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Suppose we have say $$f(2x)=\left(f(x)\right)^4,$$ with $f(0)=1$ and $f$ not being constant. How does one find out what $f$ is without guessing?

More generally, is there a systematic way of finding non-obvious functions that solve $$f(x)=g(f(h(x)))?$$

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Completely hopeless problem for general $g$ and $h$ – Norbert Apr 7 '12 at 16:40
I think you need more conditions on $f$ to have a unique solution. I doubt that there is a systematic method, so some guessing is required. In the example above, you could notice that $f(2^k) = f(1)^{4^k}$; letting $x = 2^k$ gives $f(x) = f(1)^{x^2}$, which suggests a general form. So, a function of the form $f(x) = c^{x^2}$ will work, for some $c>0$. – copper.hat Apr 7 '12 at 16:52
Of course define $f$ arbitrarily on $[1,2)$, then use the functional equation to extend it to $(0,\infty)$. – GEdgar Apr 7 '12 at 18:23
$f(x)=e^{c x^2}$ can be obtained by assuming $f(x) = e^{(a \hspace{3pt}p(x))}$ and applying $f(0) = 1$. – Kirthi Raman Apr 7 '12 at 19:22

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