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If $x \neq 0$ , find $f(x)$ if it satisfies: $f(x)f(1/x) = f(x) + f(1/x)$.

I know that the answer is $f(x) = 1 \pm x^n$ where $n \in \mathbb{R}$. I don't know how to show this.

Anon
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    Set $g(x)=f(x)-1$. Then the functional equation rearranges to $g(x^{-1})=(g(x))^{-1}$, assuming $g(x)$ nonzero. – vadim123 Apr 11 '18 at 02:58
  • @vadim123 So the equation simplifies to $g(x)g(\frac{1}{x}) = 1$. But how to solve this equation, as the only way for solving functional equations that I know of, is substituting $x = 1/y$, or $x = -y$ etc. Here no such thing helps. – Anon Apr 11 '18 at 05:10

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Solution of $g(x)g(\frac 1 x)=1$ on $\mathbb R \setminus \{0\}$: take any function $h$ on $\{x: 0<|x|<1\}$ and define $g(x)=h(x)$ for $0<|x|<1$, $g(x)=\frac 1 {h(1/x)}$ for $|x| >1$ Take $g(1)$ and $g(-1)$ to be 1 or -1. Then $g$ satisfies the given property. Conversely, given $g$ we can take $h$ to be the restriction of $g$ to $0<|x|<1$. This describes all possible solutions.

Kavi Rama Murthy
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Hint:

$$f(x)(f(1/x)-1)=f(1/x)$$ $$f(x)= \frac{f(1/x)}{f(1/x)-1}$$ $$f(x)-1 = \frac{1}{f(1/x)-1}$$

I.e. defining $g(x)=f(x)-1$, we have $g(x) = 1/g(1/x)$. Now solve this easier equation.

Tal-Botvinnik
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$f(x)f\left(\dfrac{1}{x}\right)=f(x)+f\left(\dfrac{1}{x}\right)$

$f(x)f\left(\dfrac{1}{x}\right)-f\left(\dfrac{1}{x}\right)=f(x)$

$(f(x)-1)f\left(\dfrac{1}{x}\right)=f(x)$

$f\left(\dfrac{1}{x}\right)=\dfrac{f(x)}{f(x)-1}$

$f\left(\dfrac{1}{x}\right)=\dfrac{1}{f(x)-1}+1$

$f\left(\dfrac{1}{x}\right)-1=\dfrac{1}{f(x)-1}$

Let $g(x)=f(x)-1$ ,

Then $g\left(\dfrac{1}{x}\right)=\dfrac{1}{g(x)}$

$g(x)g\left(\dfrac{1}{x}\right)=1$

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.

$\therefore g(x)=\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.

$f(x)=1\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.

doraemonpaul
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  • I think for that solution, additional constraints (like continuity or differentiability) are imposed on f(x) [hence g(x)] as otherwise Kavi Rama Murthy's answer is correct. – Anon Apr 11 '19 at 11:42