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What function satisfies $f(x)+f(−x)=f(x^2)$?

$f(x)=0$ is obviously a solution to the above functional equation.

We can assume f is continuous or differentiable or similar (if needed).

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    $\begingroup$ $f(0) = 0$ does not hold with $\log|x|$ $\endgroup$ – DeepSea May 15 '15 at 20:33
  • $\begingroup$ maybe you can consider $f(x)+f(-x)=2f(x)=f(x^2)$? $\endgroup$ – Mythomorphic May 15 '15 at 20:37
  • $\begingroup$ @hkmather802 : Where exactly did that come from? $\endgroup$ – Kevin Church May 15 '15 at 20:38
  • $\begingroup$ All even functions $f(x)$ obeys the rule $f(x)+f(-x)=2f(x)$. So we may try to find some even $f(x)$ where $2f(x)=f(x^2)$ $\endgroup$ – Mythomorphic May 15 '15 at 20:41
  • $\begingroup$ Is it given that $f$ must have domain and codomain $\mathbb R$? $\endgroup$ – user26486 May 15 '15 at 21:09
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Define $f(x)$ any way you want for $x > 0$, then define $f(-x) = f(x^2) - f(x)$ also for $x>0$. If you want continuity, make sure that $\lim_{x\to 0} f(x) = 0$.

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  • $\begingroup$ Where do you know that if the limit exists, then there is another solution other than trivial one??!! $\endgroup$ – k1.M May 15 '15 at 20:41
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    $\begingroup$ Since you can start with any continuous function on $[0,\infty)$ with $f(0)=0$ to get a continuous function on the whole real line, there are infinitely many continuous solutions. If you can prove that $f=0$ is the only solution, then the proof or the statement of the problem is wrong. $\endgroup$ – Lukas Geyer May 15 '15 at 20:43
  • $\begingroup$ For example, let $f(x)=x^2$ for $x>0$, $f(x)=x^4-x^2$ for $x<0$, and $f(0)=0$. This gives an explicit solution, and it's easy to check that it's continuous. $\endgroup$ – Noah Schweber May 15 '15 at 20:44
  • $\begingroup$ Another example: let $f(x) = x$ if $x\geq 0$ and $f(x) = x^2 + x$ if $x<0$. $\endgroup$ – David K May 15 '15 at 20:47
  • $\begingroup$ That's OK, I made a little mistake in my calculations...thanks for observations...+1 $\endgroup$ – k1.M May 15 '15 at 20:49
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Give f(x) = ln(|x|) a try in your equation

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    $\begingroup$ This was mentioned in the comments, and is not continuous at 0. (It's not even defined at 0, but the point is that no extension of the function to the whole real line is continuous.) $\endgroup$ – Noah Schweber May 15 '15 at 21:02
  • $\begingroup$ Actually the question doesn't specify the domain. So also the function $f:\{-1,1\}\to\{0,1\}$ with $f(-1)=0$ and $f(1)=1$ satisfies the given requirements. $\endgroup$ – celtschk May 15 '15 at 21:06
  • $\begingroup$ @user28111, the comment proposing $\log|x|$ was deleted (only a reply remains). And the question doesn't insist on continuity. The only thing missing here, I'd say, is to extend $f$ by defining $f(0)=0$. $\endgroup$ – Barry Cipra May 15 '15 at 21:09
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I'm going to put my comment as an answer.

$\ln|1-x|$ seems to work: $$\ln|1-x|+\ln|1+x|=\ln|1-x^2|$$ Similarly, so does $\ln|1-x^3|$ (or any odd exponent).

Also, any linear combination of these works, as you can check. Thus: $$\ln(1+x+x^2)$$ works because it's equal to $\ln|1-x^3|-\ln|1-x|$. The example of $\ln(1+x+x^2)$ is nice because it's defined, continuous, and infinitely differentiable everywhere.

As far as I know, if you want it to be defined everywhere, continuous, and infinitely differentiable, this sort of thing is the only possible solution.

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