Here's a copy which isn't behind a paywall: http://digitalcommons.plattsburgh.edu/cgi/viewcontent.cgi?article=1012&conte... On Mon, Jul 4, 2016 at 11:12 AM, Mike Stay <metaweta@gmail.com> wrote:
http://www.jstor.org/stable/2695354
On Sun, Jul 3, 2016 at 10:13 PM, Zak Seidov <math-fun@mailman.xmission.com> wrote:
My initial motivation of Q was: if one doesn't know that (trivial) soln is f(x)=tan(c*x) and EVEN doesn't know tan function at all, how s/he will solve this f.e.? =========== Ecclesiastes : "he that increaseth knowledge increaseth sorrow".
Понедельник, 4 июля 2016, 6:04 +03:00 от Eugene Salamin via math-fun <math-fun@mailman.xmission.com>:
Yes, f(x) = tan(cx) is the unique differentiable function satisfying the functional equation.
-- Gene
From: Fred Lunnon < fred.lunnon@gmail.com > To: Zak Seidov < zakseidov@mail.ru >; math-fun < math-fun@mailman.xmission.com > Sent: Sunday, July 3, 2016 4:50 PM Subject: Re: [math-fun] Solve functional equation(?)
No can do if x_i = x_j + x_k for some i, j, k, since f_i, f_j, f_k are then constrained by your equation; if only approximately true, any continuous solution would involve arbitrarily large gradients.
An obvious special solution is f(x) = tan(c x) for arbitrary c .
WFL
On 7/3/16, Zak Seidov < math-fun@mailman.xmission.com > wrote:
" any suitable initial values" means: f(x_i) = f_i, i=1..m, for any desired m, x_i and f_.
Понедельник, 4 июля 2016, 1:33 +03:00 от Zak Seidov via math-fun < math-fun@mailman.xmission.com >:
Solve functional equation, for f(x), with any suitable initial values, f (a + b) = (f (a) + f (b))/(1 - f (a)*f (b)). -- Zak Seidov _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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