Irresistible integrals: symbolics, analysis and experiments in the evaluation of integrals

Abstract

1. Introduction 2. Factorials and binomial coefficients 3. The method of partial fractions 4. A simple rational function 5. A review of power series 6. The exponential and logarithm functions 7. The trigonometric functions and pi 8. A quartic integral 9. The normal integral 10. Euler's constant 11. Eulerian integrals: the Gamma and Beta functions 12. The Riemann zeta function 13. Logarithmic integrals 14. A master formula 15. Appendix: the revolutionary WZ method.

Full text

Main CB702-Boros April 20, 2004 11:15 Char Count= 0
IRRESISTIBLE INTEGRALS
Symbolics, Analysis and Experiments in the
Evaluation of Integrals
GEORGE BOROS
Formerly of Xavier University of Lousiana
VICTOR MOLL
Tulane University
iii

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PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE
The Pitt Building, Trumpington Street, Cambridge, United Kingdom
CAMBRIDGE UNIVERSITY PRESS
The Edinburgh Building, Cambridge CB2 2RU, UK
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C

Victor Moll and George Boros 2004
This book is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without
the written permission of Cambridge University Press.
First published 2004
Printed in the United States of America
Typefaces Times Roman 10.25/13 pt. and Courier System L
A
T
E
X2
ε
[TB]
A catalog record for this book is available from the British Library.
Library of Congress Cataloging in Publication Data
Boros George, 1947–
Irresistible integrals : symbolics, analysis and experiments in the evaluation of integrals /
George Boros, Victor H. Moll.
p. cm
Includes bibliographical references and index.
ISBN 0-521-79186-3 – ISBN 0-521-79636-9 (pbk)
1. Definite integrals. I. Moll, Victor. II. Title.
QA308.B67 2004
515

.43 – dc22 2003069574
ISBN 0 521 79186 3 hardback
ISBN 0 521 79636 9 paperback
iv

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Contents
Preface page xi
1 Factorials and Binomial Coefficients 1
1.1 Introduction 1
1.2 Prime Numbers and the Factorization of n!3
1.3 The Role of Symbolic Languages 7
1.4 The Binomial Theorem 10
1.5 The Ascending Factorial Symbol 16
1.6 The Integration of Polynomials 19
2 The Method of Partial Fractions 25
2.1 Introduction 25
2.2 An Elementary Example 30
2.3 Wallis’ Formula 32
2.4 The Solution of Polynomial Equations 36
2.5 The Integration of a Biquadratic 44
3 A Simple Rational Function 48
3.1 Introduction 48
3.2 Rational Functions with a Single Multiple Pole 49
3.3 An Empirical Derivation 49
3.4 Scaling and a Recursion 51
3.5 A Symbolic Evaluation 53
3.6 A Search in Gradshteyn and Ryzhik 55
3.7 Some Consequences of the Evaluation 56
3.8 A Complicated Integral 58
4 A Review of Power Series 61
4.1 Introduction 61
4.2 Taylor Series 65
4.3 Taylor Series of Rational Functions 67
vii

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viii Contents
5 The Exponential and Logarithm Functions 73
5.1 Introduction 73
5.2 The Logarithm 74
5.3 Some Logarithmic Integrals 81
5.4 The Number e 84
5.5 Arithmetical Properties of e 89
5.6 The Exponential Function 91
5.7 Stirling’s Formula 92
5.8 Some Definite Integrals 97
5.9 Bernoulli Numbers 99
5.10 Combinations of Exponentials and Polynomials 103
6 The Trigonometric Functions and π 105
6.1 Introduction 105
6.2 The Basic Trigonometric Functions and the Existence
of π 106
6.3 Solution of Cubics and Quartics by Trigonometry 111
6.4 Quadratic Denominators and Wallis’ Formula 112
6.5 Arithmetical Properties of π 117
6.6 Some Expansions in Taylor Series 118
6.7 A Sequence of Polynomials Approximating tan
−1
x 124
6.8 The Infinite Product for sin x 126
6.9 The Cotangent and the Riemann Zeta Function 129
6.10 The Case of a General Quadratic Denominator 133
6.11 Combinations of Trigonometric Functions and
Polynomials 135
7 A Quartic Integral 137
7.1 Introduction 137
7.2 Reduction to a Polynomial 139
7.3 A Triple Sum for the Coefficients 143
7.4 The Quartic Denominators: A Crude Bound 144
7.5 Closed-Form Expressions for d
l
(m) 145
7.6 A Recursion 147
7.7 The Taylor Expansion of the Double Square Root 150
7.8 Ramanujan’s Master Theorem and a New Class of
Integrals 151
7.9 A Simplified Expression for P
m
(a) 153
7.10 The Elementary Evaluation of N
j,4
(a; m) 159
7.11 The Expansion of the Triple Square Root 160
8 The Normal Integral 162
8.1 Introduction 162
8.2 Some Evaluations of the Normal Integral 164

