I would like to solve $$f(x)=g(x)\cdot g(-x)$$ The goal is to express $g(x)$ in terms of $f(x)$. Any idea how to solve this?

  • $\begingroup$ There isn't a way to do this in general. $\endgroup$ Oct 13 at 23:53
  • $\begingroup$ Yes, there is no way unless g(-x) is the inverse of g(x) or equal to g(x) or anything similar. $\endgroup$ Oct 13 at 23:55
  • $\begingroup$ There is a solution only when $f \geq 0$ and $f(-x)=f(x)$. In this case $g(x)=\sqrt {f(x)}$ is the unique solution. $\endgroup$ Oct 14 at 0:01
  • $\begingroup$ You gave a rule to get $f$ from $g$. This can be understood as mapping $g \mapsto f$. You ask for a mapping $f \mapsto g$ (and it should be in some sense well behaved that you can write it down). However, your $g \mapsto f$ isn't even injective... $\endgroup$
    – user940347
    Oct 14 at 0:14

Assume $g$ is an arbitrary function $g:\mathbb{R} \to \mathbb{R}$ and $h^+$ is an arbitrary function $h^+:\mathbb{R^+} \to \mathbb{R}.$ Define $f:\mathbb{R} \to \mathbb{R}$ as $f(x):=g(x)g(-x)$. Now define $h:\mathbb{R} \to \mathbb{R}$ by $$h(x)=\begin{cases}h^+(x), &x>0\\g(0), &x=0\\ \frac{f(x)}{h^+(-x)},&x<0\end{cases}$$ We have $f(x)=h(x)h(-x)$, so it satisfies the same equation as $g$, but $h(x)$ is for positive $x$ an arbitrary function, so $g$ cannot be determined by $f.$


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