Bernoulli polynomials

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In mathematics, the Bernoulli polynomials occur in the study of many special functions and in particular the Riemann zeta function and the Hurwitz zeta function. This is in large part due to the fact that they are an Appell sequence, i.e. a Sheffer sequence for the ordinary derivative operator. Unlike orthogonal polynomials, the Bernoulli polynomials are remarkable in that the number of crossings of the x-axis in the unit interval does not go up as the degree of the polynomials goes up. In the limit of large degree, the Bernoulli polynomials, appropriately scaled, approach the sine and cosine functions.

Bernoulli polynomials
Bernoulli polynomials

Contents

[edit] Representations

The Bernoulli polynomials Bn admit a variety of different representations. Which among them should be taken to be the definition may depend on one's purposes.

[edit] Explicit formula

B_n(x) = \sum_{k=0}^n {n \choose k} b_k x^{n-k},

for n ≥ 0, where bk are the Bernoulli numbers.

[edit] Generating functions

The generating function for the Bernoulli polynomials is

\frac{t e^{xt}}{e^t-1}= \sum_{n=0}^\infty B_n(x) \frac{t^n}{n!}.

The generating function for the Euler polynomials is

\frac{2 e^{xt}}{e^t+1}= \sum_{n=0}^\infty E_n(x) \frac{t^n}{n!}.

[edit] Representation by a differential operator

The Bernoulli polynomials are also given by

B_n(x)={D \over e^D -1} x^n

where D = d/dx is differentiation with respect to x and the fraction is expanded as a formal power series.

[edit] Representation by an integral operator

The Bernoulli polynomials are the unique polynomials determined by

\int_x^{x+1} B_n(u)\,du = x^n.

The integral operator

(Tf)(x) = \int_x^{x+1} f(u)\,du

on polynomials f, is the same as

(Tf)(x) = {e^D - 1 \over D}f(x) = \sum_{n=0}^\infty {D^n \over (n+1)!}f(x) = f(x) + {f'(x) \over 2} + {f''(x) \over 6} + {f'''(x) \over 24} + \cdots.

[edit] Another explicit formula

An explicit formula for the Bernoulli polynomials is given by

B_m(x)= \sum_{n=0}^m \frac{1}{n+1} \sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m.

Note the remarkable similarity to the globally convergent series expression for the Hurwitz zeta function. Indeed, one has

Bn(x) = − nζ(1 − n,x)

where ζ(s,q) is the Hurwitz zeta; this, in a certain sense, the Hurwitz zeta generalizes the Bernoulli polynomials to non-integer values of n.

The inner sum may be understood to be the nth forward difference of xm; that is,

\Delta^n x^m = \sum_{k=0}^n (-1)^{n-k} {n \choose k} (x+k)^m

where Δ is the forward difference operator. Thus, one may write

B_m(x)= \sum_{n=0}^m \frac{(-1)^n}{n+1} \Delta^n x^m.

This formula may be derived from an identity appearing above as follows: since the forward difference operator Δ is equal to

\Delta = e^D - 1\,

where D is differentiation with respect to x, we have

{D \over e^D - 1} = {\log(\Delta + 1) \over \Delta} = \sum_{n=0}^\infty {(-\Delta)^n \over n+1}.

As long as this operates on an mth-degree polynomial such as xm, one may let n go from 0 only up to m.

An integral representation for the Bernoulli polynomials is given by the Nörlund-Rice integral, which follows from the expression as a finite difference.

An explicit formula for the Euler polynomials is given by

E_m(x)= \sum_{n=0}^m \frac{1}{2^n} \sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m.

This may also be written in terms of the Euler numbers Ek as

E_m(x)= \sum_{k=0}^m {m \choose k} \left(\frac{1}{2}\right)^k \left(x-\frac{1}{2}\right)^{m-k} E_k\,.

[edit] Sums of pth powers

We have

\sum_{k=0}^{x} k^p = \frac{B_{p+1}(x+1)-B_{p+1}(0)}{p+1}.

See Faulhaber's formula for more on this.

[edit] The Bernoulli and Euler numbers

The Bernoulli numbers are given by Bn = Bn(0).

The Euler numbers are given by En = 2nEn(1 / 2).

[edit] Explicit expressions for low degrees

The first few Bernoulli polynomials are:

B_0(x)=1\,
B_1(x)=x-1/2\,
B_2(x)=x^2-x+1/6\,
B_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{2}x\,
B_4(x)=x^4-2x^3+x^2-\frac{1}{30}\,
B_5(x)=x^5-\frac{5}{2}x^4+\frac{5}{3}x^3-\frac{1}{6}x\,
B_6(x)=x^6-3x^5+\frac{5}{2}x^4-\frac{1}{2}x^2+\frac{1}{42}.\,

The first few Euler polynomials are

E_0(x)=1\,
E_1(x)=x-1/2\,
E_2(x)=x^2-x\,
E_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{4}\,
E_4(x)=x^4-2x^3+x\,
E_5(x)=x^5-\frac{5}{2}x^4+\frac{5}{2}x^2-\frac{1}{2}\,
E_6(x)=x^6-3x^5+5x^3-3x.\,

[edit] Differences and derivatives

The Bernoulli and Euler polynomials obey many relations from umbral calculus:

ΔBn(x) = Bn(x + 1) − Bn(x) = nxn − 1,
ΔEn(x) = En(x + 1) + En(x) = 2xn.

