{"paper":{"title":"Identities involving Bernoulli and Euler polynomials","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CA","authors_text":"Horst Alzer, Semyon Yakubovich","submitted_at":"2017-10-19T13:18:52Z","abstract_excerpt":"We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that $$ \\sum_{k=0}^{[n/4]}(-1)^k {n\\choose 4k}\\frac{B_{n-4k}(z) }{2^{6k}} =\\frac{1}{2^{n+1}}\\sum_{k=0}^{n} (-1)^k \\frac{1+i^k}{(1+i)^k} {n\\choose k}{B_{n-k}(2z)} $$ and $$ \\sum_{k=1}^{n} 2^{2k-1} {2n\\choose 2k-1} B_{2k-1}(z) =\n\\sum_{k=1}^n k \\, 2^{2k} {2n\\choose 2k} E_{2k-1}(z). $$ Applications of our results lead to formulas for Bernoulli and Euler numbers, like, for instance, $$ n E_{n-1} =\\sum_{k=1}^{[n/2]} \\frac{2^{2k}-1}{k} (2^{2k}-2^n){n\\choose 2k-1} B_{2k}B_{n-2k}. $$"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1710.07127","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}