Bernoulli and Euler Polynomials
Güneş Çatma
Submitted to the Institute of Graduate Studies and Research in partial fulfillment of the requirements for the Degree of
Master of Science in Mathematics
Eastern Mediterranean University July 2013 Gazimağusa, North Cyprus Approval of the Institute of Graduate Studies and Research
Prof. Dr. Elvan Yılmaz Director
I certify that this thesis satisfies the requirements as a thesis for the degree of Master of Science in Mathematics.
Prof. Dr. Nazım Mahmudov Chair, Department of Mathematics
We certify that we have read this thesis and that in our opinion it is fully adequate in scope and quality as a thesis for the degree of Master of Science in Mathematics.
Assoc. Prof. Dr. Sonuç Zorlu Oğurlu Supervisor
Examining Committee
1. Prof. Dr. Nazım Mahmudov
2. Assoc. Prof. Dr. Mehmet Ali Özarslan
3. Assoc. Prof. Dr. Sonuç Zorlu Oğurlu ABSTRACT
This thesis provides an overview of Bernoulli and Euler numbers. It describes the Bernoulli and Euler polynomials and investigates the relationship between the classes of the two polynomials. It also discusses some important identities using the finite difference calculus and differentiation. The last part of this study is con- cerned with the Generalized Bernoulli and Euler polynomials. Furthermore, the properties obtained in the second chapter are also examined for the generalized
Bernoulli and Euler polynomials in this part of the thesis. The Complemen- tary Argument Theorem, the generating functions, the Multiplication and the
Euler-Maclauren Theorems are widely used in obtaining the mentioned results.
Keywords: Bernoulli -Euler Polynomials, Generalized Bernoulli -Euler Polyno- mials, Finite Difference
iii OZ¨
Bu ¸calı¸smadaBernoulli ve Euler sayıları ile Bernoulli ve Euler polinom- ları arasındaki ili¸skilerincelenmi¸stir. Bernoulli sayıları i¸cinardı¸sıklık,kapalılık ve ¨ureticilikgibi temel ¨ozellikler¸calı¸sılmı¸stır.Bunun yanında Bernoulli ve Euler
Polinomları arasındaki ili¸skilerincelenip her ikisi i¸cinde ge¸cerliolan t¨urev,inte- gral, fark ve simetri ¨ozellikleriincelenmi¸stir.Ayrıca Genelle¸stirilmi¸sBernoulli ve
Euler polinomları i¸cinde t¨urev,intregral ¨ozellikleri¸calı¸sılmı¸stır.
Anahtar Kelimeler: Bernulli-Euler Polinomları, Genelle¸stirilmi¸sBernoulli-Euler
Polinomları, Sonlu Fark
iv DEDICATION
TO MY FATHER
v ACKNOWLEDGMENTS
First of all, I would like to express my special thanks to my supervisor Assoc.
Prof. Dr Sonu¸cZorlu O˘gurlufor her patience, supervision, guidance, encourage- ment and help during the preperation of this thesis. I also would like to thank to my parents and friends specially to Noushin Ghahramanlou for their help, encouragement and love.
Finally, my special thanks go to the members of Department of Mathematics for their advices and help in completing my study at EMU. I am specially thankful to Prof.Dr.Nazım Mahmudov and Asst.Prof.Dr. Arif Akkele¸sfor their help.
vi TABLE OF CONTENTS
ABSTRACT ...... iii
OZ¨ ...... iv DEDICATION ...... v ACKNOWLEDGMENTS ...... vi
1 BERNOULLI AND EULER NUMBERS ...... 1
1.1 Bernoulli Numbers ...... 1
1.1.1 Worpitzky’s Representation for Bernoulli Numbers . . . . . 5
1.2 Euler Numbers ...... 6
1.2.1 Properties of the Euler Numbers ...... 7
1.2.2 Identities Involving Euler Numbers ...... 8
2 BERNOULLI AND EULER POLYNOMIALS ...... 10
2.1 Properties of Bernoulli and Euler Polynomials ...... 11
2.1.1 Equivalence of Relation (2.1.3) and (2.1.6) ...... 15
3 THE GENERALIZED BERNOULLI AND EULER POLYNOMIALS . 17
3.1 The ϕ Polynomials ...... 17
3.2 The β Polynomials ...... 20
3.3 Definition of Bernoulli Polynomials ...... 22
3.3.1 Fundamental Properties of Bernoulli Polynomials ...... 23
3.3.2 The Complementary Argument Theorem ...... 26
3.3.3 The Relation between Polynomials of Successive Orders . . 27
3.3.4 Relation of Bernoulli Polynomials to Factorials ...... 29
vii 3.3.4.1 The Integral of the Factorial ...... 40
3.3.4.2 Expansion of xα in Powers of x ...... 44
3.3.4.3 Expansion of xυ in Factorials ...... 46
3.3.4.4 Generating Functions of Bernoulli Numbers . . . . 48
3.3.5 Bernoulli Polynomials of the First Order ...... 51
3.3.5.1 A Summation Problem ...... 55
3.3.5.2 Bernoulli Numbers of the First Order ...... 56
3.3.5.3 The Euler-Maclaurin Theorem for Polynomials . . 60
3.3.5.4 The Multiplication Theorem ...... 63
3.3.5.5 Bernoulli Polynomials in the Interval (0, 1) . . . . 64
3.3.6 The η Polynomials ...... 66
3.3.7 Definition of Euler Polynomial ...... 68
3.3.7.1 Fundamental Properties of Euler Polynomials . . 70
3.3.7.2 The Complementary Argument Theorem . . . . . 74
3.3.7.3 Euler Polynomials of Successive Orders ...... 76
3.3.8 Euler Polynomials of the First Order ...... 77
3.3.8.1 Euler Numbers of the First Order ...... 80
3.3.8.2 Boole’s Theorem for Polynomials ...... 82
4 CONCLUSION ...... 85
REFERENCES ...... 86
viii Chapter 1
BERNOULLI AND EULER NUMBERS
1.1 Bernoulli Numbers
In Mathematics, the Bernoulli numbers Bn are a sequence of rational numbers with important relations to number theory and with many interesting arithmetic properties. We find them in a number of contexts, for example they are closely related to the values of the Riemann zeta function at negative integers and appear in the Euler-Maclaurin formula. The values of the first few Bernoulli numbers are B0 = 1, B1 = −1/2, B2 = 1/6,B3 = 0, B4 = −1/30 ( Some authors use
B1 = +1/2 and some write Bn for B2n). After B1 all Bernoulli numbers with odd index are zero and the non zero values alternate in sign. These numbers appear in the series expansions of trigonometric and are important in number theory and analysis.
The Bernoulli numbers first used in the Taylor series expansions of the tangent and hyperbolic tangent functions, in the formulas for the sum of powers of the
first positive integers, in the Euler-Maclaurin formula and in the expression for certain values of the Riemann zeta function.
Bernoulli numbers appeared in 1713 in Jacob Bernoulli’s work, who was Jo- hanns Bernoulli’s older brother, Euler’s teacher and mentor at Bassel’s university.
1 Those numbers were studied at the same time independently by Japanese math- ematician Seki Kowa. His discovery was published in 1712 in his work and in his
Arts Conjectandi of 1713. Ada Lovelace’s note G on the analytical engine from
1842 describes an algorithm for generating Bernoulli numbers with Babbage’s machine. Also, the Bernoulli numbers have the distinction of being the main topic of the first computer program.
Bernoulli numbers are still a bit misterious, they appear frequently in studying
Gamma function about Euler’s constant and people continue to discover new properties and to publish articles about them.
Definition 1.1.1 Bernoulli numbers has the following closed form expression:
n X m m m m Sm(n) = k = 1 + 2 + ... + n . (1.1.1) k=1
Note that Sm(0) = 0 for all m > 0 The equation (1.1.1) can always be written as a polynomial in n of degree m + 1.
Definition 1.1.2 The coefficients of the polynomials are related to the Bernoulli
1 Pm m+1 m+1−k numbers by Bernoulli’s formula: Sm(n) = m+1 k=0 k Bkn ,
where B1 = +1/2. Bernoulli’s formula can also be stated as
m 1 X m + 1 S (n) = (−1)k B nm+1−k. m m + 1 k k k=0
Here are some simple examples of Bernoulli numbers:
2 Example 1.1.3 Let n > 0. Taking m to be 0 and B0 = 1 gives the natural numbers 0, 1, 2, 3, ...
1 0 + 1 + 1 + ... + 1 = (B n) = n. 1 0
Example 1.1.4 Let n > 0. Taking m to be 1 and B1 = 1/2 gives the triangular numbers 0, 1, 3, 6, ...
1 1 0 + 1 + 2 + ... + n = (B n2 + 2B n1) = (n2 + n). 2 0 1 2
Example 1.1.5 Let n > 0. Taking m to be 2 and B2 = 1/6 gives the square pyramidal numbers 0, 1, 5, 14, ...
1 1 3 1 0 + 12 + 22 + ... + n2 = (B n3 + 3B n2 + 3B n1) = (n3 + n2 + n) 3 0 1 2 3 2 2
where B1 = −1/2.
There are many characterizations of the Bernoulli numbers where each can be used to introduce them. Here are most useful characterizations:
1. Recursive Definition. The recursive equation is best introduced in a
slightly more general form
n−1 X n Bk(x) B (x) = xn − . n k n − k + 1 k=0
3 For x = 0, the recursive equation becomes,
n−1 X n Bk B = [x = 0] = − n k n − k + 1 k=0
Also, when x = 1,we get the following form
n−1 X n Bk B = 1 − n k n − k + 1 k=0
2. Explicit Definition. Starting again with slightly more general formula
n k X X k(x + υ)n B (x) = (−1)υ . n υ k + 1 k=0 v=0
For x = 0, the recursive equation becomes,
n k X X k υn B = (−1)υ . n υ k + 1 k=0 υ=0
Also, when x = 1, we get the following from
n+1 k X X k − 1vn B = (−1)υ+1 . n υ − 1 k k=1 υ=1
3. Generating Function. The general formula for the generating function
is given as
∞ text X tn = B (x) . et − 1 n n! n=0
4 For x = 0, the recursive equation becomes,
∞ t X tn = B . et − 1 n n! n=0
Also, when x = 1, we get the following form
∞ t X tn = B . 1 − e−t n n! n=0
1.1.1 Worpitzky’s Representation for Bernoulli Numbers
The definition to proceed with was developed by Julius Worpitzky in 1883.
Besides elementary arithmetic only the factorial function n! and the power func- tion km are employed. The signless Worpitzky numbers are defined as
k X k! W = (−1)n+k(n + 1)n . n,k n!(k − n)! n=0
One can also express Wn,k through the Stirling numbers of the second kind as follows:
n + 1 Wn,k = k! . k + 1
A Bernoulli number is then introduced as an inclusion-exclusion sum of Worpitzky numbers weighted by the sequence 1, 1/2, 1/3, ...
n n k X Wn,k X 1 X k B = (−1)k = (−1)n(n + 1)n . n k + 1 k + 1 n k=0 k=0 n=0
5 1.2 Euler Numbers
In Combinatorics, the Eulerian number A(n, m) is the number of permutations of the numbers 1 to n in which exactly m elements are greater than the previous element. The coefficients of the Eulerian polynomials are given as follows,
n X n−m An(x) = A(n, m)x . m=0
This polynomial appears as the numerator in an expression for the generating function of the sequence 1n, 2n, 3n, ··· . Other notations for A(n, m) are E(n, m) and hinm.