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Contents ix
8.3 Formulae from Gradshteyn and Rhyzik (G & R) 171
8.4 An Integral of Laplace 171
9 Euler’s Constant 173
9.1 Existence of Euler’s Constant 173
9.2 A Second Proof of the Existence of Euler’s Constant 174
9.3 Integral Forms for Euler’s Constant 176
9.4 The Rate of Convergence to Euler’s Constant 180
9.5 Series Representations for Euler’s Constant 183
9.6 The Irrationality of γ 184
10 Eulerian Integrals: The Gamma and Beta Functions 186
10.1 Introduction 186
10.2 The Beta Function 192
10.3 Integral Representations for Gamma and Beta 193
10.4 Legendre’s Duplication Formula 195
10.5 An Example of Degree 4 198
10.6 The Expansion of the Loggamma Function 201
10.7 The Product Representation for (x) 204
10.8 Formulas from Gradshteyn and Rhyzik (G & R) 206
10.9 An Expression for the Coefficients d
l
(m) 207
10.10 Holder’s Theorem for the Gamma Function 210
10.11 The Psi Function 212
10.12 Integral Representations for ψ(x) 215
10.13 Some Explicit Evaluations 217
11 The Riemann Zeta Function 219
11.1 Introduction 219
11.2 An Integral Representation 222
11.3 Several Evaluations for ζ (2) 225
11.4 Apery’s Constant: ζ(3) 231
11.5 Apery Type Formulae 235
12 Logarithmic Integrals 237
12.1 Polynomial Examples 238
12.2 Linear Denominators 239
12.3 Some Quadratic Denominators 241
12.4 Products of Logarithms 244
12.5 The Logsine Function 245
13 A Master Formula 250
13.1 Introduction 250
13.2 Schlomilch Transformation 251
13.3 Derivation of the Master Formula 252
13.4 Applications of the Master Formula 253
13.5 Differentiation Results 257

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x Contents
13.6 The Case a = 1 258
13.7 A New Series of Examples 263
13.8 New Integrals by Integration 266
13.9 New Integrals by Differentiation 268
Appendix: The Revolutionary WZ Method 271
A.1 Introduction 271
A.2 An Introduction to WZ Methods 272
A.3 A Proof of Wallis’ Formula 273
Bibliography 276
Index 299

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1
Factorials and Binomial Coefficients
1.1. Introduction
In this chapter we discuss several properties of factorials and binomial coef-
ficients. These functions will often appear as results of evaluations of definite
integrals.
Definition 1.1.1. A function f :
N → N is said to satisfy a recurrence if
the value f (n) is determined by the values { f (1), f (2), ···, f (n − 1)}. The
recurrence is of order k if f (n) is determined by the values { f (n − 1), f (n −
2), ···, f (n − k)}, where k is a fixed positive integer. The notation f
n
is
sometimes used for f (n).
For example, the Fibonacci numbers F
n
satisfy the second-order recur-
rence
F
n
= F
n−1
+ F
n−2
. (1.1.1)
Therefore, in order to compute F
n
, one needs to know only F
1
and F
2
. In this
case F
1
= 1 and F
2
= 1. These values are called the initial conditions of the
recurrence. The Mathematica command
F[n_]:= If[n==0,1, If[n==1,1, F[n-1]+ F[n-2]]]
gives the value of F
n
. The modified command
F[n_]:= F[n]= If[n==0,1, If[n==1,1, F[n-1]+F[n-2]]]
saves the previously computed values, so at every step there is a single sum
to perform.
Exercise 1.1.1. Compare the times that it takes to evaluate
F
30
= 832040 (1.1.2)
using both versions of the function F.
1

