solving $f(x)=\frac x{1-x^2}+2(1+x)f(x^2)$ without power series expansion

Question

My question concerns the title equation $f(x)=\frac x{1-x^2}+2(1+x)f(x^2)$, which arose through the use of generating functions for a simple recurrence.
Assuming $f: R\rightarrow R$ is analytic, the solution can obviously be found by using a power series expansion $f(x)=\sum_{i\ge0}a_ix^i$, which through equating coefficients gives
$a_0=0$
$a_{2n}=2a_n$
$a_{2n+1}=2a_n+1$
with the solution $a_n=n$, which gives $f(x)=\frac x{(1-x)^2}$.
My question now is: can the functional equation be solved without using power series expansion (or repeated derivatives, for that matter)? The only thing I see is $f(0)=0$, which doesn't seem to help much.

Answer 1

Even simpler than my other answer.

Let $f(x) = \frac{x}{(1-x)^2} + \frac{h(x)}{1-x}$. Then $h$ satisfies $h(0)=0$ and $$ h(x) = 2h(x^2). $$ Then $$ h(x) = \lim_{k\to\infty} 2^k h\bigl(x^{2^k}\bigr) = \lim_{y\to0} h(y)\log(y)=0 $$ as soon as we suppose that $h'(0)$ exists.

Answer 2

Regularity assumptions are required. $f$ differentiable at $0$ is sufficient.

Setting $f(x)=\frac{x}{(1-x)^2}+h(x)$ yields the functional equation $$ h(x)=2(1+x)h(x^2) $$ with $h(0)=0$. Assume that $$ \lim_{y\to0} h(y)\log(y) = 0 . $$ Then, for $x\in(-1,1)$, iterating we get $$ h(x) = \lim_{n\to\infty} 2^k \prod_{k=0}^n \left(1+x^{2^k}\right) \, h\!\left(x^{2^{k+1}}\right) = \frac1{1-x} \lim_{n\to\infty} 2^k h\!\left(x^{2^{k+1}}\right) = 0. $$

Answer 3

Let's make use of $(-x)^2=x^2$: $$f(x)+f(-x)=4f(x^2) $$ $$f(x)-f(-x)= \frac{2 x}{1-x^2}+4xf(x^2) $$ Hence by eliminating $4f(x^2)$, $$ f(x)-f(-x)=\frac{2x}{1-x^2}+x(f(x)+f(-x))$$ and so with $g(x):=(1-x)f(x)$, $$\tag1g(x)-g(-x) = \frac{2x}{1-x^2}.$$ This suggests rewriting the original functional equation in $g$ (by multiplying both sides with $1-x$) as $$ g(x)=\frac x{1+x}+2g(x^2).$$ As $g(x^2)=g(\exp(2\ln x))=g(\exp(\exp(\ln2 + \ln\ln x)))$, we can turn squaring into addition of $1$ by letting $h(x)=g(\exp(\exp(x\ln2)))=g(\exp(2^x))$ so that $$\tag2 h(x)=\frac{\exp(2^x)}{1+\exp(2^x)}+2h(x+1).$$ Apparently, we can define arbitrarily on $[0,1)$ and use $(2)$ to extend this to $\Bbb R$. In order to achieve continuity, it is necessary and sufficient to have continuity on $[0,1]$ and $h(0)=\frac e{1+e}+2h(1)$. Now we can define $$g(x)=h\left(\frac{\ln\ln x}{\ln2}\right),$$ which works for $x>1$. Then use $(1)$ to extend $g$ to $(-\infty,-1)\cup (1,\infty)$.

We could also define $h^*(x)=g(\exp(-\exp(x\ln 2)))=g(\exp(-2^x))$ and obtain $$\tag{2$^*$}h^*(x)=\frac{\exp(-2^x)}{1+\exp(-2^x)}+2h^*(x+1)$$ and proceed as above. This gives us the general (continuous) solution for $g$ on $(-1,0)\cup(0,1)$.

Can we achieve a continuous extension of $g$ to $x=0$? As we already know that $g(0)=0$, it is necessary and sufficient to have $$\lim_{x\to +\infty}h^*(x)=0.$$

From the known particular solution $g_0$ corresponding to $f_0(x)=\frac x{(1-x)^2}$, and the corresponding $h_0^*$, we see that the genearl solution for $g$ continuous on $(-1,1)$ is obtained from $$ (h^*-h_0^*)(x)=2(h^*-h_0^*)(x+1),$$ i.e., $$ h^*(x)=h_0^*(x) +2^{-x}s(x)$$ where $s$ is a continuous function of period $1$. That makes $$ g(x)= g_0(x)+\frac{s\left(\frac{\ln(-\ln x)}{\ln 2}\right)}{\ln x},\qquad 0<x<1$$ and hence $$ f(x)= f_0(x)+\frac{s\left(\frac{\ln(-\ln x)}{\ln 2}\right)}{(1-x)\ln x},\qquad 0<x<1$$ and its extension via $(1)$ to the general continuous solution on the interval $(-1,1)$.

As $f(\pm1)$ cnnot be defined (or perhaps $f(-1)$ can?), we can pick a solution for $g$ and hence $f$ on the set $(-\infty,-1)\cup(1,\infty)$ independently, in the way described above.

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Example plot over $[-.9,.9]$ of $f$ obtained from the 1-periodic function $s(x)=\sin 2\pi x+\sin 6\pi x$: While continuous in this range and smooth away from $0$, a closer look reveals that $f'(0)$ does not exist (cf. Frederico's answer)