In Number Theory, the Euler numbers are sequence En of integers defined by the following Taylor series expansion
∞ 1 2 X En = = tn, (1.2.1) cosh t et + e−t n! n=0 where cosh t is the hyperbolic cosine. The Euler numbers appear as a special value of Euler polynomials. The odd indexed Euler numbers are all zero while the even ones have alternating signs. They also appear in the Taylor series expansions of the secant and hyperbolic secant functions. The latter is the function given in equation (1.2.1). We also play important role in Combinatorics, especially when counting the number of alternating permutations of a set with an even number of elements.
John Napier who made common use of the decimal point, was a Scottish
6 mathematician, physicist, astronomer and astrologer. He is best known as the discoverer of logarithms. In 1618, he published a work on logarithms which con- tained the first reference to the constant e. It was not until Jacob Bernoulli whom to the Bernoulli Principles named after, attempted to find the value of a compound-interest expression. Historically Euler’s number was actually ex- pressed as ”b” until Euler himself published his work Mechanica who used ”e” instead of ”b” as the variable. Eventually the letter made its way as the standard notation of Euler’s number.
To find Euler numbers A(n, m) one can use the following formula,
A(n, m) = (n − m)A(n − 1, m − 1) + (m + 1)A(n − 1, m).
Recall that,
A(n, m) = A(n, n − m − 1).
A closed form expression for A(n, m) is given by,
m X n + 1 A(n, m) = (−1)k (m + 1 − k)n. k k=0
1.2.1 Properties of the Euler Numbers
1. It is clear from the combinatorics definition that the sum of the Eulerian
numbers for a fixed value of n is the total number of permutations of the
7 numbers 1 to n, so
n−1 X A(n, m) = n! for n > 1. m=0
2. The alternating sum of the Eulerian numbers for a fixed value of n is related
to the Bernoulli number Bn+1 and
n−1 n+1 n+1 X 2 (2 − 1)Bn+1 (−1)mA(n, m) = for n 1. n + 1 > m=0
Other summation properties of the Eulerian numbers are:
n−1 X m A(n, m) (−1) n−1 = 0 for n > 2, m=0 m
n−1 X m A(n, m) (−1) n = (n + 1)Bn for n ≥ 2. m=0 m
1.2.2 Identities Involving Euler Numbers
The Euler numbers are involved in the generating function for the sequence of nth powers
∞ n X P A(n, m)xm+1 knxk = m=0 . (1 − x)n+1 k=1
8 Worpitzky’s identity expresses xn as the linear combination of Euler numbers with binomial coefficients:
n−1 X x + m xn = A(n, m) . n m=0
It follows from Worpitzky’s identity that
N n−1 X X N + 1 + m kn = A(n, m) . n + 1 k=1 m=0
Another interesting identity is given as follows,
∞ n ex X P A(n, m)xm+1 = m=0 . 1 − xe (1 − x)n+1n! n=0
9 Chapter 2
BERNOULLI AND EULER POLYNOMIALS
The classical Bernoulli polynomials Bn(x) and the classical Euler polynomi- als En(x) are usually defined by means of the following exponential generating functions:
∞ text X tn = B (x) , |t| < 2π, et − 1 n n! n=0 and
∞ 2ext X tn = E (x) , |t| < π (2.0.1) et + 1 n n! n=0 respectively. The classical Bernoulli and Euler polynomials can explicitly be defined as,
n X n B (x) = B xn−k (2.0.2) n k k k=0 and
n+1 1 X n + 1 E (x) = (2 − 2k+1) B xn+1−k, n n + 1 k k k=1
respectively, where Bk := Bk(0) is the Bernoulli number for each k = 0,1,...,n.
10 Several interesting properties and relationships involving each of these poly- nomials and numbers can be found in many books and journals ([1]- [7]) on this subject. Some of these properties are given in the following section.
2.1 Properties of Bernoulli and Euler Polynomials
The purpose of this section is to obtain interesting properties of the Bernoulli and Euler polynomials, and the relationship between these polynomials.
Recently, Cheon ([1]) obtained the results given below:
n X n B (x + 1) = B (x), n ∈ N (2.1.1) n k k 0 k=0
n X n E (x + 1) = E (x), n ∈ N (2.1.2) n k k 0 k=0
n X n B (x) = B E (x), (n ∈ ) . (2.1.3) n k k n−k N0 k=0 k6=1
Both (2.1.1) and (2.1.2) are well-known results and are obviously special cases of the following familiar addition theorems:
n X n B (x + y) = B (x)yn−k,(n ∈ ) (2.1.4) n k k N0 k=0
11 and
n X n E (x + y) = E (x)yn−k,(n ∈ ) . (2.1.5) n k k N0 k=0
Furthermore, Cheon’s main result (2.1.3) is essentially the same as the following known relationship:
n X n B (x) = 2−n B E (2x), (n ∈ ) , n k n−k k N0 k=0 or equivalently,
n x X n 2nB ( ) = B E (x), (n ∈ ) . (2.1.6) n 2 k k n−k N0 k=0
These two polynomials have many similar properties (see [1]).
The following identity will be useful in the sequel.
nn − k n j + k = . (2.1.7) k j j + k k
Theorem 2.1.1 For any integer n > 0, we have
Pn n a) Bn(x + 1) = k=0 k Bk(x)
Pn n b) En(x + 1) = k=0 k Ek(x)
Proof. a) Letting y = 1 in equations (2.1.4) and (2.1.5), one can directly obtain the equations (2.1.1) and (2.1.2) respectively. Also, applying (2.0.2) and (2.1.7),
12 we obtain
n X n B (x + 1) = B (x + 1)n−k. (2.1.8) n k k k=0
Recall the following Binomial expansion
n X n (x + y)n = xkyn−k. (2.1.9) k k=0
Applying (2.1.9) to (2.1.8), one can easily get the following formula
n n−k X n X n − k B (x + 1) = B xj. n k k j k=0 j=0
Using (2.1.7), the above equation becomes,
n n−k X X n j + k B (x + 1) = B xj. n k j + k k k=0 j=0
Expanding the last expression gives
n 0 n 1 1 B (x + 1) = B + B x + B + n 0 0 0 1 0 0 1 1 n n n n +... + B xn + B xn−1 + ... + B , n 0 0 1 1 n n which yields to
n n n X n X k X n B (x + 1) = B xk−j = B (x). n k j j k k k=0 j=0 k=0
Hence the theorem (a) is proved.
13 b) We will now give the proof of part (b) of the theorem. Replacing x with
(x + 1) in equation (2.0.1), we get,
∞ 2e(x+1)t X tn = E (x + 1) . (2.1.10) et + 1 n n! n=0
Also multiplying both sides of equation (2.0.1) by et and using the expansion
t P∞ tn e = n=0 n! ,
∞ 2extet X tn = E (x) et et + 1 n n! n=0 ∞ ∞ X tn X tn = E (x) . n n! n! n=0 n=0
Now applying Cauchy Product Formula, one can obtain the following relation
∞ n 2extet X X tk tn−k = E (x) et + 1 k k! (n − k)! n=0 k=0 ∞ n X X tk tn−k n! = E (x) , k k! (n − k)! n! n=0 k=0 ∞ n X X ntn = E (x) , k k n! n=0 k=0 which implies
n X n E (x + 1) = E (x) , n k k k=0
This gives the desired result.
14 2.1.1 Equivalence of Relation (2.1.3) and (2.1.6)
For both Bernoulli and Euler polynomials, the following Multiplication The- orems are well known :
Theorem 2.1.2 For n ∈ N0 and m ∈ N the following relations hold:
n−1 Pm−1 j a) Bn(mx) = m j=0 Bn x + m , n Pm−1 j j m j=0 (−1) En x + m , (m = 1, 3, 5, ...) b) En(mx) = 2 n Pm−1 j j − n+1 m j=0 (−1) Bn+1 x + m , (m = 2, 4, 6, ...)
The above theorem yields the following relationships between these two poly- nomials when m = 2.
2 x E (2x) = − 2n B (x) − B . n n + 1 n+1 n 2
Also letting n to be n − 1 and x to be x/2, we obtain
2n x + 1 x E (x) = B − B (2.1.11) n−1 n n 2 n 2 2 h xi = B (x) − 2nB n ∈ N n n n 2 2n x + 1 x x = 2n−1 B + B − 2nB . (2.1.12) n n 2 n 2 n 2
Also letting m = 2 in part(a), we have
1 B (2x) = 2n−1 B (x) + B x + . n n n 2
15 Using (2.1.2) and replacing 2x with x, we get
x x + 1 B (x) = 2n−1 B + B . n n 2 n 2
From (2.1.12),
2 x E (x) = B (x) − 2nB . n−1 n n n 2
1 Since B1 = − 2 , by separating the second term (k = 1) of the sum in (2.1.6), we have
x n 2nB = B E (x) n 2 1 1 n−1 n = − E (x). 2 n−1
Hence,
n x X n n 2nB = B E (x) − E (x), (n ∈ ) n 2 k k n−k 2 n−1 N0 k=0 k6=1 which in the light of the second relation in (2.1.11), immediately yields to (2.1.3).
16 Chapter 3
THE GENERALIZED BERNOULLI AND EULER
POLYNOMIALS
In this chapter we study some properties of two classes of polynomials namely
Bernoulli and Euler polynomials which play an important role in the finite cal- culus. These polynomials have been the object of much research and have been generalized in a very elegant manner by N¨orlund.
We shall here approach these polynomials by a symbolic method described by
Milne-Thomson (see [4] ) by which they arise as generalizations of the simplest polynomials, namely the powers of x. The method is applicable to whole classes of polynomials, including those of Hermite. Considerations of space must limit us to the discussion of only a few of the most interesting relations to which these polynomials give rise.
3.1 The ϕ Polynomials
(α) φ polynomials of degree n and order α are denoted as φn (x) and defined as below :
∞ X tn f (t)ext+g(t) = φ(α)(x), (3.1.1) α n! n n=0
17 where the uniformly convergent series in t on the right-hand side of (3.1.1) exists for fα(t) and g(t) in a certain range of x.
Substituting x = 0, we get
∞ X tn f (t)eg(t) = φ(α)(x), (3.1.2) α n! n n=0
(α) (α) where φn (x) = φn (0) is a φ number of order α.
Writing x + y instead of x in (3.1.1) we get obtain
∞ X tn φ(α)(x + y) = f (t)e(x+y)t+g(t), n! n α n=0 xt g(t)+yt = fα(t)e e ∞ X tn = ext φ(α)(y). (3.1.3) n! n n=0
Having put the coefficients of tn, equal on both sides, we have
n n φ(α)(x + y) = φ(α)(y) + x φ(α) (y) + ... + xn φ(α)(y) (3.1.4) n n 1 n−1 n 0 υ X k (α) = x φn−k(y). k=0
Substituting (3.1.4) and applying the Cauchy product formula to (3.1.3), we
18 have
∞ n ∞ ∞ k n−k X t X X (xt) t (α) φ(α)(x + y) = φ (y) n! n k! (n − k)! n−k n=0 n=0 k=0 ∞ ∞ k n X X n! x t (n) = φ (y) n! k!(n − k)! n−k n=0 k=0 ∞ ∞ n X X n t (n) = xk φ (y). k n! n−k n=0 k=0
Taking y = 0, we obtain
n n n φ(α)(x) = φ(α) + x φ(α) + x2 φ(α) + ... + xn φ(α), n n 1 n−1 2 n−2 n 0
(α) (α) which shows unless φ0 = 0 that φn (x) is actually of degree n.
Therefore we can write the below equality
(α) (α) n φn (x) + (φ + x) (3.1.5) where after expanding the powers, each index of φ(α) will be replaced by the corresponding suffix.