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2 Factorials and Binomial Coefficients
A recurrence can also be used to define a sequence of numbers. For instance
D
n+1
= n(D
n
+ D
n−1
), n ≥ 2 (1.1.3)
with D
1
= 0, D
2
= 1defines the derangement numbers. See Rosen (2003)
for properties of this interesting sequence.
We now give a recursive definition of the factorials.
Definition 1.1.2. The factorial of n ∈
N is defined by
n! = n · (n − 1) · (n − 2) ···3 ·2 · 1. (1.1.4)
A recursive definition is given by
1! = 1 (1.1.5)
n! = n × (n − 1)!.
The first exercise shows that the recursive definition characterizes n!. This
technique will be used throughout the book: in order to prove some iden-
tity, you check that both sides satisfy the same recursion and that the initial
conditions match.
Exercise 1.1.2. Prove that the factorial is the unique solution of the recursion
x
n
= n × x
n−1
(1.1.6)
satisfying the initial condition x
1
= 1. Hint. Let y
n
= x
n
/n! and use (1.1.5)
to produce a trivial recurrence for y
n
.
Exercise 1.1.3. Establish the formula
D
n
= n! ×
n

k=0
(−1)
k
k!
. (1.1.7)
Hint. Check that the right-hand side satisfies the same recurrence as D
n
and
then check the initial conditions.
The first values of the sequence n! are
1! = 1, 2! = 2, 3! = 6, 4! = 24, (1.1.8)
and these grow very fast. For instance
50! = 3041409320171337804361260816606476884437764156896051
2000000000000
and 1000! has 2568 digits.

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1.2. Prime Numbers and the Factorization of n!3
Mathematica 1.1.1. The Mathematica command for n!isFactorial
[n]. The reader should check the value 1000! stated above. The number
of digits of an integer can be obtained with the Mathematica command
Length[IntegerDigits[n]].
The next exercise illustrates the fact that the extension of a function from
N
to R sometimes produces unexpected results.
Exercise 1.1.4. Use Mathematica to check that

1
2

! =
√
π
2
.
The exercise is one of the instances in which the factorial is connected
to π, the fundamental constant of trigonometry. Later we will see that the
growth of n!asn →∞is related to e: the base of natural logarithms. These
issues will be discussed in Chapters 5 and 6, respectively. To get a complete
explanation for the appearance of π , the reader will have to wait until Chapter
10 where we introduce the gamma function.
1.2. Prime Numbers and the Factorization of n!
In this section we discuss the factorization of n! into prime factors.
Definition 1.2.1. An integer n ∈
N is prime if its only divisors are 1 and
itself.
The reader is refered to Hardy and Wright (1979) and Ribenboim (1989)
for more information about prime numbers. In particular, Ribenbiom’s first
chapter contains many proofs of the fact that there are infinitely many primes.
Much more information about primes can be found at the site
http://www.utm.edu/research/primes/
The set of prime numbers can be used as building blocks for all integers.
This is the content of the Fundamental Theorem of Arithmetic stated below.
Theorem 1.2.1. Every positive integer can be written as a product of prime
numbers. This factorization is unique up to the order of the prime factors.
The proof of this result appears in every introductory book in num-
ber theory. For example, see Andrews (1994), page 26, for the standard
argument.

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4 Factorials and Binomial Coefficients
Mathematica 1.2.1. The Mathematica command FactorInteger[n]
gives the complete factorization of the integer n. For example
FactorInteger[1001] gives the prime factorization 1001 = 7 ·11 · 13.
The concept of prime factorization can now be extended to rational numbers
by allowing negative exponents. For example
1001
1003
= 7 · 11 ·13 · 17
−1
· 59
−1
. (1.2.1)
The efficient complete factorization of a large integer n is one of the ba-
sic questions in computational number theory. The reader should be careful
with requesting such a factorization from a symbolic language like Mathe-
matica: the amount of time required can become very large. A safeguard is
the command
FactorInteger[n, FactorComplete -> False]
which computes the small factors of n and leaves a part unfactored. The
reader will find in Bressoud and Wagon (2000) more information about these
issues.
Definition 1.2.2. Let p be prime and r ∈
Q
+
. Then there are unique integers
a, b, not divisible by p, and m ∈
Z such that
r =
a
b
× p
m
. (1.2.2)
The p-adic valuation of r is defined by
ν
p
(r) = p
−m
. (1.2.3)
The integer m in (1.2.2) will be called the exponent of p in m and will be
denoted by µ
p
(r), that is,
ν
p
(r) = p
−µ
p
(r)
. (1.2.4)
Extra 1.2.1. The p-adic valuation of a rational number gives a new way of
measuring its size. In this context, a number is small if it is divisible by a large
power of p. This is the basic idea behind p-adic Analysis. Nice introductions
to this topic can be found in Gouvea (1997) and Hardy and Wright (1979).
Exercise 1.2.1. Prove that the valuation ν
p
satisfies
ν
p
(r
1
r
2
) = ν
p
(r
1
) × ν
p
(r
2
),
ν
p
(r
1
/r
2
) = ν
p
(r
1
)/ν
p
(r
2
),

