Utilizing the translation operator to represent Bernoulli polynomials and power sums as polynomials of Sheffer-type, we obtain concisely almost all their known properties as so as many new ones, especially new recursion relations for calculating Bernoulli polynomials and numbers, new formulae for obtaining power sums of entire and complex numbers. Then by the change of arguments from z into Z = z(z-1) and n into λ which is the 1 st order power sum we obtain the Faulhaber formula for powers sums in term of polynomials in λ having coefficients depending on Z. Practically we give tables for calculating in easiest possible manners, the Bernoulli numbers, polynomials, the general powers sums.
In many branches of mathematics the problem of Bernoulli numbers related to the millenary problem of power sums is probably the most studied since the publication of the book Ars Conjectandi by Euler in 1738 [
Concerning Bernoulli polynomials B m ( z ) , classically defined from a generating function, there had not so much properties, the most remarkable is its representation by a hyper-differential operator, the Hurwitz expansion of them into Fourier series, the Roman formula for B m ( n z ) , the Euler-McLaurin formula, etc.
As for the power sums on real and complex numbers, including the famous Faulhaber conjecture, there has no valuable formula linking them with Bernoulli polynomials until only some years ago [
Regarding the situation, we would like to perform a selection of as many as possible known and new interesting properties of Bernoulli polynomials then of Bernoulli numbers in a coherent way, i.e., by only one approach, which utilizes principally operator calculus lying on the couple of operators position and derivation, similar as the couple r → , ∇ in quantum mechanics.
In Section 2, we will treat the problem of Bernoulli polynomials, from their representation by a hyper-differential operator to almost all of their algebraic properties to the fact that B m ( n ) is equal to the primitive of the power sums of natural integers. Afterward we show that the formula giving Bernoulli polynomials of a sum of two arguments B m ( z + y ) leads to two new recurrence relations for obtaining B m ( z ) . We also give another approach for calculating without integrations, the Fourier series of Bernoulli polynomials and the Bernoulli series of functions, the relation of B m ( z ) with the Euler zeta function. Afterward we show an up-to-date procedure for obtaining B m ( z ) from and only from B m − 1 ( z ) leading to the rapid establishment of
In Section 3, we treat the problem of Bernoulli numbers B m , from its initial definition by Jakob Bernoulli in 1713 who related them by conjecture with the power sums on natural numbers. By comparison of this relation with the preceding formula linking B m ( n ) with power sums, we may identify B m with B m ( 0 ) then calculate B m by a simple matrix method side-by-side with the method, more powerful, link with l, coming from the special recurrence formula coming from B m ( z + y ) .
In Section 4, we prove by utilizing the translation operator e a ∂ z , coming from the Newtonian binomial, that the power sums on complex numbers are simply related to those on natural numbers. On the other hand, we prove that they are also related very simply to Bernoulli polynomials, from that we get again the recurrence relation between Bernoulli polynomials.
Section 5 is devoted to the Faulhaber problem regarding power sums on complex numbers. Here we show that power sums on complex numbers may be calculated from sums of entire numbers somehow by writing B 2 m ( z ) in function of the new argument Z = z ( z − 1 ) .
In 1738, Euler introduced the Bernoulli polynomials B m ( z ) via the generating function [
t e t − 1 e z t = ∑ m = 0 ∞ 1 m ! B m ( z ) t m (2.1)
which directly gives by identification
B 0 ( z ) = 1 , B 1 ( z ) = z − 1 2 , B 2 ( 0 ) = 1 6 (2.2)
Utilizing the translation operator e a ∂ z coming from the Newtonian binomial
( x + a ) m = ∑ k = 0 m ( m k ) a k x m − k = ∑ k = 0 m a k k ! ∂ z k x m = e a ∂ z x m (2.3)
and having the property
e a ∂ z f ( x ) = f ( x + a )
e ∂ z e t z = e t ( z + 1 ) = e t e t z (2.4)
∂ z e ∂ z − 1 e t z = t e t − 1 e t z (2.5)
we directly find from (2.1) that B m ( z ) is the transform of z m via a differential operator
B m ( z ) = ∂ z e ∂ z − 1 z m , m > 0 (2.6)
From (2.6) we get the famous known formulae
B ′ m ( z ) = m B m − 1 ( z ) (2.7)
B m ( z + 1 ) − B m ( z ) = ( e ∂ z − 1 ) B m ( z ) = ∂ z z m = m z m − 1 (2.8)
B m ( 1 ) − B m ( 0 ) = δ m 1 (2.9)
and the following formula which gives e i n x z as series of Bernoulli polynomials.
