Derivative equals Inverse – Lorenzo Alvarado

We are often tasked with finding the derivative of a function or finding its inverse. But what if we ask for a function that remains invariant under these two seemingly distinct operations?

The goal of this post is to find a function $f$ such that: $$ f'(x) = f^{-1}(x) $$ This inquiry will lead us to an unexpected generalization involving the Golden Ratio and the family of Metallic Means.

1. The First Case: $f'(x) = f^{-1}(x)$

For $f'$ and $f^{-1}$ to be of the same class of functions, a natural candidate is a power function of the form: $$ f(x) = c \cdot x^r $$ where $c, r \in \mathbb{R}$.

Calculating the derivative and the inverse: $$ f'(x) = c r x^{r-1} $$ $$ f^{-1}(x) = c^{-1/r} x^{1/r} $$

Equating $f'(x) = f^{-1}(x)$, we must match the powers of $x$ and the coefficients:
1) Powers: $r - 1 = \frac{1}{r} \implies r^2 - r - 1 = 0$
2) Coefficients: $cr = c^{-1/r}$

The equation $r^2 - r - 1 = 0$ is the famous equation defining the Golden Ratio $\Phi = \frac{1+\sqrt{5}}{2}$ and its conjugate $\Psi = \frac{1-\sqrt{5}}{2}$.

If we consider $r > 0$, then $r = \Phi$. Solving for the constant $c$: $$ c\Phi = c^{\Psi} \implies c = \Phi^{\frac{1}{\Psi - 1}} $$ (Using the property that $-1/\Phi = \Psi$).

Solution 1: The function satisfying $f'(x) = f^{-1}(x)$ is: $$ f(x) = \Phi^{\frac{1}{\Psi-1}} x^{\Phi} $$ This explicitly relates the derivative/inverse problem to the Golden Ratio.

2. Second Order: $f''(x) = f^{-1}(x)$

Let's take it a step further. What if the second derivative equals the inverse? $$ f''(x) = c r(r-1) x^{r-2} $$ Matching powers with the inverse $x^{1/r}$: $$ r - 2 = \frac{1}{r} \implies r^2 - 2r - 1 = 0 $$

The positive solution to this quadratic is the Silver Ratio, often denoted as $\Phi_2 = 1 + \sqrt{2}$. This suggests a pattern related to the "Metallic Means".

3. Generalizing for the $n$-th Derivative

To solve the general problem $f^{(n)}(x) = f^{-1}(x)$, we need some definitions.

Metallic Means

The Metallic Mean associated with $n \in \mathbb{N}$, denoted $\Phi_n$, is the positive solution to: $$ x^2 = nx + 1 \implies \Phi_n = \frac{n + \sqrt{n^2 + 4}}{2} $$ Its conjugate is $\Psi_n = \frac{n - \sqrt{n^2 + 4}}{2}$.
Properties: $\Psi_n = n - \Phi_n$ and $-1/\Phi_n = \Psi_n$.

Extending Factorials

Since we are dealing with derivatives of $x^r$ where $r$ is a real number (specifically, a metallic mean), we cannot use standard factorials. We use the Gamma Function $\Gamma(x)$, which satisfies $\Gamma(n+1) = n!$.

The $n$-th derivative of $f(x) = x^r$ is given by: $$ f^{(n)}(x) = \frac{\Gamma(r+1)}{\Gamma(r-n+1)} x^{r-n} $$

4. The General Theorem

We are now ready to state the general solution.

Problem: Find $f(x) = c x^r$ (with $r>0$) such that $f^{(n)}(x) = f^{-1}(x)$.

Step 1: Powers. Matching the exponents gives $r - n = 1/r \implies r^2 - nr - 1 = 0$. Since $r>0$, the solution is exactly the $n$-th metallic mean: $r = \Phi_n$.

Step 2: Coefficients. $$ c \frac{\Gamma(r+1)}{\Gamma(r-n+1)} = c^{-1/r} $$ Substituting $r = \Phi_n$ and using the property $-1/\Phi_n = \Psi_n$: $$ c \frac{\Gamma(\Phi_n+1)}{\Gamma(\Phi_n-n+1)} = c^{\Psi_n} $$ Solving for $c$ eventually yields (after some algebraic manipulation using $\Psi_n = n - \Phi_n$): $$ c = \left[ \frac{\Gamma(1-\Psi_n)}{\Gamma(1+\Phi_n)} \right]^{1-\Psi_n} $$

Theorem: The function satisfying $f^{(n)}(x) = f^{-1}(x)$ for any $n \in \mathbb{N}$ is: $$ f_1(x) = \left[ \frac{\Gamma(1-\Psi_n)}{\Gamma(1+\Phi_n)} \right]^{1-\Psi_n} x^{\Phi_n} $$ where $\Phi_n$ is the $n$-th metallic mean.

There is also a second solution $f_2(x)$ considering $r < 0$ (related to $\Psi_n$), but this only yields real solutions when $n$ is even (to avoid roots of negative numbers in the coefficient calculation).

This result beautifully connects differential operators, inverse functions, and the generalized Fibonacci sequence constants known as Metallic Means.

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