Identities involving Bernoulli and Euler polynomials

文献摘要

\sum_{k = 0}^{[n / 4]} (- 1)^{k} (\binom{n}{4 k}) \frac{B_{n - 4 k} (z)}{2^{6 k}} = \frac{1}{2^{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{1 + i^{k}}{(1 + i)^{k}} (\binom{n}{k}) B_{n - k} (2 z)

We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that $\sum_{k = 0}^{[n / 4]} (- 1)^{k} (\binom{n}{4 k}) \frac{B_{n - 4 k} (z)}{2^{6 k}} = \frac{1}{2^{n + 1}} \sum_{k = 0}^{n} (- 1)^{k} \frac{1 + i^{k}}{(1 + i)^{k}} (\binom{n}{k}) B_{n - k} (2 z)$ and $\sum_{k = 1}^{n} 2^{2 k - 1} (\binom{2 n}{2 k - 1}) B_{2 k - 1} (z) = \sum_{k = 1}^{n} k 2^{2 k} (\binom{2 n}{2 k}) E_{2 k - 1} (z) .$ Applications of our results lead to formulas for Bernoulli and Euler numbers, like, for instance, $n E_{n - 1} = \sum_{k = 1}^{[n / 2]} \frac{2^{2 k} - 1}{k} (2^{2 k} - 2^{n}) (\binom{n}{2 k - 1}) B_{2 k} B_{n - 2 k} .$

Identities involving Bernoulli and Euler polynomials

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Identities involving Bernoulli and Euler polynomials

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Identities involving Bernoulli and Euler polynomials

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讨论

|讲解博客

|AI智能校验

Identities involving Bernoulli and Euler polynomials

文献摘要

讨论

|讲解博客

|AI智能校验

讲解博客

AI智能校验

讲解博客

AI智能校验