

Fix sorne k 2:: 1, and assume that

is true. To complete the inductive step, it suffices to prove

S(k

433

+ 1), eomp1eting the

inductive step.

o

Thus, by mathematica1 induction, for eaeh n 2:: 1, Sen) holds.

(2(k+1)+1)+(2(k+1)+3)+···+(2(k+1)+2(k+1)-1) =3(k+1)2.

Beginning with the left-hand side of S(k

(2(k

Note: A direct proof is also available by rearranging terms, faetoring, and using Theorem 1.6.1 as follows:

+ 1),

+ 1) + 1) + ... + (2(k + 1) + 2(k + 1) -

3)

+ (2(k + 1) + 2(k + 1) -

1)

a + (a

+ d) + ... + (a + (n -

1) d)

=

na + [1

+ 2 + ... + (n -

1)1 d

434

na+



(n-1)n d 2

Exercise 47: See the solution to Exercise 49 when ". = 2.

o

Exercise 48: Again, this is the special case". = 3 in Exercise 49.

o

(by Thm. 1.6.1)

~[a + (n - l)d].

435

2

o

Exercise 49: Fix ". E lR, ".

2_

2

Tn - 1

2

2

+ 1)2 -

n 2[n 2 + n

(n - 1)2n 2

4

+ 1-

n 2 (4n) =-4completes the proof. An inductive proof is also available: For each n 2:: 2, let Sen) denote the statement T~ - T~_l = n 3 .

Tf = 9- 1= 23 , so S(2)

holds.

INDUCTIVE STEP (S(k) that is, assume S(k):

--->

Tf - Tf-l

= k

+ 1):

Tf+l - Tf = (k

S(k + 1):

3

rk+l - 1

r k+1(r - 1)

r-1

r-1

rk+l - 1 + rk+l(r - 1)

= k 3 + k(k + l)(k + 1) - (k - 1)k . k + 2k + 1 2 = k 3 + k[k 2 + 2k + 1 - k + k] + 2k + 1 2 = k 3 + 3k + 3k + 1,

(by Thm 1.6.1)

r-1 r k +1 _ 1 + ".k+2 _ rk+l r - 1

rk+2 - 1 ". - 1 '

+ 1), completing the inductive step.

By ]VII, for each n 2:: 2, Sen) is true.

+ 1),

---+----'----'-

(by S(k))

RHS of S(k

r-1

'--v--"

+ 1),

= k 3 + 2Tk(k + 1) - 2Tk-1k + 2k + 1

+ 1)3, the

rk+2 - 1

1 + r + r 2 + ... + r k + ".k+l = - - -

rk+l - 1 - - - +rk+l (by indo hyp.) r-1

+ 1)3

Tf = (Tk + (k + 1))2 - (n-l + k)2 2 = Tf + 2Tk(k + 1) + (k + 1)2 - TLl - 2Tk-1k + k = Tf - TLl + 2Tk(k + 1) - 2Tk-1k + 2k + 1

which is (k

(inductive hypothesis)

".-1

is also true. Beginning with the left side of S(k

follows. Starting with the LHS of S(k Tf+1 -

rk+l - 1

is true. To complete the inductive step, one needs to show that

holds. It remains to show that S(k

S(k + 1)): Fix sorne k 2:: 1 and suppose that S(k) is true,

1 + r +".2 + .. , + r n =

INDUCTIVE STEP: Let k 2:: 2 and suppose that S(k):

r-1

BASE STEP (n = 1): The statement S(l) says 1 + r = rr2~l which is true since r 2 - 1 = (r + l)(r - 1) and r f. 1. [Note, one could actually begin the induction at n = O since SeO) says 1 = ~=~.]

(n 2 - 2n + 1)]

4

BASE CASE: S(2) says Ti -

".n+l _ 1 = ---

+rn

To solve the problem, it suffices to prove that for each 71 2:: 1, Sen) holds, that is, it suffices to·consider only a = 1. For r = O, the statement Sen) becomes 1 = 1, so assume without loss of generality that r f. O. (This assumption might not be needed, but sometimes simple initial observations save headaches later.)

= [n(n + 1)]2 _ [(71 - 1)71]2

71 2(71

1, and for n 2:: 1, let Sen) be the statement

Sen) : 1 + ". + r 2 + ...

Exercise 46: The result in this exercise is obtained by simple algebra and Theorem 1.6.1:

Tn

f.

o

which is the righthand side of S(k+ 1). So S(k) step.

-->

S(k+ 1), completing the inductive


436

Hence, by the principle of MI, for aH n 2: 1, the statement Sen) holds.


o

(k - 1 + k 2k2 k + 1

Exercise 50: For n 2: 1, let Sen) be the statement 1.2 1

+

2.22

+

3.23

+ ...

k2 k +l

+ n· 2 = 2 + (n - 1)2 +1.

-+

S(k

1.2 1

S(k):

+ 1)):

+ 1)2 k + 1

+ 1,

n

n

BASE STEP (n = 1): S(l) says 1.2 1 = 2 + (1 - 1)22 , which is true. INDUCTIVE STEP (S (k)

437


+ 2.2 2 + ... + k· 2k =

2 + (k - 1)2

k

agreeing with the right side of S(k step S(k) -+ S(k + 1).

+ 1).

This completes the proof of the inductive

Therefore, by the principIe of mathematical induction, for each n 2: 1, Sen) is indeed true. O

+l

Exercise 54: For n 2: 1, let Sen) be the statement is true. To be shown is

Sen) :

12

+ 22 + 3 2 + ... + n 2 =

BASE STEP (n = 1): The statement S(l) says 12

The left side of S(k + 1) is

+ ... + k . 2k + (k + 1)2 k + 1 2 + (k - 1)2 k+ 1 + (k + 1)2 k+l (by k 2 + (k - 1 + k + 1) 2 +l 2 + (2k)2 k+l 2 + (k)2 k + 2 ,


1 .2 1 + 2.2 2

S(k»

S(k) :

-+

12

S(k

+ 1»:

Exercise 51: For each n 2: 1, let Sen) denote the statement

1 + 2·2

+ 3.2 + ... + n2

BASE STEP (n = 1): 1 = (1 - 1)20

+ 1, so

-

1

= (n

- 1)2

n

+ 22 + 3 2 + ... + k 2 =

+ ...

to be true. To be shown i8 that S (k 1 + 2·2

S(k

+ 1)

+ 1)(2k + 1)

follows. Starting with the left-hand side of S(k + 1), 12 + 22 + 32 + ...

+ k 2 + (k + 1)2 k(k + 1)(2k + 1)

(k

+ +1

)2

(by indo hyp.),

6

(k

+ 1)k(2k + 1) +6(k + 1) 6

(k

+ 1) k 1 (k)2 + + 1.

2

+ 1) 2k + k; 6k + 6)

k ) 2k ( +1

+ 1),

k 1 + 2·2 + 3 . 2 + ... + k2 k- 1 + (k + 1)2 k k = (k - 1)2 + 1 + (k + 1)2 (by S(k)) 2

k(k

:

+ k2k-1 = (k - 1)2k + 1.

follows. Starting with the left side of S(k

okay.

(k+l) [k(2k +1) +(k+1)]

S(l) holds.

+ 3.22 + '" + k2k-1 + (k + 1)2k =

lS

6

+ 1.

INDUCTIVE STEP: For sorne fixed k 2: 1, assume the inductive hypothesis S(k) 1 + 2 . 2 + 3.2 2

which


6

n

= 1(2)(3)/6

holds. Needed to be shown is that

which is the right side of S(k + 1), so S(k + 1) is also true. This completes the inductive step S(k) -+ S(k + 1). o By mathematical induction, for all n 2: 1, Sen) is true.

2

n(n + 1)(2n + 1) 6 .

(

k

+1

2

+ 7k + 6) 6

)(k+2)(2k+3) 6 '

which equals the right-hand side of S(k

+ 1).


Chapter 27. SoIutions: Identities

438

o

By MI, for every n .2: 1, Sen) is true. Exercise 56: For any n 2:: 1, let Sen) be the staternent

27.1. Solutions: Arithrnetic progressions to be true. To show that P(k

+ 1):

12 + 32 + ...

+ (2(k + 1) _

follows, begin with the left side of P( k is easy to see how to apply P(k»: BASE STEP (n = 1): S(l) says 13 INDU9TIVE STEP (S(k)

S (k):

->

S(k

=

12 +3 2 + 5 2

¡(lli2l F = 1, which holds.

+ 1»:

+ k3

= [

k(k+1)]2

13 + 23 + ...

+ k 3 + (k + 1)3

+ 1), =

[k(k; 1)

(k (k

+ 1)2

r

(:2

+ (k + 1)3 + (k +

BASE STEP (n

=

1): Since 12

=

+ 1) 2k

(2k

+ 1) 2k

(2k

+ 1) (k + 1)~2k + 3)

2 -

2

k: 6k

+;k

+ 1)

+3

+3

+ 1)(2k + 1)(2k + 3)

which is the same as the right side of P(k P(k) ---; P(k + 1).

+ 1).

This concludes the ínductive step

By the principIe of rnathernatical induction, for each n .2: 1, P(n) is true. the inductive step.

n(2n - 1)(2n + 1) + (2n - 1)2 = 3 .

Exercise 58: For n 2:: 1, denote the staternent in the exercise by S(n): BASE STEP (n

k(2k - 1)(2k 12 +3 2 + ... +(2k-1)2= 3

=

+ 42 + 62 + ... + (2n?

1): Since 22

S(k):

+ 1) .

22

=4=

= 2n(n

+ 11(2n + 1).

2(1+1~(2+1) , the statement S(l) holds.

INDUCTIVE STEP: For sorne fixed k .2: 1, assurne the inductive hypothesís

1~.3, Lhe staternent P(l) holds.

INDUCTIVE STEP: For sorne fixed k .2: 1, assurne the inductive hypothesis P(k):

(2k

1) ; 3(2k

]

3

Exercise 57: For n .2: 1, denote the proposition in the exercise by

+ ...

+ 1) k(2k -

2

Hence, by the principIe of rnathernatical induction, for each n .2: 1, the staternent Sen) holds. O

P(n):

+(2k+1)

(2k

(k

(k+1)2(k:2?,

12 + 3 2 + 52

3

(by P(k))

(by indo hyp.),

1))

+ 1), cornpleting

(adding the second last term so that it

+ (2k + 1)2

k(2k - 1)

+ 1)2 k + 4~k + 1)

which is precisely the right-hand side of S(k

+ 1)

+ 1)(2(k + 1); 1)(2(k + 1) + 1).

+. :". +J2k"':' 1)2 + (2k + 1)2

(2k+1) [

2

holds (the inductive hypothesis). To be shown is


1)2 = (k

k(2k - 11(2k + 1)

Fix sorne k .2: 1, and assurne that

3 3 1 + 2 + 33 + ...

439

22

+ 42 + 62 + ... + (2k)2 =

2k(k

+ ~(2k + 1) .

to be true. It rernains to show that S(k

+ 1):

22

+ 42 + 6 2 + ... + (2(k + 1»2 =

2(k

+ l)(k + 2)(2(k + 1) + 1) 3

O


440

follows from S(k). Starting with the left side of S(k

22 + 42 + 62 + ...

= (-1 )k+l k

2

2k(2k + 1) 3

+ (2(k + 1))2 +

(by indo hyp. S(k))

= (_1)k+2k2

(3(2(k.+1)] 3

+ 1) 2k + k: 6k + 6

2(k

+ 1) 2k



Exercise 60: Por each n ?:: 1, let Sen) denote the statement

+ 1).

This completes the inductive step

Therefore, by the principIe of mathematical induction, Sen) is true for all n ?::

O

1.

BASE STEP

+ ... + (_1)n+ln2

S(k):

= (_1)n+1(1 + 2 + 3 + ... + n).

(n = 1): The statement S(l) says 1 = (_1)1+1 (1), which is true.

(n = 1): S(l) says 12 = l(l~H) which is true. Por sorne fixed k ?:: 1, assume the inductive hypothesis

INDUCTIVE STEP:

Exercise 59: Por each n ?:: 1, denote the statement in the exercise by

BASE STEP

+ 1),

k 2 - (k _ 1)2

+ (k

_ 2)2

+ ... +

+ 1):

+ 1)2 _

k 2 + (k _ 1)2 + ... + (_1)k(1)2

+ (_1)k+lk2 + (_1)k+2(k + 1)2

(k+ 1)2 - k 2 + (k _1)2

= (_l)k+l(l

+2+3 +

...

+ k)

+ (_1)k+ (k + 1)2

(by S(k))

=(_1)k+l[(1+2+3+ ... +k)-(k+1)2] = (-ll+1 [k(k 2+ 1) - (k

+ 1)2]

(by Thm 1.6.1)

S(k

(k

2

k(k

+ 1)

=

(k + 1)(k + 2) 2

+ 1)2 (k + 1)2 _ (k

[k 2 k(k

+ ... + (_1)k(1)2 + (k - 2f + ... +

+ 1)

2

(k+1)[(k+1)-~] (k

+

1) k

+2

2'

+ 1):

(k - 1)2

-

(k + 1) 2k +22 - k = (_l)k+1k(k + 1) - 2(k + 1?

=

to be true. To see that

follows, start with the left-hand side of S(k

2

(_1)k-l(1)2

2

Por sorne fixed k ?:: 1, assume the inductive hypothesis S(k) to be true. To show that S(k + 1) follows from S(k), use two applications of Theorem 1.6.1, one with n = k and another with n = k + 1. The left side of S(k + 1) is

INDUCTIVE STEP:

1- 4 + 9 - 16 + ...

(by Thm 1.6.1),

Therefore, by the principIe of mathematical induction, for every n ?:: 1, Sen) is true. O

(k+2)(2k+3) 3 '

1- 4 + 9 -16

+ l)(k + 2) 2

+;k + 6

which agrees with the right side of S(k S(k) --. S(k + 1).

S(n):

+ 3k + 2

= (_1)k+2(1 + 2 + ... + k + (k + 1))

2(k

) ( 2k+1

4k - 2

-

2

2 2

2

2

(3(4(k + 1? 3

+

2k

= (_l)k+1 _k - 3k - 2 2

= (_1)k+2 (k

k(2k+1) ) [ ( 2k+1 3

+k -

441

2

+ (2k)2 + (2(k + l)f

2k(k + 1)(2k + 1) 3 ( k+1 )

+ 1), derive the right side:


(by S(k))

(_1)k-l(1)2]


442

which is the right-hand side of S(k

+ 1).

27.1. Solutions: Arithrnetic progressions


k(k +

Therefore, by the principIe of rnathernatical induction, for all n 2: 1, the staternent Sen) is true. D

1~{k + 2)

(k + 1)(k + 2)

443

+ (k + 1)(k + 2)

(by S(k))

(~ + 1)

Exercise 61: For each n 2: 1, denote the equality in the exercise by

13 + 3 3 + 53 + ... + (2n - 1)3

E(n): BASE STEP

(n

=

= n 2 (2n 2

-

1).

=

1): Since 13 = (1)2(2(1)2 - 1), E(1) is true.

