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This is super cool, I've never seen it before! Do you know what this is called so I can look up a proof/theorem on it?
by dvt 4y ago
This is super cool, I've never seen it before! Do you know what this is called so I can look up a proof/theorem on it?
- SonOfLilit 4y agoI'm not familiar with this result, but this comment is phrased in the language of Ordinary Differential Equations, so I'd look for a textbook on solving systems of ODEs and expect to find a technique that can prove that this is the unique solution (at least assuming differentiability of S(x) and C(x)).
- selimthegrim 4y agoYou can also consider the pair of functional equations as implicitly using a Taylor expansion
- chx 4y agoI asked https://math.stackexchange.com/q/124887/6400 https://math.stackexchange.com/q/124887/6400 I wish I had a more modern summary of the papers mentioned in the linked paper > Tannery, Fonctions d'une Variable, 1886, p. 147. Osgood, Lehrbueh der Funktionentheorie, 1912, p. 582. Van Vleck and H'Doubler, Transactions Amer. Math. Society, vol. 17 (1916), p. 30 because we spent an entire semester at the university in one class working on these two.
- dvt 4y ago> because we spent an entire semester at the university in one class working on these two. Yeah, the Math Overflow answers are a bit sparse, and I think the Euler formulation (while clever) is a bit of a red herring and might be circular. Trying to slowly go through that 1917 paper, thanks for linking!
- chx 4y agoAlso related: doi:10.1007/s00010-003-2700-z f(x)f(y)f(z) = f(x) + f(y) + f(z) sort of implies f(x) is tan(x) -- it's actually tan(kx + (1-k) pi/3).
- chx 4y agoSorry I left out the condition of x + y + z = π