In this way φ polynomials defined completely by φ numbers mentioned in (3.1.2) and also by the equality (3.1.5).
From (3.1.5), we have
d φ(α)(x) n(φ(α) + x)n−1 = nφ(α) (x) (3.1.6) dx n + n n−1
19 and
Z x (α) (α) (α) φn+1(x) − φn+1(a) φn (t)dt = . (3.1.7) a n + 1
Therefore as we already know differentiation will decrease the degree by one unit and integration will increase it one unit but in both cases we will have no changes in the order. Using 4 in equation (3.1.1), we will get
∞ X tn 4 β(α)(x) = (et − 1)f (t)ext+g(t). (3.1.8) n! n α n=0
In a similar way, using 5 in (3.1.1), we get
∞ X tn et + 1 5 φ(α)(x) = f (t)ext+g(t). (3.1.9) n! n 2 α n=0
3.2 The β Polynomials
α t −α A result from (3.1.8) is that if we take fα(t) = t (e − 1) in (3.1.1) where α is any integer (either positive, negative or zero), we get a particularly simple class of φ polynomials which are called β polynomials and are written as follows
∞ tα X tn ext+g(t) = β(α)(x) (3.2.1) (et − 1)α n! n n=0
20 so that from (3.1.8)
∞ ∞ X tn X tn 4 β(α)(x) = t β(α−1)(x) n! n n! n n=0 n=0 ∞ X tn = β(α)(x + 1) − β(α)(x) n! n n n=0 ∞ ∞ X tn X tn = β(α)(x + 1) − β(α)(x) n! n n! n n=0 n=0 α t h (x+1)t+g(t) xt+g(t) i = e − e (et − 1)α ∞ X tn = t β(α−1)(x), n! n n=0 where
tα tα β(α)(x + 1) − β(α)(x) = e(x+1)t+g(t) − ext+g(t) n n (et − 1)α (et − 1)α tα = e(x+1)t+g(t) − ext+g(t) (et − 1)α ttα−1 = ext+g(t) (et − 1)α−1 ∞ X tn = t β(α−1)(x). n! n n=0
Replacing n + 1 with n,
∞ n X t (α−1) = β (x) (n − 1)! n−1 n=0 ∞ n X n t (α−1) = β (x) n (n − 1)! n−1 n=0 ∞ n X t (α−1) = n β (x) n! n−1 n=0 (α−1) = nβn−1 (x).
21 Therefore we have
(α) (α−1) 4βn (x) = nβn−1 (x). (3.2.3)
It is clear that 4 decreases the order and the degree both by unit one .
Using (3.1.5), we can rewrite (3.2.3) as follows
(α) n (α) n (α−1) n−1 β + x + 1 − β + x + n β + x .
Substituting x = 0, we obtain
(α) n (α) (α−1) β + 1 − βn + nβn−1 (3.2.4) which results a one to one relation between the β numbers of orders α and α − 1.
3.3 Definition of Bernoulli Polynomials
The function ext+g(t) generates the β polynomials of order zero, where if we put g(t) = 0 we will obtain the simplest polynomials of this kind ext. These β polynomials are known as Bernoulli polynomials of order zero which are defined as follows :
(0) n Bn (x) = x . (3.3.1)
22 Therefore we have
∞ ∞ ∞ X tn X (xt)n X tn ext = B(0)(x) = = B(0)(x). n! n n! n! n n=0 n=0 n=0
Using (3.2.1), we can expand this definition to Bernoulli polynomials of order
α given by the identity
∞ tαext X tn = B(α)(x). (3.3.2) (et − 1)α n! n n=0
Putting x = 0, we will obtain
∞ tα X tn = B(α). (et − 1)α n! n n=0
3.3.1 Fundamental Properties of Bernoulli Polynomials
Since β polynomials are φ polynomials, so Bernoulli polynomials are also
φ polynomials. Here are some properties of Generalized Bernoulli polynomials:
(α) (α) n Bn (x) = Bn + x . (3.3.3)
d B(α)(x) = nB(α) (x). (3.3.4) dx n n−1
Z x (α) 1 h (α) (α) i Bn (t)dt = βn+1(x) − βn+1(a) . (3.3.5) a n + 1
23 (α) (α−1) 4Bn (x) = nBn−1 (x). (3.3.6)
(α) n (α) (α−1) B + 1 − Bn + nBn−1 . (3.3.7)
Properties (3.3.3), (3.3.4), and (3.3.5) are common in an φ polynomials and properties (3.3.6) and (3.3.7) are shared in all β polynomials (see [4]).
For n ≥ α , repeated applications of property (3.3.6) will give the relation
α (α) n−α 4 Bn (x) = n(n − 1)(n − 2)...(n − α + 1)x .
Let us prove the first property. From (3.3.6)
α−1 (α) α−1 (α−1) 4 4 Bn (x) = 4 nBn−1 (x)
α−2 (α−1) α−2 (α−1) 4 4 nBn−1 (x) = 4 n(n − 1)Bn−1 (x) .
Applying 4, α times yields to
α (α) (0) 4 Bn (x) = n(n − 1)(n − 2)...(n − α + 1)Bn−α(x)
= n(n − 1)(n − 2)...(n − α + 1)xn−α.
(0) n Here note that Bn (x) = x . Now if n < α, the right-hand will vanish, since we can not have a Bernoulli polynomial of negative degree. Norlund’s theory of
Bernoulli polynomials arised from relations (3.3.6) and (3.3.7).
There are some useful results gained from (3.3.6) such as the following theo-
24 rem.
Theorem 3.3.1 For any integer n, α > 0, we have
(α) (α) (α−1) Bn (x + 1) = Bn (x) + nBn−1 (x). (3.3.8)
Proof. Putting x = 0 in equation (3.3.8) yields to
(α) (α) (α−1) Bn (1) = Bn + nBn−1 .
Using (3.3.5) and (3.3.6), we get
Z x+1 (α) 1 (α) (α−1) Bn (t)dt = 4 Bn+1(x) = Bn (x). (3.3.9) x n + 1
Replacing x + 1 by x and x by a in equation (3.3.5)
Z x+1 (α) 1 h (α) (α) i Bn (t)dt = βn+1(x + 1) − βn+1(x) x n + 1 1 = 4 B(α) (x) n + 1 n+1
(α−1) = Bn (x) and in particular
Z 1 (α) (α−1) Bn (t)dt = Bn . (3.3.10) 0
25 3.3.2 The Complementary Argument Theorem
Theorem 3.3.2 If the arguments x and α − x are complementary, then
(α) n (α) Bn (α − x) = (−1) Bn (x). (3.3.11)
Proof. Considering (3.3.2),
∞ X tn tαe(α−x)t B(α)(α − x) = n! n (et − 1)α n=0 tαe(α−x)t e−αt = (et − 1)α e−αt tαe−xt = (et − 1)α(e−t)α tαe−xt = (1 − e−t)α (−t)αe−xt = (1 − e−t)α ∞ X (−t)n = B(α)(x), n! n n=0 equating the coefficients of tn on both sides of (3.2.1), we prove the theorem.
The equation (3.2.1) is called the complementary argument theorem. The theorem holds for any β polynomial with an even function as its generating func-
(α) (α) tion. Taking x = 0, n = 2µ in (3.3.11), B2µ (α) = B2µ , which results x = α,
(α) (α) x = 0 as zeros of B2µ (x) − B2µ .
1 In a similar way putting x = 2 α, n = 2µ + 1in (3.3.11), we have
26 1 1 B(α) (α − α) = (−1)2µ+1B(α) ( α) 2µ+1 2 2µ+1 2 1 1 B(α) ( α) = −B(α) ( α). 2µ+1 2 2µ+1 2
Resulting that
1 B(n) ( n) = 0. (3.3.12) 2µ+1 2
3.3.3 The Relation between Polynomials of Successive Orders
The following theorem gives the relation between the polynomials of suc- cessive orders.
Theorem 3.3.3 For any integer n, α > 0, we have
n x B(α+1)(x) = 1 − B(α)(x) + n − 1 B(α) (x). n α n α n−1
Proof. Differentiating both sides of the equality below
∞ X tn tαext B(α)(x) = (3.3.13a) n! n (et − 1)α n=0
27 and then multiplying it by t, we get
∞ X tn B(α)(x) (n − 1)! n n=0 h i t [(αtα−1ext + tαxext)(et − 1)α] − ttαextα (et − 1)α−1 et = (et − 1)2α h i [αtαext (et − 1)α] + [tα+1xext (et − 1)α] − tα+1et(x+1)α (et − 1)α−1 = (et − 1)2α αtαext (et − 1)α [tα+1xext (et − 1)α] tα+1e(x+1)tα (et − 1)α−1 = + − (et − 1)2α (et − 1)2α (et − 1)2α αtαext xtα+1ext αtα+1e(x+1)t = α + α − (et − 1) (et − 1) (et − 1)α+1 ∞ ∞ ∞ X tn X tn X tn = α B(α)(x) + xt B(α)(x) − α B(α+1)(x + 1). n! n n! n n! n n=0 n=0 n=0
By equating the coefficients of tn we have
(α) (α) (α) (α+1) nBn (x) = αBn (x) + xnBn−1(x) − αBn (x + 1). (3.3.14)
Using (3.3.8) we obtain
(α+1) (α+1) (α) Bn (x + 1) = Bn (x) + nBn−1(x). (3.3.15)
Therefore we get
n x B(α+1)(x) = 1 − B(α)(x) + n − 1 B(α) (x) (3.3.16) n α n α n−1 which demonstrates a relation between Bernoulli polynomials of order α and α+1, as required.
28 Now letting x = 0, we also have
n B(α+1) = 1 − B(α) − nB(α) (x). n α n n−1
Again putting x = 0 in equation (3.3.14), we get
n B(α+1)(1) = 1 − B(α), (3.3.18) n α n
(α) (α) (α+1) nBn (0) = αBn (0) − αBn (1)
(α) (α) (α+1) nBn = αBn − αBn (1)
(α+1) (α) (α) αBn (1) = αBn − nBn (α − n)B(α) B(α+1)(1) = n n α n = 1 − B(α), α n replacing α by n + α in (3.3.18), we have
α B(n+α+1)(1) = B(n+α). (3.3.19) n n + α n
3.3.4 Relation of Bernoulli Polynomials to Factorials
Letting n = α in (3.3.16), we obtain
(α+1) (α) (α−1) Bn (x) = (x − α)Bα−1(x) = (x − α)(x − α + 1)Bα−2 (x) = ...
(1) = (x − α)(x − α + 1) ... (x − 2)(x − 1)B0 (x).
29 Putting α for n in equation (3.3.16)
α x B(α+1)(x) = 1 − B(α)(x) + α − 1 B(α) (x) α α α α α−1
(α) = (x − α)Bα−1(x).
Now putting α − 1 for α at right side,
(α+1) (α−1) Bα (x) = (x − α)(x − α + 1)Bα−2 (x) = ...
(1) = (x − α)(x − α + 1)...(x − 2)(x − 1)B0 (x).
