Is there a function whose inverse is exactly the reciprocal of the function, that is $f^{-1} = \frac{1}{f}$?

Is there a function whose inverse is exactly the reciprocal of the function? That is $f^{-1} = \frac{1}{f}$.

We know that the inverse of a function is not necessarily equal to its reciprocal in general.

Solution 1:

Assuming that the function maps $\Bbb{R} \to \Bbb{R}$ and is differentiable, the answer is no.

Here's why. In order for a continuous function to be invertible, it must be strictly monotonic (monotone increasing or monotone decreasing). In fact:

CLAIM 1: If $f$ is an increasing function, then $f^{-1}$ is too.

PROOF: This is a straightforward, standard calculation using the chain rule and the fact that $f(f^{-1}(x)) = x$ for all $x$: $$ \frac{d}{dx} f(f^{-1}(x)) = f'(f^{-1}(x)) \cdot (f^{-1})'(x) \quad\text{and}\quad \frac{d}{dx} x = 1 $$ so $$ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. $$ According to this last expression, the signs of $(f^{-1})'$ and $f'$ are the same.

CLAIM 2: If $f$ is an increasing function, then $(1/f)(x)$ is decreasing.

PROOF: Calculate the derivative using the chain rule: $$ \frac{d}{dx} \frac{1}{f(x)} = -\frac{f'(x)}{(f(x))^2}. $$ Since the denominator is positive, the signs of $(1/f)'$ and $f'$ must be opposite.

Solution 2:

It is possible for this to be satisfied in $\Bbb{R} \to \Bbb{R}$ everywhere except $ 0$.

$$ f(x) = \begin{cases} -x & x \geq 0 \\ -{x}^{-1} & x \lt 0 \\ \end{cases} $$

EDIT: User Slade asked whether it is possible in $\mathbb{R}^+ \to\mathbb{R}^+$. I've constructed a solution.

Let $ g(x) $ be a bijective function from $ \left(0,\frac{1}{2}\right] $ to $ \left(\frac{1}{2},1\right) $.

Then the following function satisfies $ f^{-1} = \dfrac{1}{f} $:

$$ f(x) = \begin{cases} g(x) & 0 \lt x \leq \frac{1}{2} \\ \dfrac{1} {g^{-1}(x)} & \frac{1}{2} \lt x \lt 1 \\ 1 & x = 1 \\ g^{-1}(\frac{1}{x}) & 1 \lt x \lt 2 \\ \dfrac{1} {g\left(\frac{1}{x}\right)} & 2 \leq x \\ \end{cases} $$

EDIT 2: Here's an example of such a $g$:

$$ g(x) = \begin{cases} \dfrac 1 2 + \dfrac 1 {n+1} & x = \dfrac 1 n, n \in \Bbb{N} \\ \dfrac 1 2 + x & \text{otherwise} \\ \end{cases} $$

Solution 3:

This is by no means a complete answer, but it is a bit long for a comment.

If $f\circ 1/f = 1/f \circ f = x$, we have the equations:

$$f(1/f(x)) = x$$

$$f(f(x)) = 1/x$$

In particular, $f$ is a half-iterate of $1/x$. Note that there cannot possibly be a solution defined at $x=0$.

One observation is that if $f$ satisfies the above equations, $1/f$ does as well. So if we could somehow prove that there was at most one solution to the above equations, then we could conclude that $f=1/f$ and arrive at a contradiction.

This problem deserves a very clear specification about what domain $f$ should be defined on, whether it should be continuous, and so forth. For example, the answer $x^i$ works as a function on the complex plane, but it can only be a continuous function away from a branch cut. And the much more interesting question of whether there exists such a function $\mathbb{R}^+ \to\mathbb{R}^+$ is still open.

Solution 4:

The complex function $f(x) = x^{\pm i}$ satisfies this property.

Its inverse is $f^{-1}(x) = x^{\mp i}$ since $f^{-1}\circ f(x) = \left(x^{\pm i}\right)^{\mp i} = x$.

One can see that $1/f(x) = x^{\mp i} = f^{-1}(x)$.

Solution 5:

For definiteness, let $X$ and $Y$ denote non-empty sets of complex numbers and let $f:X \to Y$ be a bijection whose inverse $f^{-1}:Y \to X$ is the reciprocal $1/f:X \to 1/Y$. (Here, "$1/Y$" denotes the set of reciprocals of elements of $Y$. Consequently, $X = Y = 1/Y$: The domain and target of $f$ are the same set, and are closed under taking reciprocals, and $0$ is not an element of $X$.)

At the level of elements, for all $x$ in $X$ and $y$ in $Y$, $y = f(x)$ if and only if $$ x = f^{-1}(y) = \frac{1}{f(y)}. \tag{1} $$

Substituting $y = f(x)$ in (1) and taking reciprocals, $$ f^{[2]}(x) := f\bigl(f(x)\bigr) = \frac{1}{x}\quad\text{for all $x$ in $X$.} \tag{2} $$ Applying $f$ to both sides and using associativity of composition, $$ f^{[3]}(x) = f\left(\frac{1}{x}\right) = \frac{1}{f(x)}\quad\text{for all $x$ in $X$.} \tag{3} $$ One more application of $f$ gives $$ f^{[4]}(x) = f^{[2]}\left(\frac{1}{x}\right) = x\quad\text{for all $x$ in $X$.} \tag{4} $$ That is, $f$ generates an action of the cyclic group of order $4$ on $X$. Further, each point $x$ in $X$ "generates" a square in which the solid arrows denote application of $f$, and the dashed arrows represent application of the reciprocal function ($x \mapsto 1/x$).

Such functions certainly exist (modulo the axiom of choice, depending how large you want the domain $X$ to be); just pick distinct complex numbers $a$ and $b$ other than $0$, $1$, and $-1$, place $a$, $b$, $1/a$, and $1/b$ cyclically around the corners of such a square, and define $f$ according to the diagram. (You can also consistently define $f(1) = 1$ and/or $f(-1) = -1$, or $f(1) = -1$ and $f(-1) = 1$. These numbers are special because they're fixed by the reciprocal function itself.)

The question remains: Does any such function have an explicit algebraic formula? Offhand I don't know.

The function proposed in the comments, $f(x) = x^{i} = \exp(i\log x)$ with $\log$ denoting the principal branch, does not work without qualification: For $x > 0$ real, $$ f\bigl(f(x)\bigr) = \exp\bigl(i\log(\exp i\log x)\bigr) \neq \exp(-\log x) = \frac{1}{x} $$ unless $\log(\exp i\log x) = i\log x$, which happens if and only if $\exp(i\log x)$ has absolute argument less than $\pi$, if and only if $e^{-\pi} < x < e^{\pi}$.

The formula $f(x) = \exp(i\log x)$ does work on any reciprocal-invariant, $f$-invariant set. No real interval has this property, however, since $f$ is not real-valued on any real interval. Every open disk with $a > 0$ and $1/a$ lying on a diameter is reciprocal-invariant, but a bit of numerical experimentation suggests no such disk is $f$-invariant. (I haven't carefully checked.)