# Bernoulli polynomials

In

mathematics , the**Bernoulli polynomials**occur in the study of manyspecial functions and in particular theRiemann zeta function and theHurwitz zeta function . This is in large part because they are anAppell sequence , i.e. aSheffer sequence for the ordinaryderivative operator. Unlikeorthogonal polynomials , the Bernoulli polynomials are remarkable in that the number of crossings of the "x"-axis in theunit 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.**Representations**The Bernoulli polynomials "B"

_{"n"}admit a variety of different representations. Which among them should be taken to be the definition may depend on one's purposes.**Explicit formula**:$B\_n(x)\; =\; sum\_\{k=0\}^n\; \{n\; choose\; k\}\; b\_k\; x^\{n-k\},$

for "n" ≥ 0, where "b"

_{"k"}are theBernoulli number s.**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 :$frac\{2\; e^\{xt\{e^t+1\}=\; sum\_\{n=0\}^infty\; E\_n(x)\; frac\{t^n\}\{n!\}.$

**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 .**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

:$egin\{align\}(Tf)(x)\; =\; \{e^D\; -\; 1\; over\; D\}f(x)\; \{\}\; =\; sum\_\{n=0\}^infty\; \{D^n\; over\; (n+1)!\}f(x)\; \backslash \; \{\}\; =\; f(x)\; +\; \{f\text{\'}(x)\; over\; 2\}\; +\; \{f"(x)\; over\; 6\}\; +\; \{f"\text{\'}(x)\; over\; 24\}\; +\; cdots.end\{align\}$

**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:$B\_n(x)\; =\; -n\; zeta(1-n,x)$

where $zeta(s,q)$ is the Hurwitz zeta; thus, 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 "n"th

forward difference of $x^m$; 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 "m"th-degree polynomial such as "x"

^{"m"}, 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 number s $E\_k$ as:$E\_m(x)=\; sum\_\{k=0\}^m\; \{m\; choose\; k\}left(frac\{1\}\{2\}\; ight)^k\; left(x-frac\{1\}\{2\}\; ight)^\{m-k\}\; E\_k,.$**ums of "p"th 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.**The Bernoulli and Euler numbers**The

Bernoulli number s are given by $B\_n=B\_n(0).$The

Euler number s are given by $E\_n=2^nE\_n(1/2).$**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:$E\_0(x)=1,$:$E\_1(x)=x-1/2,$:$E\_2(x)=x^2-x,$:$E\_3(x)=x^3-frac\{3\}\{2\}x^2+frac\{1\}\{4\},$:$E\_4(x)=x^4-2x^3+x,$:$E\_5(x)=x^5-frac\{5\}\{2\}x^4+frac\{5\}\{2\}x^2-frac\{1\}\{2\},$:$E\_6(x)=x^6-3x^5+5x^3-3x.,$

**Differences and derivatives**The Bernoulli and Euler polynomials obey many relations from

umbral calculus ::$Delta\; B\_n(x)\; =\; B\_n(x+1)-B\_n(x)=nx^\{n-1\},$

:$Delta\; E\_n(x)\; =\; E\_n(x+1)-E\_n(x)=2x^n.$

(Δ is the

forward difference operator ).These

polynomial sequence s areAppell sequence s::$B\_n\text{'}(x)=nB\_\{n-1\}(x),,$

:$E\_n\text{'}(x)=nE\_\{n-1\}(x).,$

**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 sequence s. (Hermite polynomials are another example.)**ymmetries**:$B\_n(1-x)=(-1)^nB\_n(x),quad\; n\; ge\; 0,$

:$E\_n(1-x)=(-1)^n\; E\_n(x),$

:$(-1)^n\; B\_n(-x)\; =\; B\_n(x)\; +\; nx^\{n-1\},$

:$(-1)^n\; E\_n(-x)\; =\; -E\_n(x)\; +\; 2x^n,$

Zhi-Wei Sun and Hao Pan [*http://arxiv.org/abs/math/0409035*] 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

:$[s,t;x,y]\; \_n=sum\_\{k=0\}^n(-1)^k\{s\; choose\; k\}\{tchoose\; \{n-kB\_\{n-k\}(x)B\_k(y).$

**Fourier series**The

Fourier series of the Bernoulli polynomials is also aDirichlet series , given by the expansion:$B\_n(x)\; =\; -frac\{n!\}\{(2pi\; i)^n\}sum\_\{k\; ot=0\; \}frac\{e^\{2pi\; i\; x\{k^n\}.$

This is a special case of the analogous form for the

Hurwitz zeta function :$B\_n(x)\; =\; -Gamma(n+1)\; sum\_\{k=1\}^infty\; frac\{\; exp\; (2pi\; ikx)\; +\; e^\{ipi\; n\}\; exp\; (2pi\; ik(1-x))\; \}\; \{\; (2pi\; ik)^n\; \}.$

This expansion is valid only for 0 ≤ "x" ≤ 1 when "n" ≥ 2 and is valid for 0 < "x" < 1 when "n" = 1.