Therefore
(α+1) (α) Bα (x) = (x − 1)(x − 2) ... (x − α) = (x − 1) . (3.3.20)
Putting x + 1 for x in equation (3.3.20),
(α+1) α Bα (x + 1) = x(x − 1)(x − 2) ... (x − α + 1) = x . (3.3.21)
Taking integral from 0 to 1 in both of the above equations and considering
(3.3.10), we get
Z 1 (α) (x − 1)(x − 2)...(x − α)dx = Bα . 0
Using relation between (3.3.10) and (3.3.21) and from (3.3.15)
Z 1 (α) 1 (α−1) (x − 1)(x − 2) ... (x − α + 1)dx = Bα (1) = − Bα . 0 α − 1
30 Note that the above equation will be the same we put α = −1 in (3.3.19).
Using (3.3.4) and differentiating (3.3.20) α − n times (α > n), we obtain,
dα−n α(α − 1) ... (α − α + n + 1)B(α+1)(x) = (x − 1)α n dxα−n
(α+1) which represents an expression for Bn (x) as below
n! dα−n B(α+1)(x) = [(x − 1)(x − 2) ... (x − α)] . n α! dxα−n
In Stirling’s and Bessel’s interpolation formula, we have the following coefficients:
p+s 1. a2s+1(p) = 2s+1
p p+s−1 2. a2s(p) = 2s 2s−1
1 p− 2 p+s−1 3. b2s+1(p) = (2s+1) 2s
p+s−1 4. b2s(p) = 2s
1 (2s+2) 5. a2s+1(p) = (2s+1)! B2s+1 (p + s + 1)
We will now prove the above properties. To prove the property on a2s(p),we
31 rewrite equation (3.3.20) as
(α+1) Bα (x) = (x − 1)(x − 2)...(x − α) (x − 1)(x − 2) ... (x − α)(x − α − 1) ... 3.2.1 = (x − α − 1)! (x − 1)! = (x − α − 1)! x − 1 = α! . (3.3.22) α
Substitute x = p + s + 1, α = 2s + 1, α + 1 = 2s + 2, in (3.3.22) to get
p + s + 1 − 1 B(2s+2)(p + s + 1) = (2s + 1)! 2s+1 2s + 1 p + s = (2s + 1)! . 2s + 1
Now, substitute α = 2s − 1, α + 1 = 2s, x = p + s in (3.3.22) to get
p + s − 1 B(2s) (p + s) = (2s − 1)! . 2s−1 2s − 1
Use the above relation below to obtain the desired property
p p + s − 1 a (p) = (2s − 1)! 2s (2s)! 2s − 1 p p + s − 1 = . 2s 2s − 1
Let us prove the property (3) on b2s+1(p).Substitute x = p + s, n = 2s, n + 1 =
2s + 1in equation (3.3.22),
p + s − 1 B(2s+1)(p + s) = (2s)! . 2s 2s
32 Then, use the above relation below,
p − 1 b (p) = 2 B(2s+1)(p + s) 2s+1 (2s + 1)! 2s p − 1 p + s − 1 2 (2s)! (2s + 1)! 2s p − 1 p + s − 1 = 2 (2s + 1) 2s which proves the desired property.
Next, we prove the property on b2s(p). Using the relation
1 b (p) = B(2s+1)(p + s) (3.3.23) 2s (2s)! 2s and substituting x = p + s, α = 2s, α + 1 = 2s + 1 in equation (3.3.22), we easily obtain
p + s − 1 B(2s+1)(p + s) = (2s)! . 2s 2s
Using the above result in (3.3.23),
1 p + s − 1 b (p) = (2s)! 2s (2s)! 2s p + s − 1 = . 2s
Hence the property is obtained.
Differentiating each of these coefficients m times with respect to p and
1 then putting p = 0 in a2s+1(p) and a2s(p) and also putting p = 2 in b2s+1(p) and
33 b2s(p), we get
1 Dma (0) = B(2s+2) (s + 1). (3.3.24) 2s+1 (2s − m + 1)! 2s−m+1
From (3.3.4),
d d 1 a (p) = B(2s+2)(p + s + 1) , dp 2s+1 dp (2s + 1)! 2s+1 2s + 1 D (a (p)) = B(2s+2)(p + s + 1) 2s+1 (2s + 1)! 2s 1 = B(2s+2)(p + s + 1). (2s)! 2s
Differentiating once more,
1 D2 (a (p)) = D B(2s+2)(p + s + 1) 2s+1 (2s)! 2s 2s = B(2s+2)(p + s + 1) (2s)! 2s−1 1 = B(2s+2)(p + s + 1). (2s − 1)! 2s−1
Repeatedly, we obtain
1 Dm (a (p)) = B(2s+2) (p + s + 1). (3.3.25) 2s+1 (2s − m + 1)! 2s−m+1
Now putting p = 0 in (3.3.25),
1 Dm (a (0)) = B(2s+2) (s + 1). (3.3.26) 2s+1 (2s − m + 1)! 2s−m+1
34 Now, let us obtain the relation,
m Dma (0) = B(2s) (s). (3.3.27) 2s 2s(2s − m)! 2s−m
From (3.3.4),
d d 1 (a (p)) = pB(2s) (p + s) , dp 2s dp (2s)! 2s−1
thus
1 h i D(a (p)) = B(2s) (p + s) + p(2s − 1)B(2s) (p + s) . 2s (2s)! 2s−1 2s−2 1 D2(a (p)) = 2s (2s)!
h (2s) (2s) × (2s − 1)B2s−2(p + s) + (2s − 1)B2s−2(p + s)
(2s) i + p(2s − 1)(2s − 2)B2s−3(p + s) 1 h i = 2(2s − 1)B(2s) (p + s) + p(2s − 1)(2s − 2)B(2s) (p + s) (2s)! 2s−2 2s−3 1 h i = (2s − 1) 2(B(2s) (p + s)) + p(2s − 2)B(2s) (p + s) (2s)! 2s−2 2s−3 1 h i = 2(B(2s) (p + s)) + p(2s − 2)B(2s) (p + s) . (2s)(2s − 2)! 2s−2 2s−3
Repeatedly,
1 h i Dm(a (p)) = m(B(2s) (p + s)) + p(2s − 2)B(2s) (p + s) . 2s (2s)(2s − m)! 2s−2 2s−3
35 Now place p = 0 in the last equation
m Dm(a (0)) = B(2s) (s). 2s (2s)(2s − m)! 2s−2
Whence the result. The next result is about the mth derivative of the coefficient
1 b2s+1(p) at p = 2 which is,
1 m 1 Dmb ( ) = B(2s+1) (s + ). (3.3.28) 2s+1 2 (2s + 1)(2s − m + 1)! 2s−m+1 2
From property (3)
d d p − 1 b (p) = 2 B(2s+1)(p + s) dp 2s+1 dp (2s + 1)! 2s 1 1 D (b (p)) = B(2s+1)(p + s) + p − (2s)B(2s+1)(p + s) 2s+1 (2s + 1)! 2s 2 2s−1 1 h i D2 (b (p)) = (2s)B(2s+1)(p + s) + (2s)B(2s+1)(p + s) 2s+1 (2s + 1)! 2s−1 2s−1 1 1 + p − (2s)(2s − 1)B(2s+1)(p + s) (2s + 1)! 2 2s−2 1 1 = (2s) 2B(2s+1)(p + s) + p − (2s − 1)B(2s+1)(p + s) (2s + 1)! 2s−1 2 2s−2 1 1 = 2B(2s+1)(p + s) + p − (2s − 1)B(2s+1)(p + s) (2s + 1)(2s − 1)! 2s−1 2 2s−2
Repeatedly,
1 Dm (b (p)) = 2s+1 (2s + 1)(2s − m + 1)! 1 × mB(2s+1) (p + s) + p − (2s − m + 1)B(2s+1)(p + s) . 2s−m+1 2 2s−m
36 1 We will put p = 2 in last equation to get,
1 1 Dm b ( ) = 2s+1 2 (2s + 1)(2s − m + 1)! 1 1 1 1 × mB(2s+1) ( + s) + − (2s − m + 1)B(2s+1)( + s) 2s−m+1 2 2 2 2s−m 2 m 1 = B(2s+1) ( + s). (2s + 1)(2s − m + 1)! 2s−m+1 2
To prove the relation below
1 1 1 Dmb ( ) = B(2s+1)(s + ), (3.3.29) 2s 2 (2s − m)! 2s−m 2 we use equation (3.3.23) to have,
d d 1 (b (p)) = B(2s+1)(p + s) dp 2s dp (2s)! 2s 1 D (b (p)) = (2s)B(2s+1)(p + s) 2s (2s)! 2s−1 1 h i D2 (b (p)) = (2s)(2s − 1)B(2s+1)(p + s) 2s (2s)! 2s−2 (2s)(2s − 1) = B(2s+1)(p + s) (2s)(2s − 1)(2s − 2)! 2s−2 1 = B(2s+1)(p + s). (2s − 2)! 2s−2
Repeatedly,
1 Dm (b (p)) = B(2s+1)(p + s). 2s (2s − m)! 2s−m
1 We will now put p = 2 in last equation to obtain
1 1 1 Dm b ( ) = B(2s+1)( + s), 2s 2 (2s − m)! 2s−m 2
37 which gives the desired relation.
From equation (3.3.26),
1 Dma (0) = B(2s+2) (s + 1) 2s+1 (2s − m + 1)! 2s−m+1 and
1 D2ma (0) = B(2s+2) (s + 1). 2s+1 (2s − 2m + 1)! 2s−2m+1
Using (3.3.12)
2m D a2s+1(0) = 0.
From equation (3.3.27),
m Dma (0) = B(2s) (s) 2s 2s(2s − m)! s−m and
2m + 1 D2m+1a (0) = B(2s) (s). 2s 2s(2s − 2m − 1)! 2s−2m−1
Using (3.3.12), we obtain
2m+1 D a2s(0) = 0.
38 m 1 2m 1 By equation (3.3.28), D b2s+1( 2 ) and D b2s+1( 2 ) become,
1 m 1 Dmb ( ) = B(2s+1) (s + ) 2s+1 2 (2s + 1)(2s − m + 1)! 2s−m+1 2 and
1 2m 1 D2mb ( ) = B(2s+1) (s + ), 2s+1 2 (2s + 1)(2s − 2m + 1)! 2s−2m+1 2 respectively.
Now from equation (3.3.12),
1 D2mb ( ) = 0. 2s+1 2
From (3.3.29),
1 1 1 Dmb ( ) = B(2s+1)(s + ) 2s 2 (2s − m)! 2s−m 2 and
1 1 1 D2m+1b ( ) = B(2s+1) (s + ). 2s 2 (2s − 2m + 1)! 2s−2m+1 2
Using (3.3.12), we get
1 D2m+1b ( ) = 0. 2s 2
39 3.3.4.1 The Integral of the Factorial
An important function in the theory of numerical integration is the χ(x) function :
Z x χ(x) = (y − 1)(y − 2)...(y − 2n + 1)dy (3.3.30) 1−k Z x (2n) = B2n−1(y)dy. (3.3.31) 1−k
The above integral results with the following expression,
Z x (2n) 1 h (2n) (2n) i B2n−1(y)dy = B2n (x) − B2n (1 − k) , (3.3.32) 1−k 2n where k is either zero or unit. Considering the Complemantary Argument Theo- rem, substituting n = 2n, α = 2n in (3.3.11) and using (3.3.30), we have
(2n) 2n (2n) B2n (2n − (−k + 1)) = (−1) B2n (1 − k)
(2n) = B2n (1 − k).
Also, substituting the above equalities in (3.3.30), we have
B(2n)(2n + k − 1) − B(2n)(1 − k) χ(2n + k − 1) = 2n 2n 2n B(2n)(1 − k) − B(2n)(1 − k) = 2n 2n 2n = 0.
Resulting that χ(2n + k − 1) = χ(1 − k) = 0.
40 Using Complementary Argument Theorem mentioned in (3.3.11) and letting x = 1 − k, we also have the following relation
(2n+1) 2n+1 (2n+1) B2n+1 (2n + 1 − x) = (−1) B2n+1 (1 − k)
(2n+1) = −B2n+1 (1 − k).
