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

Question

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?

Answer 1

Problem Clarification - Solving a Simpler Related Problem

Given ( with $f^{-1}=\frac{1}{f}$ ) as $f^{-1}$ is the fraction (and not the inverse, as there are two possible conventions for $f^{-1}$ ):

$$f^{-1} = f^{'} \tag{ Eq. 1}$$

Then solve for $f$.

Solution in the case that $f^{-1}=\frac{1}{f}$, and $f^{'}$ is the derivative, the easier case

The convention of $f^{-1}=\text{ Inverse (}f\text{)}$ can be solved later.

Multiply the left and right sides of Equation 1 by $f$ so: $$ 1=f f^{'} \tag{Eq. 2}$$ Now integrate both sides of Equation 2 using the identities that $\frac{d}{df}\frac{1}{2} \left( f \right)^2 = f f' $ and $\frac{d}{df} f =1$, then, with $C/2$ an arbitrary integration constant: $$ f+C/2=\frac{1}{2}f^2 \tag{Eq. 3}$$ Rearrange Equation 3: $$ f=\frac{1}{2}f^2 - C/2 \\ \tag{Eqs. 4}$$ Consider the function $f=\sqrt{a x}=\sqrt{2 x}$ so $f'=\frac{a}{2}\frac{1}{\sqrt{ a x}}$ and then: $$ f^{'}f=\left( \frac{a}{2}\frac{1}{\sqrt{ a x}} \right) \sqrt{a x} \\ \text{And with }a=2 \\ f^{'}f=\left( \frac{2}{2}\frac{1}{\sqrt{ 2 x}} \right) =1 \tag{Eqs. 5} $$ Generally, to figure out the integration constant, consider functions again $f^{'}f=1$ and $f=\sqrt{a x}+C$, and $a x \ne 0$ so (not dividing by zero): $$ f' f=\left(\frac{a}{2}\frac{1}{\sqrt{a x}}\right) \left( \sqrt{a x} + C\right) \\ f' f=\left( \frac{a}{2} \right) + \left( \frac{a C}{2}\frac{1}{\sqrt{a x}} \right) = 1 \\ \underset{\text{implies}}{\implies} \frac{a}{2}=1 \text{ and }C=0 \tag{Eqs.6}$$ So the Solution to Equation 1 is (when just considering $f^{-1}=\frac{1}{f}$ from the title: $$\boxed{f(x)=\sqrt{2 x}}\tag{Eqs.7}$$

Another Possible Angle Towards the Problem (Detailed Problem Statement and Not Just the Title)

Sometimes $f^{-1}$ represents an inverse function, a different convention for the inverse $-1$ sign. (After all, the title for this question just had the $-1$ sign without additional explanation, so both possibilities need to be covered.)

To start with, take the function $f(x)$ from Equation 7. $$ f(x)=\sqrt{2 x} \text{ function } \\ f^{-1}(x)=g(x)=\frac{1}{2}\left(g(x)\right)^2 \text{ inverse function } \tag{Eq. 8}$$ In this concrete example: $$ f^{-1}(x)=\frac{1}{2}\left(x\right)^2 \\ f(x) = \sqrt{ 2 x} \\ f^{'}(x)=\frac{1}{\sqrt{2 x}} \\ \frac{f^{-1}(x)}{f^{'}(x)}= \frac{\left( \frac{1}{2}\left(x\right)^2 \right)} {\left( \frac{1}{\sqrt{2 x}} \right)} = \frac{ \left(x\right)^{\frac{5}{2}}} {\sqrt{2}} \tag{Eqs. 9}$$ It does not equal $\frac{f^{-1}(x)}{f^{'}(x)}=1$ as to be desired, but it is a starting point, since one idea is to represent the function in terms of powers (and possibly sums of x).

Another Example with $f(x)=a x^n$ and $g(x)=f^{-1}(x)=\left(\frac{x}{a}\right)^{\frac{1}{n}}$

Equations 9 seem to suggest that the representation of the function in terms of a polynomial might be the right direction for a solution, inserting some additional degrees of freedom.