(Δ is the forward difference operator).

These polynomial sequences are Appell sequences:

B_n'(x)=nB_{n-1}(x),\,
E_n'(x)=nE_{n-1}(x).\,

[edit] Translations

B_n(x+y)=\sum_{k=0}^n {n \choose k} B_k(x) y^{n-k}
E_n(x+y)=\sum_{k=0}^n {n \choose k} E_k(x) y^{n-k}

These identities are also equivalent to saying that these polynomial sequences are Appell sequences. (Hermite polynomials are another example.)

[edit] Symmetries

B_n(1-x)=(-1)^n B_n(x)\,
E_n(1-x)=(-1)^n E_n(x)\,
(-1)^n B_n(-x) = B_n(x) + nx^{n-1}\,
(-1)^n E_n(-x) = -E_n(x) + 2x^n\,

Zhi-Wei Sun and Hao Pan [1] established the following surprising symmetric relation: If r + s + t = n and x + y + z = 1, then

r[s,t;x,y]n + s[t,r;y,z]n + t[r,s;z,x]n = 0,

where

[s,t;x,y]_n=\sum_{k=0}^n(-1)^k{s \choose k}{t\choose {n-k}} B_{n-k}(x)B_k(y).

[edit] Fourier series

The Fourier series of the Bernoulli polynomials is also a Dirichlet series and is a special case of the Hurwitz zeta function

B_n(x) = -\Gamma(n+1) \sum_{k=1}^\infty \frac{ \exp (2\pi ikx) + (-1)^n \exp (2\pi ik(1-x)) } { (2\pi ik)^n }.

This expansion is valid only for 0\leq x\leq 1 when n \geq 2 and is valid for 0 < x < 1 when n = 1.

Defining the functions

C_\nu(x) = \sum_{k=0}^\infty \frac {\cos((2k+1)\pi x)} {(2k+1)^\nu}

and

S_\nu(x) = \sum_{k=0}^\infty \frac {\sin((2k+1)\pi x)} {(2k+1)^\nu}

for ν > 1, the Euler polynomial has the Fourier series

C_{2n}(x) = \frac{(-1)^n}{4(2n-1)!} \pi^{2n} E_{2n-1} (x)

and

S_{2n+1}(x) = \frac{(-1)^n}{4(2n)!} \pi^{2n+1} E_{2n} (x)

Note that the Cν and Sν are odd and even, respectively:

Cν(x) = − Cν(1 − x)

and

Sν(x) = Sν(1 − x)

They are related to the Legendre chi function χν as

Cν(x) = Reχν(eix)

and

Sν(x) = Imχν(eix)

[edit] Inversion

The Bernoulli polynomials may be inverted to express the monomial in terms of the polynomials. Specifically, one has

x^n = \frac {1}{n+1} \sum_{k=0}^n {n+1 \choose k} B_k (x).

[edit] Relation to falling factorial

The Bernoulli polynomials may be expanded in terms of the falling factorial (x)k as

B_{n+1}(x) = B_{n+1} + \sum_{k=0}^n \frac{n+1}{k+1} \left\{ \begin{matrix} n \\ k \end{matrix} \right\} (x)_{k+1}

where Bn = Bn(0) and

\left\{ \begin{matrix} n \\ k \end{matrix} \right\} = S(n,k)

denotes the Stirling number of the second kind. The above may be inverted to express the falling factorial in terms of the Bernoulli polynomials:

(x)_{n+1} = \sum_{k=0}^n \frac{n+1}{k+1} \left[ \begin{matrix} n \\ k \end{matrix} \right] \left(B_{k+1}(x) - B_{k+1} \right)

where

\left[ \begin{matrix} n \\ k \end{matrix} \right] = s(n,k)

denotes the Stirling number of the first kind.

[edit] Multiplication theorems

The multiplication theorems were given by Joeseph Ludwig Raabe in 1851:

B_n(mx)= m^{n-1} \sum_{k=0}^{m-1} B_n \left(x+\frac{k}{m}\right)
E_n(mx)= m^n \sum_{k=0}^{m-1} (-1)^k E_n \left(x+\frac{k}{m}\right) \quad \mbox{ for } m=1,3,\dots
E_n(mx)= \frac{-2}{n+1} m^n \sum_{k=0}^{m-1} (-1)^k B_{n+1} \left(x+\frac{k}{m}\right) \quad \mbox{ for } m=2,4,\dots

[edit] Integrals

Indefinite integrals

\int_a^x B_n(t)\,dt = \frac{B_{n+1}(x)-B_{n+1}(a)}{n+1}
\int_a^x E_n(t)\,dt = \frac{E_{n+1}(x)-E_{n+1}(a)}{n+1}

Definite integrals

\int_0^1 B_n(t) B_m(t)\,dt = (-1)^{n-1} \frac{m! n!}{(m+n)!} B_{n+m} \quad \mbox { for } m,n \ge 1
\int_0^1 E_n(t) E_m(t)\,dt = (-1)^{n} 4 (2^{m+n+2}-1)\frac{m! n!}{(m+n+2)!} B_{n+m+2}

[edit] References

  • Tom M. Apostol Introduction to Analytic Number Theory, (1976) Springer-Verlag, New York. (See Chapter 12.11)
  • Djurdje Cvijović and Jacek Klinowski, "New formulae for the Bernoulli and Euler polynomials at rational arguments", Proceedings of the American Mathematical Society'123 (1995), 1527-1535.
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