I was looking for examples of real valued functions $f$ such that $f(x)+f(-x)=f(x)f(-x)$. Preferably, I'd like them to be continuous, differentiable, etc.

Of course, there are the constant functions $f(x)=0$ and $f(x)=2$. I also showed that $1+b^x$, where $b>0$, is another solution. Are there any other nice ones?


From your solution, I thought to consider the change of variable $f(x) = 1 + g(x)$: your functional equation then becomes

$$ g(x) g(-x) = 1 $$

and now the entire solution space becomes obvious: you can pick $g(x)$ to be any function on the positive reals that is nowhere zero, and then the values at the negative reals are determined by the functional equation. And at zero, you can pick either $g(0) = 1$ or $g(0) = -1$.

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    $\begingroup$ Awesome, thanks! $\endgroup$ – Nishant Jul 12 '14 at 4:19
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    $\begingroup$ Neat, very neat. $\endgroup$ – Lubin Jul 12 '14 at 4:19

Hurkyl's solution is nice, but the change of variables obscures the inherent closed orbit property, which is the crucial part of the problem.

Observe that the functional equation only involves $x$ and $ -x$. In particular, if $h(x) = -x$, then the orbit of $x \neq 0$ is $\{x, -x\}$ and the orbit of $0$ is $\{0\}$. As such, the function is uniquely defined by the non-negative part.

For $x = 0$, we have $2 f(0) = f(0)^2 $, which means $ f(0) = 0$ or $2$.
For $x \neq 0$, we have $ f(-x) = \frac{ f(x) } { f(x) - 1}$, if $ f(x) \neq 1$.
Note that if $ f(x) = 1$, then there is no possible value for $ f(-x)$.

This is a necessary and sufficient condition.


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