∑ m = 0 ∞ t m B m ( z ) m ! = ∂ z e ∂ z − 1 e t z = t e t − 1 e t z (2.10)
From (2.8) we get the formula given by Roman [
B m + 1 ( z y + N ) − B m + 1 ( z y ) = ( m + 1 ) ∑ n = 0 N − 1 ( z y + n ) m (2.11)
From (2.10) we get the formulae on relations of Bernoulli polynomials versus trigonometric functions, especially the Castellanos formula [
∑ m = 2 ∞ ( 2 i x ) m B m ( 0 ) m ! = x cos x sin x − 1 (2.12)
The formulae (2.7) and (2.8) give the important formulae
∫ 0 1 B m − 1 ( z ) d z = 1 m ( B m ( 1 ) − B m ( 0 ) ) = δ m 1 (2.13)
B m ( n ) − B m ( 0 ) = m ( 0 m − 1 + 1 m − 1 + ⋯ + ( n − 1 ) m − 1 ) (2.14)
and the Taylor expansion
B m ( z ) = B m ( a ) + ⋯ + ( m k ) ( z − a ) k B m − k ( a ) + ⋯ + ( z − a ) m B 0 ( a )
which may be put under symbolic form
B m ( z + a ) = : ( B ( a ) + z ) m (2.15)
where undefined symbols B k ( a ) are to be replaced with B k ( a ) .
Exploring now the inter-relations between Bernoulli polynomials.
From (2.4) and (2.7) we get the complementary of (2.15)
B m ( z + a ) = e a ∂ z B m ( z ) = ( 1 + ⋯ + a k k ! ∂ k + ⋯ + a m m ! ∂ m ) B m ( z ) = B m ( z ) + ⋯ + ( m k ) B m − k ( z ) a k + ⋯ + B 0 ( z ) a m = : ( B ( z ) + a ) m , 0 0 = 1 (2.16)
From (2.13)
∫ z z + 1 B m ( y ) d y = ∫ 0 1 B m ( z + y ) d y = z m ∫ 0 1 B 0 ( y ) d y = z m
∫ 0 1 B m ( y ) d y = 0 m
∫ 0 n B m ( y ) d y = 0 m + 1 m + ⋯ + ( n − 1 ) m (2.17)
i.e.,
“The sum of powers of order m of n first entire numbers from 0 to ( n − 1 ) , denoted by S m ( n ) , is equal to the simple primitive (without constant of integration) of the Bernoulli polynomial B m ( n ) ” and vice-versa,
“The Bernoulli polynomial B m ( n ) is equal to the derivative of the power sums S m ( n ) ”
As for B m ( − z ) we see that
B m ( − z ) = − ∂ z 1 − e − ∂ z ( − z ) m = ( − 1 ) m e ∂ z ∂ z e ∂ z − 1 z m = ( − 1 ) m B m ( z + 1 ) = : ( − 1 ) m ( B ( z ) + 1 ) m (2.18)
which leads to
B m ( − z + 1 2 ) = ( − 1 ) m B m ( z + 1 2 ) (2.19)
i.e., to the theorem
“The graph of a Bernoulli polynomial is symmetric with respect to the axis z = 1 2 if m is pair and anti-symmetric if m is impair”.
Joint (2.19) with (2.9) we get the famous property [
B 2 m + 1 ( 1 ) = − B 2 m + 1 ( 0 ) = 1 2 δ m 0 (2.20)
Now, by replacing in (2.6) z with z n so that ∂ z is with n ∂ z we get
B m ( z n ) = n ∂ z e n ∂ z − 1 ( z n ) m = n ∂ z ( e ∂ z − 1 ) ( 1 + e ∂ z + e 2 ∂ z + ⋯ + e ( n − 1 ) ∂ z ) ( z n ) m
and the formula
∑ k = 0 n − 1 B m ( z + k n ) = n 1 − m B m ( z ) (2.21)
saying that
B m ( z ) is n m − 1 times the sum of B m ( z + k n ) , k < n
For examples:
2 1 − m B m ( z ) = B m ( z 2 ) + B m ( z + 1 2 )
B 2 m + 1 ( 1 3 ) + B 2 m + 1 ( 2 3 ) = 0 = ( 3 − 2 m − 1 ) B 2 m + 1 ( 1 )
By replacing in (2.6) z with nz and ∂ z with 1 n ∂ z we find again the formula given by Raabe [
B m ( n z ) = n m − 1 ( B m ( z ) + B m ( z + 1 n ) + ⋯ + B m ( z + n − 1 n ) ) (2.22)
saying that
“ B m ( n z ) is n m − 1 times the sum of B m ( z + k n ) , k < n .”