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1.2. Prime Numbers and the Factorization of n!5
and
ν
p
(r
1
+r
2
) ≤ Max

ν
p
(r
1
),ν
p
(r
2
)

,
with equality unless ν
p
(r
1
) = ν
p
(r
2
).
Extra 1.2.2. The p-adic numbers have many surprising properties. For in-
stance, a series converges p-adically if and only if the general term converges
to 0.
Definition 1.2.3. The floor of x ∈
R, denoted by

x

, is the smallest integer
less or equal than x. The Mathematica command is Floor[x].
We now show that the factorization of n! can be obtained without actually
computing its value. This is useful considering that n! grows very fast—for
instance 10000! has 35660 digits.
Theorem 1.2.2. Let p be prime and n ∈
N. The exponent of p in n! is given
by
µ
p
(n!) =
∞

k=1

n
p
k

. (1.2.5)
Proof. In the product defining n! one can divide out every multiple of p, and
there are

n/ p

such numbers. The remaining factor might still be divisible
by p and there are

n/ p
2

such terms. Now continue with higher powers of
p.

Note that the sum in (1.2.5) is finite, ending as soon as p
k
> n. Also, this
sum allows the fast factorization of n!. The next exercise illustrates how to
do it.
Exercise 1.2.2. Count the number of divisions required to obtain
50! = 2
47
·3
22
·5
12
·7
8
·11
4
·13
3
·17
2
·19
2
·23
2
·29 ·31 ·37 ·41 ·43 ·47,
using (1.2.5).
Exercise 1.2.3. Prove that every prime p ≤ n appears in the prime factoriza-
tion of n! and that every prime p > n/2 appears to the first power.

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6 Factorials and Binomial Coefficients
There are many expressions for the function µ
p
(n). We present a proof
of one due to Legendre (1830). The result depends on the expansion of
an integer in base p. The next exercise describes how to obtain such
expansion.
Exercise 1.2.4. Let n, p ∈
N. Prove that there are integers n
0
, n
1
, ···, n
r
such that
n = n
0
+ n
1
p + n
2
p
2
+···+n
r
p
r
(1.2.6)
where 0 ≤ n
i
< p for 0 ≤ i ≤ r . Hint. Recall the division algorithm: given
a, b ∈ N there are integers q, r, with 0 ≤ r < b such that a = qb +r.To
obtain the coefficients n
i
first divide n by p.
Theorem 1.2.3. The exponent of p in n! is given by
µ
p
(n!) =
n − s
p
(n)
p − 1
, (1.2.7)
where s
p
(n) = n
0
+ n
1
+···+n
r
is the sum of the base- p digits of n. In
particular,
µ
2
(n!) = n − s
2
(n). (1.2.8)
Proof. Write n in base p as in (1.2.6). Then
µ
p
(n!) =
∞

k=1

n
p
k

= (n
1
+ n
2
p +···+n
r
p
r−1
)
+(n
2
+ n
3
p +···+n
r
p
r−2
) +···+n
r
,
so that
µ
p
(n!) = n
1
+n
2
(1 + p) +n
3
(1 + p + p
2
) +···+n
r
(1 + p + ··· +p
r−1
)
=
1
p − 1

n
1
( p − 1) +n
2
( p
2
− 1) +···+n
r
( p
r
− 1)

=
n − s
p
(n)
p − 1
.

Corollary 1.2.1. The exponent of p in n! satisfies
µ
p
(n!) ≤
n − 1
p − 1
, (1.2.9)
with equality if and only if n is a power of p.

