I try to get a series expansion (not necessarily Taylor) or even an integral or any nice formula, for a special continuous function satisfying a peculiar, simple functional equation:
$f(x^a) = f(1+x) - f(1-x)$.
Here $a$ is a negative constant. Let $x_0$ be a solution of $x_0^a = 1+x_0$ or $x_0^a =1 - x_0$ and let $x_1 = x_0^a$ and $x_2 = 2-x_0$.
It is easy to obtain a simple relationship between the derivatives of $x$ taken at $x=x_1$ and $x=x_2$. While this is not enough to get a Taylor series around $x=x_1$ or $x=x_2$, is it enough to get a series expansion for $f$?
Note that if $x_0^a=1+x_0 = x_1$, we have:
$f(x_1) = f(x_1) - f(x_2)\\ ax_0^{a-1} f'(x_1) = f'(x_1) - f'(x_2)\\ ax_0^{a-1}f''(x_1) + a(a-1)x_0^{a-2}f'(x_1) = f''(x_1) - f''(x_2).$
It is easy to iteratively differentiate the above expression, but does it lead to something interesting, in terms of a nice expression for the function $f$ -- an expression that allows you to easily tabulate $f$?