That is,
Z 2n+k−1 Z 2n+k−1 1 · χ(x)dx = − x(x − 1)(x − 2) ... (x − 2n + 1)dx 1−k 1−k B(2n+1)(2 − k) + B(2n+1)(1 − k) = 2n+1 2n+1 . (3.3.33) 2n + 1
Rewriting the left hand side of the above equation leads to the following integral,
Z 2n+k−1 Z 2n+k−1 Z x 1 · χ(x)dx = (y − 1)(y − 2)...(y − 2n + 1)dydx. 1−k 1−k 1−k
Letting u = χ(x) and dv = 1dx and using integration by parts method we have,
R 2n+k−1 h R x 2n+k−1 1−k 1 · χ(x)dx = x · 1−k(y − 1)(y − 2) ... (y − 2n + 1)dy |1−k R 2n+k−1 − 1−k x(x − 1)(x − 2) ... (x − 2n + 1)dx
2n+k−1 R 2n+k−1 = [x · χ(x)] |1−k − 1−k x(x − 1)(x − 2) ... (x − 2n + 1)dx
= [(2n + k − 1) · χ (2n + k − 1) · χ (1 − k)]
R 2n+k−1 − 1−k x(x − 1)(x − 2) ... (x − 2n + 1)dx.
Since χ(2n + k − 1) = χ(1 − k) = 0,
Z 2n+k−1 Z 2n+k−1 1 · χ(x)dx = − x(x − 1)(x − 2) ... (x − 2n + 1)dx. 1−k 1−k
41 Substituting 2n instead of n in (3.3.21), the above integral will be equal to the integral given below
Z 2n+k−1 Z 2n+k−1 (2n+1) 1.χ(x)dx = − B2n (x + 1)dx. 1−k 1−k
From (3.3.5),
Z 2n+k−1 (2n+1) 1 h (2n+1) (2n+1) i − B2n (x + 1)dx = − B2n+1 (2n + k) − B2n+1 (2 − k) , 1−k 2n + 1 which proves the relation (3.3.33).
(2n−1) R 1 It is easy to verify that B2n−1 = 0 (y − 1)(y − 2)...(y − 2n + 1)dy is negative, while
Z 2 (2n−1) B2n−1 (1) = (y − 1)(y − 2)...(y − 2n + 1)dy 1 is positive.
Continuing in this way, when υ is an integer, 0 ≤ υ ≤ 2n,
υ+1 (2n−1) (−1) B2n−1 (υ) > 0.
At this point we will prove the following inequality,
υ (2n−1) υ+1 (2n−1) (−1) B2n−1 (υ − 1) > (−1) B2n−1 (υ), 1 ≤ υ ≤ n − 1. (3.3.35)
42 From (3.3.9), we have
Z υ (2n−1) B2n−1 (υ − 1) = (y − 1)(y − 2) ... (y − 2n + 1)dy, (3.3.36) υ−1 and
Z υ+1 (2n−1) B2n−1 (υ) = (y − 1)(y − 2) ... (y − 2n + 1)dy υ Z υ = y(y − 1)(y − 2) ... (y − 2n + 2)dy υ−1 Z υ (2n−1) y B2n−1 (υ) = − (y − 1)(y − 2) ... (y − 2n + 1)dy. (3.3.37) υ−1 2n − y − 1
1 Since υ−1 ≤ y ≤ υ and υ ≤ n−1, we have y ≤ n− 2 . Thus y/(2n−y−1) is positive and less than unity. Considering (3.3.36) and (3.3.37) it is clear that the absolute value of the integrand of (3.3.37) is less than the absolute value of the integrand of (3.3.36), which results (3.3.35).According to the above explanations we will show that the function χ(x) has a fixed sign whenever
1 − k ≤ x ≤ 2n + k − 1.
Let υ − 1 ≤ x ≤ υ. If υ − 1 ≤ y ≤ υ, then the sign of the integrand of (3.3.30) will not change. Hence,
Z υ−1 Z υ (y−1)(y−2) ... (y−2n+1)dy ≤ χ(x) ≤ (y−1)(y−2) ... (y−2n+1)dy. 1−k 1−k
If we rewrite each integral using the intervals (1 − k, 2 − k), (2 − k, 3 − k),..., we
43 will realize that χ(x) is between the following two sums.
(2n−1) (2n−1) (2n−1) B2n−1 (1 − k) + B2n−1 (2 − k) + ... + B2n−1 (υ − 2),
(2n−1) (2n−1) (2n−1) B2n−1 (1 − k) + B2n−1 (2 − k) + ... + B2n−1 (υ − 1).
According to the Complementary Argument theorem, we will consider υ ≤ n, when the terms satisfy this condition.
Considering (3.3.35) the absolute magnitude of the terms in the above sums are in descending order and their signs alternate. Thus the sign of each sum
(2n−1) is the same as the sign of the first term, namely, B2n−1 (1 − k).
(2n) (2n) Following the above applications we proved that B2n (x) − B2n has no zeros in 0 ≤ x ≤ 2n, and
(2n−1) (2n−1) B2n−1 (1 − k) ≥ B2n−1 (2 − k) , for k = 0 or 1.
3.3.4.2 Expansion of xα in Powers of x
Having differentiated (3.3.21) p times, we have
dp x(α) = n(p)B(α+1)(x + 1). dxp α−p
44 To prove the above relation, we differentiate (3.3.21) with respect to x, that is
(α+1) (α) Bα (x + 1) = x(x − 1)(x − 2) ... (x − α + 1) = x
(α+1) (α+1) D Bα (x + 1) = αBα−1 (x + 1)
2 (α+1) (α+1) D = D αBα−1 (x + 1) = α(α − 1)Bα−2 (x + 1)
3 2 (α+1) (α+1) D = D α(α − 1)Bα−2 (x + 1) = α(α − 1)(α − 2)Bα−3 (x + 1).
Repeatedly,
p p−1 (α+1) (α+1) D = D α(α − 1)Bα−2 (x + 1) = α(α−1)(α−2) ... (α−p+1)Bα−p (x+1).
Hence,
dp x(α) = α(p)B(α+1)(x + 1). dxp α−p
Now, letting x = 0, using (3.3.18) and (3.3.21), we obtain
p d (α) (p) (α+1) p x = α Bα−p (1) dx x=0 α(α − 1)(α − 2) ... (α − p + 1).(α − p)! = B(α+1)(1) (α − p)! α−p α! = B(α+1)(1) (α − p)! α−p α! p = B(α) . (α − p)! α α−p
By applying Maclaurin’s Theorem on x(α), we have
α p α X x α! p (α) X α − 1 (α) x(α) = B = xpB . p! (α − p)! α α−p p − 1 α−p p=0 p=0
45 3.3.4.3 Expansion of xυ in Factorials
By Newton’s Interpolation formula which is
p 2 p 3 p n−1 p n (n) f(x) = f(a)+p4f(a)+ 2 4 f(a)+ 3 4 f(a)+...+ n−1 4 f(a)+ n ω f (ξ)
(α) and knowing that Bn (x + h) is a polynomial of degree n from (3.3.6), we get
υ X x(s) B(α)(x + h) = B(α)(h) + 4s B(α)(h) (3.3.38) n n s! n s=1 n X n (s) (α−s) = s x Bn−s (h). s=0
(α) By putting h = 0 in (3.3.38), we have a factorial series for Bn (x),
n (α) X n (s) (α−s) Bn (x) = s x Bn−s . s=0
(0) One can easily see that, having n = 0 gives Bn (x) = x which yields to the following required expansion
n n X n (s) (−s) x = s x Bn−s . (3.3.40) s=0
Operating 4α on (3.3.40) (α ≤ n), we will obtain the differences of zeros having taken x = 0,
n! 4α0n = B(−α). (n − α)! n−α
Taking α = n + 1, replacing h by h + 1 in (3.3.38) and using (3.3.21) leads to,
n n X n (s) (n−s) (x + h) = s x h s=0
46 which will be used in follows
n (n+1) X n (s) (n+1−s) Bn (x + h + 1) = s x Bn−s (h + 1) s=0 n X n (s) (n−s+1) = s x Bn−s (h + 1) s=0 n X n (s) (n−s) = s x h . s=0
The above relation is the well known Vandermonde’s theorem in factorials and demonstrates an analogou to the Binomial Theorem as below
n n X n s n−s (x + h) = s x h . s=0
Also interchanging x and h in (3.3.38), we get
(α) (α) n Bn (x + h) − Bn (x) X (α−s) = n(h − 1)(s−1)B (x). (3.3.41) h s n−s s=1
(α) Taking limit when h → 0 we will get the derivative of Bn (x) on the left-hand side
n (α) X n (s−1) (α−s) nBn−1(x) = s (−1) (s − 1)!Bn−s (x) s=1
47 we will use (3.3.4) and (3.3.41) and the definition of derivative,
(α) (α) n Bn (x + h) − Bn (x) X (α−s) = n(h − 1)(s−1)B (x) h s n−s s=1 n d X (α−s) B(α)(x) = n(h − 1)(s−1)B (x) dx n s n−s s=1 n (α) X n (s−1) (α−s) nBn−1(x) = s (−1) (s − 1)!Bn−s (x). s=1
Specially, when x = 0
n (α) X n (s−1) (α−s) nBn−1 = s (−1) (s − 1)!Bn−s . s=1
3.3.4.4 Generating Functions of Bernoulli Numbers
Using Binomial Theorem, we have
x−1 ∞ X (x − 1)(x − 2) ... (x − n) X tn (1 + t)x−1 = tn = B(n+1)(x) n! n! n n=0 n=0 x−1 X x−1 n α−n = n t (1) n=0 x−1 x−1 X X (x − 1)! = x−1tn = tn n n!(x − 1 − n)! n=0 n=0 x−1 X (x − 1)(x − 2) ... (x − n)(x − 1 − n)! = tn n!(x − 1 − n)! n=0 x−1 X (x − 1)(x − 2) ... (x − n) = tn n! n=0 x−1 X tn = B(n+1)(x). n! n n=0
48 Having differentiated the above equation α times with respect to x, we get
∞ n α X t (n+1) (1 + t)x−1 [log(1 + t)] = B (x). (α − n)! n−α n=α
Letting x = 1 and then dividing by tα, we obtain
α ∞ log(1 + t) X tn = B(α+n+1)(1) (3.3.42) t n! n n=0 ∞ X tn α = B(α+n). (3.3.43) n! α + n n n=0
∞ n α X t (n+1) (1 + t)x−1 [log(1 + t)] = B (x) (n − α)! n−α n=α n (n+1) 1−1 α ∞ t (1 + t) [log(1 + t)] X (n−α)! Bn−α (1) = tα tα n=α α ∞ n log(1 + t) X t (n+1) = B (1) t tn(n − α)! n−α υ=0 ∞ n−α X t (n+1) = B (1) (n − α)! n−α n=0 ∞ n+n−n X t (n+α+1) = B (1) (n + α − α)! n+α−α n=0 ∞ X tn = B(n+α+1)(1) n! n n=0 ∞ X tn α = B(α+n). n! α + n n n=0
In particular, for α = 1,
∞ log(1 + t) X tn = B(n+1). t (n + 1)! n n=0
49 Integrating (1 + t)x−1 α times with respect to x from x to x + 1 and using
(3.3.9), we have
∞ (1 + t)x−1tα X tn = B(n−α+1)(x). [log(1 + t)]α n! n n=0
Taking x = 0, we get
∞ tα X tn = B(n−α+1). (1 + t) [log(1 + t)]α n! n n=0
(α) In a particular case where α = 1, we get the generating function of Bn numbers as below
∞ t X tn = B(n). (1 + t) log(1 + t) n! n n=0
Now putting x = 1, we obtain
α ∞ t X tn = B(n−α+1)(1) log(1 + t) n! n n=0 resulting that the equation (3.3.42) holds for negative α’s as well.
