Let $f:\Bbb{R}^+\rightarrow\Bbb{R}^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^{-1}$ (where $f^{-1}$ denotes the inverse of $f$). Find $f$.

This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^{-1}$, but in the comments, there is a counterexample given: $$g(x)=\varphi^{1-\varphi}x^{\varphi}$$ where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio. It's straightforward to check that $g'=g^{-1}$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?

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    $\begingroup$ See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question. $\endgroup$ Jul 27, 2018 at 7:23
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    $\begingroup$ If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice. $\endgroup$ Jul 27, 2018 at 7:27
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    $\begingroup$ I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^{-1}(x)= x^{1/a}$. At the same time, $f_a'(x) = a x^{a - 1}$. If $f' = f^{-1}$, then we need something like $a - 1 = 1/a$. It is known that $\varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^\varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^{-1}$ $\endgroup$
    – Antoine
    Jul 27, 2018 at 7:51
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    $\begingroup$ mathoverflow.net/questions/34052/function-satisfying-f-1-f $\endgroup$ Jul 27, 2018 at 8:11
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    $\begingroup$ To "derive" $g$ just make the ansatz $g(x) = c x^\alpha$ and find what $c$ and $\alpha$ must be. $\endgroup$
    – md2perpe
    Jul 27, 2018 at 8:19

1 Answer 1


Here is an idea. Something along those lines should work.

Suppose we are trying to find out what the solutions looks like locally. Formally then, we expand in Taylor series \begin{equation} f(x)=\sum_{n\geq 0}c_n x^n\implies f^\prime(x)=\sum_{n\geq 1}nc_n x^{n-1}. \end{equation} The condition we need satisfied is \begin{equation} f(f^\prime)(x)=x, \end{equation} or in other words \begin{equation} \begin{aligned} c_0 &+ c_1\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)\\ &+ c_2\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^2\\ &+ c_3\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^3\\ &+ c_4\left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^4\\ &+ c_5 \left(c_1+2c_2 x+3c_3 x^2+4c_4 x^4+\dots\right)^5\\ & \dots\\ = &x \end{aligned} \end{equation} Let, for simplicity, the vector $C=(c_0,c_1,c_2,c_3,c_4,\dots)$. Then, the above relation is equivalent to \begin{equation} \begin{aligned} & C^T\cdot(1,c_1,c_1^2,c_1^3,\dots)=0, \\ & C^T\cdot (0,2c_2,0,0,\dots)=2c_1c_2=1,\\ & C^T\cdot (0,3c_3,(2c_2)^2,0,0,\dots)=0,\\ & C^T \cdot (0,4c_4,(3c_3)^2,(2c_2)^3,0,\dots,)=0,\\ & \dots. \end{aligned} \end{equation} We begine from the second equation above. We note that if we choose $c_1,c_2$, in the second equation above, then the rest $c_3,c_4,c_5$ are chosen inductively.

However, there is only one choice of $c_1,c_2$ such that the first equation is satisfied. Now since you have already found a solution, this solution is unique.

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    $\begingroup$ This expansion makes no sense. You cannot Maclaurin expand a function that is defined only on the positive real numbers. $\endgroup$
    – Angel
    Dec 8, 2021 at 15:28
  • $\begingroup$ I suggest this video : youtu.be/rNUfiQgj6ZI $\endgroup$ Sep 29, 2023 at 8:01

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