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How do I solve the following functional equation: $$f(x)+12f^{-1}(x)=\frac{1}{x}f(x)$$ I've been doing a lot of functional equations, but I haven't done one yet that has the function and its inverse together. All I've done so far is figure out that $f(x)$ has a fixed point at $x=\frac{1}{13}$ and that $f(0)$ starts a cycle of orbit $2$.

Thanks! All help is appreciated!

Franklin Pezzuti Dyer
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    Does $f$ need to have real domain and codomain? – orlp May 22 '17 at 13:39
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    You can trade the inverse function for a nesting with

    $$f(f^{-1}(x))=x=f\left(\frac{1-x}{12x}f(x)\right)$$ or $$f(f(x))+12x=\frac{f(f(x))}{f(x)}.$$

     –  May 22 '17 at 15:44

2 Answers2

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We have

$$ y=f(x) $$

and:

$$ x=f^{-1}(y)=g(y) $$

Thus

$$ (\frac{1}{x}-1)f(x)=12f^{-1}(x)\\ x=f(f^{-1}(x))=f(\frac{1}{12}(\frac{1}{x}-1)f(x))\\ $$

Thus, we can formulate a recursive numeric search on $f$:

$$ min_{df}\left|x-f\left(\frac{1}{12}(\frac{1}{x}-1)f(x)+df(x)\right)+df\left(\frac{1}{12}(\frac{1}{x}-1)f(x)\right)\right|\\ $$

Brethlosze
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  • What exactly is a recursive numeric search? – Franklin Pezzuti Dyer May 21 '17 at 21:31
  • I've been unable to make it converge though... Perhaps this problem dont have a clear solution, but also i was unable to prove that too. I discretized a vector $f$ over several testing range and applied a Newton Raphson gradient search over the function $f$ defined on that way, with different iterating rates, considering the indicated expression to be minimized for both the vector $x$, the current recursion for $f$ and the new step $df$ evaluated at the indicated function value. But i didnt have any results, yet. – Brethlosze May 21 '17 at 22:16
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This is not a full answer but an alternate approach in addition to hypfco's. We have $y = f(x)$ and $x = f^{-1}(y)$.

Writing your equation as $(\frac{1}{x} - 1)f(x) = 12f^{-1}(x)$ we get

$$(\frac{1}{x} - 1)y = 12f^{-1}(f^{-1}(y))$$ $$(\frac{1}{f^{-1}(y)} - 1)y = 12f^{-1}(f^{-1}(y))$$ $$y = \frac{12}{\frac{1}{f^{-1}(y)} - 1}f^{-1}(f^{-1}(y))$$ $$y = -12\frac{f^{-1}(y)}{f^{-1}(y) - 1}f^{-1}(f^{-1}(y))$$

If we just write $g(x) = f^{-1}(x)$, we can solve the problem by finding a $g$ such that:

$$-12\frac{g(x)}{g(x)-1}g(g(x)) = x$$

orlp
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