English

Identities for combinatorial sums involving trigonometric functions

Classical Analysis and ODEs 2023-01-02 v1

Abstract

Let Am,n(a)=j=0m(4)j(m+j2j)k=0n1sin(a+2kπ/n)cos2j(a+2kπ/n) A_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j\choose 2j}\sum_{k=0}^{n-1} \sin(a+2k\pi/n) \cos^{2j}(a+2k\pi/n) and Bm,n(a)=j=0m(4)j(m+j+12j+1)k=0n1sin(a+2kπ/n)cos2j+1(a+2kπ/n), B_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j+1\choose 2j+1}\sum_{k=0}^{n-1} \sin(a+2k\pi/n) \cos^{2j+1}(a+2k\pi/n), where m0m\geq 0 and n1n\geq 1 are integers and aa is a real number. We present two proofs for the following results: (i) If 2m+10(\mboxmodn)2m+1 \equiv 0 \, (\mbox{mod} \, n), then Am,n(a)=(1)mnsin((2m+1)a). A_{m,n}(a)=(-1)^m n \sin((2m+1)a). (ii) If 2m+1≢0(\mboxmodn)2m+1 \not\equiv 0 \, (\mbox{mod} \, n), then Am,n(a)=0A_{m,n}(a)=0. (iii) If 2(m+1)0(\mboxmodn)2(m+1) \equiv 0 \, (\mbox{mod} \, n), then Bm,n(a)=(1)mn2sin(2(m+1)a). B_{m,n}(a)=(-1)^m \frac{n}{2} \sin(2(m+1)a). (iv) If 2(m+1)≢0(\mboxmodn)2(m+1) \not\equiv 0 \, (\mbox{mod} \, n), then Bm,n(a)=0B_{m,n}(a)=0.

Keywords

Cite

@article{arxiv.2212.14841,
  title  = {Identities for combinatorial sums involving trigonometric functions},
  author = {Horst Alzer and Semyon Yakubovich},
  journal= {arXiv preprint arXiv:2212.14841},
  year   = {2023}
}
R2 v1 2026-06-28T07:57:32.817Z