In particular for α = 1, we will obtain the generating function of the
(n) numbers Bn (1),
∞ t X tn = B(n)(1). (3.3.44) log(1 + t) n! n n=0
50 Using (3.3.19) on (3.3.44), we get
∞ n (n−1) t 1 X t Bn = 1 + t − . log(1 + t) 2 n! n − 1 n=2
(n) Below the first ten of the Bn numbers are listed :
1 19087 B(1) = − ,B(6) = , 1 2 6 84 5 36799 B(2) = ,B(7) = − , 2 6 7 24 9 1070017 B(3) = − ,B(8) = , 3 4 8 90 251 2082753 B(4) = ,B(9) = − , 4 30 9 20 475 134211265 B(5) = − ,B(10) = . 5 12 10 132
3.3.5 Bernoulli Polynomials of the First Order
(1) In the rest of the thesis we will write Bn(x) instead of Bn (x), having in mind that the order is one. Therefore from (3.3.2), we get the below function as the generating function of the Bernoulli polynomials :
∞ text X tn = B (x). (3.3.45) et − 1 n! n n=0
The generating function of Bernoulli numbers, Bn of the first order are shown as below
∞ t X tn = B . (3.3.46) et − 1 n! n n=0
51 Using the properties given in the subsection (3.2.1), the following properties are satisfied
n Bn(x) + (B + x) . (3.3.47)
Putting α = 0 in equation (3.3.3)
(0) (0) n Bn (x) = B + x , hence
n Bn(x) + (B + x) .
Now, we will prove the following property.
n (B + 1) − Bn + 0, n = 2, 3, 4,... . (3.3.48)
We put x = 0, α = 0 in equation (3.3.15),
(0+1) (0+1) (0) Bn (0 + 1) = Bn (0) + nBn−1(0)
(0) Bn(1) = Bn + nBn−1
(1) (1) Bn (1) − Bn = 0
n (B + 1) − Bn + 0, n = 2, 3, 4,... .
52 Another property is related with differentiation.
d B (x) = nB (x). (3.3.49) dx n n−1
To prove this, we will differentiate both sides of equation (3.3.45) with respect to x.
That is,
∞ ! d text d X tn = B (x) dx et − 1 dx n! n n=0 ∞ X tn d ttext (B (x)) = n! dx n et − 1 n=0 ∞ X tn = t B (x) n! n n=0 ∞ X tn+1 = B (x) n! n n=0 ∞ X tn+1 = (n + 1) B (x) (n + 1)! n n=0 ∞ X tn = n B (x). (n)! n−1 n=0
Thus
d (B (x)) = nB (x). dx n n−1
The following is the integral representation of Bn(x)
Z x 1 Bn(t)dt = [Bn+1(x) − Bn+1(a)] . (3.3.50) a n + 1
53 By integrating both sides of equation (3.3.45), we have
∞ ! Z x X tn Z x tety B (t) dt = dy n! n et − 1 a n=0 a ∞ X tn Z x ext eat B (t)dt = − n! n et − 1 et − 1 n=0 a ∞ ∞ X tn−1 X tn−1 = B (x) − B (a) n! n n! n n=0 n=0 ∞ ∞ X tn X tn = B (x) − B (a) (n + 1)! n+1 (n + 1)! n+1 n=0 n=0 ∞ X tn 1 = (B (x) − B (a)) . n! n + 1 n+1 n+1 n=0
Thus,
Z x 1 Bn(t)dt = [Bn+1(x) − Bn+1(a)] . a n + 1
Next, consider the difference operator.
n−1 4Bn(x) = nx . (3.3.51)
From (3.3.15),
(α+1) (α+1) (α+1) Bn (x + 1) = Bn (x) + nBn (x)
(α+1) (α+1) (α+1) Bn (x + 1) − Bn (x) = nBn (x)
(α+1) (α+1) 4Bn (x) = nBn (x).
54 Putting α = 0,
(1) (1) 4Bn (x) = nBn (x)
n−1 4Bn(x) = nx .
Below the first seven polynomials are listed
B0(x) = 1, 1 B (x) = x − , 1 2 1 B (x) = x2 − x + , 2 6 3 1 B (x) = x(x − 1)(x − 2) = x3 − x2 + x, 3 2 2 1 B (x) = x4 − 2x3 + x2 − , 4 30 1 1 5 5 1 B (x) = x(x − 1)(x − )(x2 − x − ) = x5 − x4 + x3 − x, 5 2 3 2 3 6 5 1 1 B (x) = x6 − 3x5 + x4 − x2 + . 6 2 2 42
Also the values for the first seven numbers are :
B0 B1 B2 B3 B4 B5 B6 . 1 1 1 1 1 − 2 6 0 − 30 0 42
3.3.5.1 A Summation Problem
By (3.3.50) and (3.3.51), we have
Z s+1 1 Bn(x)dx = [Bn+1(s + 1) − Bn+1(s)] s n + 1 = sn.
55 One can easily show the obove relation as follows:
Z s+1 1 Bn(x)dx = [Bn+1(s + 1) − Bn+1(s)] s n + 1 1 = 4 B (s) n + 1 n+1 1 = (n + 1)sn n + 1 = sn.
Also,
α X Z α+1 1 sn = B (x)dx = [B (α + 1) − B ] . n n + 1 n+1 n+1 s=1 0
As an example for n = 3, we have
α X 1 s3 = [B (α + 1) − B ] 4 4 4 s=1 1 1 1 = (α + 1)4 − 2(α + 1)3 + (α + 1) − − − 4 30 30 1 = (α + 1)4 − 2(α + 1)3 + (α + 1) 4 1 2 = α(α + 1) . 2
3.3.5.2 Bernoulli Numbers of the First Order
(3.3.46) results
∞ t X tn t et + 1 + B = · (3.3.52) 2 n! n 2 et − 1 n=0
56 we will use (3.3.46)
∞ t X tn t t + B = + 2 n! n 2 et − 1 n=0 t(et − 1) + 2t = 2(et − 1) t(et − 1 + 2) = 2(et − 1) t et + 1 = · . 2 et − 1
Since changing t to −t does not make change the function on the right-hand side, the function on the right is even. Therefore the expansion above does not contain any odd powers of t and hence
B2µ+1 = 0, µ > 0, (3.3.53)
1 B = − . 1 2
Replacing t by 2t in (3.3.52) we get
∞ 2t X (2t)n 2t e2t + 1 + B = · 2 n! n 2 e2t − 1 n=0 e2t + 1 = t · e2t − 1 22t2 23t3 = 1 + t + 2tB + B + B + .... 1 2! 2 3! 3
From (3.3.53)
∞ X 2ntn 22t2 24t4 t + B = 1 + t + B + B .... n! n 2! 2 4! 4 n=0
57 Writing it instead of t we get
22t2 24t4 t coth = 1 − B + B − ... . (3.3.54) 2! 2 4! 4
Similarly, one can easily obtain expansions for csc t and tan t as follows.
1 csc t = cot t − cot t, 2 tan t = cot t − 2 cot 2t
∞ X 22n(22n − 1) = (−1)n−1 B t2n−1. (2n)! 2n n=1
We have another expansion in partial fractions
∞ X 1 πt coth πt = 1 + 2t2 . (3.3.55) t2 − n2 n=1
Rearranging these series and comparing with the coefficients of t2p in (3.3.55), and also considering the series for πt coth πt in (3.3.54),
p ∞ (2π) B2p X 1 (−1)p−1 = . (3.3.56) 2(2p)! α2p α=1
It is clear to see that the summation on the right hand side of (3.3.56) lies between 1 and 2. Thus, as p increases, B2p increases rapidly and also the Bernoulli numbers alternate in sign. Furthermore, we have
p−1 (−1) B2p > 0.
58 In order to express the Bernoulli numbers using determinants, we use (3.3.48)
1 B + 1 = 0, 2! 1! 1 1 B 1 B + 1 + 2 = 0, 3! 2! 1! 1! 2! ...... 1 1 B 1 B 1 B B + 1 + 2 + ... + n−1 + α = 0, (α + 1)! α! 1! (α − 1)! 2! 2! (α − 1)! α!
α (−1) Bα Solving the above equations, for α! , we have the following determinant
:
1 1 0 0 ··· 0 2!
1 1 1 0 ··· 0 3! 2!
1 1 1 0 ··· 0 . 4! 3!
...... ··· .
1 1 1 1 ··· 1 (α+1)! α! (α−1)! (α−2)! 2!
We have
1 1 B (x) + nxn−1 (x + B)n + nxn−1 n 2 + 2
n n n−2 n n−4 = x + 2 x B2 + 4 x B4 + ...,
1 n−1 so that Bn(x) + 2 nx is an even function when n is even and odd function whenever n is odd, since B2µ+1 = 0, µ > 0.
59 3.3.5.3 The Euler-Maclaurin Theorem for Polynomials
Suppose that P (x) is an polynomial of degree α.
Using (3.3.48) and (3.3.51) we have,
n−1 nx = 4Bn(x)
= Bn(x + 1) − Bn(x)
= (B + x + 1)n − (B + x)n,
resulting that
0 P (x) + P (x + B + 1) − P (x + B), (3.3.57) and therefore
0 P (x + y) + P (x + y + B + 1) − P (x + y + B) (3.3.58)
+ P (x + 1 + B(y)) − P (x + B(y)).
Applying Taylor’s Theorem, we have
1 1 P (x+B(y)) P (x)+B (y)P 0(x)+ B (y)P 00(x)+...+ B (y)P (α)(x). (3.3.59) + 1 2! 2 α! α
Substituting equation in (3.3.58) and (3.3.59), we will get the Euler-Maclaurin
60 Theorem for polynomials as follows:
0 P (x + y) + P (x + 1 + B(y)) 1 1 − P (x) + B (y)P 0(x) + B (y)P 00(x) + ... + B (y)P (α)(x) 1 2! 2 α! α 1 = P (x + 1) + B (y)P 0(x + 1) + B (y)P 00(x + 1) 1 2! 2 1 + ... + B (y)P (α)(x + 1) α! α 1 1 − P (x) + B (y)P 0(x) + B (y)P 00(x) + ... + B (y)P (α)(x) 1 2! 2 α! α
0 0 = [P (x + 1) − P (x)] + B1(y)[P (x + 1) − P (x)] 1 + B (y)[P 00(x + 1) − P 00(x)] + ... + 2! 2 1 + B (y) P (α)(x + 1) − P (α)(x) α! α 1 1 = 4P (x) + B (y) 4 P 0(x) + B (y) 4 P 00(x) + ... + B (y) 4 P (α)(x) 1 2! 2 α! α which gives us the Euler-Maclaurin theorem for polynomials.
In the special case when y = 0,
1 1 1 P 0(x) = 4P (x)+B 4P 0(x)+ B 4P 00(x)+ B 4P (iv)(x)+...+ B 4P (α)(x), 1 2! 2 4! 4 α! α (3.3.60) where B3,B5,B7,... all vanish. Now considering P (x) as :
Z a P (x) = φ(t)dt. x
Integrating both sides of equation (3.3.60) we get,
Z x+1 1 0 1 000 φ(x) = φ(t)dt + B1 4 φ(x) + B2 4 φ (x) + B4 4 φ (x) + ... (3.3.61) x 2! 4!
61 1 Since B1 = − 2 , we have the following equation for any polynomial φ(x)
Z x+1 1 1 0 1 000 φ(t)dt = [φ(x + 1) + φ(x)] − B2 4 φ (x) − B4 4 φ (x) − ... . x 2 2! 4!
