# Find a function such that $f^{-1}=f'$

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?

• See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question. Jul 27, 2018 at 7:23
• 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. Jul 27, 2018 at 7:27
• 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 $f' = f^{-1}$ Jul 27, 2018 at 7:51
• mathoverflow.net/questions/34052/function-satisfying-f-1-f Jul 27, 2018 at 8:11
• To "derive" $g$ just make the ansatz $g(x) = c x^\alpha$ and find what $c$ and $\alpha$ must be. Jul 27, 2018 at 8:19

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