13 + 3 3 + 53 + ... + (2k - 1)3 + (2(k + 1) - 1)3 13 + 33 + 53 + ... + (2k - 1)3 + (2k + 1)3 -

4

2

1)

+ (2k + 1)3

+ 1)2(2(k + 1)2 -

BASE STEP

A(k):

1),

O

1·2+2·3+···+n(n+1)=

n(n + l)(n 3

+ 2)

1·2+2·3+···+k(k+1)=

1.3+2.4+ ... +(k-1)(k+1)= (k-1)(k¿c

2k

k(k

A(k + 1): 1·3 + 2.4 + ... + (k - l)(k

+ 1) + k(k +

2) = (k)(k + ¿(2k

+ 1) + k(k + 2)

(k - 1)(kj(2k + 5) + k(k + 2) (by indo hyp.) k [(k

-1)~2k + 5) + (k + 2)]

+ 1)(k + 2) 3

k [

1·2 + 2 . 3 + ... + k(k + 1) + (k + 1)(k + 2) =

(k + 1)(k + 2)(k + 3) 3

2k2 +3k 6

5+--6k + 12J

.

+ 1),

1·2 + 2 ·3+ .. , + k(k + 1) + (k + 1)(k + 2)

+ 5)

holds. To see that

1·3 + 2·4 + .. , + (k - 1)(k

Let k 2: 1, and suppose that the inductive hypothesis


.

For sorne k 2: 2 assurne that

.

is true. Yet to be proved is

S(k + 1):

6

holds, start with the left side:

The staternent S(n says 1 ·2= 1(2~(3), which is true.

S(k):

(n - 1)(n)(2n + 5)

(n = 2): A(2) says l· 3 = (1)(~)(9) which is true.

INDUCTIVE STEP:

Exercise 63: For each n 2: 1, define the staternent

INDUCTION STEP:

1·3+2·4+···+(n-1)(n+1)=

2

3

By the principie of rnathematical induction, for each n 2: 1, E(n) is true.

BASE STEP:

Hence, by the principIe of rnathematical induction, for every n 2: 1, Sen) holds. D

A(n):

(by E(k))

and so 13 + 33 + 53 + ... + (2k - 1)3 + (2(k + 1) - 1)3 = (k + 1)2(2(k + 1)2 - 1), which is E(k + 1). This concludes the inductive step E(k) ---+ E(k + 1).

Sen):

+ 1), cornpleting the inductive step.

Exercise 64: For each n 2: 1, let A(n) be the assertion

2k - k + 8k + 12k + 6k + 1 2k 4 + 4k 3 + k'2 + 4k 3 + 8k 2 + 2k + 2k'2 + 4k + 1 k 2 (2k 2 + 4k + 1) + 2k(2k 2 + 4k + 1) + 2k 2 + 4k + 1 (k 2 + 2k + 1)(2k 2 + 4k + 1) (k

k+3

+ 1)(k + 2)-3-'


INDUCTIVE STEP: For sorne fixed k 2: 1, assurne the inductive hypothesis E(k) to be true. To see that E(k + 1) follows,

k 2 (2k 2

(k

k(k

+ 1)(2k + 7) 6

6

+ 7)

444


which is the right-hand side of A(k + 1). So A(k + 1) foHows from A(k), completing the inductive step. Thus by mathematica1 induction, for each n 2: 2, A(n) ho1ds.

27.1. SoIutions: Arithmetic progressions

445

INDUCTIVE STEP: For some fixed k 2: 1, assume the inductive hypothesis k

o

S(k): ¿ j ( j

+ 1)(j + 2)(j + 3) = k(k + 1)(k + 2)(k + 3)(k + 4) 5

j=l

Exercise 65: For each n 2: 1, let Sen) denote the statement

1·2·3 + 2 . 3·4 + 3·4·5 + ... + n(n + 1)(n + 2)

1

= -n(n

4

to be true. It remains to show that

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

k+1

S(k

+ 1): ¿j(j + 1)(j + 2)(j + 3) = (k + 1)(k + 2)(k + 3)(k + 4)(k + 5)

follows. Starting with the 1eft side of S(k + 1) and separating the 1ast term,

INDUCTIVE STEP: For some fixed k 2: 1, assume the inductive hypothesis S(k):

1·2·3 + 2 . 3 ·4+ ... + k(k + 1)(k + 2)

5

j=l

BASE STEP (n = 1): Since 1·2·3 = 6 = ~1 ·2·3·4, the base case S(1) is true.

k+1

1

= :¡k(k

+ 1)(k + 2)(k + 3).

¿j(j + 1)(j + 2)(j + 3) j=l

to be true. To be shown is that S(k+1):

(t,jU + l)U +

, 1 1·2·3+2·3·4+···+(k+l)(k+2)(k+3) = :¡(k+1)(k+2)(k+3)(k+4).

follows. Beginning with the 1eft side of S(k term inserted for clarity)

+ 1),

+(k k(k

(k

[~k + 1]

+ 1)(k + 2)(k + 3)

(k

+ 1)(k + 2)(k + 3) [l(k + 4)] ,

=

which is indeed the right side of S(k S(k + 1).

+ 1),

concluding the inductive step S(k)

---4

Therefore, by the principIe of mathematica1 induction, for each n 2: 1, the statement Sen) is true. O

.

~ j(j + 1)(j + 2)(j + 3) ~

+ 1)(n + 2)(n + 3)(n + 4).

+ 1)(k + 2)(k + 3)(k + 4)

which agrees with the right side of S(k S(k + 1).

[~+ 1]

+ 1),

comp1eting the inductive step S(k)

BASE STEP (n = 1): The 1eft side of S(1) is 1 ·2·3·4 = 24, and the right side of S(1) is 1'2,~A.5 = 24 as well, proving the base case.

---+

Exercise 67: Fix sorne k E Z+, and for each n 2: 1, denote the statement

t,j(j +

1)··· U + k - 1)

~ (k +(~) +:~~ 1)1·

5

j=1

(by S(k»

Therefore, by the principIe of mathematica1 induction, for aH n 2: 1, Sen) is true. O

S(n) , = n(n

+ 1)(k + 2)(k + 3)(k + 4)

(k+1)(k+2)(k+3)(k+4)k+5, 5

Exercise 66: For each n 2: 1, let Sen) denote the statement S(n).

+ 1)(k + 2)(k + 3)(k + 4)

+ 1)(k + 2)(k + 3)(k + 4) 5 +(k

+ 1)(k + 2)(k + 3) + (k + 1)(k + 2)(k + 3) (by indo hyp.)

(k

3))

(rewritten with the penultimate

1·2·3 + 2 ·3·4 + ... + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3) 1 :¡k(k

2)(j +

BASE STEP (n = 1): The staternent S(1) says (where j sumrnand on the 1eft) 1.2 ... k=

k

(k+1)! . + 1 . (O!)'

=

1 produces the only


446

since (k + 1)! = (k + l)(k!) and O! = 1, the two sides agree, cúmpleting the prúof of the base case S(l). INDUCTIVE STEP: For so me fixed m 2: 1, assume the inductive hypothesis m

S(m):

+ 1)··· (J + k -

1) = (k

+ 1) . (m -

n

1)!·

S(n):

¿)2k - 1)(2k

)=1

+ 1):

?= J (J + 1) ... (J +

m+1 . .

.

k - 1)

=

(k (k

(2·1- 1)(2·1 + 1)(2·1

3

+ 8n2 + 7n -

2).

1(2(1 3 )

=

+ 8(2 2 ) + 7(1) -

2).

15, showing that indeed S(l) holds.

INDUCTIVE STEP: For some fixed t 2: 1, assume the inductive hypothesis

m+1

+ 1) ... (j + k

+ 3) =

This simplifies to 1· 3·5 = 2 +8+ 7 - 2, or 15

follows. Beginning with the left side of S(m + 1) and separating the last summand, j (j

= n(2n

BASE STEP (n = 1): The statement S(l) says (where k = 1 gives the only surnmand on the left)

+ m + 1)! + 1) . m!

)=1

¿

+ 1)(2k + 3)

k=1

to be true. [Note: One shouldn't use S(k) here, because the statement needed to prove already has a k i!l it, so employ a new variable m in the inductive step.j It remains to show that S (m

447

Exercise 68: This identity was mentioned by Peter Ross in [465], a media review of the article A LISP prover for induction formulae, where he says that an induction proof of this identity "is tedious, although straightforward." For each n 2: 1, denote the statement in the exercise by

(k+m)!

. . .

¿J(J


t

- 1)

S(t):

j=1

(~j0 + 1) ... (j + k -1)) + (m + l)(m + 2)··

¿)2k - 1)(2k

+ 1)(2k + 3) = t(2t 3 + 8t 2 + 7t -

2)

k=1

(m + k)

tú be true. It remains to show that t+1

(k + m)' (k + 1) . Cm ~ 1)!

S(t + 1):

+ (m + l)(m + 2)··· (m + k)

¿(2k - 1)(2k + 1)(2k + 3) = (t + 1)(2(t + 1)3 + 8(t + 1)2 + 7(t

(by S(m))

follows. Beginning with the left side of S(t

(k + m)! (k + 1) . (m - 1)! (k + m)!m (k + 1) . (m)! (k

+ (m + k)!

(k

+

+ m)! [~ + m!

(k+m)! m!

k+1

2)

+ 1),

t+l

m!

¿)2k - 1)(2k

+ 1)(2k + 3)

k=1

+ m)! m!

[p'2k - 1)(2k + l)(2k

1]

1

+ 3) + (21 + 1)(21 + 3) (21 + 5)

+ 8t 2 + 7t - 2) + (2t + 1)(2t + 3)(2t + 5) (by S(t)) 2t + 8t 3 + 7t 2 - 2t + 8t 3 + 36t2 + 46t + 15 4 2t + 16t 3 + 43t 2 + 44t + 15 (t + 1)(2t 3 + 14t 2 + 25t + 15) (by polynomial division) (t + 1)(2(t + 1)3 + 8(t + 1)2 + 7(t + 1) - 2), t(2t 3 4

m+k+1 k+ 1

which reduces to the right side of S(m + 1) as desired. This concludes the inductive step S(m) - t S(m + 1). Therefore, by the principIe of mathematical induction, for all n 2: 1, Sen) is h~.

+ 1) -

k=l

O

where the last equality is easiest to see by multiplying out the second expressiún in the last lineo This completes the derivation of S(t + 1) using S(t), concluding the inductive step.


448

k+1 k+2'

Therefore, by the principIe of mathematical índuction, for all n ? 1, Sen) is O true.

which proves S(k+ 1) from the truth of S(k), thereby completíng the inductive step.

Exercise 70: For each n 2: 1, Iet Sen) denote the statement O . 01

+ 1 . 11 + 2·21 + 3·31 + ... + n

BASE STEP (n = 1): S(l) says 0·01 sides equal 1.

+ 1 . 11

. nl = (n

+ 1)1 -

1.

21 - 1, which is correct, since both

=

INDUCTIVE STEP: For sorne fixed k ? 1, assurne the inductive hypo~hesis S(k) to be true. To see that S(k + 1) follows, the left side of S(k + 1) is 0·01

Therefore, by the principie of mathematical induction, for all n ? 1, Sen) is true. O

Another solution ta Exercíse 71: Proving the equaIity in the exercise can be done directIy with an old trick: the partial fraction identity n(n1+l = ~ - n~l' The sum can then be seen to teIescope, a phenomenon that itself can ~e proved by induction (see Exercise 137).

+ 1 . 1! + 2· 2! + ... + k· kl + (k + 1) . (k + 1)! (k + 1)! - 1 + (k + 1) . (k + 1)! (by S (k )) (1 + (k + 1)) . (k + 1)! - 1 (k + 2) . (k + 1)1 - 1

1

1

/2

=

n

1

1

1 1

k(k+2)+1 (k+1)(k+2) k 2 + 2k + 1 (k+1)(k+2) _ (k+1)2 - (k + l)(k + 2)

71

? 1, let S (n) denote the statement 1

~ (71 + í)(71 + i + 1) =

l¡l' a true statement.

+ (k+1)(k+2)

o

This completes the second proof.

n-l

1-2 + 2·3 + ... + k(k + 1) + (k + l)(k + 2) k = k+1

71+1

+1

INDUCTIVE STEP: For sorne fixed k ? 1, assume the inductive hypothesis S(k) to be true. To be shown is that S(k + 1) follows: 1

71+1

71

n

1

+ 3·4 + ... + n(n + 1)

BASE STEP (n = 1): S(l) merely says

1

+ (~ _ ~) + (~ _ ~) + ... + (~ _ _ 1 ) ( ~1 _ ~) 2 2 3 3 4 n n+1

Exercise 72: For each

Exercise 71: For each n? 1, let Sen) denote the statement 1 2·3

1

1 1---

which equals the right side of S(k + 1). This completes the inductive step S(k) -+ S(k + 1). Therefore, by the principie of mathematical induction, for al! n ? 1, Sen) is O true.

1

1

+ - + - + ... +-;---. 1·22·33·4 n(n+1)

(k+2)!-1,

"1.2+

449


271'

The truth of Sen) for aH 71 ? 1 follows from Exercise 71, sínce the sum in Sen) is the sum of roughly the last half of a much longer series beginning at rather than at n(n1+l)' These calculations are given first, followed by an inductive proof of Sen). Direct proaf of Exercise 72:

¿

n-l

(by S(k))

1

1

1

2n-2

~ (71 + i)(n + i + 1) = ~

(i

2n - 1

2n

n-2

+ 1)(i + 2) 71 - 1 271

271 - 1- 2n 2n

-

1

~ (i + l)(i + 2) (by Exercise 71)

+2

1

2n

o


450

Inductive proof of Exercise 72: BASE STEP (n = 1): In S(l), the sum on the left is the trivial sum


451

BASE STEP: S(l) says 1.~.3 = l2~3 which is correct, since both sides are equal to ~. (1+0)(i+O+l)'

whereas the right-hand side is ~; these two are equal, so S(l) is true.

INDUCTIVE STEP: For sorne fixed k 2:: 1, assume the inductive hypothesis

INDUCTIVE STEP: For sorne fixed k 2:: 1, assume the inductive hypothesis 1

k-l

L

S(k):

k. S( ).

1

1

k

L

1

(k + 1 + i)(k + i + 2)

z=O

=

S k

1

L

(k + 1 + i)(k + 2 + í)

z=O

= =

tt k+l

1

i(i + l)(i + 2) =

L (k + j)(k + j + 1) 1

1

1

1

=

~

1 1 1 k(k + 1) + 2k(2k + 1) + (2k + 1)(2k + 2)

~

2k + 2 + 2k k(k + 1) + 2k(2k + 1)(2k + 2)

1 2k

(by S(k))

1

1

i(i + l)(í + 2) + (k + l)(k + 2)(k + 3)

(by S(k))

k 3 + 6k2 + 9k + 4 4(k + l)(k + 2)(k + 3)

1

(k + 1)2(k + 4) 4(k + l)(k + 2)(k + 3)

1

k(k + 1) + 2k(k + 1)

_ (k ~ 4(k

k+l~2+1

2k(k

+ 1)

+ 1).


Therefore, by the principIe of mathematical induction, for each n 2:: 1, Sen) is

+ 1)'

O

~oo.

completing the proof of S(k

+ 1)

and hence thc inductive step.

Therefore, by the principIe of mathematical induction, for all n 2:: 1, Sen) is O true.