The simplest polynomial $f(x)=a x^n$ results in the inverse function $g(x)=\left(\frac{x}{a}\right)^{\frac{1}{n}}$, so from Equation 1 $\frac{g(x)}{f^{'}(x)}$ is: $$ \frac{g(x)}{f^{'}(x)}= \frac{\left( \frac{x}{a} \right)^{\frac{1}{n}} } {\left( a\text{ }n \text{ }x^{n-1} \right)} =\left( \frac{1}{a} \right)^{\left( 1+ \frac{1}{n} \right)} \left( \frac{1}{n} \right) \left( x \right)^{\left( \frac{1}{n} + 1 - n \right)}=1 \tag{Eqs. 10}$$ From Equations 10, one approach is to have $x$ raised to the zero power so that $ \frac{g(x)}{f^{'}(x)} $ is independent of the value of $x$, since the solution $ \frac{g(x)}{f^{'}(x)} = 1$ is a constant that does not vary with $x$. Then: $$ \left( \frac{1}{n} + 1 - n \right)=0 \underset{\text{implies}}{\implies}n^2 -n -1=0 \underset{\text{implies}}{\implies} \left( n - \frac{1}{2}\right)^2 =1+\frac{1}{4} \\ \underset{\text{implies}}{\implies} n_{\pm} = \frac{1}{2} \pm \sqrt{\left(1+\frac{1}{4}\right)} \\ \underset{\text{implies}}{\implies} n_{\pm} = \frac{1}{2} \left( 1 \pm \sqrt{ 5 } \right) \ne 0 \tag{Eqs. 11}$$ So it is now apparent that there is are at least 2 solutions to the question referenced in the text, by setting the constants at the left of $\left( x \right)^{\left( \frac{1}{n_\pm} + 1 - n_\pm \right)}$ to $1$ as (inserting the already solved values for $n_{\pm}$) as follows:

$$ \left( \frac{1}{a_\pm} \right)^{\left( 1+ \frac{1}{n_\pm} \right)} \left( \frac{1}{n_\pm} \right)=1 \underset{\text{implies}}{\implies} \left( \frac{1}{a_\pm} \right)^{\left( 1+ \frac{1}{n_\pm} \right)} ={n_\pm} $$ $$ \underset{\text{implies}}{\implies} \left( a_\pm \right)^{\left( 1+ \frac{1}{n_\pm} \right)} =\frac{1}{n_\pm} \underset{\text{implies}}{\implies} \left( a_\pm \right)=\left(\frac{1}{n_\pm}\right)^\frac{1}{\left( 1+ \frac{1}{n_\pm} \right)} \tag{Eqs. 12}$$

Notice that with the notation here for $a_\pm$, it is not referring to a value of $a$ that is positive or negative; rather it refferes to the corresponding solution for which $n$ is positive or negative, either $n=n_{+}$ or $n=n_{-}$. From Equation 12 it is obvious how to handle the case of $n=n_{+}$. From Equations 10, in the case $n=n_{-}$, this $f=a\text{ }x^n$ is not necessarily a real function because of the (possibly complex) roots of a negative number involved in Equations 12. Thus for the purpose of this question, only $n=n_{+}$ is selected since the solution for $a=a_{+}$ is real in this case.

Finally, there is a solution for Equation 1 in the real domain, namely $f^{-1} = f^{'} $ of the form of Real $f(x)=a_{+} x^n$ and the inverse $g(x)=f^{-1}(X)$ as follows: $$ \boxed{f^{-1} = f^{'} \text{ with }f(x)= a_{+} x^{n_{+}} \text{ and with } \left( a_{+} \right)= \left( \frac{1}{n_{+}} \right)^ {\frac{1} {\left( 1+ \frac{1}{n_{+}} \right)} } \text{ and with }n_{+} = \frac{1}{2} \left(1 + \sqrt{ 5 } \right) } $$ $$ \boxed{ f(x)=a_+ x^{n_+} = \left( \frac{2}{1+\sqrt{5}} \right )^ {\left( \frac{\displaystyle 1+\sqrt{5}} {\displaystyle 3+\sqrt{5}} \right)} \left( x \right)^ {\left(\frac{\displaystyle 1 + \sqrt{ 5 }}{\displaystyle 2} \right)} } \tag{Eqs. 13}$$ $$ g(x)=f^{'}(x)= \left( \frac{2}{1+\sqrt{5}} \right )^ {\left( \frac{\displaystyle 1+\sqrt{5}} {\displaystyle 3+\sqrt{5}} \right)} \left(\frac{\displaystyle 1 + \sqrt{ 5 }}{\displaystyle 2} \right) \left( x \right)^ {\left(\frac{\displaystyle -1 + \sqrt{ 5 }}{\displaystyle 2} \right)} $$ $$ \boxed{ g(x) = \left( \frac{1+\sqrt{5}}{2} \right )^ {\left( \frac{\displaystyle 2} {\displaystyle 3+\sqrt{5}} \right)} \left( x \right)^ {\left(\frac{\displaystyle -1 + \sqrt{ 5 }}{\displaystyle 2} \right)} } \tag{Eq. 14}$$ The problem statement was to "come up with a way to derive the function g or more functions satisfying this property?" which was derived in this answer.

As to the question part, "Also IS this really the unique solution?" As far as I can tell, since a function can be expanded around a point using a Taylor Series, this result seems unique in terms of a finite expansion, because of this link on an Inverse Function for a polynomial, which seems to suggest that for other Polynomials that the inverse is an infinite series, making it very difficult to tract, since the series does not usually terminate, except for this case here of $f(x)=a_{+} x^n$.

Answer 2

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.