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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.

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    $\begingroup$ @Mr.MBB, you should edit your question to include that functional equation. $\endgroup$ – Sammy Black Dec 22 '15 at 10:07
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    $\begingroup$ $f(x)=x^i=e^{i\log x}$. $\endgroup$ – A.S. Dec 22 '15 at 10:17
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    $\begingroup$ It should be $f(x) = x^i = e^{iln x}, x>0$. Good example $\endgroup$ – Mr. MBB Dec 22 '15 at 10:24
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    $\begingroup$ $\ln x = \log x$ for any proper mathematician unless specified otherwise. $\endgroup$ – Zain Patel Dec 22 '15 at 14:04
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    $\begingroup$ A great article on exactly this topic: When Does $f^{-1} = 1/f$? by R. Cheng, A. Dasgupta, B. R. Ebanks, L. F. Kinch, L. M. Larson, and R. B. McFadden, in the American Mathematical Monthly, Vol. 105, Number 8 in October 1998. $\endgroup$ – Xoque55 Dec 23 '15 at 4:07
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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 $1/f(x) = x^{\mp i} = f^{-1}(x)$.

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    $\begingroup$ Formally this certainly works, but "$x^{i}$" isn't a (single-valued) function, and the law of exponents $(x^{a})^{b} = x^{ab}$ isn't unconditionally true for complex exponentiation. $\endgroup$ – Andrew D. Hwang Dec 22 '15 at 16:01
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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.


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    $\begingroup$ +1. As a side note, the claim 1 is also true for non differentiable functions, whereas the claim 2 is true for non differentiable functions whose sign is constant. Of course, the initial claim, that an invertible function is monotonic is true for continuous functions, but not for non continuous functions. $\endgroup$ – Taladris Dec 23 '15 at 1:04
  • $\begingroup$ Thanks for the note. I couldn't see how to relax the differentiability in the second claim, and the proof of the first claim is straightforward with that hypothesis. $\endgroup$ – Sammy Black Dec 23 '15 at 5:28
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    $\begingroup$ The second claim is elementary: the reciprocal function $g(x)=\frac{1}{x}$ is decreasing on $(-\infty,0)$ on $(0,\infty)$, so, if $f$ is increasing and has a constant sign, then the composition $1/f=g\circ f$ is decreasing. (To prove this last claim, remember that "increasing=order-preserving" and "decreasing=order-reversing". $\endgroup$ – Taladris Dec 23 '15 at 9:03
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    $\begingroup$ There is also an elementary proof of the first claim: assume that $y_1<y_2$ are two elements of the range of $f$. Then $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1$ and $x_2$ in the domain of $f$. We cannot have $x_1=x_2$ since $f$ is one-to-one. If $x_1> x_2$, then $y_1=f(x_1)> f(x_2)=y_2$ since $f$ is increasing. It is a contradiction with $y_1<y_2$, so $f^{-1}(y_1)=x_1<x_2=f^{-1}(y_2)$. $\endgroup$ – Taladris Dec 23 '15 at 9:07
  • $\begingroup$ Your proof certainly works but there can be no function defined on point $0$ that satisfies this equation, because if it was possible then the inverse function will equal to $0$ at some point which is not possible by equation. So the domain of our function cannot be all real numbers. $\endgroup$ – Юрій Ярош Dec 30 '18 at 17:40
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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} $:

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

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  • $\begingroup$ Excellent !!! I like this example. $\endgroup$ – Mr. MBB Dec 22 '15 at 22:44
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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.

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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 ($x \mapsto 1/x$).

Pointwise action of a map whose reciprocal is its inverse

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.)

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Let $$ g(x)= \frac{I}{1!} x -\frac{1 + I}{2!} x^{2} +\frac{3 + I}{3!} x^{3} -\frac{10}{4!} x^{4} +\frac{40 - 10I}{5!} x^{5} -\frac{190 - 90I}{6!} x^{6}\\ +\frac{1050 - 730I}{7!} x^{7} -\frac{6620 - 6160I}{8!} x^{8} +\frac{46800 - 55900I}{9!} x^{9} -\frac{365300 - 549900I}{10!} x^{10} \\ +\frac{3103100 - 5864300I}{11!} x^{11} -\frac{28269800 - 67610400I}{12!} x^{12} \\ +\frac{271627200 - 839594600I}{13!} x^{13} -\frac{2691559000 - 11186357000I}{14!} x^{14} \\ +\frac{26495469000 - 159300557000I}{15!} x^{15}\\ + O(x^{16}) $$ and $f(x) = g(x-1)+1$ then $$f(f(x)) = \frac{1}{x} \\ f(x)=\frac{1}{f°^{-1}(x)} \\ f°^{-1}(x) = \frac{1}{f(x)} $$ for $x-1$ in the range of convergence of $g(x)$.

(The range of convergence can be extended by Eulersummation as far as the resulting signs in $g(x)$ alternate, that means effectively for $f(x)$ with $x \gt -1$)


The function $g(x)$ has been constructed by the idea that $$ g°^2(x) = \exp(-1 \cdot \log(1+x))-1 \\ g(x) = \exp( i \cdot \log(1+x))-1 \\ $$ which has also already been suggested by earlier answers as well.
One can use Pari/GP with the function-call serlaplace(exp(I*log(1+x))-1) to get the coefficients for the function $g(x)$

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