1

1

1

~ + 2·3·4 + 3·4·5 + ... + n(n + l)(n + 2)

= 4(n

Exercise 74: For each n 2:: 1, let Sen) denote the statement

111 1 1 1 1 - - + - ~ - + ... + - - - - = - 2

Exercise 73: For each n 2:: 1, let Sen) denote the statement

1

+ l)(k + 4) + 2)(k + 3)'

one arrives at the right-hand side of S(k

1 - 2(k

+ 1),

1

1

= 2k ~

(k+1)(k+4)

k(k + 3)(k + 3) 4 = 4(k + l)(k + 2)(k + 3) + 4(k + l)(k + 2)(k + 3)

L (k + j)(k + j + 1) ~ k(k + 1) + 2k(2k + 1) + (2k + 1)(2k + 2) 1

_

k(k + 3) 4 = 4(k + l)(k + 2) + 4(k + l)(k + 2)(k + 3)

J=O

= 2k

tt k

J=l

k-l

1

k+l

tri(i+1)(i+2) - 4(k+2)(k+3)

follows. Starting with the left-hand side of S(k

1

k+l

l.

( + ).

2(k + 1)

follows, starting with the left-hand side, k

k(k + 3) 4(k + l)(k + 2)

to be true. It remains to show that

to be true. To see that S(k + 1):

z=l

(k + i)(k + i + 1) = 2k

t=O

~

1

k

L i(i + 1)(i + 2) -

n(n + 3) + l)(n + 2)"

3

4

2n

~

1

2n

n

+1

+ -1- + n+2

1 ... + - . 2n

The expression on the left constitutes the first 2n - 1 terms of what is called the "alternating harmonic series" .


452

BASE STEP (n = 1): Note that the left side of Sen) has denominators which range from 1 to 2n, whereas the denominators on the right range from n + 1 to 2n. So, for n = 1, the denominators on the left range from 1 to 2, whereas on the right, they range from 1 + 1 = 2 to 2 . 1 = 2, that is, on the right, there is only one termo Hence, S(I) says 1 - ~ = ~, which is true. INDUCTIVE STEP: For sorne fixed k :2: 1, assume the inductive hypothesis S(k): 111 1 1 1 1 1 - - + - - - + ... + - - - - = - - + - 2 3 4 2k - 1 2k k +1 k +2 to be true. It remains to prove that S(k 1-

+ ... +

1

-. 2k

+ 1):

111 1 1 -+ ... + - - - - 2 3 4 2k + 1 2k + 2

-+- -

=

1 1 - - + --+ k+ 2 k+ 3

+ 1)

follows. Beginning with the left side of S(k terms)

1

... + - 2k + 2

(and filling two more penultimate


INDUCTIVE STEP: For sorne fixed k :2: 1, assume the inductive hypothesis P(k): _1_+_1_+_1_+ ... + 1 _ _k_ 1·33·55·7 (2k-1)(2k+1) 2k + 1 to be true. The consequence to be proved is P(k

1 = k+1

The express ion to the left of the equal sign in P(k + 1) (written with the second last summand explicit) is equal to 1

1

+

1 =

k

+ 2 + ... +

1 2k

1

1

+ 2k + 1 + k + 1

2k+l

1 k + 2 + ...

1

+ 2k

1

+ 2k

2'

agreeing with the right side of S(k+ 1). (This sequence of equalities turned out to be far easier than one might have thought at the onset!) This completes the inductive step S(k) -+ S(k + 1). Therefore, by the principle of mathematical induction, for all n :2: 1, Sen) is k~. O Exercise 75: (This exercise is the subject of an exposition in [433, Vol 1, p. 112114].) For each n :2: 1, let P(n) be the proposition 1 1 1 1·33·55·7

- + - + - + ... + BASE STEP: P(l) says

¿

= 2(1)+1'

1 n =-(2n-1)(2n+1) 2n+1·

which is true.

1 + -:-----,----

(by S(k))

(2k+1)(2k+3)

2k 2 + 3k + 1 (2k + 1)(2k + 3)

+2

1

+ 1 + 2k +

1

(byS(k))

(2k + l)(k + 1) (2k + 1)(2k + 3)

11121 = k + 2 + ... + 2k + 2k + 1 + 2k + 2 - 2k + 2 =

1

_ k(2k + 3) + 1 - (2k + 1)(2k + 3)

1

- 2k

1

- + - + - + ... + +..,.------1·33·55·7 (2k-1)(2k+l) (2k+1)(2k+3)

k

1 111 k+2+···+2k+2k+1-2k+2

+ 1):

1 1 1 1 k+l - + - + - + ... + =-1·33·55·7 (2k+1)(2k+3) 2k+3'

= --

111 1 1 1 1 1 - 2" + "3 - 4 + ... + 2k - 1 - 2k + 2k + 1 - 2k + 2

453

k+l - 2k + 3' which equals the express ion on the right side of P(k + 1). This completes the proof of P(k + 1), and hence the inductive step P(k) -+ P(k + 1). By mathematical induction, oue concludes that for all n :2: 1, P(n) is true. Exercise 77: For n :2: 1, denote the statement in the exercise by S(n):

_1_

1·3

BASE STEP: Since

¿

=

+ _1_ + ... + __1_ _ 2·4

(n)(n+2)

n(3n + 5) 4(n

+

l)(n

+ 2)"

1.1.;, S(1) is true.

INDUCTIVE STEP: For sorne fixed m :2: 1, assume that 111 S(m) : + + ... + .,.----1·3 2·4 (m)(m+2)

m(3m + 5) 4(m+1)(m+2)'

O


454


k

is true. It rernains to show that

S(m

+ 1):

1 1.3

+

1 2.4

1 (m + l)(m

+ ... +

foHows. Starting with the left-hand side of S(m 1 1· 3

+

1 2·4

+ ... +

1 m(m + 2)

_ m(3m + 5) - 4(m+1)(m+2)

+

+

= 4k

(m + 1)(3m + 8) 4(m + 2)(m + 3)

+ 3)

(4k + 1)(k + 1) (4k + 1)(4k + 5)

(by S(m))

m(3m + 5)(m + 3) + 4(m + 2) 4(m + 1)(m + 2)(m + 3) m3

+ 14m2 + 19m + 8

4(m

+ 1)(m + 2)(m + 3)

=

Therefore, by the principIe of rnathematieaI induction, for aH n 2: 1, Q(n) is true. D

Exercise 79: This appeared in [499, Probo 14]' for exarnpIe. Por n 2: 1, let P(n) be the proposition

3m 2 + 11m+8 4(m + 2)(m + 3)

proving S(m

+ 1)

k+1 4k + 5'

proving Q(k + 1), and henee eompleting the induetive step Q(k) ~ Q(k + 1).

(3m 2 + 11m + 8)(m + 1) 4(m + 1)(m + 2)(m + 3)

_ (m + 1) (3m - 4(m + 2)(m

(sinee m

+ 1 i- O)

12

22

+ 8)

BASE STEP: The left side of P(1) is so P(1) is true. also

+ 3)'

1,

as desired, thereby eoncluding the induetive step. D

k

1 1 1 1 1.5+5.9+9.13+···+(4n-3)(4n+1)

n 4n+1

I:5 + 1

z=1

i2 1)(2i

+

1 9.13

+ ... +

1 (4k + 1)(4k + 5)

1

N + 5·9 + 9 . 13 + ... + (4k -

1·

3)(4k

k(k

+ 1)

= 2(2k

i2

k+1

is ~:~, whieh is

+ 1) + 1)

P(k

+ 1):

I: (2i _ 1)(2i + 1) = z=1

(k

+ 1)(k + 2) 2(2k + 3)

follows; this is aecornplished by:

k+l 4k + 5

follows. Beginning with the expression on the left side of the equality in Q( k 1

!' and the right side of P(1)

is true. To be proved is that

¿

Q(k + 1):

I: (2i -

P(k):

BASE STEP: It is trivial that = 41+1' and so Q(1) is true. INDUCTIVE STEP: Suppose that for sorne k 2: 1, Q(k) holds. It suffiees to show that 1 5.9

n(n + 1) 2(2n+1)

=.

INDUCTIVE STEP: Let k 2: 1 be fixed and suppose that

Exercise 78: Por n 2: 1, denote the equality in the exercise by

1

n2

32

+ - + - + ... + (2n-1)(2n+1) 1·33·55·7

By MI, for eaeh n 2: 1, Sen) is true.

Q(n):

(by Q(k))

4k 2 + 5k + 1 (4k + 1)(4k + 5)

+ 3)

1 . (m+1)(m+3)

1

+ 1 + (4k + 1) (4k + 5)

k(4k+5)+1 (4k + 1)(4k + 5)

+ 1),

1 (m + l)(m

455

k+1

+ 1),

[k

i2

f; (2i - 1)(2i + 1)

=

1

+ 1 + (4k + 1)(4k + 5)

=

tt

i2

(2i - 1)(2i + 1)

k(k+1) 2(2k + 1)

1

+ (2(k + 1)

(k+1)2 - 1)(2(k + 1)

(k+1)2

+ (2k + 1)(2k + 3)

(by P(k))

+ 1)


456

k(k + 1)(2k + 3) + 2(k + 1)2 2(2k + 1)(2k + 3)

27.1. Solutions: Arithmetic progressions INDUCTIVE STEP: For sorne m

~

457

1, suppose that

1 m 2 t ; k +3k+2=2(m+2) m

(k + 1)[k(2k + 3) + 2(k + 1)] 2(2k + 1)(2k + 3)

Q(m):

is true. Prove the statement

(k + 1)[2k 2 + 5k + 2] 2(2k + 1)(2k + 3)

Q(m (k + 1)(2k + l)(k + 2) 2(2k + 1)(2k + 3)

(k proves P(k

+ 1).

~

m+l

~

D

1, that P(n) is true.

Exercise 80: For each n

~

1, let Q(n) denote the statement 1

n

L

k=l

n

k 2 + 3k + 2 = 2( n + 2) .

Proaf using Exercise 71: Notice that the denominators in each summand factors: n

1

L

k 2 + 3k + 2

k=l

1

n

= k=l L

(k + 1) (k + 2)

=

L

j(j

J=2

+ 1)

=L ·C + 1) j=l J J n+ 1 n+2 2n

2"

1 2

+1

k 2 + 3k + 2 =

~ k 2 + 3k + 2 +

(m + 1)2 + 3(m + 1) + 2

m 1 +----",---2(m+2) m 2 +5m+6 m 1 = + -:-----,---,,...----....,.. 2(m + 2) (m + 2)(m + 3) m(m+3) +2 2(m + 2)(m + 3)

(by Q(k))

=

m 2 +3m+2 2(m + 2)(m + 3) (m + l)(m + 2) 2(m + 2)(m + 3) 1

+ 3)"

This completes the inductive step Q(m) --7 Q(m + 1). By mathematical induction, for any n 2' 2, Q(n) is true.

1

-

m

~ k 2 +3k+2 = 2(m+3)

m I l

2(m

1

n+l

1

m+

1

n+l

1

m+l

as foHows:

+ l)(k +2) 2(2k + 3)


By MI, for aH n

+ 1):

o

Exercise 81: This probIem oeeurs in an old Canadian grade sehool text [385] without solution. For n 2' 1, denote the statement of the exereise by

(by Exercise 71)

+2-

(n + 2) 2(n + 2)

S(n):

~

i

~1+i2+i4

=

n(n + 1) . 2(n2+n+1)

n

BASE STEP: S(l) says l+~+1 = 2(1!-;+ll) whieh is true.

2(n+2)'

as desired. Inductive proaf af Q(n): BASE CASE: Q(l) says (ll2+j(ll+2 = 2(1~2l' which reduces to

D

i = i, and so is true.

INDUCTIVE STEP: Fix m ~ 1 and suppose that S(m) is true. S(m + 1), observe that

1 + (m + 1)2 + (m + 1)4

=

To prepare for

(m 2 + m + 1)(m2 + 3m + 3).

458

Chapter 27. SoIutions: Identities

To prove S(m i

m+1

¿

1=1

+ 1), start

1 + i2

with the LHS of S(m i

m

+ i4 =

¿

1=1

1 +i2

m +1 + i 4 + 1 + (m + 1)2 + (m + 1)4

m(m+1)

m+1

+ m + 1) + 1 + (m + 1)2 + (m + 1)4 m(m + 1) m+1 2(m2+m+1) + (m 2 +m+1)(m2 +3m+3) +1 1 ] m 2 + m + 1 2" + m 2 + 3m + 3

=

m

(by S(m))

[m

m(m2 +3m+3)+2 m2 + m + 1 2(m 2 + 3m + 3) m +1 m 3 + 3m 2 + 3m + 2 2 m + m + 1 . 2((m + 1)2 + (m + 1) + 1)) m+1 (m2+m+1)(m+2) m 2 + m + 1 . 2((m + 1)2 + (m + 1) + 1)) (m+1)(m+2) 2«m + 1)2 + (m + 1) + 1))' m+1

= =

which is the right side of S(m

+ 1),

which proves S(k

By the principIe of mathematical induction, for aH n 2 2, the statement Sen) is

D

Exercise 84: For each n 2 2, let Sen) be the statement

or using product notation,

Exercise 82: For each n 2 1, let Sen) be the statement 1


tr~

By the principIe of mathematical induction, for each n 2 1, Sen) is true. [In fact, SeO) is also true.] O

1

+ 1).

rr 1) 1

concluding the inductive step.

1

++ -7·-10+ ... + (3n 1·4 4·7

459

k(3k + 4) + 1 (3k + 1)(3k + 4) 3k 2 + 4k + 1 (3k + 1)(3k + 4) (k + 1)(3k + 1) (3k + 1)(3k + 4) k+1 3k +4 k+1 3(k+1)+1'

+ 1):

= 2(m 2 =


n

i=2

(

1- -:~

=-.

BASE STEP: When n = 2, the equality says 1 -

n

! = !' which is true.

INDUCTIVE STEP: For sorne k 2 2, suppose that S(k) is true. Then

n

1

- 2)(3n

+ 1)

=-3n + 1 '

or written in sigma notation, n

~ (3i BASE STEP: When n = 1, S(l) says

(by S(k))

n

1

2)(3i + 1)

=

3n + 1" k

(1)1(4)

=

L a true statement.

INDUCTIVE STEP: For sorne fixed k 2 1, suppose that S(k) is true. To see that S(k + 1) is true,

E

(3i - 2)1(3i

+ 1)

~ [t (3i - 2)\3i + 1)1+ (3(k + 1) - 2)1(3(k + 1) + 1) = [3k:1]

+ (3k+1;(3k+4)

(byS(k))

k+ 1 1

k

proves that S(k

+ 1)

+ l'

follows. This completes the inductive step.

By mathematical induction, for all n 2 2, Sen) is true. Exercise 85: For each n 2 2, let Sen) be the statement

o


460

follows. Starting with the left side of S(k

or in product notation,

rr (1 _~) n

z2

~. 2n

=

i=2

BASE STEP (n

S(2) says 1 - ~

= 2):

= k(k

1-

[f! (

~ ) 1(1 -

1-

=

k+1 (

--;¡¡;-

=

k

1 - (k

1)

(k:

1)' )

+1

k 2+ 2k

2k

(k

+ 1)

+ l)(k + 2)

_ k(k

+ 1)(k + 2)

_

+

6

+

(k

2

(k

2 3(k

+ l)(k + 2)

+ 6 k(k + l)(k + 2) + 3(k + 1)(k + 2) 6

Exercise 89: For by

+ 1).

71, ;:::

Solutions: Sums with binomial coefficients

L

-+

= 2-

71,

+ 1·

2

P(l) is true.

