Solve for f(x) if f(x)f(-x)=g(x) [closed]

Question

Solve for $f(x)$ if $f(x)\cdot f(-x)=g(x)$.

I have been having trouble figuring this out. I asked ChatGPT, and its answers don't work. I then looked online (I googled it), and was surprised to find nothing about it.

I was playing around in Desmos and found that you can express sin(x) and cos(x) by using factorials, like so:

$$ \left(-\frac{x+\frac{\pi}{2}}{\frac{x+\frac{\pi}{2}}{\pi}!\cdot-\left(-\frac{x+\frac{\pi}{2}}{\pi}\right)!},-\frac{x}{\frac{x}{\pi}!\cdot-\left(-\frac{x}{\pi}\right)!}\right) $$

$\cos(x)$ can be derived by adding $\frac\pi2$ to every instance of $x$.

I was trying to figure out if it is possible to express $x!$ in terms of $\sin(x)$ (without using any factorials or series (plural) or sums or whatever beyond that of the definition of the sine function itself), and got as far as the following:

$$ \pi x\frac{1}{\sin\left(\pi x\right)}=x!\cdot\left(-x\right)! $$

It can be simplified to $\pi x\csc\left(\pi x\right)$.

Given that $f\left(x\right)=x!$ and $g\left(x\right)=\pi x\csc\left(\pi x\right)$, I want to find $f(x)$ in terms of $g(x)$.

Even better if the problem can be solved regardless of equation. Thanks!

Edit: I also asked Wolfram Alpha, and it worked up until $f(x)\cdot f(x)=g(x)$, but not for $f(x)\cdot f(-x)=g(x)$. Also, if the problem is impossible, I would like to know why. Thanks again!

Answer 1

First, I will clarify some conventions: the function $x!$ is only initially defined for natural numbers, so by $x!$ for a real-valued $x$, we mean the Gamma function $\Gamma(x+1)$. In this case your equality $$ x!\cdot (-x)!=\Gamma(x+1)\cdot\Gamma(-x+1)=\pi x\csc(\pi x) $$ does hold.

Now, as for your question: I very highly suspect that without using series, there will not be a good answer to your problem. The reason for my suspicion is this: given a function $g(x)$ like yours, a function $f(x)$ satisfying $$ g(x)=f(x)\cdot f(-x) $$ need not be unique. For a simple example, let $f$ be any function satisfying the above equality. Then one can define $$ \tilde f(x)= f(x)\cdot2^x $$ which is a completely different function which also satisfies $\tilde f(x)\cdot \tilde f(-x) = g(x)$, since $$ \tilde f(x)\cdot \tilde f(-x) = f(x)2^x\cdot f(-x)2^{-x} = f(x)\cdot f(-x) = g(x) $$ So the function $f$ should not necessarily be derivable from just the equality $f(x)\cdot f(-x)=g(x)$ and information about $g$, without knowing something else about $f$ in advance (such as its Taylor series), since it is not unique.

As noted in the comments, whenever the function $g(x)$ is even (which it must be) and non-negatively valued, you may take $f(x)=\sqrt{g(x)}$ to be one solution to this problem. Unfortunately in your case, the graph of $g(x)=\pi x\csc(\pi x)$ looks like this:

In particular, it is non-negatively valued precisely when $\sin(\pi x)$ is positively valued, and also at $0$ (since the factor of $x$ allows there to be no pole at $0$). This happens on the interval $(-1,1)$ and on intervals of the form $(2n,2n+1)$ and $(-2n-1,-2n)$ for $n\in\mathbb{N}$. So this solution is only valid on intervals of this form. However, this is not the same as $x!$, which is immediately clear by a quick comparison of their graphs:

This is again a consequence of the function $x!$ not being unique with the property which you listed.