Editing Bernoulli polynomials (section)

==Another explicit formula==

An explicit formula for the Bernoulli polynomials is given by
<math display="block">
B_n(x) = \sum_{k=0}^n \biggl[ \frac{1}{k + 1}
                       \sum_{\ell=0}^k (-1)^\ell { k \choose \ell } (x + \ell)^n \biggr].</math>

That is similar to the series expression for the [[Hurwitz zeta function]] in the complex plane. Indeed, there is the relationship
<math display="block">B_n(x) = -n \zeta(1 - n,\,x)</math>
where <math>\zeta(s,\,q)</math> is the [[Hurwitz zeta function]]. The latter generalizes the Bernoulli polynomials, allowing for non-integer values {{nobr|of {{mvar|n}}.}}

The inner sum may be understood to be the {{mvar|n}}th [[forward difference]] of <math>x^m,</math> that is,
<math display="block">\Delta^n x^m = \sum_{k=0}^n (-1)^{n - k}{n \choose k}(x + k)^m</math>
where <math>\Delta</math> is the [[forward difference operator]]. Thus, one may write
<math display="block">B_n(x) = \sum_{k=0}^n \frac{(-1)^k}{k + 1}\Delta^k x^n.</math>

This formula may be derived from an identity appearing above as follows. Since the forward difference operator {{math|Δ}} equals
<math display="block">\Delta = e^D - 1</math>
where {{mvar|D}} is differentiation with respect to {{mvar|x}}, we have, from the [[Mercator series]],
<math display="block">\frac{ D }{e^D - 1} = \frac{\log(\Delta + 1)}{\Delta} = \sum_{n=0}^\infty \frac{(-\Delta)^n }{n + 1}.</math>

As long as this operates on an {{mvar|m}}th-degree polynomial such as <math>x^m,</math> one may let {{mvar|n}} go from {{math|0}} only up {{nobr|to {{mvar|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
<math display="block">E_n(x) = \sum_{k=0}^n \left[ \frac{1}{2^k}\sum_{\ell=0}^n (-1)^\ell {k \choose \ell}(x + \ell)^n \right] .</math>

The above follows analogously, using the fact that
<math display="block">\frac{2}{e^D + 1} = \frac{1}{1 + \tfrac12 \Delta} = \sum_{n = 0}^\infty \bigl( {-\tfrac{1}{2}} \Delta \bigr)^n .</math>