Approximation with special partitions of unity

Question

Question:
what can be recommended for calculating $f(x)$ that solves $\frac{f(x)}{f(x)+f(1-x)}\approx g(x)$ for $x\in[0,1]$?

I have tried comparing Taylor series, but they look intimidating and I would appreciate suggestions for better solutions

Answer 1

I don't know what significance the approximate equality has, so I'll just replace $\approx$ by $=$. Any function $g(x)$ on $[0,1]$ can be split into its even $g_e(x) = (g(x)+g(1-x))/2$ and odd $g_o(x) = (g(x)-g(1-x))/2$ parts, $g(x) = g_e(x) + g_o(x)$. So the desired equation takes the following form, with an immediate implication: $$ \frac{f_e(x)+f_o(x)}{2f_e(x)} = g_e(x) + g_o(x) \quad \implies \quad \frac{f_e(x)}{2f_e(x)} = \frac{1}{2} = g_e(x) . $$ So the equation will have no solution at all, unless $g_e(x) = 1/2$. But if that holds, the odd part of the equation simply states that: $$ \frac{f_o(x)}{2f_e(x)} = g_o(x) . $$ That is, the most general solution is parametrized by a choice of $f_e(x) \ne 0$, with $f_o(x) = 2 g_o(x) f_e(x)$, or $f(x) = f_e(x) (1+2g_o(x)) = 2 f_e(x) g(x)$.