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Find all functions $f(x)$ such for a given fixed $a\in \mathbb{R}$ such that the following functional equation holds

$$f(x)^{2}=f(x/a)$$

I'm not sure how to solve this equation other then using the method of power series, any tips?

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  • $\begingroup$ How would you use power series here? It is not even given $\;f\;$ is continuous, leave along differentiable. $\endgroup$ – Timbuc Nov 19 '14 at 20:52
  • $\begingroup$ @Timbuc A solution can proceed by finding a solution of certain form and then later proving that the solution is unique. $\endgroup$ – user192614 Nov 19 '14 at 20:58
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    $\begingroup$ True but In general if you have no idea if the function is continuous or not it can be a good idea to see if you can guess a solution a candidate solution using power series to get a recurrence relation with the coefficients then show that this candidate solution does in fact hold. $\endgroup$ – lance wellton Nov 19 '14 at 20:58
  • $\begingroup$ $f(x) = c\cdot \exp(\frac{x}{2a})$ for any $c$ is one family of solutions. $\endgroup$ – Solomonoff's Secret Nov 19 '14 at 20:59
  • $\begingroup$ @jef That is not a solution, RHS is $exp(\frac{x}{a})$ the LHS is $exp(\frac{x}{2a^{2}})$ these are not the same. The equation is nonlinear you do not have that if $f(x)$ is a solution then so is $cf(x)$ unless $c=1$ $\endgroup$ – lance wellton Nov 19 '14 at 21:03
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I will describe 2 trivial functions and one nontrivial family of functions satisfying the above relation for each $a$. I will give this answer even though I am not finding all such functions because nobody else has provided an answer in a day.

First let $c = \frac{1}{a}$. I think it's more natural to think about the equation as $f(x)^2 = f(cx)$ than as in the OP. Clearly $f(x) = 0$ and $f(x) = 1$ are solutions.

More interestingly, the family of functions $$f(x) = p^{(qx)^{\log_c(2)}}$$ for any $p, q > 0$ satisfies the inequality, which we can see as follows:

$$f(x)^2 = p^{2\cdot (qx)^{log_c(2)}} = 2^{c^{\log_c(2)} (qx)^{\log_c(2)}} = p^{(q(cx))^{\log_c(2)}} = f(cx).$$

This approach immediately generalizes to solve equations of the form $$f(x)^n = f(cx),$$ with solution $$f(x) = p^{(qx)^{\log_c(n)}}.$$

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