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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