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Nov 26, 2022 at 21:37 history edited Todd Trimble CC BY-SA 4.0
cot --> coth
Dec 1, 2015 at 12:07 comment added Fedor Petrov oh, I did not know this trick with polynomial
Dec 1, 2015 at 11:32 comment added Pietro Majer (no: I've just found Herglotz trick on line)
Dec 1, 2015 at 11:24 comment added Pietro Majer Just to complete the picture, the expansion of the cotangent can be obtained by logarithmic differentiation from the infinite product for $\sin(\pi x)$, and the infinite product for $\sin(\pi x)$ can be obtained as dominated limit on the (easily computed) complete factorization of the polynomials $(1+iz/n)^n-(1-iz/n)^n$, that converge to $e^{iz}-e^{-iz}=2i\sin(z)$ (Is this Herglotz' proof?)
Dec 1, 2015 at 7:48 history edited Fedor Petrov CC BY-SA 3.0
added 965 characters in body
Dec 1, 2015 at 5:59 comment added Fedor Petrov Indeed, but this works for all rational functions.
Dec 1, 2015 at 0:14 comment added Alexandre Eremenko Yes, but he asked to sum $f(n)$ with a more general class of functions $f$.
S Nov 30, 2015 at 22:52 history suggested CommunityBot CC BY-SA 3.0
Corrected two typos.
Nov 30, 2015 at 22:45 review Suggested edits
S Nov 30, 2015 at 22:52
Nov 30, 2015 at 21:58 history answered Fedor Petrov CC BY-SA 3.0