For examples
B m ( 2 z ) = 2 m − 1 ( B m ( z ) + B m ( z + 1 2 ) )
B m ( 1 2 ) = ( 2 1 − m − 1 ) B m ( 0 )
B 1 ( 3 z ) = B 1 ( z ) + B 1 ( z + 1 3 ) + B 1 ( z + 2 3 )
5 − m B m ( 0 ) = ( 1 + ( − 1 ) m ) ( B m ( 1 5 ) + B m ( 2 5 ) )
From the following property of operators that we characterize fundamental [
f ( ∂ z ) g ( z ) ≡ g ( z ) f ( ∂ z ) + 1 1 ! g ′ ( z ) f ′ ( ∂ z ) + 1 2 ! g ″ ( z ) f ″ ( ∂ z ) + ⋯ (2.23)
we get
B m + 1 ( z ) = ∂ z e ∂ z − 1 z z m = z ∂ z e ∂ z − 1 z m + ( 1 e ∂ z − 1 − ∂ z e ∂ z ( e ∂ z − 1 ) 2 ) z m = z B m ( z ) + 1 e ∂ z − 1 z m − e ∂ z ∂ z ( e ∂ z − 1 ) 2 z m
∂ z B m + 1 ( z ) = ∂ z z B m ( z ) + B m ( z ) − e ∂ z ∂ z 2 ( e ∂ z − 1 ) 2 z m
( m − 1 ) B m ( z ) = z ∂ z B m ( z ) − e ∂ z ∂ z e ∂ z − 1 B m ( z )
Now, because
∂ z + y f ( z + y ) = ∂ z f ( z + y ) = ∂ y f ( z + y ) (2.24)
( m − 1 ) B m ( z + y ) = m ( z + y ) B m − 1 ( z + y ) − e ∂ y ∂ y e ∂ y − 1 B m ( z + y ) = : m ( z + y ) B m − 1 ( z + y ) − ( B ( z ) + B ( y + 1 ) ) m (2.25)
The above recurrence formula is to be compare with that given by Weisstein [
( 1 − m ) B m ( z + y ) + m ( z + y − 1 ) B m − 1 ( z + y ) = : ( B ( z ) + B ( y ) ) m
From (2.25) and knowing that B k ( 1 ) = ( − 1 ) k B k ( 0 ) we obtain another type of recurrence formula for Bernoulli polynomials
( m − 1 ) B m ( z ) = m z B m − 1 ( z ) − ( B ( z ) + B ( 1 ) ) m (2.26)
B m ( z ) = B 1 ( z ) B m − 1 ( z ) − 1 m ∑ k = 2 m ( − 1 ) k ( m k ) B k ( 0 ) B m − k ( z ) (2.27)
For examples, with B 1 ( z ) = z − 1 2 , B 2 ( 0 ) = 1 6 ,
B 2 ( z ) = B 1 ( z ) B 1 ( z ) − 1 2 B 2 ( 0 ) B 0 ( z ) = ( z − 1 2 ) 2 − 1 12 = z 2 − z + 1 6
B 3 ( z ) = B 1 ( z ) B 2 ( z ) − 1 3 3 B 2 ( 0 ) B 1 ( z ) = ( z − 1 2 ) ( z 2 − z + 1 6 − 1 6 ) = z 3 − 3 2 z 2 + 1 2 z
B 4 ( z ) = B 1 ( z ) B 3 ( z ) − 1 4 ( B 2 ( z ) + B 4 ( 0 ) )
By successive integrations by parts and utilizing the formula (2.13) for n , m ≥ 1 we get, knowing (2.9),
∫ 0 1 B n ( z ) B m ( z ) d z = 1 m + 1 ∫ 0 1 B n ( z ) B ′ m + 1 ( z ) d z = 1 m + 1 ( B n ( z ) B m + 1 ( z ) ) | 0 1 − n m + 1 ∫ 0 1 B n − 1 ( z ) B m + 1 ( z ) d z = ( − 1 ) n − 1 n ! m ! ( m + n ) ! ( B 1 ( z ) B m + n ( z ) ) | 0 1 = ( − 1 ) n − 1 n ! m ! ( m + n ) ! B m + n ( 0 ) (2.28)
Because of the factor ( − 1 ) n − 1 we may conclude that
B 2 n + 1 ( 0 ) = 0 for n > 0 and B 2 n + 2 ( 0 ) has opposite sign with respect to B 2 n ( 0 ) .