One can easily prove the above equation by substituting
Z x+1 1 0 1 000 φ(t)dt = − B1 4 φ(x) + B2 4 φ (x) + B4 4 φ (x) + ... x 2! 4! 1 1 = −B 4 φ(x) − B 4 φ0(x) − B 4 φ000(x) − ... 1 2! 2 4! 4 1 1 1 = [φ(x + 1) − φ(x)] − B 4 φ0(x) − B 4 φ000(x) − .... 2 2! 2 4! 4
After a finite number of terms, the series on the right-hand side will terminate.
The equation (3.3.57) shows that
u(x) + P (x + B) + P (B(x)) (3.3.62) is the polynomial solution of the difference equation below :
4u(x) = P 0(x). (3.3.63)
Thus, as an example consider
1 u(x) = B (x) − B (x) + B (x) + c, 4 4 3 1
4u(x) = x3 − 3x2 + 1,
62 having c as an arbitrary constant.
We use (3.3.63), (3.3.62) and (3.3.47)
4u(x) == P 0(x) = x3 − 3x2 + 1.
We then integrate both sides to get
x4 P (x) = − x3 + x + c 4 (x + B)4 P (x + B) = u(x) = − (x + B)3 + (x + B) + c 4 1 = B (x) − B (x) + B (x) + c. 4 4 3 1
Replacing c by an arbitrary periodic function $(x), we will get the general solu- tion as :
$(x + 1) = $(x).
3.3.5.4 The Multiplication Theorem
Considering m as a positive integer in (3.3.45), we have,
∞ n m−1 m−1 (x+ s )t X t X s X te m B x + = n! n m et − 1 n=0 s=0 m=0 t t xt t mx m te (e − 1) m m e = t = t (et − 1)(e m − 1) e m − 1 ∞ X tn = m. B (mx). mn n n=0
63 Thus,
m−1 X s B (mx) = mn−1 B x + . n n m s=0
The above result is the well-known multiplication theorem for Bernoulli poly- nomials of order one.
Letting x = 0, we get
m−1 X s 1 B = − 1 − B . n m mn−1 n s=1
Therefore, if m = 2,
1 1 B = − 1 − B , n = 1, 2,.... n 2 2n−1 n
3.3.5.5 Bernoulli Polynomials in the Interval (0, 1)
Recall that
B2n(1 − x) = B2n(x) and
B2n+1(1 − x) = −B2n+1(x). (3.3.64)
64 n We use the relation Bn(1 − x) = (−1) Bn(x) and write 2n + 1 for n to get,
2n+1 B2n+1(1 − x) = (−1) B2n+1(x)
= −B2n+1(x).
So, the zeros of B2n(x) − B2n are 0 and 1. We must prove that they are the only zeros in [0, 1].
1 1 Letting x = 2 in (3.3.64), we get B2n+1 2 = 0 and also using (3.3.64)
1 we realize that B2n+1(x) is symmetric around x = 2 , thus for n < 0, B2n+1(x)
1 has the zeros 0, 2 , 1. Our aim is to prove that these are the only zeros in [0, 1].
Therefore, including n = µ > 0, we suppose that both statements are true.
B2µ+2(x) − B2µ+2, vanishes at x = 0, x = 1. Also for 0 < x < 1, its’
1 minimum or maximum occurs only at x = 2 .
Since
D [B2µ+2(x) − B2µ+2] = (2µ + 2) B2µ+1(x). (3.3.65)
Therefore, it cannot vanish in (0, 1).
In a similar way,
DB2µ+3(x) = (2µ + 3) [B2µ+2(x) − B2µ+2] + (2µ + 3) B2µ+2
1 where in the interval 0 < x < 2 the above equation can vanish at most once.
65 1 So B2µ+3(x) cannot vanish in 0 < x < 2 and consequently can not vanish in
1 2 < x < 1 by (3.3.64).
The properties will now follow by induction.
n+1 n+1 Recall that (−1) B2n > 0. (−1) B2n+1(x) will have the same sign as its derivative whenever x is sufficiently small and positive which means that it will
n+1 n+1 have the same sign as (−1) B2n, which also has the same sign as (−1) B2n and is positive. Hence
1 (−1)n+1B (x) > 0, 0 < x < . 2n+1 2
1 µ+1 As x increases from 0 to 2 , from (3.3.65) (−1) (B2µ+2(x) − B2µ+2) will also exceed 0, and therefore is positive. Since the expression above vanishes only at 0 and 1, we get
µ+1 (−1) (B2µ+2(x) − B2µ+2) > 0, 0 < x < 1.
3.3.6 The η Polynomials
(3.1.9) is another method for generalizing polynomials. Writing fα(t) =
2α(et + 1)−α a new class of polynomials will be formed which are named η poly- nomials and are denoted by
∞ 2αext+g(t) X tn = η(α)(x), (3.3.66) (et + 1)α n! n n=0
66 so that
∞ ∞ X tn X tn ∇η(α)(x) = η(α−1)(x). (3.3.67) n! n n! n n=0 n=0
α t −α Replacing η for φ and writing fα(t) = 2 (e + 1) in equation (3.1.9)
∞ X tn et + 1 2α ∇η(α)(x) = ext+g(t) n! n 2 (et + 1)n n=0 2α−1 = ext+g(t). (et + 1)α−1
From (3.3.66),
∞ X tn 2α−1 ∇η(α)(x) = ext+g(t) n! n (et + 1)α−1 n=0 ∞ X tn = η(α−1)(x). n! n n=0
Therefore we have
(α) (α−1) ∇ηn (x) = ηn (x). (3.3.68)
As we can see, ∇ decreases the order by one unit and makes no changes in the degree.
Using (3.1.5), we get
(α) n (α) n (α−1) n η + x + 1 + (η + x) + 2 η + x (3.3.69)
67 where η satisfies the recurrence relation below
(α) n (α) (α−1) η + 1 + ηn + 2ηn . (3.3.70)
3.3.7 Definition of Euler Polynomial
Letting g(t) = 0 and α = 0 in the generating function, we get the simplest
η polynomials with ext as their generating function. These η polynomials are the powers of x. These polynomials are also called Euler polynomials of order zero.
Therefore
(0) n En (x) = x (3.3.71) and
∞ X tn ext = E(0)(x), n! n n=0 which is simply obtained as follows
∞ X (xt)n ext = n! n=0 ∞ X xntn = n! n=0 ∞ X tn = E(0)(x). n! n n=0
(0) Here En (x) is the Euler polynomial of degree n and order zero.
68 The Euler polynomials of order α are defined as below using (3.3.66),
∞ 2αext X tn = E(α)(x). (3.3.72) (et + 1)α n! n n=0
(α) Generally Euler numbers are the values of En (0). In order to avoid confu- sion with N¨orlund’snotation for polynomials, we will use the notations below as
N¨orlunddid well
(α) −n (α) En (0) = 2 Cn . (3.3.73)
Thus we give the generating function for the C numbers
∞ 2α X tn 1 = C(α) (et + 1)α n! 2n n n=0 ∞ X tn = E(α)(x). n! n n=0
We put x = 0 in equation (3.3.72) and (3.3.73),
∞ 2αe0.t X tn = E(α)(0) (et + 1)α n! n n=0 ∞ 2α X tn 1 = C(α). (et + 1)α n! 2n n n=0
1 n (α) (α) Substituting x = 2 α in 2 En (x), we get the Euler numbers of order α, En as,
1 E(α) = 2nE(α)( α). n n 2
Euler numbers with an odd suffix vanish as we shown in (3.3.73).
69 3.3.7.1 Fundamental Properties of Euler Polynomials
As we know Euler polynomials are η polynomials and hence φ polynomi- als. Therefore
1 n E(α) C(α) + x (3.3.74) n + 2 and
d E(α)(x) = nE(α) (x). (3.3.75) dx n n−1
We differentiate both sides of equation (3.3.72),
∞ X tn d d 2αext E(α)(x) = n! dx n dx (et + 1)α n=0 t2αext = (et + 1)α ∞ X tn = t E(α)(x) n! n n=0 ∞ X tn+1 = E(α)(x). n! n n=0
We put (n + 1) instead of n to obtain the relation (3.3.75),
∞ ∞ X tn d X tn+1 E(α)(x) = (n + 1) E(α)(x) n! dx n (n + 1)! n n=0 n=0 ∞ n X t (α) = n E (x). n! n−1 n=0
70 Let us prove the following equation representing the integral representation for generalized Euler polynomials
Z x (α) 1 h (α) (α) i En (t)dt = En+1(x) − En+1(a) . (3.3.76) a n + 1
We will integrate both sides of equation (3.3.72)
∞ X tn Z x Z x 2αeyt E(α)(t)dt = dy n! n (et + 1)α n=0 a a 2α 1 = ext − eat (et + 1)α t 1 2αext 1 2αeat = − t (et + 1)α t (et + 1)α ∞ ∞ 1 X tn 1 X tn = E(α)(x) − E(α)(a) t n! n t n! n n=0 n=0 ∞ ∞ X tn−1 X tn−1 = E(α)(x) − E(α)(a). n! n n! n n=0 n=0
We replace n − 1 with n,
∞ n ∞ n X t (α) X t (α) E (x) − E (a) (n + 1)! n+1 (n + 1)! n+1 n=0 n=0 " ∞ n ∞ n # 1 X t (α) X t (α) = E (x) − E (a) , n + 1 n! n+1 n! n+1 n=0 n=0 ∞ n Z x " ∞ n ∞ n # X t 1 X t (α) X t (α) E(α)(t)dt = E (x) − E (a) . n! n n + 1 n! n+1 n! n+1 n=0 a n=0 n=0
Thus, from (3.3.69) and (3.3.73),
(α) (α−1) ∇En (x) = En (x) (3.3.77)
71 and
1 n 1 2 C(α) + 1 + C(α) C(α−1). 2 2n n + 2n n
Replacing E for η in equation (3.3.70) proves (3.3.74).
(α) n (α) (α−1) Theorem 3.3.4 a) E + 1 + En + 2En
(α) (α) (α−1) b) En (1) + En = 2En
1 (α) n 1 (α) 2 (α−1) c) 2 C + 1 + 2n Cn + 2n Cn .
Using (3.3.73) and (3.1.5),
(α) n (α) (α−1) C + 2 + Cn + 2Cn . (3.3.78)
Using (3.3.77) repeatedly,
α (α) n ∇ = En (x) = x (3.3.79)
(0) n since En (x) = x .
Applying ∇ to both sides of equation (3.3.77) once more,
2 (α) (α−1) ∇ En (x) = ∇En (x)
(α−2) = En (x).
72 Applying difference operator consecutively gives,
α (α) (α−α) ∇ En (x) = En (x)
(0) = En (x)
= xn.
Again using (3.3.77)
(α) (α−1) (α) En (x + 1) = 2En (x) − En (x), (3.3.80)
(α−1) (α) En (x) = ∇En (x) E(α)(x + 1) + E(α)(x) = n n , 2
(α−1) (α) (α) 2En (x) = En (x + 1) + En (x).
Since
α 1 αn E(α) = 2nE(α) C(α) + 2n, (3.3.81) n n 2 + 2 2 using (3.3.74) we will get
(n) (n) E + n + C .
73 Equation (3.3.81) yields to the following,
1 αn 1 n E(α) = 2n C(α) + = 2n C(α) + α n 2 2 2 1 n = 2n C(α) + α 2n n = C(α) + n .