=

[t (i~l) 1+ 2

l=l

2

INDuCTrON STEP (S(k) hypothesis

2

(i+l)

l=l

2

(~)

=

= 2- k

(1)(~)(3) , so S(1) holds.

(k!2) 2

1

(by P(k))

+ 1 + (k~2) 2

2

=2---+-----

S(k+1)): Fix sorne k;::: 1 and suppose that the induction

k

+1

(k

+ 2)(k + 1)

=2- [1- k:2] _2__2_ [kk+2 + 1] k!l

-

is true. Next, show that

( ). (2) + (3) + (4) + ... +

Sk+1.

2

2

2

(k

+ 2

1)

= (k

+ 1)(k + 2)(k + 3) 6

k+1

=2- _2_ k +2'

proving P(k

+ 1),

+ 1)

is

o

1, Sen) holds.

1

n

t=l

S(k

1, denote the first proposition in the statement of the exercise

P(n):

~ (i~l) = 1): The statement S(1) says

~

-+

INDUCTIVE STEP: For sorne fixed k ;::: 1, assume that P(k) holds. Then

Exercise 88: For each n;::: 1, let Sen) be the statement

(71,

71,

the inductive step S(k)

BASE STEP: Since

Exercise 86: See [266, pp. 125-6] for solution.

BASE STEP

(by S(k))

+ 1)(k + 2)

By mathematical induction, for all

follows, completing the inductive step.

2)

6

Therefore, by mathematical induction, for all n ;::: 2, the statement Sen) holds true. D

27.2

+

which yields the right side of S(k completed.

+ 1)2

k+2 2(k+1)'

shows that S(k

_ k(k -

(by S(k))

+ 1)2

+ 1)(k + 2) 6

INDUCTION STEP: For sorne fixed k ;::: 2, suppose that S(k) is true. Then

TI ( ~)

+ 1),

G) + ... + (k; 1) + (k; 2)

22:i, which is correct, both sides being i·

=

461

27.2. Solutions: Sums with binomial coefficients

and completing the inductive step.


462

By rnathernatical induction, for aH n 2:: 1, P(n) is true.

¿

i=1

1

n

1

2

i=1

2

(i+l) ~ 2 = J~~¿ (i+l)

= J~~ 2 - n

+1 =

k+2-r = k:; =

2 2.

o See [402] for a related diagram. It turns out that aH of this machinery was unnecessary, since

~ - i";1' and so the sum L~l (,!l) telescopes to 2 -

k+ 2 (k +r 1)

=

An infinite series converges iff the limit of the partial sums exists. In this case, the Hmit exists because the partial surns are increasing and each is less than 2. Rence, condude that this lirnit exists (and is at rnost 2). Indeed, one can calculate the limit directly: 00


d)

=

=

n!l'

Exercise 90 (Pascal's identity): Two proofs are presented; the first is by induction and is rather curnbersome, whereas the second is direct and very simple. This demonstrates that induction is not always the preferred proof, but the inductive method is presented here anyway, if only to show its utility.

e

~ r [(~) + ~ 1)]

2

1-

463

(by P(k))

1) + k + 1k+1 - (r - 1) (kr-1 + 1)]

k+ [k + r (k + k+2-r k+1 r

11) + kk ++ 2(kr +- 1) (k + 2)(k + 1- r) (k + 1) + _1_ (k + 1) + (k + 1) (k+2-r)(k+1) r k+1 r-1 r-1 (k (k

+ 2)(k + r) (k + + 2 - r)(k + 1) r

1

(k+2)(k+1-r)(k+1) + 2 - r)(k + 1) r

r

+ (k + l)(k -

= (k

= (k+2)(k+1-r)+r(k+1)

(k+2-r)(k+1)

1

r

r + 2)

(k+1) r

+

(k+1) r- 1

+ (k+l) r-1

Proof by induction: Fix r, and for aH n 2:: r, let P(n) be the proposition

which is the right-hand side of P(k BASE STEP (n = r): The statement P(r) says (r~l) = (~) 1 + r, and so P(r) is true.

+ C~l)'

that is, r

+1 =

INDUCTION STEP: Fix sorne k 2:: r and suppose that

+ 1),


Consequently, by rnathematical induction, for aH n 2:: O, P(n) holds.

O

Direct proof: Let S be a set with n + 1 elernents, and consider sorne fixed x E S. 1 There are (n~ ) r-subsets of S-count them according to whether or not they contain x: there are (~) not containing x, (each forrned by choosing r of the rernaining n elernents in S\{x}), and there are Cr~1) r-sets containing x, (each formed by selecting an additional r - 1 elements in S\{x}). O

is true. Consider the staternent

Beginning with the left side of P(k + 1), apply Lernma 9.6.1 four times, in the first, third and sixth Hnes below (in the first Hne, use m = k + 2 and s = r; in the third line, use m = k + 1 and s = r and s = r - 1 respectively; in the sixth line, use m = k + 1 and s = r):

Exercise 92 [Pascal]: For each n 2:: 2 let S(n) be the assertion that for aH k satisfying 1 ~ k ~ n - 1, (~) k+1 (k:l) n-k' The proof is by induction on n, and in the inductive step, Lernma 9.6.1 is applied. BASE STEP (n

= 2): When n = 2, the only choice for k is k JIL=2=1+1 2) 2-1' (1+1

= 1. In this case,


464

and so S(2) is true. INDUCTIVE STEP (S(m) --> S(m + 1)): Fix m 2: 2, and suppose that S(m) is true, that is, for any k satisfying 1 -:::; k -:::; m - 1,

k+1 m-k

465


INDUCTIVE STEP (S(m) -> S(m + 1)): Fix m 2: 2, and suppose that S(m) is true, that is, for any k satisfying 1 ~ k -:::; m - 1,

is true. It remains to prove that for any 1

~

k

~

m,

is true. To be shown is that for any 1 -:::; k -:::; m,

For m

= k,

the desired equality is true, so assume that k < m. Then

(m:l) (m+1) k+l

m+l (m) m+l-k k m+l ( m) m-k k+l m-k e;:) m+1- k . (k~1) m-k k+1 m+1-k m-k k+1 m+1-k'

For m = k, the desired equality is true, so assume that k hypothesis, k + 1 _ (7:) m - k - (k~l)'

(by Lemma 9.6.1, twice) and simple algebra shows

l+~ 1 + k+1 m-k

m

k+l +1- k

l+~ 1 + m-k k+l

k

k+l

m+ 1- k Multiplying out, one arrives at

+1

n-k'

Proceed by induction on n, and use Pascal's identity: BASE STEP (n = 2): When n = 2, the only choice for k is k

and so S(2) is true.

k+1 +1- k

1 + (k'.:\)

(r;:)

Exercise 93 [Pascal]: The proof provided here is far less intuitive than that in Exercise 92, and was arrived at by working backwards from what was required. For each n 2: 2 let Sen) be the claim that for aH k satisfying 1 ~ k -:::; n - 1,

(~) (k:l)

m

By inductive hypotheses Sk-l(m) and Sk(m),

Hence, by mathematical induction, for each n 2: 2, the statement Sen) holds true. O

(i)

k+l k

m -

Thus,

as desired. This completes the inductive step.

2) (1+1

< m. By the inductive

Applying Pascal's identity,

= 1.

In this case,

= 2 = 1+1

2-1'

whence the desired equality foHows, completing the inductive step. So, by mathematical induction, for all n 2: 2, Sen) is true.

o


466


467

Exercise 95 [Euler]: For t 2': 0, let 8(t) be the statement that for any non-negative integers m and n with m + n = t, and any p 2': 0, finishing the proof of the desired equality, and hence the proof of S(k completes the inductive step. BASE STEP

(t

=

O): When t

=

0, only m = n =

°

is possible. In this case, 8(0) says

If p = 0, the left side of 8(0) is equal to 1, and the right side has only one summand, namely (g) (o~o), also equal to 1. If p 2': 1, the left side is equal to 0, and every summand on the right-hand side will have a factor of the form (?) where i > 0, and so every summand on the right is also O. Thus, 8(0) holds.

Suppose that for sorne k 2': 0, 8(k) holds, that is, for every m 2': with m + n = k, and any p 2': 0,

holds. To show that S(k + 1) holds, show that for any m 2': m + n = k, and any p 2': 0,

°

°

and n 2':

and n 2':

°

Therefore, by mathematical induction, for all t 2: 0, S(t) holds.

=

t (7) (p ~ i) + ~ (7) (p - ~ - J 1-0

°

with

(Pascal 's id.)

BASE STEP (n = 0, 1): The statement SeO) says (g) = (g) 2, which is true since both sides are equaI to 1. To start the induction step for k 2': 1, also check the case n = 1: the statement S(l) says (i) = (g)2 + (~)2, and upon evaluating, says 2 = 1 + 1, which is true.

(S(k)

~

S(k + 1)): Fix sorne k 2': 1 and assume that the inductive

hypothesis

is true. It remains to prove

(ind. hyp.)

1-0

Starting with the 1eft side of S(k

=~(7) [V-i)+C,-~-i)l+(;) =

[~(7)(;~~)l +(;W~l)

o

Exercise 96 [Lagrange]: This is a very interesting equality, because proving it directly by induction seems far more difficult than proving the more general statement of Theorem 9.6.2. One attempt to prove this by induction was given in [350, Ex. 55], however the carefu1 reader will spot that this proof is not "pure1y inductive". In fact, in the inductive step of this proof, another spedal case of Theorem 9.6.2 is invoked; this seems rather pointless, since the Corollary 9.6.3 follows directly in one step from another special case of Theorem 9.6.2! Nevertheless, there is value in examining such an attempt. The presentation given here differs slightly from that in [350], mostly in the order of equalities used; the main idea is the same.

INDUCTIVE STEP

=

This

Proof of Corollary 9.6.3: For each integer n 2': 0, let Sen) denote the statement

If p = 0, then both sides are equal to 1, so assume that p 2': 1. Beginning with the left side of the aboye equality,

(m+;+ 1) (m;n) + (:~~)

+ 1).

2)

2k + ( k+1 (Pascal's id.)

+ 1)

and applying Pascal's identity twice,


468

where the last line used ( 2k ) = (k2~¡). In applying the inductive hypothesis to the k+l . . . dI d? N . first summand in the bracket m the last lme, how lS the last term han e. otlce that by Theorem 9.6.2,


469

It seems that Lagrange's identity is more difficult to prove than Eu1er's equa1ity because upon using Pasca1's identity, certain variables are reduced only by oue, and then the inductive hypothesis is of no help because there are binomial coefficients which are uot of the form It is entirely possible, however, that a purely inductive proof exists, though it seems as if it would be either messy 01' intricate.

e:).

Exercise 97: This exercise occurs in [350, Ex. 59J, however there strong induction is used, although not needed. The proof is rather simple. Por each n 2: 0, let Sen) be the statement that for any m 2: 0,

2k) (k) ( k ) (k-l =~ i k+l-i' k+l

and so, by also applying the inductive hypothesis,

+ 2) (k)2 (k) ( k ) (k)2 (2kk+1 =~ i +2~ i k+1-i +[; j k+l

k

=~ k

k

BASE STEP: When n sides are equal to 1.

(k)i +2~ (k)í (k+1-i k ) +[; (k)j 2 2

k

=

0, SeO) says

i=l

(~)

2

+

t (G)

G~!D, which is true since both

k

Suppose that for some k 2: 0, S(k) is true. To be shown is that is true, that is, for any m 2: 0,

INDUCTION STEP:

S(k

+ 1)

_ k (k) 2 k (k) ( k) k+ 1 ( -'" +2'" . k1) 2 ~ z. ~ z. .z- 1 +¿ . zi=O

L?=o (m~i) =

+

e: ¡))

~

¿

¡=l

i=O

2

+

G)

(m + i) = (mm+1 + + 2) . m

k

Starting with the 1eft side of this equation,

2

(Pascal's id.) (by S(k)) (by Pascal's id.), This completes the inductive step. Therefore, by induction, for each n 2: 0, the statement Sen) is true.

o

Note: The technique used in [350] to begin the sequence of equalities aboye was to use the identity = r:;(n;.::}) (which follows by direct computation or by Lemma 9.6.1 twice) as follows:

C;)

+ 2) = 2k + 2(2k + 1) = 2(2k + 1) , (2kk+1 k+1 k k and then apply Pascal's identity. This technique does not seem as natural as the one employed aboye.

finishing the proof of S(k

+ 1)

and hence the inductive step.

Therefore, by the principIe of mathematicaI induction, for al! n 2: O, the statement Sen) is truco O Exercise 98: This exercise occurs in [350, Ex. 60], where strong induction is used, however unnecessarily so. The equality in this exercise is the very same as that in Exercise 97, because binomial coefficients are symmetric, that is, for M < N (n) = ( n ). so in this case (m+i) = (m+i) and (m+n+l) = (m+n+l) '1 S(k + 1).

+

1).

This finishes the ínductive step

By the principIe of mathematical induction, for all n 2: 0, the statement Sen) is true. O Exercise 103: For each n 2: 1, let Sen) be the statement

INDUCTIVE STEP: Fix some k 2: 5 and suppose that

BASE STEP: S(l) says (1

+ X)l

= (~)

+ (i)x

which is valido


476

+ x)k+l

(1

+ x)k(l + x)

[(~) + (~)x + ... + (~)xk] (1 + x)

(~)

+

(:)x + ... + (:)x

(x

(by S(k))

+ y)k+l (x + y)k(x + y)

[t G)xk-iyi] [t G)xHu'] x [t G)xk-iu']

k

(x

+(~)x + (:)x 2 + ... + (~)xk+l

mt, [G) +

=

(

k

+

o


+

1) + (k + 1) 1

+ 1), and

(i ~ 1)1 + G) Xi

X1

+ (k + 2

1)

x2

477

sum, however due to space constraints, sigma notation is used throughout. [If the notation is at all confusing, the reader might write out the sigma expressions over two lines, lining up like terms.] Beginning with the left side of S(k + 1),

INDUCTrVE STEP: Suppose that for sorne k 2:: 1, S(k) is true. Then

(1


+ y)

(by S(k))

+

I

y

Xk+

1)

k+l + ... + (kk++x1 '

hence the inductive step.

o

Therefore, by MI, S (n) holds for all n 2:: 1.

Exercise 104: The letters x and y are variables indicating non-zero values for which xO = 1 and yO = 1 make sense-these values can be taken from any number field, for example. Por each n 2:: 1, let Sen) be the statement

k+ ( O BASE STEP: S(l) says

(x + y)l

=

(6)x + G)Y

+ 1)

Y°+ ~ ~ (k +. 1) x k+l-j y k

x

k+l

j=l

J

j

1)

+ (kk + 1 +

Yk+l (Pase. id.)

which is valido

INDUCTrVE STEP: Por sorne k 2: 1, suppose that S(k) is true. To complete the inductive step, it suffices to prove that

S(k

1)

:

holds. To prove S(k+ 1), the same trick is used as in the inductive step for Exercise 103, namely break off one factor, apply the inductive hypothesis, use distributivity, combine like terms (that is, ones with the same xayb's), and use Pascal's equality. Just how to collect like terms is most easily seen by writing out a few terms in each

k+l

=

¿:(k+l)xk+l-jyj, j=O

J


+ 1), and hence the inductive step.

Therefore, by MI, Sen) holds for all n 2:: 1. Exercise 106: Let Sen) be the statement

o


478


479

BASE STEP (P(n, 1»: When n 2: 2, P(n, 1) is simply equation (9.7).

