I'm trying to solve this task:
Find all $f : (-1, 1) \to \mathbb{R}$ continuous at $x = 0$ and satisfying $f(x^{2}) = \frac{f(x)}{1 + x}$.
I tried substituting $x$ with $-1, 0, 1$, but it doesn't really do anything.
I thought that maybe I should use limits here since I know that function is continuous at $x = 0$, so I also tried to decrease the power:
$f(x) = \frac{f(x^{\frac{1}{2}})}{1 + x^{\frac{1}{2}}}$
$f(x^{\frac{1}{2}}) = \frac{f(x^{\frac{1}{4}})}{1 + x^{\frac{1}{4}}}$
$...$
I am not sure what to do next. Could somebody please give a hint?
Thanks in advance.
We have $$f(x)=(1+x)(1+x^2)\ldots (1+x^{2^n})f(x^{2^{n+1}})$$ As $f(x^{2^{n+1}})$ tends to $f(0)$ we get $$f(x)=\lim_n\prod_{k=0}^n (1+x^{2^k})\, f(0) $$ Observe that (by induction) $$(1-x)\prod_{k=0}^n (1+x^{2^k})=1-x^{2^{n+1}}\to 1$$ Hence $$f(x)={f(0)\over 1-x}$$
Remark Once we know the solution we may argue as follows. The equation is equivalent to $$(1-x^2)f(x^2)=(1-x)f(x)$$ By iteration we get $$(1-x)f(x)=(1-x^{2^n})f(x^{2^n})\to f(0)$$