Euler polynomials

Polynomials of the form

$$ E _ {n} ( x) = \sum _ { k=0}^ { n } \left ( \begin{array}{c} n \\ k \end{array} \right ) \frac{E _ k}{2 ^ {k}} \left ( x - \frac{1}{2} \right ) ^ {n-k} , $$

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

$$ E _ {n} ( x) + \sum _ { s=0} ^ { n } \left ( \begin{array}{c} n \\ s \end{array} \right ) E _ {s} ( x) = 2 x ^ {n} . $$

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

$$ \frac{d}{dx} E _ {n} ( x) = n E _ {n-} 1 ( x) . $$

The generating function of the Euler polynomials is

$$ \frac{2 e ^ {xt} }{e ^ {t} + 1 } = \ \sum _ { n=0}^ \infty \frac{E _ {n} ( x) }{n!} t ^ {n} . $$

The Euler polynomials admit the Fourier expansion

$$ \tag{* } E _ {n} ( x) = \frac{n!} {\pi ^ {n+ 1 }} \sum _ { k=0} ^ \infty \frac{\cos [ ( 2 k + 1 ) \pi x + ( n+ 1) \pi / 2 ] }{( 2 k + 1 ) ^ {n+1} } , $$

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

They satisfy the relations

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

$$ E _ {n} ( mx) = m ^ {n} \sum _ { k=0} ^ { m-1} ( - 1 ) ^ {k} E _ {n} \left ( x + \frac{k}{m} \right ) $$

if $ m $ is odd,

$$ E _ {n} ( mx) = - \frac{2 m ^ {n} }{n+1} \sum _ { k=0} ^ { m-1} ( - 1 ) ^ {k} B _ {n+1} \left ( x + \frac{k}{m} \right ) $$

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.

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Comments[edit]

The Euler polynomials satisfy in addition the identities

$$ E _ {n} ( x+ h) = $$

$$ = \ E _ {n} ( x) + \left ( \begin{array}{c} n \\ 1 \end{array} \right ) h E _ {n-1} ( x) + \dots + \left ( \begin{array}{c} n \\ n- 1 \end{array} \right ) h ^ {n-1} E _ {1} ( x) + E _ {0} ( x), $$

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) $,

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