BASE STEP: When n = 1,

INDUCTIVE STEP: ([P(n, i -1) A P(n -1, i - l)J --t P(n, i»: Fix i 2: 2 and let n > i. Suppose that both P(n, i - 1) and P(n - 1, i - 1) are true. Then

~ (~)(-lt-kki

and so S(l) is true. INDUCTIVE STEP: For sorne fixed k 2:: 1, assume that S(k) is true, that is,

Next to show is that S(k

(~G)(_1)"-1ki) +ni

(~;;(~ =:)t _1)"-kki ) + ni

+ 1)

is true. To accomplish this, perform a trick of adding and subtracting an extra term:

Xk+ 1

_

xyk

x(xk _ yk)

x(." - y)

+ xyk

+ (x

_ yk+1

-v yV-1)

(~ (~=:) (-I)"-kki-l) + ni

n

(~ [(~) -

(n

~ 1)1(_I)"-kk i1 )

n(~ (~) (_lj"-kk

_ y)yk

(~Xk

n

+ (x -

n

y)y" (by S(k))

t

(by Lemma 9.6.1)

i

-

1

)

+

+ ni

(by Pascal's id.)

(-1)t) n n~ (n ~ 1) (_1)" i

_

kk

i1

(~)(_l)n-kki-l_n~ (n~ l)(_l)n-k k i-l

k=O

k=O

n· O-n· O

(by P(n, i - 1) and P(n - 1, i - 1»,

shows that P(n, i) holds, completing the inductive step. By mathematical induction, for aH j 2:: 1, and any n > j, P(n, j) is true.

D

Exercise 108 [Abel identity 1]: Let a E :IR. and for each n 2:: 1, let Sen) denote the statement

t (~)x(x

- ka)k-l(y

+ ka)n-k

=

(x

+ y)n.

k=O

thereby proving S(k

+ 1).

Mathematical induction proves that for all n 2: 1, the statement Sen) is true.

D

Exercise 107: For each n 2:: 2 and O < j < n, let P(n,j) be the statement that 'L~=o G) (-1 )n-k k j = o. As suggested in the exercise, one proof is by induction on j. Note that when j = O, if one interprets 0° as 1, then P(n, O) is true by equation (9.5).

It seems quite difficult to prove Sen) by a standard inductive argument, so a trick contained in the following lemma (which needs a bit of calculus for its proof) is used. Lernrna 27.2.1. Polynomials p(x, y) and q(x, y) agree if and only if both p(x, -x) = q(x, -x), and

ap

aq

ay

8y·


480

Proof outline: If P and q are differentiable functions with domain JR2, the conditions imply that for every fixed Xo, the restriction of P and q to single variable functions f(y) = p(xo, y) and g(y) = q(xo, y) is identical becausc they agree at the point y = -xo and their derivatives are the same. (This is a standard result in firstyear calculus, following easily from the Mean Value Theorem.) Then use the fact that if two polynomials agree everywhere, they must be the same polynomial. O Let Pn(x, y) be the left-hand side of Sen) and let qn(x, y) be the right side. The conditions in Lemma 27.2.1 are proved separately; denote these two statements by T(n):

Pn(x, -x)

apn = aqn ay ay'

Statement T(n) is proved directly, and U(n) is proved by induction.

x

t (~)(_1)n-k(x

- ka)n-l = O.

k=O

Indeed, by expanding the term (x - ka)n-l and reinterpreting the order of summation, find an express ion equal to zero inside:

x

t (~)

Proof of D(n): BASE STEP: For n = 1, it is straightforward to check that both partials equal l. INDUCTIVE STEP: For sorne fixed m 2: 1, assume that U(m) holds. Then by T(m), S (m) holds. To prove that U (m + 1) follows from S (m), apm+l

---¡j]J

=

a ay [Pm+l (x, y)]

(-1t- k

k=O

I:(

(m + 1)

a [m+l ay {;

k

x(x - ka)k-l(y

a [ x(x - (m + 1)a)m ay

a [m

ay {;

Proof of T(n): First rewrite T(n):

481

+ ka)m+l-k

1

= qn(x, ~x),

and . U( n. )


f

(m + 1)

(m:

k

1)

(m

+

t; (m + 1) k

m

x(x - ka)k-l(y

+

1-

x(x - ka)k-l(y

1

+ ka)m+l-k

1

+ ka)m+l-k

k)x(x _ ka)k-l(y

+ ka)m-k

k=O

(m

+ 1) ~ (7)x(x

- ka)k-l(y

+ ka)m-k

j -ka)jxn - (by binomial thm)

J=O

(m + 1)(x

+ y)m

(by S(m))

a

ay [(x + y)m+l], which is the partial with respect to y of qm+l(x, y) as desired. This completes the inductive step U(m) ---+ U(m + 1).

xI: ax j

j=O

O as desired.

n- j

(n -:- 1) (-1)i . O (by Exercise 107) J

Hence, by MI, for every n 2: 1, U(n) is true, and so it follows that for every n 2: 1, Sen) is true. Exercise 109 [Abel identity 2]: A proof appeared in [354, Probo l.44(a)]. Here is one outline of a proof. For each n 2: 1, let Sen) denote the statement

t (~)x(x k=O

+ k)k-l(y + n -

k)n-k = (x

+ y + n)n.


482

A direct proof of Sen) is available by Exercise 108 with a = -1 and replacing y with y + n. The inductive proof can be found by imitating the solution of Exercise 108 using these replacements. D

27.3. Solutions: Trigonometry

483

By the principIe of mathematical induction, (or as referred to in Section 3.4, "alternative mathematical induction") for aH n 2': 1, Sen) holds. O Exercise 114: Let Sen) be the statement

Exercise 110: For m 2': O, let A(m) be the assertion that Sm(n) is a polynomíal in n of degree m + 1 wíth constant term O. BASE STEP: So(n) = 1° + 2° + ... + nO = n, which is a polynomial of degree l. INDUCTIVE STEP: Fix p 2': 0, and assume that for O -::; j < p that A(j) holds, that is, each of So(n), . .. ,Sp(n) is a polynomial of appropriate degree with constant ter m O. Using m = p in (9.9),

(p+p 1) Sp(n) =

(n

+ l)P+l

_

~ (p ~ 1) Sj(n). j=O

(27.1)

S(k):

Isin«k +

l)x)1 ::; (k + 1)1 sin(x)1

Isin(kx +

l)x)1

x)1

-::;

I sin(kx) cos(x) , + 'cos(kx) sin(x)' (triangle ineq.)

Isin(kx)1 . 'cos(x)' + ::; Isin(kx)' + 'sin(x)1

-1, S(l) holds. Since cos(2n)

= 1=

(_1)2,

::;

kl sin(x)' +, sin(x)'

::;

(k + 1)1 sin(x)l,

'cos(kx)I'1 sín(x)' (bccause I cos(e), ::; 1) (by indo hyp.)

the right-hand side of S(k + 1), completing the inductive step. By mathematical induction, for aH n 2': 1, the statement Sen) is true.

O

Exercise 115 (De Moivre's Theorem): Let D(n) be the statement

D(n):

[cosCO) + isin(e)¡n = cos(ne) + isin(ne).

BASE STEP: D(l) is triviaHy true because it says only [cos(e)+isin(eW = cos(e) +

cos(mn + 2n) cos(mn) (by aboye identity with

Isín«k +

Isin(kx) cos(x) + cos(kx) sin(x)1

INDUCTIVE STEP (S(m) ---t S(m + 2)): Suppose, for so me m 2': 1, that S(m) holds. RecaH the identity cos(e + 2n) = cos(e). Starting with the left si de of S(m + 2), cos( (m + 2)n)

Isín(kx)f ::; kl sin(x)1

holds. Beginníng with the left-hand side of S(k + 1),

Solutions: Trigonometry

=

ni sin(x)l.

holds. To be proved is that

S(k + 1):

Exercise 113: This exercise appeared, e.g., in [582, Probo 39]. There are many proofs; one simple proof is given here with two base cases and an inductive step that jumps by two (similar to the proof of Theorem 3.4.1, where there are three base cases). Let Sen) denote the proposition cos(mr) = (-l)n.

::;

INDUCTIVE STEP: Fix some k 2': 1 and assume that

By mathematical induction, for each m 2': O, Sm(n) is a polYllomial in n of degree D m + 1 alld with constant term O.

BASE STEPS (n = 1,2): Since cos(n) and so S(2) also holds.

Isin(nx)1

BASE STEP: The statement S(l) says Isin(x)1 ::; Isin(x)j, which is triviaHy true.

J

By induction hypothesis, for each j < p, Sj(n) is a polynomial in n of degree j + 1 with constant ter m O, and so the right-hand side of (27.1) is a polynomial in n of degree p + 1 with constant term O. Thus Sp( n) is a polyllomial in n of degree p + 1 with constant term O, proving A(p), completing the inductive step.

27.3

for any real x,

S(n):

e=

i sín(e).

mn)

INDUCTIVE STEP: Let k 2': 1 be fixed and assume that D(k) is true. Then

(_l)m (_1)m+2, which is the right side of S(m + 2). This completes the inductive step S(m) S(m + 2).

[cos( e) + i sineO) ]k+l ---t

(cos(e) + isin(e))[cos(e) + isin(e)t (cos(e) + isin(e))(cos(ke) + isin(ke))

(by D(k»)


484


485

cosCO) cos(kO) - sin(B) sin(kO) + i(sin(O) cos(kO) + cosCO) sin(kO)) cos(B + kB) + isin(O + kO)

sin(e) cos(n7f) + cos(B) . O sin(B)( _l)n.

(by eq'ns (9.12) and (9.11))

cos((k + 1)61) + isin((k + 1)61) proves D(k + 1), completing the inductive step. By MI, for each n 2 1, D(n) is true, completing the inductive proof of De Moivre's theorem (also called De Moivre's formula). O

Exercise 117: This exercise appeared in, for example, [550]. For each n 2 1, let

E(n):

cos(e

Exercise 116: For any n 2 1 let

S(n):

sin(e +mr) = (-ltsin(e)

denote the statement in the exercise. In fact, Sen) is defined for any integer n-see remark following the proof. The identity sin(a + 7f) = - sin(a) is relied on, which can be proved either by noticing that the angles a and a+7f correspond to antipodal points on the unit circle, or directly by equation (9.11) as follows: sin(a + 7f)

= sin(a) cos(7f) + cosCa) sin(7f) = sin(a)( -1) + cos(a) . O = - sin(a).

(-lt cos(e)

BASE STEP: By the identity mentioned aboye, cos(B + 7f) = - cos(e), and so E(l) is true. (Notice that E(O) is also true, so this could have been the base case.) INDUCTIVE STEP: For sorne fixed k 2 1, assume that E(k) is true. Then cosCO

+ (k + 1)7f)

+ k7f) + 7f) cosCO + k7f)

cos((e -

-( _l)k cosCO)

(by E(k))

(_l)k+l cosCO)

sin((B + k7f) + 7f) proves E(k

- sin(e + k7f) -( _l)k sin (e)

=

denote the equality in the exercise. Use the identity cosCa + 7f) = - cosCa) which follows because the angles a + 7f and a correspond to antipodal points on the unit circle; one can also derive this identity by applying equation (9.12) as follows: cos(a+ 7f) = cos( a) cos( 7f) - sin (a) sine 7f) = cos(a)( -1) - sin(a) . O = - cos( a).

BASE STEP: Since sin(O + 7f) = - sin(e), Sen) is true. [Notice that SeO) is true as well, so the base case could have been n = O.] INDUCTIVE STEP: For sorne fixed k 2 1, assume that S(k) is true. Then sin (e + (k + 1)7f)

+ n7f)

(by S(k))

+ 1),

completing the inductive step E(k)

->

E(k + 1).

Therefore, by MI, for all n 2 1 the statement E(n) is true.

o

(_l)k+l sin(e)

proves S(k + 1), completing the inductive step S(k)

->

S(k + 1).

Therefore, by MI, for all n 2 1 the statement Sen) is true.

o

Remark: Notice that since SeO) is also true, one could have concluded that Sen) is true for all n 2 O. In fact, Sen) is true for all n E Z; this can be shown in a number of ways. One could prove this by induction for the negative integers using sin (a - 7f) = - sin(a), and imitating the proof above, showing sin(O - n7f) = (-l)n sin(e) fOl' n 2 O. One can also see that sin(O - n7f) = (-l)n sin(O) follows directly from Sen) by

sin(e - n7f) Lastly, sin(e - n7f) Exercise 113:

= sin(O +

n7f - n27f)

= sin(e + n7f).

Remark: Notice that since E(O) is also true, one could conclude that E(n) is true for all n 2 O. Furthermore, just as in Exercise 116, one can show that E(n) is true for all n E Z.

Exercise 118: This exercise appeared in many places, for example, [550] and [499, Probo 33]. For each n 2 1, denote the statement in the exercise by n : S()

sin(e) cos( -n7f) + cos(e) sine -n7f)

~

'61

Slll J

=

sin(_n~_10)sin(!l...2e) --,---,,,--:--:-;--,-.:....!!...:....

sin(O/2)'

BASE STEP: The statement S(l) says

( -1) n sin (O) follows from S (n) using eq uation (9.11) and

sin(B - n7f)

~. j=l

sin(O) = which is true.

sine+1e) sin(Q) 2 2 sin(O /2)


486

S(k):

487

proves that S(k + 1) follows, completing the inductive step.

INDUCTIVE STEP: For sorne fixed k 2: 1, suppose that

o

By MI, for every n 2: 1, Sen) holds.

sine k!1 O) sine ~) Lsm)O = sin(Oj2)

k

27.3. Solutions: Thigonornetry

..

Exercise 119: This prablem appeared in, for example, [550]. For each integer n 2: 1, let the statement Sen) denote the statement

J=1

is true. It rernains to prove that

cos(n+lo) sin(n8) cosB+cos(20)+···+cos(nB)=

k+1 ~.. S(k + 1):

. (k+20) . (Ck+l)e) sm -2sm 2 L-sm)O = sin(Oj2)

J=1

BASE STEP: Since

.

is true. To streamIíne the proof of S(k + 1), use the identity 2cos

k+1) sin(Oj2) = sin (k+2) . -2-0 - sm(k8j2), (-2-8

which follows fram using A = k!18 and B = ~(J in the following identity sin(A + B) - sin(A - B) sin Acos B + cos Asin B - (sin A cos( -B) + cos Asin( -B))

S(l) is true.

(27.2)

In a naive attempt to prove the inductive step, one sees that a rather strange identity is necessary; one which boils down to the following:

cosCA + B) sin(A) + cos(2A + 2B) sin(B) = cosCA + 2B) sin(A + B).

(27.3)

The following sequence of identities shows one way to derive equation (27.3), though there rnight be other sírnpler ways: Beginning with sin A(l - sin 2 B) = sin A cos 2 B,

2cosAsinB.

and adding cos A cos B sin B to each side gives

Beginning with the left side of S(k + 1),

sin A + cos A cos B sin B - sin A sin B sin B

k+l

L sin (j8)

[t.

2.

cosct!lB) sin(~) sin(B/2) = cos(B),

sin A cos B + cos A sin B - (sin A cos B - cos A sin B)

j=l

s~n(Bj2)

= sin A cos B cos B + cos A sin B cos B.

