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Euler Polynomial


A Polynomial $E_n(x)$ given by the sum

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

Euler polynomials are related to the Bernoulli Numbers by
$\displaystyle E_{n-1}(x)$ $\textstyle =$ $\displaystyle {2^n\over n} \left[{B_n\left({x+1\over 2}\right)-B_n\left({x\over 2}\right)}\right]$ (2)
  $\textstyle =$ $\displaystyle {2\over n}\left[{B_n(x)-2^nB_n\left({x\over 2}\right)}\right]$ (3)
$\displaystyle E_{n-2}(x)$ $\textstyle =$ $\displaystyle 2{n\choose 2}^{-1} \sum_{k=0}^{n-2} {n\choose k}[(2^{n-k}-1)B_{n-k}B_k(x)],$  

where ${n\choose k}$ is a Binomial Coefficient. Setting $x=1/2$ and normalizing by $2^n$ gives the Euler Number
E_n=2^nE_n({\textstyle{1\over 2}}).
\end{displaymath} (5)

Call $E_n'=E_n(0)$, then the first few terms are $-1/2$, 0, 1/4, $-1/2$, 0, 17/8, 0, 31/2, 0, .... The terms are the same but with the Signs reversed if $x=1$. These values can be computed using the double sum
E_n(0)=2^{-n}\sum_{j=1}^n\left[{(-1)^{j+n+1}j^k\sum_{k=0}^{n-j}{n+1\choose k}}\right].
\end{displaymath} (6)

The Bernoulli Numbers $B_n$ for $n>1$ can be expressed in terms of the $E_n'$ by
B_n= -{nE'_{n-1}\over 2(2^n-1)}.
\end{displaymath} (7)

See also Bernoulli Polynomial, Euler Number, Genocchi Number


Abramowitz, M. and Stegun, C. A. (Eds.). ``Bernoulli and Euler Polynomials and the Euler-Maclaurin Formula.'' §23.1 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 804-806, 1972.

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, 1979.

Spanier, J. and Oldham, K. B. ``The Euler Polynomials $E_n(x)$.'' Ch. 20 in An Atlas of Functions. Washington, DC: Hemisphere, pp. 175-181, 1987.

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© 1996-9 Eric W. Weisstein