How to solve this functional equation: $f(x^2) = f(x) f^{-1}(x)$?

Question

We know that the equation admits the particular solutions:

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

But how to get these results? The method of guessing a function and substituting it into the equation is very limited and we found no other possible solutions. Is there any powerful method to solve these types of equations involving inverse and composite functions? I'm finding the same difficulty to solve the similar problem already posted here: $f(2x) = f(x) + f^{-1}(x)$

If I were you, I would start with $f(1^2)=f(1)f^{-1}(1)$, meaning that $f^{-1}(1) = 1$ or $f(1) = 0$. — Dominique, Jul 23 '23 at 15:46
If the solution is in the form $f(x)=ax^p+b$. Then, it follows that $f^{-1}(x)=\sqrt[p]{\frac{x-b}{a}}$. Since equation is valid for all $x$ in $\mathbb R$, it comes that $p=1$. Then it is easy to conclude that $a=1$ and $b=0$ or $-1$.
Now, if the solution is polynomial, I think we can show that it of the previous form. — elmas, Jul 23 '23 at 17:34
Another solution is $f\left( x \right) = x^{-1}$. Wolfram|Alpha says True. — Kevin Dietrich, Jul 31 '23 at 14:10

score 0 · Answer 1 · answered Jul 23 '23 at 18:20

They set $f(x) = ax+b$ and found which $a$ and $b$ works: $$ax^2+b=(ax+b)\left(\frac{x-b}{a}\right) = x^2-bx+\frac{bx}{a}-\frac{b^2}{a}.$$ If you set the coëfficiënts of $x^2$ equal to each other, you find $a=1$. If you then set the constant terms equal to each other, you get $b=-b^2\Longrightarrow b=0$ or $b=-1$, meaning the only linear polynomials that satisfy this equation are $f(x) = x$ and $f(x)=x-1$.

Finding all the solutions seems pretty hard (probably even impossible), since plugging values in for x can only give you the value of the function at certain points, but you can't use those to make clever substitutions and find all the solutions, like you normally can with functional equations with more unknowns.

So you will need to exploit something very specific to the functional equation to find all the solutions, but again: I don't think this is possible here. The equation gives too little information about the function.

How to solve this functional equation: $f(x^2) = f(x) f^{-1}(x)$?

1 Answers1