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


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$f(x) = p^{(qx)^{\log_c(n)}}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.