The same method also gives
∫ 0 1 B m ( z ) e − 2 i k π z d z = − 1 2 π i k ∫ 0 1 B m ( z ) ( e − 2 i k π z ) ′ d z = − 1 2 π i k δ m 1 − − m 2 π i k ∫ 0 1 B m − 1 ( z ) e − 2 i π k z d z = ⋯ = − m ! ( 2 π i k ) m (2.29)
which provides us the following formula on Fourier series of B m ( z ) proven by Hurwitz in 1890 by another method [
B m ( z ) = ∑ k ∈ Z , k ≠ 0 ∞ ( ∫ 0 1 B m ( z ) e − 2 i k π z d z ) e i 2 π k z = − m ! ( 2 i π ) m ∑ k ∈ Z , k ≠ 0 ∞ 1 k m e i 2 π k z , 0 ≤ z ≤ 1 (2.30)
Let f ( z ) be a periodic function defined on an interval a ≤ z < b and has the period P = b − a . For expanding f ( z ) into a Fourier series of exponentials
f ( z ) = ∑ n ∈ Z c ( n ) e i 2 π n z P , a ≤ z < b = a + P (2.31)
we firstly write
∫ a b − e − i 2 π n 0 z P f ( z ) d z P = ∑ n ∈ Z c ( n ) ∫ a b − e i 2 π ( n − n 0 ) z P d z P
and see that the second member is equal uniquely to c ( n 0 ) so that
c ( n 0 ) = 1 P ∫ a b − e − i 2 π n 0 z P f ( z ) d z (2.32)
The Fourier series of a function, if it exists, is then
f ( z ) = 1 P ∑ n ∈ Z ∞ e i 2 π n z P ∫ a b − e − i 2 π n z P f ( z ) d z (2.33)
To avoid integrations in the calculation, we may utilize the method of integrations by parts and get
c ( n ) = 1 P ∫ a b − f ( z ) e − i 2 π n z P d z
P c ( n ) = − P 2 i π n ( f ( b − ) e − i 2 π n b − P − f ( a ) e − i 2 π n a P ) + P 2 i π n ∫ a b − f ′ ( z ) e − i 2 π n z P d z
P c ( n ) = − ∑ k = 0 ∞ ( P 2 i π n ) k + 1 ( f ( k ) ( b − ) e − i 2 π n b − P − f ( k ) ( a ) e − i 2 π n a P ) − 0 ( k )
so that we may write down the Fourier series formula
f ( z ) = 1 P ∫ a b − f ( z ) d z − 1 P ∑ k = 0 ∞ f ( k ) ( z ) | a b − ∑ n ∈ Z , n ≠ 0 ( P 2 i π n ) k + 1 e i 2 π n z − a P (2.34)
In the case 0 ≤ z < 1 , jointed the preceding formula written under the form
f ( z ) = ∫ 0 1 − f ( z ) d z − ∑ n ∈ Z , n ≠ 0 ∑ k = 0 ∞ f ( k ) ( z ) | 0 1 − ( 1 2 i π n ) k + 1 e i 2 π n z
