Euler Polynomials

Polynomials of the form

$E_{n} (x) = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) \frac{E_{k}}{2^{k}} {(x - \frac{1}{2})}^{n - k},$

where $ E _ {k} $ are the Euler numbers. The Euler polynomials can be computed successively by means of the formula

$E_{n} (x) + \sum_{s = 0}^{n} (\begin{matrix} n \\ s \end{matrix}) E_{s} (x) = 2 x^{n} .$

In particular,

$E_{0} (x) = 1, E_{1} (x) = x - \frac{1}{2}, E_{2} (x) = x (x - 1) .$

The Euler polynomials satisfy the difference equation

$E_{n} (x + 1) + E_{n} (x) = 2 x^{n}$

and belong to the class of Appell polynomials, that is, they satisfy

$\frac{d}{d x} E_{n} (x) = n E_{n -} 1 (x) .$

The generating function of the Euler polynomials is

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

The Euler polynomials admit the Fourier expansion

$\begin{matrix} (*) & E_{n} (x) = \frac{n!}{π^{n + 1}} \sum_{k = 0}^{\infty} \frac{\cos [(2 k + 1) π x + (n + 1) π / 2]}{(2 k + 1)^{n + 1}}, \end{matrix}$

$0 \leq x \leq 1, n \geq 1 .$

They satisfy the relations

$E_{n} (1 - x) = (- 1)^{n} E_{n} (x),$

$E_{n} (m x) = m^{n} \sum_{k = 0}^{m - 1} (- 1)^{k} E_{n} (x + \frac{k}{m})$

if $ m $ is odd,

$E_{n} (m x) = - \frac{2 m^{n}}{n + 1} \sum_{k = 0}^{m - 1} (- 1)^{k} B_{n + 1} (x + \frac{k}{m})$

if $ m $ is even. Here $B_{n+1}$ is a Bernoulli polynomial (cf. Bernoulli polynomials). The periodic functions coinciding with the right-hand side of (*) are extremal in the Kolmogorov inequality and in a number of other extremal problems in function theory. Generalized Euler polynomials have also been considered.

References[edit]

[1]	L. Euler, "Opera omnia: series prima: opera mathematica: institutiones calculi differentialis" , Teubner (1980) (Translated from Latin)
[2]	N.E. Nörlund, "Volesungen über Differenzenrechnung" , Springer (1924)

Comments[edit]

The Euler polynomials satisfy in addition the identities

$E_{n} (x + h) =$

$= E_{n} (x) + (\begin{matrix} n \\ 1 \end{matrix}) h E_{n - 1} (x) + \dots + (\begin{matrix} n \\ n - 1 \end{matrix}) h^{n - 1} E_{1} (x) + E_{0} (x),$

written symbolically as

$E_{n} (x + h) = {E (x) + h}^{n} .$

Here the right-hand side should be read as follows: first expand the right-hand side into sums of expressions $ ( {} _ {i} ^ {n} ) \{ E ( x) \} ^ {i} h ^ {n-i} $ and then replace $ \{ E ( x) \} ^ {i} $ with $ E _ {i} ( x) $.

Using the same symbolic notation one has for every polynomial $ p( x) $,

$p (E (x) + 1) + p (E (x)) = 2 p (x) .$