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1.3. The Role of Symbolic Languages 7
Mathematica 1.2.2. The command IntegerDigits[n,p] gives the
list of numbers n
i
in Exercise 1.2.4.
Exercise 1.2.5. Define
A
1
(m) = (2m + 1)
m

k=1
(4k − 1) −
m

k=1
(4k + 1). (1.2.10)
Prove that, for any prime p = 2,
µ
p
(A
1
(m)) ≥ µ
p
(m!). (1.2.11)
Hint. Let a
m
=

m
k=1
(4k − 1) and b
m
=

m
k=1
(4k + 1) so that a
m
is the prod-
uct of the least m positive integers congruent to 1 modulo 4. Observe that for
p ≥ 3 prime and k ∈ N
, exactly one of the first p
k
positive integers congruent
to 3 modulo 4 is divisible by p
k
and the same is true for integers congruent
to 1 modulo 4. Conclude that A
1
(m) is divisible by the odd part of m!. For
instance,
A
1
(30)
30!
=
359937762656357407018337533
2
24
. (1.2.12)
The products in (1.2.10) will be considered in detail in Section 10.9.
1.3. The Role of Symbolic Languages
In this section we discuss how to use Mathematica to conjecture general
closed form formulas. A simple example will illustrate the point.
Exercise 1.2.3 shows that n! is divisible by a large number of consecutive
prime numbers. We now turn this information around to empirically suggest
closed-form formulas. Assume that in the middle of a calculation we have
obtained the numbers
x
1
= 5356234211328000
x
2
= 102793666719744000
x
3
= 2074369080655872000
x
4
= 43913881247588352000
x
5
= 973160803270656000000,
and one hopes that these numbers obey a simple rule. The goal is to obtain a
function x :
N → N that interpolates the given values, that is, x(i) = x
i
for
1 ≤ i ≤ 5. Naturally this question admits more than one solution, and we will

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8 Factorials and Binomial Coefficients
use Mathematica to find one. The prime factorization of the data is
x
1
= 2
23
· 3
6
· 5
3
· 7
2
· 11 · 13
x
2
= 2
15
· 3
6
· 5
3
· 7
2
· 11 · 13 · 17
2
x
3
= 2
18
· 3
12
· 5
3
· 7
2
· 11 · 13 · 17
x
4
= 2
16
· 3
8
· 5
3
· 7
2
· 11 · 13 · 17 ·19
3
x
5
= 2
22
· 3
8
· 5
6
· 7
2
· 11 · 13 · 17 ·19
and a moment of reflection reveals that x
i
contains all primes less than i + 15.
This is also true for (i +15)!, leading to the consideration of y
i
= x
i
/
(i + 15)!. We find that
y
1
= 256
y
2
= 289
y
3
= 324
y
4
= 361
y
5
= 400,
so that y
i
= (i + 15)
2
. Thus x
i
= (i + 15)
2
× (i + 15)! is one of the possible
rules for x
i
. This can be then tested against more data, and if the rule still
holds, we have produced the conjecture
z
i
= i
2
× i!, (1.3.1)
where z
i
= x
i+15
.
Definition 1.3.1. Given a sequence of numbers {a
k
: k ∈
N
}, the function
T (x) =
∞

k=0
a
k
x
k
(1.3.2)
is the generating function of the sequence. If the sequence is finite, then we
obtain a generating polynomial
T
n
(x) =
n

k=0
a
k
x
k
. (1.3.3)
The generating function is one of the forms in which the sequence {a
k
:
0 ≤ k ≤ n} can be incorporated into an analytic object. Usually this makes it
easier to perform calculations with them. Mathematica knows a large number

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1.3. The Role of Symbolic Languages 9
of polynomials, so if {a
k
} is part of a known family, then a symbolic search
will produce an expression for T
n
.
Exercise 1.3.1. Obtain a closed-form for the generating function of the
Fibonacci numbers. Hint. Let f (x) =

∞
n=0
F
n
x
n
be the generating func-
tion. Multiply the recurrence (1.1.1) by x
n
and sum from n = 1to∞.In
order to manipulate the resulting series observe that
∞

n=1
F
n+1
x
n
=
∞

n=2
F
n
x
n−1
=
1
x
(
f (x) − F
0
− F
1
x
)
.
The answer is f (x) = x/(1 − x − x
2
). The Mathematica command to gen-
erate the first n terms of this is
list[n_]:= CoefficientList
[Normal[Series[ x/(1-x-x^{2}), {x,0,n-1}]],x]
For example, f[10] gives {0, 1, 1, 2, 3, 5, 8, 13, 21, 34}.
It is often the case that the answer is expressed in terms of more complicated
functions. For example, Mathematica evaluates the polynomial
G
n
(x) =
n

k=0
k!x
k
(1.3.4)
as
G
n
(x) =−
e
−1/x
x

(0, −
1
x
) + (−1)
n
(n + 2)(−1 − n, −
1
x
)

, (1.3.5)
where e
u
is the usual exponential function,
(x) =

∞
0
t
x−1
e
−t
dt (1.3.6)
is the gamma function, and
(a, x) =

∞
x
t
a−1
e
−t
dt (1.3.7)
is the incomplete gamma function. The exponential function will be dis-
cussed in Chapter 5, the gamma function in Chapter 10, and the study of
(a, x) is postponed until Volume 2.