with the Hurwitz formula we get the new and precious formula on expansion of derivable functions into series of Bernoulli polynomials
f ( z ) = ∫ 0 1 f ( z ) d z + ∑ k = 0 ∞ [ f ( k ) ( 1 ) − f ( k ) ( 0 ) ] 1 ( k + 1 ) ! B k + 1 ( z ) (2.35)
or
f ( z ) = ∫ 0 1 f ( z ) d z + ∑ k = 0 N f ( k ) ( z ) | 0 1 B k + 1 ( z ) ( k + 1 ) ! − ∑ n ∈ Z n ≠ 0 ∑ k = N + 1 ∞ f ( k ) ( z ) | 0 1 ( 1 2 i π n ) k + 1 e i 2 π n z (2.36)
For examples, under matrix form
( f ( z ) 1 z z 2 ⋮ z m ) = ( [ ∫ f ( z ) ] 0 1 [ f ( z ) ] 0 1 [ f ′ ( z ) ] 0 1 [ f ″ ( z ) ] 0 1 ⋯ [ f ( m − 1 ) ( z ) ] 0 1 1 1 / 2 1 1 / 3 1 2 ⋮ ⋮ ⋮ ⋮ ⋱ 1 / ( m + 1 ) 1 m m ( m − 1 ) ⋯ m ! ) ( 1 B 1 ( z ) / 1 ! B 2 ( z ) / 2 ! B 3 ( z ) / 3 ! ⋮ B m ( z ) / m ! ) (2.37)
to be compared with
( f ( z ) 1 z z 2 ⋮ z m ) = ( [ ∫ f ( z ) ] 0 1 [ f ( z ) ] 0 1 [ f ′ ( z ) ] 0 1 [ f ″ ( z ) ] 0 1 ⋯ [ f ( m − 1 ) ( z ) ] 0 1 1 1 / 2 1 1 / 3 1 2 ⋮ ⋮ ⋮ ⋮ ⋱ 1 / ( m + 1 ) 1 m m ( m − 1 ) ⋯ m ! ) ( 1 ∑ n ≠ 0 e 2 i π n z / ( 2 i π n ) ∑ n ≠ 0 e 2 i π n z / ( 2 i π n ) 2 ∑ n ≠ 0 e 2 i π n z / ( 2 i π n ) 3 ⋮ ∑ n ≠ 0 e 2 i π n z / ( 2 i π n ) m ) (2.38)
Formula (2.36) leads also to
f ( 0 ) = ∫ 0 1 f ( z ) d z + ∑ k = 0 ∞ [ f ( k ) ( 1 ) − f ( k ) ( 0 ) ] 1 ( k + 1 ) ! B k + 1 ( 0 ) (2.39)
f ′ ( z ) = ∑ k = 1 ∞ [ f ( k ) ( 1 ) − f ( k ) ( 0 ) ] 1 ( k + 1 ) ! B k + 1 ( z ) (2.40)
As first interesting applications
e z = ( e − 1 ) + ( e − 1 ) ∑ k = 0 ∞ B k + 1 ( z ) ( k + 1 ) ! 0 ≤ z < 1
e − 2 e − 1 = − ∑ k = 0 ∞ B k + 1 ( 0 ) ( k + 1 ) ! (2.41)
By (2.36) we also obtain a precious recurrence formula of Bernoulli polynomials
z m = ∫ 0 1 z m d z + ∑ k = 1 m ( m k − 1 ) B k ( z ) k 0 ≤ z < 1 (2.42)
i.e., under matrix form
( z z 2 z 3 z 4 ⋮ ) = ( 2 − 1 3 − 1 4 − 1 5 − 1 ⋮ ) + ( 1 ⋯ 1 2 ⋯ 1 3 3 ⋯ 1 4 6 4 ⋯ ⋮ ⋮ ⋮ ⋮ ⋱ ) ( B 1 ( z ) / 1 B 2 ( z ) / 2 B 3 ( z ) / 3 B 4 ( z ) / 4 ⋮ ) (2.43)
which may be resolved for B m ( z ) and B m ( 0 ) my matrix calculus.
