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A function whose inverse is equal to its derivative

Here's a problem: find a function \(f\) whose inverse \(f^{-1}\) is equal to its derivative \(f'\).

A possible approach

Assume the function \(f\) is a polynomial and can be written as

\begin{equation*} f(x) = ax^b, \end{equation*}

where \(a\) and \(b\) are constants we wish to determine. We know the derivative \(f'\) is given by

\begin{equation*} f'(x) = abx^{b-1}, \end{equation*}

and the inverse \(f^{-1}\) by

\begin{equation*} f^{-1}(x) = \left ( \frac{x}{a} \right )^{\frac{1}{b}} = a^{-\frac{1}{b}} x^{\frac{1}{b}}, \end{equation*}

and so we obtain the following system of equations:

\begin{equation*} \begin{cases} ab = a^{-\frac{1}{b}}, \\ b - 1 = \frac{1}{b}. \end{cases} \end{equation*}

The second equation looks familiar, let's expand on that. Rewriting gives us

\begin{equation*} b^2 - b - 1 = 0, \end{equation*}

which may be solved using the quadratic formula,

\begin{equation*} b = \frac{1 \pm \sqrt{5}}{2} = \phi. \end{equation*}

The golden ratio. You know when you see the golden ratio pop up somewhere, things are getting more interesting.

Now let's solve for \(a\), we know \(\frac{1}{b} = b - 1\), so

\begin{equation*} b = a^{-\frac{1}{b} - 1} = a^{-b}, \end{equation*}

and

\begin{equation*} a = b^{-\frac{1}{b}} = b^{1-b}. \end{equation*}

Finally we have,

\begin{equation*} f(x) = \phi^{1-\phi} x^{\phi}. \end{equation*}

Are there more?

If so, how many? Infinite? How can we find them?

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