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10 Factorials and Binomial Coefficients
1.4. The Binomial Theorem
The goal of this section is to recall the binomial theorem and use it to find
closed-form expressions for a class of sums involving binomial coeffici-
ents.
Definition 1.4.1. The binomial coefficient is

n
k

:=
n!
k!(n − k)!
, 0 ≤ k ≤ n. (1.4.1)
Theorem 1.4.1. Let a, b ∈ R and n ∈ N. Then
(a + b)
n
=
n

k=0

n
k

a
n−k
b
k
. (1.4.2)
Proof. We use induction. The identity (a + b)
n
= (a + b) × (a + b)
n−1
and
the induction hypothesis yield
(a + b)
n
=
n−1

k=0

n − 1
k

a
n−k
b
k
+
n−1

k=0

n − 1
k

a
n−k−1
b
k+1
= a
n
+
n−1

k=1

n − 1
k

+

n − 1
k − 1

a
n−k
b
k
+ b
n
.
The result now follows from the identity

n
k

=

n − 1
k

+

n − 1
k − 1

, (1.4.3)
that admits a direct proof using (1.4.1).

Exercise 1.4.1. Check the details.
Note 1.4.1. The binomial theorem
(1 + x)
n
=
n

k=0

n
k

x
k
(1.4.4)
shows that (1 + x)
n
is the generating function of the binomial coefficients

n
k

:0≤ k ≤ n

.

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1.4. The Binomial Theorem 11
In principle it is difficult to predict if a given sequence will have a simple
generating function. Compare (1.3.5) with (1.4.4).
We now present a different proof of the binomial theorem in which we
illustrate a general procedure that will be employed throughout the text. The
goal is to find and prove an expression for (a + b)
n
.
a) Scaling. The first step is to write
(a + b)
n
= b
n
(1 + x)
n
(1.4.5)
with x = a/b, so that it suffices to establish (1.4.2) for a = 1.
b) Guessing the structure. The second step is to formulate an educated
guess on the form of (1 + x)
n
. Expanding (1 + x)
n
(for any specific n) shows
that it is a polynomial in x of degree n, with positive integer coefficients,
that is,
(1 + x)
n
=
n

k=0
b
n,k
x
k
(1.4.6)
for some undetermined b
n,k
∈ N
. Observe that x = 0 yields b
n,0
= 1.
c) The next step is to find a way to understand the coefficients b
n,k
.
Exercise 1.4.2. Differentiate (1.4.6) to produce the recurrence
b
n,k+1
=
n
k + 1
b
n−1,k
0 ≤ k ≤ n − 1. (1.4.7)
Conclude that the numbers b
n,k
are determined from (1.4.7) and initial con-
dition b
n,0
= 1.
We now guess the solution to (1.4.7) by studying the list of coeffi-
cients
L[n]:={b
n,k
:0≤ k ≤ n}. (1.4.8)
The list L[n] can be generated symbolically by the command
term[n_,k_]:=If[n==0,1, If[ k==0, 1,
n∗term[n-1,k-1]/k]];
L[n_]:= Table[ term[n,k], {k,0,n}];

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12 Factorials and Binomial Coefficients
that produces a list of the coefficients b
n,k
from (1.4.7). For instance,
L[0] ={1}
L[1] ={1, 1}
L[2] ={1, 2, 1}
L[3] ={1, 3, 3, 1}
L[4] ={1, 4, 6, 4, 1}
L[5] ={1, 5, 10, 10, 5, 1}
L[6] ={1, 6, 15, 20, 15, 6, 1 }. (1.4.9)
The reader may now recognize the binomial coefficients (1.4.1) from the
list (1.4.9) and conjecture the formula
b
n,k
=