Integrating two times as followed the Hurwitz formula on Fourier series of Bernoulli polynomials we get
∫ 0 x B m ( z ) d z = − m ! ∑ n ∈ Z n ≠ 0 ∞ ( 1 2 i π n ) m + 1 ( e i 2 π n z − 1 ) = 1 m + 1 B m + 1 ( z ) + m ! ∑ n ∈ Z n ≠ 0 ∞ ( 1 2 i π n ) m + 1
∫ 0 1 d z ∫ 0 z B m ( x ) d x = m ! ∑ n ∈ Z n ≠ 0 ∞ ( 1 2 i π n ) m + 1
B m + 1 ( z ) = ( m + 1 ) ∫ 0 z B m ( x ) d x − ( m + 1 ) ∫ 0 1 d z ∫ 0 z B m ( x ) d x (2.44)
i.e.,
B m + 1 ( z ) is equal to ( m + 1 ) times the primitive of B m ( z ) minus the double primitive of B m ( z ) calculated for z = 1 . The second term is so equal to B m ( 0 ) = ( − 1 ) m B m ( 1 ) . (2.45)
This new algorithm for obtaining B m + 1 ( z ) from B m ( z ) and B m ( 0 ) is very easy to perform and may be utilized to establish
For examples:
B 0 ( x ) = 1
B 1 ( x ) = x − x 2 2 | x = 1 = x − 1 2
B 2 ( x ) = 2 ( x 2 2 − x 2 ) − 2 ( x 3 6 − x 2 4 ) | x = 1 = x 2 − x + 1 6
B 3 ( x ) = 3 ( x 3 3 − x 2 2 + x 6 ) − 3 ( 1 12 − 1 6 + 1 12 ) = x 3 − 3 2 x 2 + x 2
B 4 ( x ) = 4 ( x 4 4 − x 3 2 + x 2 4 ) − 4 ( 1 20 − 1 8 + 1 12 ) = x 4 − 2 x 3 + x 2 − 1 30
B 5 ( x ) = 5 ( x 5 5 − x 4 2 + x 3 3 − x 30 ) = x 5 − 5 2 x 4 + 5 3 x 3 − 1 6 x
B 6 ( x ) = 6 ( x 6 6 − x 5 2 + 5 x 4 12 − x 2 12 ) + 6 ( 1 6.7 − 1 2.4 + 5 12.4 − 1 12.3 ) = x 6 − 3 x 5 + 5 x 4 2 − x 2 2 + 1 42
B 7 ( x ) = 7 ( x 7 7 − x 6 2 + x 5 2 − x 3 6 + x 42 ) + 0 = x 7 − 7 2 x 6 + 7 x 5 2 − 7 x 3 6 + x 6
B 8 ( x ) = 8 ( x 8 8 − x 7 2 + 7 x 6 12 − 7 x 4 24 + x 2 12 ) − 8 ( 1 8.9 − 1 2.8 + 7 8.6 − 7 24.5 + 1 12.3 ) = x 8 − 4 x 7 + 14 3 x 6 − 7 3 x 4 + 2 3 x 2 − 1 30
B 9 ( x ) = 9 ( x 9 9 − x 8 2 + 2 x 7 3 − 7 x 5 15 + 2 x 3 9 − x 30 ) + 0 = x 9 − 9 x 8 2 + 6 x 7 − 21 x 5 5 + 2 x 3 − 3 x 10
This method for establishing a table of Bernoulli polynomials is extremely easier if we utilize the list of fifty Bernoulli numbers B m ( 0 ) conscientiously established by Coen [
B 10 ( x ) = x 10 − 5 x 9 + 15 2 x 8 − 7 x 6 + 5 x 4 − 3 2 x 2 + 5 66
B 11 ( x ) = x 11 − 11 2 x 10 + 55 6 x 9 − 11 x 7 + 11 x 5 − 11 2 x 3 + 5 6 x
B 12 ( x ) = x 12 − 6 x 11 + 11 x 10 − 33 2 x 8 + 22 x 6 − 33 2 x 4 + 5 x 2 − 691 2730 (2.46)
From the Hurwitz formula
1 k ! B k ( z ) = − 1 ( 2 i π ) k ∑ n ∈ Z , n ≠ 0 ∞ 1 n k e i 2 π n z 0 ≤ z ≤ 1
we get the Euler zeta function one may find references in Coen [
ζ ( 2 m ) = ∑ k = 1 ∞ 1 k 2 m = ( − 1 ) m + 1 1 ( 2 m ) ! 2 ( 2 π ) 2 m B 2 m ( 0 ) (2.47)
as so as
( 2 π ) 2 m = ( − 1 ) m + 1 ( 2 m ) ! 2 B 2 m ( z ) ∑ k = 1 ∞ 1 k 2 m cos 2 π k z (2.48)
( 2 π ) 2 m + 1 = ( − 1 ) m + 1 ( 2 m + 1 ) ! 2 B 2 m + 1 ( z ) ∑ k = 1 ∞ 1 k 2 m + 1 sin 2 π k z (2.49)
Moreover, by taking z = 0 , 1 8 , 1 6 , 1 4 , 1 3 , 1 2 , 2 3 , 1 in these formulae we get the known property
B 2 m + 1 ( 0 ) = B 2 m + 1 ( 1 ) = − ( 2 m + 1 ) ! ∑ k ∈ Z k ≠ 0 ∞ ( 1 2 i π k ) 2 m + 1 = 0 for m > 0 (2.50)
and the powers of pi.
