English

Identities involving Bernoulli and Euler polynomials

Classical Analysis and ODEs 2017-10-20 v1

Abstract

We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that k=0[n/4](1)k(n4k)Bn4k(z)26k=12n+1k=0n(1)k1+ik(1+i)k(nk)Bnk(2z) \sum_{k=0}^{[n/4]}(-1)^k {n\choose 4k}\frac{B_{n-4k}(z) }{2^{6k}} =\frac{1}{2^{n+1}}\sum_{k=0}^{n} (-1)^k \frac{1+i^k}{(1+i)^k} {n\choose k}{B_{n-k}(2z)} and k=1n22k1(2n2k1)B2k1(z)=k=1nk22k(2n2k)E2k1(z). \sum_{k=1}^{n} 2^{2k-1} {2n\choose 2k-1} B_{2k-1}(z) = \sum_{k=1}^n k \, 2^{2k} {2n\choose 2k} E_{2k-1}(z). Applications of our results lead to formulas for Bernoulli and Euler numbers, like, for instance, nEn1=k=1[n/2]22k1k(22k2n)(n2k1)B2kBn2k. n E_{n-1} =\sum_{k=1}^{[n/2]} \frac{2^{2k}-1}{k} (2^{2k}-2^n){n\choose 2k-1} B_{2k}B_{n-2k}.

Keywords

Cite

@article{arxiv.1710.07127,
  title  = {Identities involving Bernoulli and Euler polynomials},
  author = {Horst Alzer and Semyon Yakubovich},
  journal= {arXiv preprint arXiv:1710.07127},
  year   = {2017}
}
R2 v1 2026-06-22T22:19:19.328Z