First use the identities (9.12) and (9.11) and then rnultiply each side by cos(A+B), giving

1

cosCA + B)[sin A

,in(j6) Hin( (k+ 1)6)

· ( l l i 8) . (kO) sm 2 sm '2 + sin«k + 1)8) sin(8j2)

B) cos B.

Multiplying out the left síde, (by S(k)

cosCA + B) sin A

sin(k+1e)sin(~) . (k+l k+l) 2 2 +SIn --8+ -8 sin(ej2) 2 2

(k+2e) c o(k+28) s--

sin(k!18)sin(~) + 2 SIn . ---'---'''---':-'---'--'''-sin(8j2) 2

+ cosCA + B) sin B] = cosCA + B) sin(A +

2

· (k + 1 ) sin(k8j2) + 2cos (~8) sin(8j2) SIn -2- 8 sin(8j2) · (k + 1 ) sin(k8j2) + sin (~8) - sin(k8j2) SIn -2- 8 sin(8j2) · (k + 1 ) sinsin(8j2)' (~8) SIn -2- 8

+ cos2 (A + B) sinB =

cosCA + B) sin(A + B) cos B.

Subtracting sin 2 (A + B) sinB fram each side gives (by eqn (9.11)

cosCA =

+ B) sin A + [cos 2 (A + [cosCA

+ B) cosB -

B) - sin2 (A + B)] sinB

sin(A + B) sinB] sin (A + E).

Finally, applying identity (9.12) twice, on the left with o: = A right with o: = A + B and (3 = B, arrive at (byeqn (27.2))

+B

= (3, and on the

cosCA + B) sin(A) + cos(2A + 2B) sin(B) = cosCA + 2B) sin(A + B), finishing the proof of equation (27.3)


488

INDUCTIVE STEP: For sorne fixed k 2: 1, assume that k

S(k):

2

2

2

2

. «(k+l)/:J) COS (k+2B) -2- sm -2sin(Bj2)

.

L...cosJB=

INDUCTIVE STEP: For sorne fixed k 2: 1, suppose that

)=1

P(k):

follows. Beginning with the 1eft side of S(k

[t

L cosue) j=1

2

sin2 A.

k+1 ~

k+l

+ 1),

~

sm(Oj2)

"2

+ cos«k + 1)0)

P(k

+ 1):

j=l

[t

k+1

Lsin«(2j - 1)0) j=l

P( n): sine O) + sin(3B) + ... +

2

sin (kB)

o-o

1 )0)]

+ s;n( (2k + 1)0)

+ sin«2k + 1)8)

(by P(k))

+ sine (2k + 1)0) sine 8)

sine B) sin 2 «k + 1)0) sine B) where the last equality follows fram equation (27.4) using A = (k + 1)B and B This finishes the proof of P(k + 1), and hence the inductive step.

sin 2 (nB) sine (2n - 1)8) = sine O) .

Therefore, by MI, for all n 2: 1, P(n) is true.

s~~(W' P(l) is clearly true.

= k8.

o

Exercise 121: This exercise appeared, for example, in [550]. Fix sorne angle e which is not a multiple of 7f. For n 2: 1, denote the statement in the exercise by

In the inductive step, the following identity is used:

+ sin2 (B) =

s;n( (2j

sin 2 (kB) sin(O)

O

Exercise 120: This prob1em appeared in, for example, [550]. For n 2: 1, let the statement in the exercise be denoted by

sin 2 (A).

(27.4)

Sen) :

+ sin2 (B)

+ cos Asin B)(sin Acos( -B) + cos Asin( -B)) + sin 2 B + cos A sin B) (sin A cos B - cos A sin B) + sin2 B

t

j=l

The proof of equation (27.4) is fairly straightforward:

(sin A cos B

. 2 ) sm (~k + 1 O) sm(8)

follows. Indeed,

By MI, for all n 2: 1 the statement Sen) is true.

(sin Acos B

sm(B)

k+l L sin«2j _ l)B) =

where the 1ast equality follows from equation (27.3) with A = kBj2 and B = oj2. This completes the proof of S(k + 1) and hence the inductive step.

sin(A + B) sin(A - B)

si~2(k8)

(by S(k))

cos( ~B) sine ~) + cos( (k + 1)0) sine ~) sin(8j2) «(k+1)/:J) k+2B) . ( cos:2 sm - 2 - (b e 'n 27.3) sin(Oj2) y q

sin(A + B) sin(A - B)

sin«2j - 1)0) =

is true. It remains to show that

1+ cos((k+ 1)0)

SI

t

j=1

cos(jO)

cos(k+1 B) ·n( k/:J)

BASE STEP: Since sin(B) =

+ sin 2 B

+ (1 - cos A) sin 2 B sin A cos B + sin 2 A sin 2 B sin 2 A( cos 2 B + sin2 B)

is true. It remains to prove that

S(k+1):

2

sin A cos B

sin(Oj2)

f;COSJB =

489

sin A cos 2 B - cos 2 A sin 2 B

cos(kt 10 ) sin(~)

.


cos (2j - 1)0 = sin(2nO) . 2sin(8)

BASE STEP: S(l) says cos( B) = sin(28) 2 sin(8)' which is true because sin(20) = 2 cosCO) sin(o).


490

k

.

¿cos«2J -1)0)

=

491

o

By MI, for every n 2: 1, Sen) is true.

INDUCTIVE STEP: For sorne fixed k 2: 1, assurne that

S(k) :

27.3. Solutions: Trigonornetry

Exercise 122: This exercise appeared in e.g., [550J. It is given that So = O, SI = 1, and for n 2: 2,

sin(2kO) 2sin(0)

)=1

Sn

is true. To be proved ís

S(k+ 1):

~

)0) _ sin(2(k + 1)8) L...,.,cos 2J -1 2sin(0) .

Sen) :

To thís end, use the following identity:

sin(2kO) + 2cos«2k + 1)0) sin(O) sin(2kO) + cos«2k + 1)0) sín(O)

+ cos«2k + 1)0) sin (O) sin(2kO) + cos(2kO + O) sine O) + cos( (2k + 1)0) sine O) sin(2kO) + [cos(2kO) cos( O) - sin(2kO) sine O)) sine O) + cos( (2k + 1)0) sine O) sin(2kO) - sin(2kO) sin 2 (O) + cos(2kO) sine O) cos( O) + cos( (2k + 1)0) sine O) sin(2kO) cos 2 (0) + cos(2kO) sin(O) cosCO) + cos«2k + 1)0) sin (O) [sin(2kO) cosCO) + cos(2kO) sin(O)] cos(O) + cos«2k + 1)0) sin(O) [sin(2kO + O)] cosCO) + cos«2k + 1)0) sin(O) [sin«2k + 1)0)] cosCO) + cos«2k + 1)0) sin(8) sin«2k + 1)0 + O) sin«2k + 2)0).

Now that the tools are assernbled, he re is the proof of S(k

cos (2j - 1)0

+ 1):

[~COS«2j - 1)0)] + cos«2k + 1)9)

j=1

sin(2kO) . (O)

2sm sin(2kO)

+ cos«2k + 1)8) (by S(k)) + 2 cos( (2k + 1)0) sine O) 2 sine O)

sin(2(k + 1)0) (b '(275) 2sin(O) yeqn .. This proves S(k

+ 1),

and hence finishes the inductive step.

C(n):

(27.5)

A proof of equation (27.5) rnight go as follows:

¿

Sn

=

sin(n8) sín(8) ,

and for each n 2: 1,

+ 2 cos«2k + 1)0) sín(O) = sin(2(k + 1)0).

k+l

Sn-2·

To be proved is that for each n 2: O the staternent

«.

)=1

sin(2kO)

= 2cos(0)Sn_1 -

cos(nO) = cos(O)sn - Sn-1

are true. Both Sen) and C(n) are proved by simple induction, inducting on each at the sarne time. BASE STEP SeO), S(l), AND C(l): Since Sn is defined by two previous values, one would think that there should be two base cases for Sen). In fact, the only reason that two base cases are needed for Sen) is to have S(l), since the induction is started on C(n) at n = 1 as well; the actual induction is simple (as opposed to strong). SeO) says that So = s;~f~~r) which is true since both sides are O. S(1) says SI = sin(I·B) . sin(B) , a1so true smce each SI'de equa1s 1. F'mally, C( 1) says cos(l· O) = cos( O)SI - So, which is true since SI = 1 and So

= O.

INDUCTIVE STEP: For sorne fixed k 2: 1, assurne that both S(k) and C(k) are true. The inductive step is accornplished in two stages, one for S(k + 1), and the other for C(k + 1). First show

S(k

+ 1)

:

sk+l

=

sin«k + 1)0) sin (O) .

Here is the derivation: 2 cos( O)Sk - Sk-1

(def'n of sn)

cOS(8)Sk + cOS(O)Sk - Sk-l cOS(8)Sk + cos(kO) (by C(k)) sin(kO) cos(O) sin(O) + cos(kO) (by S(k)) cos(O) sin(kO)

+ cos(kO) sin(8)

sin(O) sin(kO + O) sin(O) sin«k + 1)0) sin(O)


492

completing the proof of S(k + 1). To finish the inductive step, one needs to prove

C(k + 1) : cos«k Having both S(k) and S(k

cos«k

+ 1)

+ 1)0) = cOS(e)Sk+l

- Sk.

A(k

Sk+l

completing the proof of C(k + 1). Since the implications S(k) and S(k) 1\ S(k + 1) -+ C(k + 1), are proved, the implication 1\

C(k)]

-+

[S(k

Sk+l = cos«k + 1)0)

1\

C(k)

-+

2cos(e)cos(ke) - cos«k -1)0) (by A(k) and A(k - 1)) 2 cosCO) cos(kO) - cos(kO - O) 2 cos(e) cos(ke) - [cos(ke) cos( -e) - sin(ke) sine -e)]

2 cos(e) cos(ke) - [cos(ke) cos(e)

+ sin(ke) sin(e)]

cos(e) cos(ke) - sin(ke) sin(e)

cos(ka + e) (by identity (9.12)) + 1)61),

cos«k

as desired. This completes the inductive step. By MI, for all n 2 1, A(n) is true.

o

Exercise 124: Use Exercise 194 with x replaced by cos 2 (x) and y replaced by 2 sin (x). Then x + y = 1, and the result foHows directly. O

S(k

+ 1)

Exercise 125: This exercise appears in many places, for example, [499, Probo 31], where a brief solution outline is given. For n 2 0, let A(n) be the assertion that cos(a) cos(2a) cos(4a) ... cos(2na) =

+ 1) 1\ C(k + 1)]

is proved, completing the inductive step. By MI, for aH n 2 1, Sen) and C(n) hold. Together with the base case SeO), all that was required is proved. O Exercise 123: This exercise appears in [499, Probo 32], together with a sketch of the solution. Write SI = cosCO), S2 = cos(20) and for n > 2,

Sn = 2 cos(e)Sn-l - Sn-2' For every n 2:: 1, denote the assertion in the exercise by

A(n):

+ 1):

follows. Indeed,

in hand,

cos(O) cos(ke) - sin(8) sin(k8) cos(e) cos(ke) sin(e) - sin2 (8) sin(ke) sin(e) cosca) cos(ke) sin(e) - (1 - cos 2 (8)) sin(ke) sin (e) cos(8) cos(k8) sin(O) + cos 2(e) sin(ke) - sin(kO) sin (e) cos(O)[cos(kO) sin(O) + cosCO) sin(kO)] - sin(ke) sin (e) cosCO) sin«k + 1)61) - sin(ke) sine O) cos(e) sin«k + 1)0) sin(ke) sin(a) - sin(O) cOS(O)Sk+l - Sk (by S(k + 1) and S(k)),

[S(k)

493

INDUCTIVE STEP: For some fixed k 2 2, assume that both A(k - 1) and A(k) are true. To be shown is that

cos( e + ke)

+ 1)0)


Sn = cos(nO).

BASE STEP: Both A(I) and A(2) are true by definition of SI and S2.

sin(2n+la) . l' . 2n+ sm(a)

BASE STEP: Since cos(a) = ;i~~~~~ (just expand the numerator using identity (9.11)), A(I) is seen to be true. INDUCTIVE STEP: For sorne k 2 Osuppose that

A(k):

cos(a) cos(2a)··· cos(2ka) =

sin(2 k +la) k 1 . ( ) 2 + sm a

is true. Then using the identity sin(A) cosCA) = sin(2A)j2, cos(a) cos(2a) ... cos(2ka) cos(2k+la) sin(2 k+la) 2Ml sin(Q) cos(2k+la) sin(2 k+ 1 a + 2k +la)j2 2k+l sin(a)

(by A(k))


494


sin(2k+2a) 2k+2 sin(a)

sine (2k+ 1 )t )

. (t)

+ cos«k + l)t) (by (P(k))

2sm 2

shows that A(k + 1) is true. This completes the inductive step.

sin(~) + 2cos«k + 1)t)sin(!)

o

By MI, for every n ?: O, A(n) is true.

2 sin(~)

sin(~) + sin«k + l)t + ~) - sin«k + 1)t - ~)

Exercise 126: (Dirichlet kernel) This exercise appears in rnany places, for exarnple, [499, Probo 34]. Let be an angle which is not an integer rnultiple of 27r; for each n ?: 1, denote the statement in the exercise by

2sin(~)

e

P(n):

495

sin(~)

1 ~ . sin«2n + 1)tj2) "2 + ~ cos(Jt) = 2sin(tj2) .

2sin(!)

)=1

(by eqn (27.6))

,

this proves P(k + 1), and hence finishes the inductive step.

BASE STEP: To prove P(l), use the identity

By rnathernatical induction, for al! n ?: 1, P(n) is true.

sin(A + B) - sin(A - B) = 2cosAsinB,

(27.6)

o

Exercise 127: (Fejér kernel) Using the notation from Exercise 126 put

which was proved in the solution of Exercise 118. Using A = t and B = tj2, equation (27.6) yields sin(3tj2) - sin(tj2) = 2 cos(t) sin(tj2); hence

1 sin(tj2) + 2cos(t) sin(tj2) _ sin(3tj2) "2+cos(t)= 2sin(tj2) -2sin(tj2)'

For N ?: O, let denote the staternent to be proved by

SeN) :

which proves P(1).

K

2

(t) = sin «N + 1)tj2) . N 2(N + 1) sin 2 (tj2)

INDUCTIVE STEP: For sorne fixed k ?: 1 assume that 1

P(k):

k

f:r

- + '"' cos(jt) = 2

sin«2k+l)t) . 2sm 2

(t)

BASE STEP: When N = O, there is only one sumrnand in Do, namely when n = O; in this case, Ka = and

!'

sin 2 «O + 1)tj2) 1 2(0 + 1) sin 2(tj2) = "2

is true. To be proved is

P(k+ 1):

1 k+1 sine (2k+3)t) "2 + LcoS(jt) = 2sin(1) )=1

Starting with the left side of P(k + 1), 1

"2

2

as well, so SeO) is true. Instead of proving Sen) by induction, a sirnpler but equivalent statement is proved. Since by Exercise 126, K (t) = _1_ ~ D (t) = _1_ ~ sin«2n + 1)tj2) N N +1~ n N +1~ 2sin(tj2) ,

k+1

+ L cos(jt) J=1

SeN) says 1

k

1

- + "cos(jt) + cos«k + 1)t) 2~ J=1

N

N

+1~

sin (2n2+1t) _

sin 2

(Ni 1t )

2 sin(tj2) - 2(N + 1) sin 2(tj2)'


496

Cancelling the ter m 2(N+1{sin(t/2) shows that SeN) is equivalent to *

497

Let x be a real number which is not an integer multiple of 21f. For each n 2: 1 let S (n) be the statement

(2n+1) = sin2(N;¡-lt) sin(t/2) .