For examples
( 2 π ) 2 m = ( − 1 ) m + 1 ( 2 m ) ! 2 B 2 m ( 1 / 2 ) ∑ k = 1 ∞ ( − 1 ) k k 2 m (2.51)
π 2 m = ( − 1 ) m + 1 ( 2 m ) ! 2 2 m − 1 ( 1 − 2 2 m − 1 ) B 2 m ∑ k = 1 ∞ ( − 1 ) k 1 k 2 m (2.52)
( 2 π ) 2 m = ( − 1 ) m + 1 ( 2 m ) ! 2 B 2 m ( 1 / 6 ) ∑ k = 1 ∞ cos ( k π 3 ) 1 k 2 m (2.53)
( 2 π ) 2 m + 1 = ( − 1 ) m + 1 ( 2 m + 1 ) ! 2 B 2 m + 1 ( 1 / 4 ) ∑ k = 1 ∞ sin ( k π 2 ) 1 k 2 m + 1 (2.54)
and
π 4 = 1 1 − 1 3 + 1 5 − 1 7 + ⋯
( π 4 ) 2 = 1 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯
π 2 = 36 ( 1 2 1 1 2 − 1 2 1 2 2 − 1 3 2 − 1 2 1 4 2 + 1 2 1 5 2 + 1 6 2 ) + ( 1 2 1 7 2 − 1 2 1 8 2 ) + ⋯
π 3 = 32 ( 1 − 1 3 3 + 1 5 3 − 1 7 3 + ⋯ )
etc.
In 1713, according to Jacob Bernoulli (1655-1705), was published the list of ten first sums of powers of entire numbers [
∑ n m = 1 m + 2 m + ⋯ + n m (3.1)
in terms of the numbers B k which are conjectured to be the same for all m
∑ n m = 1 m + 1 ∑ k = 0 m ( − 1 ) k ( m + 1 k ) B k n m + 1 − k . (3.2)
Afterward, the B k were baptized Bernoulli numbers.
By comparison of the relation coming from (3.2)
∂ n ∑ n m = ∑ k = 0 m ( − 1 ) k m ! ( m − k ) ! k ! B k n m − k = : ( B − n ) m = B 0 n m − m B 1 n m − 1 + m ( m − 1 ) 2 B 2 n m − 2 + ⋯ + ( − 1 ) m B m (3.3)
with the formula coming from (2.16), (2.17)
∂ n ( 1 m + ⋯ + n m − 1 ) + ∂ n n m = B m ( n ) + ∂ n n m = : ( B ( 0 ) + n ) m + m n m − 1 = B 0 ( 0 ) n m + m ( 1 + B 1 ( 0 ) ) n m − 1 + ( m 2 ) B 2 ( 0 ) n m − 2 + ⋯ + B m ( 0 ) (3.4)
we get, combining with (2.20),
B 0 = B 0 ( 0 )
B 1 = − B 1 ( 0 ) − 1 = − 1 2
B 2 m = B 2 m ( 0 )
B 2 m + 1 = − B 2 m + 1 ( 0 ) = B 2 m + 1 ( 1 ) = 1 2 δ m 0
B m = B m ( 0 ) (3.5)
i.e.
“The Bernoulli numbers B m are equal to the values at origin of the Bernoulli polynomial B m ( z ) ”.
The above formula (3.5) and the recurrence formula for Bernoulli polynomials (2.43) corresponding to z = 0
0 m = 1 m + 1 + ∑ k = 1 m ( m k − 1 ) B k ( 0 ) k (3.6)
lead to that for Bernoulli numbers
1 m + 1 B 0 + ( m 0 ) B 1 1 + ( m 1 ) B 2 2 + ( m 2 ) B 3 3 + ⋯ + ( m m − 1 ) B m m = 0 , m > 0 (3.7)
which, knowing B 0 = B m ( 0 ) = 1 , gives B 1 , B 2 , B 4 , ⋯ , B m according to following
This matrix equation may be resolved by doing linear combinations over lines from the second one in order to replace them with lines containing only some non-zero numbers.