n
k

=
n!
k!(n − k)!
(1.4.10)
from this data. Naturally this requires a priori knowledge of the binomial
coefficients. An alternative is to employ the procedure described in Section
1.3 to conjecture (1.4.10) from the data in the list L[n].
The guessing of a closed-form formula from data is sometimes obscured
by dealing with small numbers. Mathematica can be used to generate terms
in the middle part of L[100]. The command
t:= Table[ L[100][[i]],{i,45,49}]
chooses the elements in positions 45 to 49 in L[100]:
L[100][[45]] = 49378235797073715747364762200
L[100][[46]] = 61448471214136179596720592960
L[100][[47]] = 73470998190814997343905056800
L[100][[48]] = 84413487283064039501507937600
L[100][[49]] = 93206558875049876949581681100, (1.4.11)
and, as before, we examine their prime factorizations to find a pattern.
The prime factorization of n = L[100][[45]] is
n = 2
3
·3
3
·5
2
·7 ·19 ·23 ·29 ·31 ·47 ·59 ·61 ·67 ·71 ·73 ·79 ·83 ·89 ·97,
suggesting the evaluation of n/97!. It turns out that this the reciprocal of an

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1.4. The Binomial Theorem 13
integer of 124 digits. Its factorization
97!
n
= 2
91
· 3
43
· 5
20
· 7
13
· 11
8
· 13
7
· 17
5
· 19
4
· 23
3
· 29
2
· 31
2
· 37
2
· 41
2
·43
2
· 47 · 53
leads to the consideration of
97!
n × 53!
= 2
42
· 3
20
· 5
8
· 7
5
· 11
4
· 13
3
· 17
2
· 19
2
· 23 · 29 · 31 ·37 · 41 ·43,
and then of
97!
n × 53! ×43!
=
2
3
× 3 × 11
5 × 7
. (1.4.12)
The numbers 97, 53 and 43 now have to be slightly adjusted to produce
n =
100!
56! × 44!
=

100
56

. (1.4.13)
Repeating this procedure with the other elements in the list (1.4.11) leads to
the conjecture (1.4.10).
Exercise 1.4.3. Use the method described above to suggest an analytic ex-
pression for
t
1
= 33422213193503283445319840060700101890113888695441
601636800,
t
2
= 4786578310918590163752805570088320851200018220954
9068756000,
t
3
= 63273506018045330510555274728827082768779925144537
753208000,
t
4
= 77218653725969794800710549093404104300057079699419
429079500.
d) Recurrences. Finally, in order to prove that our guess is correct, define
a
n,k
:=

n
k

−1
b
n,k
(1.4.14)
and show that (1.4.7) becomes
a
n,k+1
= a
n−1,k
, n ≥ 1, 0 ≤ k ≤ n − 1, (1.4.15)
so that a
n,k
≡ 1.

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14 Factorials and Binomial Coefficients
Exercise 1.4.4. Check that b
n,k
=

n
k

by verifying that

n
k

satisfies (1.4.7)
and that this recurrence admits a unique solution with b
n,0
= 1.
Note 1.4.2. The sequence of binomial coefficients has many interesting prop-
erties. We provide some of them in the next exercises. The reader will find
much more in
http://www.dms.umontreal.ca/~andrew/Binomial/
index.htlm
Exercise 1.4.5. Prove that the exponent of the central binomial coefficients
C
n
=

2n
n

satisfies
µ
p
(C
n
) =
2s
p
(n) − s
p
(2n)
p − 1
. (1.4.16)
Hint. Let n = a
0
+ a
1
p +···+a
r
p
r
be the expansion of n in base p.Define
λ
j
by 2a
j
= λ
j
p + ν
j
, where 0 ≤ ν
j
≤ p − 1. Check that λ
j
is either 0 or 1
and confirm the formula
µ
p
(C
n
) =
r

j=0
λ
j
. (1.4.17)
In particular µ
p
(C
n
) ≤ r + 1. Check that C
n
is always even. When is C
n
/2
odd?
The binomial theorem yields the evaluation of many finite sums involving
binomial coefficients. The discussion on binomial sums presented in this
book is nonsystematic; we see them as results of evaluations of some definite
integrals. The reader will find in Koepf (1998) and Petkovsek et al. (1996).
a more complete analysis of these ideas.
Exercise 1.4.6. Let n ∈
N.
a) Establish the identities
n

k=0

n
k

= 2
n
n

k=0
k

n
k

= 2
n−1
n
n

k=0
k
2

n
k

= 2
n−2
n(n + 1). (1.4.18)