N.


S (N) : ¿SIn - 2 - t

~ .. (.)

~JSlllJX

n=O

=

(n + 1) sin(nx) - nsin((n + l)x) 2

4 sin (x/2)

j=l

Proceed with the induction using S* rather than S. (The base case for S*(N) is proved by S(O).) INDUCTIVE STEP: For some fixed k 2: O, assume that S*(k):

k

(2n + 1 ) -2-t

.

¿Slll

2 _ sin (k!lt) sin(l)

n=O

BASE STEP: S(l) says sin (x) = 2S;~~2(;~ngx). Why is this true? To begin with, a standard identity is arrived at by applying identity (9.12) with o: 2 2 cos(x) = cos (x/2) - sin (x/2)

-

2

~,

1 - sin2(x/2) - sin 2(x/2) = 1 - 2sin 2(x/2).

4sin 2(x/2) = 2 - 2cos(x).

S*(k+ 1):

k+1 . ¿sm

(2n- 2+ -1 t )

_ sin2 (k!2t) sin(l) . 2

2sin(x) - sin(2x) = 2sin(x) - 2sin(x) cos(x) 4sin 2(x/2) 2 - 2cos(x)

Starting with the left side of S* (k + 1),

(2n- 2+-1t)

[

n=O

=

(2n

k .~sin - 2+-1t )

sin2(~t) sin(~)

1+sin (-2k2+-3t)

+sin(2k+3 t) 2

(byS*(k))

S(k) :

(k + 1) sin(kx) - ksin(k + l)x)

= --'-----'---'-------'---;;.,-,-------.:...:--~

4sin 2 (x/2)

S(k + 1):

I:j

sin(jx) = (k + 2) sin«k +

j=1

1)~)

- (k + 1) sin(k + 2)x) 2 4sm (x/2)

foUows. To prove S(k + 1), use the following identity sin(kx) - 2 sin«k + 1)x) cos(x) = - sin«k + 2)x),

which is precisely S*(k + 1). Since this is equivalent to S(k + 1), the inductive step is completed. O

Note: The statement S*(N) is equivalent to that in Exercise 120, using the replacement t = 28. Exercise 128: This exercise occurs in [499, Probo 35), for example.

~

is truco It remains to show that

_ sin2 (k!2t) . (l) , Slll 2

Rence, by mathematical induction, for aU n 2: O, SeN) is true.

~ J.. (.) sm JX j=l

Using sin 2 B + sin(A + B) sin(A - B) = sin 2 A (this is equation (27.4), proved in Exercise 120) with A = k!2t and B = k!lt in the last line aboye, obtain

n=O

)

INDUCTIVE STEP: Fix k 2: 1, and suppose that

sine ~)

(2n + 1 ) ¿sm - 2 - t

= sin(x

shows that SU) is indeed true.

sin 2 (~t) ++sin(~t)sin(~)

k+1 .

(27.7)

Also, identity (9.11) with ex = (3 = x yields sin(2x) = 2sin(x)cos(x). Using these identities,

-

n=O

k+1

=

= (3 =

It foUows that

is true. To be proved is

¿sin

.

(27.8)

which is perhaps most easily seen by using the trick kx = (k + 1)x - x and expanding sin(kx) using identity (9.11) as follows: sin(kx) - 2 sine (k

+ l)x) cos(x)

sin«k + l)x - x) - 2sin(k + 1)x) cos(x) sine (k

+ 1)x) cos( -x) + cos( (k + l)x sine -x)

- 2 sine (k

+ l)x) cos(x)

sin«k + l)x) cos(x) - cos«k + l)x sin(x) - 2 sine (k + l)x) cos(x) - sine (k + l)x) cos(x) - cos( (k + 1)x sin (x )

Chapter 27. Solutions: Identíties

498

=

-sin((k

27.3. Solutions: Trigonometry verifies that S(l) indeed holds.

+ 2)x).

Starting with the left side of S(k

INDUCTIVE STEP: Let k

+ 1),

S(k):

k+l

2..: j sin(jx) j=l

~

1 be fixed and assume that

k

Lcos(jx) = (k

+ l)cos(kx) -

kcos«k

+ l)x) -1

2

4sin (x/2)

j=l

[t,j

is true. To complete the inductive step, one must prove

Sin(jX)]

+ (k + 1) sin((k + l)x) S(k

+ 1):

(k + 1) sin(k~) ~ kt~«k + l)x) + (k + 1) sin«k + 1)x) (by S(k)) 4sm x 2

(k + 1) cos(kx) - 2(k

(by eq'n (27.7))

4sin2 (x/2) (by eq'n (27.8)),

4sin 2 (x/2) which is the right si de of S(k

+ 1).

+ 1) cos(x) cos«k + l)x) = -(k + 1) cos«k + 2)x).

(27.9)

cos(kx) - 2 cos(x) cos( (k + l)x) [cos«k + l)x) cos(x) + sin«k + l)x) sin(x)] - 2cos(x) cos«k + l)x) = -[cos«k + l)x)cos(x) - sin«k + l)x)sin(x)] = - cos«k + 2)x).

(k + 1) sin«k + l)x) + (k + l)[sin(kx) - 2 sin«k + l)x) cos(x)]

+ 1) sin(k + 2)x)

+ 2)x) - 1

The sequence of steps in proving equation (27.9) is very similar to that used to prove equation (27.8), so only an outline is given:

(k + 1) sin«k + l)x) + (k + 1) sin(kx) - 2(k + 1) sin«k + l)x) cos(x) 4sin2 (x/2)

- (k

+ 1) cos«k 2 4sin (x/2)

(k + 2) cos«k + l)x) - (k

Similar to that used in Exercise 128, the following identity is used to accomplish the proof of S(k + 1):

4sin2 (x/2)

+ 2) sin«(k + l)x)

k+l L cos(jx) = j=l

(k + 1) sin(kx) - ksin«k + 1)x) + 2(k + 1) sin«k + 1)x)(1 - cos(x))

(k

499

Multiplication throughout by (k + 1) finishes the proof of (27.9). To prove S(k + 1),


By MI, for all n ~ 1, Sen) is true, completing the solution to Exercise 128.

k+l

O

¿jcos(jx) j=l

Exercise 129: This exercise occurs in e.g., [499, Probo 36]. Let x E lR be fixed which is not an integer multiple of 21f. Por each n ~ 1 let Sen) be the statement

.;¡-.,

(') =

~COS]x

j=l

(n

BASE STEP: S(l) says cos(x) =

+ 1) cos(nx) - ncos«n + l)x) 2

2 cos(x) - cos(2x) - 1 .

2

cos(2x) = cos 2 (x) - sin2 (x) and equation (27.7) (which says

4sin 2 (x/2) =

=

(k

1

1+

,os(jx)

(k

.

. Usmg

+ 1) cos(kx) - kcos«k + l)x) 4sin 2 (x/2)

1

+ (k + l)cos«k + l)x)

(by S(k)

(k + 1) cos(kx) - kcos«k + l)x) - 1 + 2(1 - cos(x»)(k + 1) cos«k + l)x) 4 sin2 (x/2) (by eq'n 27.7)

2

2 cos (x) - 1,

(k

+ 2) cos«k + l)x) + (k + 1) cos(kx) -

(k

+ 2) cos«k + l)x)

(k 4sin2 (x/2)

2 - 2cos(x), proved in Exercise 128], 2

+ 1) cos((k + 1)x)

.

4sin (x/2)

4sm (x/2)

[t,j

2cos(x) - cos(2x) - 1 = 2cos(x) - (2cos (x) - 1) - 1 = cos(x) 4sin 2 (x/2) 2 - 2cos(x) ,

+ 1) cos(x) cos«k + l)x) -

- (k + 1) cos«k + 2)x) - 1 4sin 2 (x/2)

(by eq'n (27.9),

1


500


+ 1),

27.3. Solutions: Trigonometry was actually proved, or equivalently,

and hence the inductive step.

By mathematical induction, for all n 2:: 1, Sen) is true, completing the solution to Exercise 129. D Exercise 130: This exercise occurred in, for example, [499, Probo 37]. Suppose that x E lR is not an integer multiple of TI. Por each n 2:: 1, let T(n) be the statement 1 tan -----:- = - 1 cot " n -----:- cot(x). L 2J 2J 2n 2n

(X)

(X)

j=l

cot(A) In proving T(k

+ 1),

k+1

X

sin(~)

2

2

2cos(~)

1

2k

This proves T(k

INDUCTIVE STEP: Por sorne fixed k 2:: 1, suppose that

z= k

;j tan

(~)

1

( x )

1

=

2k cot ( ; ) - cot(x)

j=l

is true. I t remains to prove

+ 1):

(27.10)

tan

[~cot Ck:1) ]

2k~1 cot (2 k:

- cot(x)

(2:+

1)

(by T(k))

- cot(x)

(by eq'n 27.10)

1 ) - cot(x).

+ 1), completing

the inductive step.

By MI, for every n 2:: 1, T(n) is true, completing the solution to Exercise 130.

Thus T(l) is true.

T(k

+ 2k~1

~ 2k+1 2k [cot (-=-) 2k + ~2 tan (~)]

=

-cot(x).

~cot (4)·

1

1

2sin(~)cos(~)

~cot(~)

=

j=l

sin 2( ~) 2 sine ~) cos( ~) cos2(~) - [cos2(~) - sin2(~)] cos( ~) cos(x) sine ~) - sin(x)

(4)

use this with A = {k, vastly simplifying calculations:

2k cot ( ; ) - cot(x)

1

T(k):

+ ~tan

"-----:-tan (~) L 2J 2J

BASE STEP: Applying identity (9.12) with CIó = i3 = x/2, one gets cos(x) = cos 2 (x/2) - sin 2(x/2). Similarly, using equation (9.11), sin(x) = 2cos(~)sin(~). . . . . () cos(A) d (B) _ sin(B) Usmg these, together wIth the defimtlOns cot A = sin(A)' an tan - cos(B) ' one proof of T(l) is: - tan( -)

501

k+1 z=

2j tan 2j

1

= 2k+1 cot

(X) 2k+1

Exercise 131: This exercise occurs in, for example, [499, Probo 38]. It serves as a good workout for understanding of inverse trigonometric functions (though all steps are found to be simple). Recall that if tan(e) = y, then the tan inverse function tan- 1 is defined by tan- 1 (y) = e. The reciprocal of tan(e) is (tan(e))-l = cot(e), though remember that tan -1 and cot are different functions (one is the inverse with respect to functions, and the other is the inverse with respect to multiplication). Por each n 2:: 1 let Sen) be the statement

- cot(x).

j=l

To save a little work, notice that in proving the base case, for any angle A which is not a multiple of TI, the identity.

~tan (4)

=

~cot (4)

-cot(A),

O

BASE STEP: S(l) says

Chapter 27. Solutíons: Identitíes

502

To prove S(l), take the tangent of each side. For the 1eft side, 1et cot(O) = 3; then tan(B) = and O = cot- 1 (3), so tan(cot- 1 (3)) = tan(B) = For the right side, an

l,

27.3. Solutions: Trigonornetry This can be checked by taking tangents of each side:

!.

tan(cot- 1 (2m

application of (9.15) gives tan(tan- 1 (2)) - tan(tan- 1 (1)) 1 + tan(tan 1(2)) tan(tan 1(1))

503

+ 3)) =

1 --,

2m+3

and

(2) - (1) 1 + (2)(1)

tan ( tan-

1

agree, and so S(m

3'

m+2) ~-1 _1 1 1 ) ( m + 1 - tan- (1) = 1 ~+~(1) = 2%:~ = 2m + 3

1

m+l

+ 1)

m+l

is true. This concludes the inductive step.

o

By MI, for each n 2: 1, Sen) is true.

which agrees with the tangent of the left side. Since tangent is defined so that it is one-to-one, conclude that since the tangents of both sides of S(l) agree, then S(l) itself is true. INDUCTIVE STEP: Fix sorne m 2: 1 and assurne that

Exercise 132: This exercise occurs in Trirn's calculus book [534, Ex. 40, p. A-6]it received a three star rating (out of three). For n 2: 0, let P( n) be the proposition that there exist constants ao, al, ... ) a n , and bo, b1 , .. . ,bn so that n

sinn(x)

= L[arcos(rx) + brsin(rx)]. r=O

°

is true. It rernains to show that S(m

+ 1):

L

m+1

cot- 1 (2k

+ 1)

L

m+1

=

k=l

tan-

T 1) -

1 ( .

+

BASE STEP: The exercise asks to prove P(n) for n 2': 2, however, it appears to be true for even n 2': O. If x is not a rnu1tiple of 7f, sin (x) =J. 0, and then the resu1t is true even for n = O, since sinO(x) = 1 = 1· cos(O· x) + sin(O· x). When n = 1, pick ao = 0, al = 0, bo can be anything, and b1 = lo For n = 2, equations (9.12) and (9.13) yield cos(2x) = 1 - 2sin2 (x), and henee

(m

+ 1) tan

-1

(1)

J=l

is true. Beginning with the left side of S(k

sin 2

+ 1), so ao = ~, anything for bo, al

{[t, [t.

tan-'(i;

tan-'

1

(2m + 3)

(m

= bl = b2 = O and a2

+

1) ,.,,-'(1) +tan~'(I) I wt-'(2m + 3)

+ br sin(rx)].

r=O

To be shown is that P(k+ 1) is true, that is, that there exist constants and bo , ,bk~l, so that

b;, ...

k+l

sink+1(x) = L[ar cos(rx) So, to finish the proof of S(m eoC 1 (2m

+ 1), it suffiees to prove that

+ 3)

= -./ show that 8(2) is true.

k

sink(x) = L[a r cos(rx)

1)] - m tan-'(l)} + oot-' (2m + 3) (by S(m))

e; 1)] -

22'

INDUCTIVE STEP: Let k ;::: 2 be fixed, and suppose that P(k) is true, that is, assurne for sorne fixed constants ao, al, ... , ak and bo , b1 , ... , bk

k=l

{~cot-l(2k + 1)} + cot-

= ~ - ~ cos(2x)

= tan -1

(m + 2) --

m+1

-

) tan -1( 1.

+ br sin(rx)].

r=O

Starting with the left-hand side of this aboye equation,

ao, al, . .. ,ak+1


504

Applying P(k) to the factor sink(x) aboye, there exist ai's and bi's so that

sink+l(x) =

sin(xl

(t.1a,

4sin 3 (x) = 3sin(x) - sin(3x),

cos(rx)

+ b, sin(rx ll )

I::[a r sin(x) cos(rx) + br sin(x) sin(rx)J. r=O

It suffices to express, for each r = O, 1, ... , k, each of sin (x ) cos( rx) and sine x) sine rx) as linear combinations of cos(O), cos(x), ... , cos«k + 1)x), and sin(O), sin(x), ... , sin«k + l)x). This is done via the two identities

sin(x) sin(rx)

= ~ sin«1 + 1

= "2 cos«1

~ sin«l -

r)x) +

r)x),

which ca~ be easily verified. One also notices that the di 's depend only on previous bj's and bi's depend only on previous aj's, and since for n = 2, alI bi's are zero, by induction, if n is even, sinn(x) is expressible as a linear combination of only cos(rx)'s and if n is odd, sinn(x) is expressible only as a linear combination of sin(rx)'s. It might be interesting to try and give a formula which would describe the coefficients explicitly for any n. Perhaps this has been done. Exercise 133: This appeared in, e.g., [161, 8.19, pp. 208, 215]. Fix x and a so that x + ~ = 2 cos( a). Fór every n ?: 1, denote the equality in the exercise by

1

E(n) :

2 cos«l + r)x).

