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Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that $$f^{-1}(x)=f'(x)$$

I think one such function would be of the form $f(x)=ax^b$. But then $b$ would be irrational and when $x\lt0$ this causes problems. So I guess letting $f(x)=-a(-x)^b$ for $x<0$ might work. But that's only one possibble solution, what are all the solutions?

Edit: With the comment of Joey Zou, the domain and range has been changed to $(0,\infty)$.

Edit: I already know the solution of the form $ax^b$. However, what I'm really asking is whether it is the unique solution.

Yuxiao Xie
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  • Hint: Use the Inverse Function Theorem – Moya Sep 06 '16 at 04:33
  • @Moya Can you please elaborate on that? – Yuxiao Xie Sep 06 '16 at 04:47
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    If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable and $f'$ takes on both positive and negative values, then $f$ can't be invertible. So if an invertible differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfied $f^{-1}(x) = f'(x)$, then $f'$ must take on both positive and negative values as the range of $f^{-1}$ is all of $\mathbb{R}$. This is a contradiction. If you change your domain and range to $f:(0,\infty)\rightarrow(0,\infty)$ you can circumvent this issue. – Joey Zou Sep 06 '16 at 04:56
  • @JoeyZou I see. Let's change the domain and range to $(0,\infty)$. – Yuxiao Xie Sep 06 '16 at 05:11
  • Related quetsion here –  Sep 06 '16 at 07:42
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    See this question at MO: http://mathoverflow.net/questions/34052/function-satisfying-f-1-f/34095#34095 – Christian Blatter Sep 06 '16 at 08:02
  • I think there is no harm in assuming that the domain range is $\mathbb R$. See here – polfosol Sep 06 '16 at 17:03

2 Answers2

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Partial answer, in order to confirm the Omnomnomnom's assumption that some solutions on the form $f(x)=ax^b$ do exist.

$$f^{-1}(x)=f'(x)\qquad \text{on }x>0$$ Search of particular solutions on the form $\quad f(x)=ax^b$ :

$f=ax^b \quad\to\quad x=\left(\frac{f}{a}\right)^{1/b}$

Hence the inverse fonction of $f(x)$ expressed as a function of $x$ is : $$f^{-1}(x)=\left(\frac{x}{a}\right)^{1/b}$$ to be not confused with $x$ expressed as a function of $f$.

If a solution of the form $f(x)=ax^b$ exists, it must satisfy the equation : $$f^{-1}(x)=f'(x)=\left(\frac{x}{a}\right)^{1/b}=abx^{b-1}$$ First condition : $x^{1/b}=x^{b-1}\quad\to\quad b^2-b-1=0$ $$b=\frac{1\pm\sqrt{5}}{2}$$ Second condition : $\left(\frac{1}{a}\right)^{1/b}=ab \quad\to\quad a=b^{-\frac{b+1}{b}}$

So, two different particular solutions are obtained :

First : $\quad f(x)=\left(\frac{2\:x}{1+\sqrt{5}}\right)^{\frac{1+\sqrt{5}}{2}}$

Second : $\quad f(x)=\left(\frac{2\:x}{1-\sqrt{5}}\right)^{\frac{1-\sqrt{5}}{2}}$

JJacquelin
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  • How do you deal with the irrational power of $-1$ in your second partial answer? – Yuxiao Xie Sep 06 '16 at 09:20
  • OK. If restricted to $x>0$ AND $f(x)$ real, reject the second solution. If restricted to $x<0$ AND $f(x)$ real, reject the first solution. If $f(x)$ not restricted to real (so, can be complex), both solutions are convenient. – JJacquelin Sep 06 '16 at 09:28
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Make the substitution $y = f(x)$ to rewrite this equation as $$ y = f'(f(y)) $$ Rewrite this as $$ y\,f'(y) = f'(f(y)) \,f'(y) $$ Now, integrate both sides with respect to $y$. Using integration by parts, note that $$ \int y\,f'(y) \,dy = y\,f(y) - \int f'(y)\,dy = (y - 1)f(y) + C $$ Using substitution, note that $$ \int f'(f(y)) \,f'(y)\,dy = \int f'(u)\,du = f(u) + C = f(f(y)) + C $$ Conclude that any $f$ satisfying our original equation must satisfy $$ (y - 1)f(y) = f(f(y)) + C $$ In terms of $x$, this says $$ (f^{-1}(x) - 1)x = f(x) + C $$ This isn't quite a solution, but at least we don't have any derivatives here.

Things to be said from here:

Note that with $y = 1$, we have $$ 0 \cdot f(1) = f(f(1)) + C \implies C = -f(f(1)) $$ Note also that if $f(y) = 0$ for any $y$, that $y$ must satisfy $$ f(f(y)) = f(f(1)) $$ but if $f$ is "invertible" and therefore $1$ to $1$, then this should never happen.

Ben Grossmann
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