Therefore
1 1 n E(α)(x) x − α + E(α) , (3.3.82) n + 2 2 α + x x + E(α) n E(n) . n 2 + 2
Letting x = 1 and x = −1 in terns and then adding them up, we have
n n α + 1 α − 1 E(α) + 1 + E(α) − 1 2nE(α) + 2nE(α) + n 2 n 2 α − 1 2n+1∇E(α) + n 2 α − 1 2n+1E(α−1) + n 2
(α−1) + 2En .
3.3.7.2 The Complementary Argument Theorem
As we know x and n − x are called complementary. We will now show that the following equation holds
(α) n (α) En (α − x) = (−1) En (x). (3.3.83)
74 Using (3.3.72) and putting α − x for x we have
∞ X tn 2ne(α−x)t E(α)(α − x) = n! n (et + 1)n n=0 2αeαte−xt e−αt = · (et + 1)α e−αt 2αe−xt = (et + 1)α (e−t)α 2αe−xt = (1 + e−t)α ∞ X (−t)n = E(α)(x). n! n n=0
Equating the coefficients of tn, (3.3.83) will be proved. Equation (3.3.83) is the
Complementary Argument Theorem. This theorem holds for any η polynomial with an even generating function g(t).
Taking x = 0 in (3.3.83) for n = 2µ, we get
(α) (α) −2µ (α) E2µ (α) = E2µ (0) = 2 C2µ .
Putting x = 0 and n = 2µ in equation (3.3.83) and using equation (3.3.73) leads to the following
(α) 2µ (α) E2µ (α) = (−1) E2µ (0)
(α) = E2µ (0)
−2µ (α) = 2 C2µ .
(α) −2µ (α) Therefore at x = 0 and x = α, E2µ (x) − 2 C2µ has zeros.
75 1 Again letting x = 2 α and n = 2µ + 1 in (3.3.83), we get
(α) (α) E2µ+1 = −E2µ+1,
(α) which will result E2µ+1 = 0. Thus Euler numbers with odd suffixes vanish.
3.3.7.3 Euler Polynomials of Successive Orders
Define,
∞ X tn 2αext E(α)(x) = . (3.3.84) n! n (et + 1)α n=0
Differentiating both sides with respect to t and then multiplying by t, we obtain
h i ∞ α xt t α t α−1 t xt X tn 2 xte ((e + 1) ) − α (e + 1) e te E(α)(x) = (n − 1)! n t 2α n=0 (e + 1) 2αxtext 2ααtet(x+1) = α − (et + 1) (et + 1)α+1 ∞ n ∞ n X t (α) 1 X t (α+1) = x E (x) − α E (x + 1). (n − 1)! n−1 2 (n − 1)! n−1 n=0 n=0
By equating the coefficients of tn+1, we have
1 E(α) (x) = xE(α)(x) − αE(α+1)(x + 1). n+1 n 2 n−1
Using (3.3.80)
(α+1) (α) (α+1) En (x + 1) = 2En (x) − En (x).
76 Thus,
2 2 E(α+1)(x) = E(α) (x) + (α − x)E(α)(x). (3.3.85) n α n+1 α n
Taking x = 0 in equation (3.3.85), we get the relation,
1 C(α+1) = C(α) + 2C(α). n α n+1 n
3.3.8 Euler Polynomials of the First Order
(1) For order one in Euler numbers we will use En(x) instead of En (x). Let us recall the following definitions.
Consider,
∞ ∞ 2ext X tn 2 X tn 1 = E (x), = C . (3.3.86) et + 1 n! n et + 1 n! 2n n n=0 υ=0
1 Putting x = 2 and α = 1 in equation (3.3.86),
1 t ∞ n 2e 2 X t En = (3.3.87) et + 1 n! 2n n=0 1 t ∞ n 2e 2 X t 1 = E . et + 1 n! n 2 n=0
Substituting α = 1, x = 0 in (3.3.82), we have
1 E n E n 2 + 2 E n . + 2n
77 1 En Replacing En 2 with 2n in the above equation
1 n E (x) = + x ,(C + 2)n + C 0, n > 0. n 2 n +
Putting α = 1 in equation (3.3.78),
n (0) (C + 2) + Cn + 2Cn
n n + 2.2 .0
+ 0.
Also,
n ∇En(x) = x (3.3.88)
DEn(x) = nEn−1(x) (3.3.89)
n En(1 − x) = (−1) En(x). (3.3.90)
78 Below, the first seven Euler polynomial are fixed.
E0(x) = 1, 1 E (x) = x − , 1 2
E2(x) = x(x − 1), 1 1 E (x) = x − x2 − x − , 3 2 2
2 E4(x) = x(x − 1) x − x − 1 , 1 E (x) = x − x4 − 2x3 − x2 + 2x + 1 , 5 2
4 3 2 E5(x) = x(x − 1) x − 2x − 2x + 3x + 3 .
Also,
E0 E2 E4 E6 E8 E10 E12 . 1 −1 5 −61 1385 −50521 2702765
Pα s n Example 3.3.5 Use equality (3.3.90) to evaluate s=1(−1) s .
α α X s n X s (−1) s = (−1) ∇En(s) s=1 s=1 α 1 X = (−1)s [E (s + 1) + E (s)] 2 n n s=1 1 = [−E (2) − E (1) + E (3) + E (2) + ... + (−1)αE (α + 1) + (−1)αE (α)] 2 n n n n n n 1 1 = (−1)αE (α + 1) − E (1). 2 n 2 n
79 3.3.8.1 Euler Numbers of the First Order
Substituting t by 2t in (3.3.87), we have
∞ 2 X tn = E . et + e−t n! n n=0
Putting 2t for t
∞ X tn 2et E = n! n e2t + 1 n=0 et (2) = et (et + e−t) 2 = . et + e−t
Therefore
∞ X tn t2 t4 sec ht = E = 1 + E + E + ..., n! n 2! 2 4! n=0 taking it instead of t, we have
t2 t4 sec t = 1 − E + E − ... . (3.3.91) 2! 2 4!
Rearranging the expansion
∞ π X (−1)n(2n + 1) = 4 cos πx (2n + 1)2 − x2 2 n=0
80 and equating the coefficient of x2p in the above equation with the coefficient of
2p 1 πx x in series for 4 π sec 2 obtained from (3.3.91), we get
E 1 1 1 (−1)p = 2p π2p+1 = 1 − + − + .... 22p+2(2p)! 32p+1 52p+1 72p+1
It is easy to see that Euler numbers increase and alternate in sign.
Using the method given in the subsection “Bernoulli Numbers of the First
α Order”, we obtain the determinant below for (−1) E2α / (2α)!
1 1 0 0 ··· 0 2! 1 1 1 0 ··· 0 4! 2! 1 1 1 1 ··· 0 . 6! 4! 2! . . . . . . . . . ··· . 1 1 1 1 1 (2α)! (2α−2)! (2α−4)! (2α−6)! ... α!
Regarding the numbers Cn and using (3.3.86), we have the odd function below
∞ n t X t Cn e − 1 1 − 1 = = − tanh t. n! 2n et + 1 2 n=0
Therefore all the numbers C2µ = 0, µ > 0. Writing 2t for t, we get
t3 t5 t7 tanh t = t − C − C − C − ..., 3! 3 5! 5 7! 7
81 again as we did before, substituting t by it
t3 t5 t7 tan t = t + C − C + C − .... 3! 3 5! 5 7! 7
Equating the corresponding coefficients in the above series and the series in the subsection “Bernoulli Numbers of the first order”, we get
22n(22n − 1) C = − B . 2n−1 2n 2n
Since for µ > 0, C2µ = 0 we have
1 n E (x) − xn x + C − xn n + 2 xn−1 xn−3 = n C + n C + ..., 1 2 1 3 23 3
n resulting that En(x) − x is an even function when n is odd and is odd when n is even.
3.3.8.2 Boole’s Theorem for Polynomials
Putting α = 1 in equation (3.3.82), we have
1 1n E (x) x + E − . n + 2 2
Thus
1 1n 1 1n 2xn = 2∇E (x) x + 1 + E − + x + E − , n + 2 2 2 2
82 and if P (x) is a polynomial
1 1 1 1 2P (x) P x + 1 + E − + P x + E − . (3.3.92) + 2 2 2 2
Replacing x by x + y,
1 1 1 1 2P (x + y) P x + y + 1 + E − + P x + y + E − + 2 2 2 2
+ P (x + y + 1 + E(y)) + P (x + E(y)) .
Now applying Taylor’s Theorem
1 00 P (x + E(y)) P (x) + E (y)P 0(x) + E (y)P (x) + .... + 1 2! 2
Hence
1 00 P (x + y) = ∇P (x) + E (y)∇P 0(x) + E (y)∇P (x) + ..., (3.3.93) 1 2! 2
which leads to Boole’s Theorem. Letting x = 0, we get an expansion for P (y) in terms of Euler’s polynomials.
From the subsection “The Complementary Argument Theorem” ,
−2s E2s(1) = E2s(0) = 2 C2s = 0
83 and
−2s E2s+1(1) = −E2s+1(0) = −2 C2s.
Taking y = 1 in (3.3.93), we obtain
1 00 1 v P (x+1)−P (x) = −C ∇P 0(x)− C ∇P (x)− C ∇P (x)−.... (3.3.94) 1 3!22 3 5!24 5
Using (3.3.92), we get a solution for
∇u(x) = P (x) (3.3.95)
as below
1 1 u(x) P x + E − P (E(x)) . + 2 2 +
For example the equation
∇u(x) = x3 + 2x2 + 1
has the solution below
u(x) = E3(x) + 2E2(x) + 1.
In order to get a general solution we can add the above solution with an arbitrary periodic function π(x) such that π(x + 1) = −π(x).
84 Chapter 4
CONCLUSION
In this thesis an overview of Bernoulli and Euler numbers is provided. In
Chapter 2, the Bernoulli and Euler polynomials are described and the relationship between these classes of polynomials are investigated. Some important identities involving differentiation, integration, summation are discussed using the basics of
Finite Difference Calculus and Differentiation. The last part of this study is con- cerned with the Generalized Bernoulli and Euler polynomials. Furthermore, the properties obtained in the second chapter are also examined for the Generalized
Bernoulli and Euler polynomials in this part of the thesis. The Complementary
Argument Theorem, the generating functions, the Multiplication and the Euler-
Maclauren Theorems are widely used in obtaining the results given in Chapter 2 and Chapter 3.
85 REFERENCES
[1] Cheon, G. S., A Note on the Bernoulli and Euler Polynomials, Applied
Mathematics Letters, Vol. 16, 365-368, 2003.
[2] Pan, H., and Sun, Z.W., New identities involving Bernoulli and Euler
polynomials, Journal of Combinatorial Theory, Series A 113 (2006) 156-175.
[3] Chen, K. W., Algorithms for Bernouli numbers and Euler Numbers, Journal
of Integer Sequences, Vol. 4(2001), Article 01.1.6.
[4] Milne-Thomson, L. M., The Calculus of Finite Differences, Macmillan and
Co., Limited St. Martin’s Street, London, 1933.
[5] Srivastava, H. M., and Pinter, A., Remarks on Some Relationships Between
the Bernoulli and Euler Polynomials, Applied Mathematics Letters, Vol 17.
375-380, 2004.
[6] Sun, Z. W., Introduction to Bernoulli and Euler Polynomials, A lecture
given in Taiwan on June 6, 2002.
86 [7] http://en.wikipedia.org/wiki/Bernoulli number.
[8] http:/en.wikipedia.org/wiki/Eulerian number.
[9] http://en.wikipedia.org/wiki/Bernoulli polynomials.
[10] http://oeis.org/wiki/Eulerian polynomials.
[11] http://dlmf.nist.gow/24.
87