- r)x) -

(The proofs ofthese are simple: for the first, expand each ofsin(A+B) and sin(A-B) using equation (9.11) and add the two equations; for the second, expand cos(A+ B) and cosCA - B) using equation (9.12) and subtract the two equations.) Notice also that when r > 1, one replaces sin«(1 - r)x) with - sin«r - l)x) and cos«(1 - r)x) with cos«r - l)x). So, using these replacements, sink+l(x) is indeed expressible as linear combinatioIls of cos(O), cos(x), ... , cos«k + l)x), and sin(O),sin(x), ... ,sin«k+ l)x). This completes the proof of P(k+ 1) and henee the inductive step. By MI, for all n 2: 2, the statement is true (in fact, for all n 2: 1, and if sin(x) for all n ?: O).

1= 0,

BASE STEP:

a

br ~al ao - ~a2

3

~bl ~b2 ~(b3 - bd ~(b4 - b2 )

k-1 k k+1

~(bk - bk-2) -~bk-l -~bk

r

1 2

sin (x) =

4

E(l) is given. To see E(2), x2 +

-}z

=

(x + 1/x)2 - 2 = 4cos 2 (a) - 2 =

INDUCTIVE STEP: Suppose that for sorne k 2: 2, both E(k) and E(k - 1) hold. To see that E(k + 1) follows, the identity cosCA

+ E) + cosCA -

B) = 2cos(A) cos(E)

(27.11)

helps. [This follows from expanding cos(A+B) and cos(A- B) using equation (9.12) and adding the two equations.] Then

(

X

+ -xl) (xk

+~) xk

_

xk-l _ _ 1 xk-1

4 cos(o:) cos(ko:) - 2 cos«k - 1)0:) (by E(l), E(k), E(k - 1»

which shows that E(k + 1) is true, completing the induetive step. By induetion, for eaeh n 2: 1, the statement E(n) is proved.

j(ak-2 - ak) ~ak-l

~ sin(x) - ~ sin(3x)

4

= 2 cos(na).

2cos«k + 1)0:) + 2cos«k -1)0:) - 2eos«k - 1)0:) (byeqn (27.11» 2eos«k + 1)0:),

°

Actually, if one were to develop a recursion, perhaps bo = should be declared in every case. Using the chart aboye recursively, one arrives at 3

+~ xn

O

i(al - a3) 1(a4 - a2)

~ak

xn

2cos(2a).

Comments on Exercise 132: In the inductive step aboye, what are the di's and b~ 's explicitly? 1 think that they work out as follows: r O

505

or equivalently,

k

sin (x) cos(rx)

27.4. Solutions: Miscellaneous identities

27.4

O

Solutions: Miscellaneous identities

Exercise 134: (Outline) Let Sen) be the statement in the exercise. For n = 1, the result is clear. For n = 2, if Xl + x2 = O and say, Xl > 0, then X2 < O, contrary to x2 being non-negative, so S(2) hold.


506

For the inductive step, use 8(2) and 8(k) to prove 8(k+ 1) as follows: Xl + ... + Xk + Xk+l = O implies by 8(2) that both Xl + ... + Xk = O and Xk+1 = O. Applying 8(k) to the first expression shows that Xl = ... = Xk = O as well. O Exercise 135: Fix X and b. Use the fact that for every y E IR+, logb(xy) = logb(x) + 10gb(Y). [The proof of this is quite simple, since brb s = br +s .] Here is the proof of the result by induction on n:

27.4. Solutions: MisccHaneous identities follows. Beginning with the left si de of P(k

[t

k+l

.L(ai i=1

+ bd

=

k logb(x). To be shown

+ logb(X k ) logb(x) + klogb(x) (k + 1) logb(x),

bj ]

+ (aH1 + bH

(t~) +ak+1 + (t

(by fact aboye with y = x k )

(~~) + (~bj)

(by induction hypothesis)

(by Def'n 2.5.6)

¡) (by indo hyp. P(k))

bJ ) +bk+l)

(by Dcf'n 256),

the right-hand side of P(k + 1), completing the inductive step.

concluding the inductive step. Hence, by MI, the result is true for aH n E Z+.

D

Exercise 136: Let al, a2, ... and bl , b2 , ... be real nurnbers, and let P( n) denote the proposition n

.L(ai

P(n):

+ 1),

+ b')] + (",+1 + b'+l) +

logb(X . x k ) logb(x)

(a,

[t a, t

BASE STEP: For n = 1, the statement is trivial. INDUCTIVE STEP: Suppose that for so me k ~ 1, 10gb(X k ) is that logb(x k + l ) = (k + 1) logb(x). Then

507

+ bd

=

i=l

n

n

i=l

j=l

Hence, by the principie of mathematical induction, for aH n ~ 1, P( n) is true.

O

Exercise 137: (Telescoping sum) Let al, a2, a3, ... be a sequence of real nurnbers. For each positive integer n ~ 1, let T(n) be the claim that

.L ai + .L b

n

j .

.L(ai - ai+¡) =

al -

an+l.

i=l

BASE STEP (n = 1): Since bl , the statement

2:}=1 bi =

2:;=1 (ai + bi )

=

1 P(I):

.L(ai

i=1

+ bd

=

al

+ b1, and

1

1

both

2:;=1 ai

=

al and

BASE STEP: When i

.L ai + .L b

k

j

i=l

j=l

For some fixed k

~

k

k

i=l

j=l

To prove T(k -4

P(k

+ 1)): k

P(k):

.L(ai i=l

+ bd

=

k+1 ¿)ai - ai+¡) = al - ak+2.

.L ai + .L b

i=l

j

Beginning with the left side of this equation, k+l

k+l

+ 1):

+

.L(a; - ai+¡) = al - ak+l· i=l 1), one needs to show

1, assume that

holds. To be proved is that P( k

1, there is only one summand and so T(1) is triviaHy true.

INDUCTION STEP: For sorne fixed k ~ 1, assume that T(k) is true, that is, assume that

follows. INDUCTIVE STEP (P(k)

=

k+l

k+1

.L (ai + bi ) .L ai + .L b =

i=l

j

i=1

j=l

.L(ai - ai+¡)

i=l

[tIa, ~ a'+l)] + (a,+! ~ aH2) al - ak+l

+ (ak+1

- ak+2)

(by T(k))


508

I1

¡ I J as desired. This completes the inductive step T(k)

->

T(k

,

+ 1).

o

Hence, by MI, for all n E Z+, the statement T(n) is true.

¡ 1

Exercise 138: This exercise appeared in, e.g., (582, Probo 22]. For n 2: 1, let S (n) denote the statement

-\

¡

n

2 2::)3i -j+2) =n(n +n+2).

S(n):

1

2

+ 1 + 2),

k

S(k):

2::)3j2 - j

+ 2)

= k(k

2+k

+ 2).

~

Therefore, by the principIe of mathematical induction, for all n

1, Sen) is

O

Exercise 140: Outline: Observe that 1 = ~, so for the purpose of this exercise, it suffices to consider 1 as a prime. Now suppose that % E Q is sorne given rational number. Let p be the largest prime in the factorization of either a or b; without loss, suppose that for sorne m ~ 1, %= ~ (where p does not divide k). Use the induction assumption that for aH fractions ~ with x and y having all prime factors smaller than p, that ~ is expressible as desired. Then, for example, when m = 1,

a

pk

b= b

which

INDUCTIVE STEP: Por sorne fixed k ~ 1, assume the inductive hypothesis

509

tr~

I

j=l

BASE STEP (n = 1): Checking S(l), it says 3(1 2) - 1 + 2 = 1(1 boils down to 4 = 4, so S (1) is true.


p!k = (p - l)!b'

and k, p - 1, and b aH have factorizations using primes smaller than p, so apply the induction hypothesis to (p-\)!b. When m > 1, a similar argument applies. O Exercise 141: This exercise appeared in (499, Problem 17]. Induct on n to show that for every n ~ 1, the proposition P(n):

j=l

1

---+ x(x + 1)

To be shown is that

1

(x

+ l)(x + 2)

+ ... +

n

1

(x

+ n - 1)(x + n)

---- x(x + n)

k+l

S(k

+ 1):

2:)3j2 - j

+ 2) =

+ l)«k + 1)2 + (k + 1) + 2).

(k

j=l

follows. Beginning with the left side of S(k

j=l

(t(3

j2 - j

k(k 2 + k

BASE STEP: P(1) says X(X~l) = x(x~l)' whích is clearly true.

+ 1),

INDUCTION STEP: Por sorne fixed k that P(k + 1):

+ 2}) + 3(k + 1}2 -

+ 2) + 3(k + 1)2 -

(k

(k

+ l) + 2

+ 1) + 2

2

(k + 1)[k + 3(k + 1) + 1] (k+1)[k 2 +3k+4] (k + 1)(k2 + 2k + 1 + k + 1 + 2]'

+ 1).

~

1, assume that P(k) is true. To be shown is

1 1 -1 -- + + ... + - - - - - - x(x + 1) (x + 1)(x + 2) (x + k)(x + k + 1)

k+1 x(x+k+l)

is true. Beginning with the left side, (writing in the second last term so that it is clear how to apply the inductive hypothesis)

(by indo hyp.)

k(k(k + 1) + 2) + 3(k + 1)2 - (k + 1) + 2 k 2(k + 1) + 2k + 3(k + 1)2 - (k + 1) + 2 k 2(k + 1) + 3(k + 1? + k + 1

which is precisely the right side of S(k S(k) - 4 S(k + 1).

is true.

This concludes the inductive step

1 1 1 -- + ... + + ..,.---:---,----x(x + 1) (x + k - l)(x + k) (x + k)(x + k + 1) k

x(x

1

+ k) + (x + k)(x + k + 1)

(by P(k))

k(x+k+1) x --;---'--:-----'--+------x(x + k)(x + k + 1) x(x + k)(x + k + 1) k(x+k)+k+x x(x + k)(x + k + 1)

510



(k+1)(x+k) x(x + k)(x + k + 1)

+ 1),

n.

BASE STEP: S(l) states 1 - x = (_1)1 xil which is true. 1

INDUCTIVE STEP: Let k 2: 1 and assume that S(k) is true; to be shown is S(k


1_ ~ 1!

o

Therefore, by MI, for aH n 2: 1 the statement P( n) is true.

+ x(x -

1) _ ...

2n

2

E(n) : - - + - - + ... + - - 1+x 1 + x2 1 + x2n

1

= -X -

1

+

+

2n +l

1 _ x2n+l

BASE STEP'. E(l) says _1_ l+x + ~ l+x = -L x- 1 + ~4 -x ; putting everything over a common 4 denominator of 1 - x indeed shows that this is an equality. (Details are left to the reader.) INDUCTION STEP: For some fixed k 2: 1, assume that E(k) is true. To prove E(k+1), use sigma notation for brevity: 2j

k

2 k +l

+ 1)

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

By the inductive hypothesis, the left side of S(k + 1) is equal to

(-l)k (x - l)(x - 2) ... (x - k) k!

-

J=O

+ (_l)k+l x(x -1) ... (x (k + 1)!

k)

+ l)(x - l)(x - 2).·· (x - k) (k+1)! (k+1)! -1 k+l (x - l)(x - 2)··· (x - k)(x - k - 1) ( ) (k+l)!

= (_l)k+l x(x - 1) ... (x - k) _ (_l)k+l (k

_

+ -1-+-x-=2~k+-;-1

-1-+-x-·2-J

1) ... (x - k k!

+ 1):

(_l)k+l (x -l)(x - 2)··· (x - k)(x - k - 1) (k + 1)!

-~--;-;-

is true.

L

+ (_l)k X(X -

2!

Exercise 142: This problem appeared in, e.g., [499, Probo 21]. To be shown is that for n 2: 1, the expression 1

(-l)n (x - l)(x - ~) ... (x - n).

=

k+1 x(x+k+1)' arrive at the right side of P(k

511

that which was desired. This completes the ínductive step. By mathematícal induction, for all n 2: 1, the statement Sen) holds. Exercise 144:

o

For each n 2: 0, denote the equality in the exercise by

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

= (2n + 1)1. 2 n n1

[Note: "C is the lower case Greek letter pronounced "ksee".] 1 x - 1

giving the truth of E(k

X2k+2 ,

Exercise 143: This problem has appeared many places, e.g., [499, Probo 23]. For n 2: 1, let Sen) denote the statement x

11 +

x(x-1) nx(x-1)···(x-n+1) 2! _···+(-1) n!

or =

O, ~(O) is indeed

INDUCTIVE STEP: For so me fixed k 2: O, assume that

+ 1), completing the inductive step.

By mathematical induction, for n 2: 1, the express ion E(n) holds.

1-

BASE STEP: For n = O, ~(n) reads (1) = ~; sínce 2° = 1 and an equality.

2k+2

+1_

o

~(k):

(1)(3)(5)··· (2k

+ 1) =

(2k k+ 1)! 2 k!

is true. It remains to show that

~(k + 1): (1)(3)(5)· .. (2(k + 1) + 1)

=

(2~k; 1) + 1)1 2 + (k + 1)1


512

follows from the truth of ~(k). Rewriting the 1eft side of ~(k+ 1) with the penultimate factor for clarity,

(1)(3)(5)··· (2k + 1)(2(k + 1) + 1)

=

(2k + 1)' 2k k!' (2k + 3)


513

Exercise 146: This problem occurs in (at least) [462, p. 282). For each n 2 1, let Sen) denote the statement

(by ~(k»

1

L

Sen):

-O =n.

0;iSQn]

(2k + 3)(2k + 2)(2k + 1)! (2k + 2)2 k k!

BASE STEP: When n = 1, the set [1] has only one non-empty subset, namely {l}, in which case the sum in the left side of S(1) is simply = 1; so S(l) is true.

t

INDUCTIVE STEP: Fix sorne k 2 1 and assume that

(2k + 3)! (2)2 k (k + l)k! _

(2k

~(k

+ 1)

+ 3)! + 1)!'

~(n)

is true. To complete the inductive step, it suffices to show that

1 + n(n

+ l)(n + 2)(n + 3)

= [en

follows. Startíng with the left side of S(k

L

0;iSQk+l] TIsEs s

+ l)(m + 2)(m + 3)(m + 4) + l)(m + 2)(m + 3) + 4(m + l)(m + 2)(m + 3) [(m + l)(m + 2) - 1)2 + 4(m + l)(m + 2)(m + 3) [m 2 + 3m + 1]2 + 4(m 2 + 3m + 2)(m + 3) m 4 + 9m 2 + 1 + 6m 3 + 2m 2 + 6m + 4m 3 + 24m 2 + 44m + 24 m 4 + 10m3 + 35m 2 + 50m+ 25

= = =

TIsEs s

+ 1),

L

_1_+

0;iS

HANDBOOKOF MATHEMATICAL INDUCTION o

HANDBOOKOF MATHEMATICAL INDUCTION o

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