The Fourier series of the Euler polynomials may also be calculated. Defining the functions

:$C\_\; u(x)\; =\; sum\_\{k=0\}^infty\; frac\; \{cos((2k+1)pi\; x)\}\; \{(2k+1)^\; u\}$

and

:$S\_\; u(x)\; =\; sum\_\{k=0\}^infty\; frac\; \{sin((2k+1)pi\; x)\}\; \{(2k+1)^\; u\}$

for $u\; >\; 1$, 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 $C\_\; u$ and $S\_\; u$ are odd and even, respectively:

:$C\_\; u(x)\; =\; -C\_\; u(1-x)$

and

:$S\_\; u(x)\; =\; S\_\; u(1-x).$

They are related to the

Legendre chi function $chi\_\; u$ as:$C\_\; u(x)\; =\; mbox\{Re\}\; chi\_\; u\; (e^\{ix\})$

and

:$S\_\; u(x)\; =\; mbox\{Im\}\; chi\_\; u\; (e^\{ix\}).$

**Inversion**The Bernoulli polynomials may be inverted to express the

monomial in terms of the polynomials. Specifically, one has:$x^n\; =\; frac\; \{1\}\{n+1\}\; sum\_\{k=0\}^n\; \{n+1\; choose\; k\}\; B\_k\; (x).$

**Relation to falling factorial**The Bernoulli polynomials may be expanded in terms of the

falling factorial $(x)\_k$ as:$B\_\{n+1\}(x)\; =\; B\_\{n+1\}\; +\; sum\_\{k=0\}^nfrac\{n+1\}\{k+1\}left\{\; egin\{matrix\}\; n\; \backslash \; k\; end\{matrix\}\; ight\}(x)\_\{k+1\}$where $B\_n=B\_n(0)$ and

:$left\{\; egin\{matrix\}\; n\; \backslash \; k\; end\{matrix\}\; ight\}\; =\; S(n,k)$

denotes the

Stirling number of the second kind . The above may be inverted to express the falling factorial in terms of the Bernoulli polynomials::$(x)\_\{n+1\}\; =\; sum\_\{k=0\}^n\; frac\{n+1\}\{k+1\}left\; [\; egin\{matrix\}\; n\; \backslash \; k\; end\{matrix\}\; ight]\; left(B\_\{k+1\}(x)\; -\; B\_\{k+1\}\; ight)$

where :$left\; [\; egin\{matrix\}\; n\; \backslash \; k\; end\{matrix\}\; ight]\; =\; s(n,k)$

denotes the

Stirling number of the first kind .**Multiplication theorems**The

multiplication theorem s were given byJoseph Ludwig Raabe in1851 ::$B\_n(mx)=\; m^\{n-1\}\; sum\_\{k=0\}^\{m-1\}\; B\_n\; left(x+frac\{k\}\{m\}\; ight)$

:$E\_n(mx)=\; m^n\; sum\_\{k=0\}^\{m-1\}\; (-1)^k\; E\_n\; left(x+frac\{k\}\{m\}\; ight)quad\; mbox\{\; for\; \}\; m=1,3,dots$

:$E\_n(mx)=\; frac\{-2\}\{n+1\}\; m^n\; sum\_\{k=0\}^\{m-1\}\; (-1)^k\; B\_\{n+1\}\; left(x+frac\{k\}\{m\}\; ight)quad\; mbox\{\; for\; \}\; m=2,4,dots$

**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:$int\_0^1\; B\_n(t)\; B\_m(t),dt\; =\; (-1)^\{n-1\}\; frac\{m!\; n!\}\{(m+n)!\}\; B\_\{n+m\}quad\; mbox\; \{\; for\; \}\; m,n\; ge\; 1$

:$int\_0^1\; E\_n(t)\; E\_m(t),dt\; =\; (-1)^\{n\}\; 4\; (2^\{m+n+2\}-1)frac\{m!\; n!\}\{(m+n+2)!\}\; B\_\{n+m+2\}$

**Periodic Bernoulli polynomials**A

**periodic Bernoulli polynomial**"P"_{"n"}("x") is a Bernoulli polynomial evalated at thefractional part of the argument "x". These functions are used to provide theremainder term in theEuler–Maclaurin formula relating sums to integrals. The first polynomial is a sawtooth function.**References*** Milton Abramowitz and Irene A. Stegun, eds. "

Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables", (1972) Dover, New York. "(See [*http://www.math.sfu.ca/~cbm/aands/page_804.htm Chapter 23*] )"* See chapter 12.11 of Apostol IANT

* Djurdje Cvijović and Jacek Klinowski, "New formulae for the Bernoulli and Euler polynomials at rational arguments", Proceedings of the American Mathematical Society123"' (1995), 1527-1535.

* Jesus Guillera and Jonathan Sondow, " [

*http://arxiv.org/abs/math.NT/0506319 Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent*] " (2005) "(Reviews relationship to the Hurwitz zeta function and Lerch transcendent.)"* [

*http://upload.wikimedia.org/wikipedia/en/9/9f/Bernoulli_Polynomials_and_Their_Applications.pdf Kurtulan, Ali Burak "Bernoulli Polynomials and Their Applications"*]*

*Wikimedia Foundation.
2010.*

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