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Let's say $f:D\to R$ is an injective function on some domain where it is also differentiable. For a real function, i.e. $D\subset\mathbb R, R\subset\mathbb R$, is it possible that $f'(x)\equiv f^{-1}(x)$?

Intuitively speaking, I suspect that this is not possible, but I can't provide a reasonable proof since I know very little nothing about functional analysis. Can anyone provide a (counter)example or prove that such function does not exist?

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  • $\begingroup$ Can you be clearer about the domains? Is $f$ supposed to be a bijection $D\to R$? If so, then shouldn't $D=R$, since $D$ is the domain of $f'$ and $R$ is the domain of $f^{-1}$? And is $D$ required to be an interval? $\endgroup$ – Eric Wofsey Aug 31 '16 at 6:17
  • $\begingroup$ @EricWofsey Yes it should. I stated the problem in general form. But this assumption restricts $R$ to be equal to $D$ $\endgroup$ – polfosol Aug 31 '16 at 6:22
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    $\begingroup$ "Functional analysis" has a specific meaning which is different from what you had in mind, I don't think it has any relevance to questions like this. $\endgroup$ – Meni Rosenfeld Aug 31 '16 at 15:01
  • $\begingroup$ If you pick $g$ so that $f(g(x)) = g(x+1)$, then the problem reduces to solving the delay differential equation $g(x-1) g'(x) = g'(x+1)$. Unfortunately I have no good ideas for that one. $\endgroup$ – Hurkyl Sep 1 '16 at 6:56
  • $\begingroup$ @Hurkyl Does the Greg's answer fit into your equation? I wasn't able to verify that $\endgroup$ – polfosol Sep 1 '16 at 7:03
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It is possible! Here is an example on the domain $D=[0,\infty)$: $$ f(x) = \bigg(\frac{\sqrt{5}-1}{2}\bigg)^{(\sqrt5-1)/2} x^{(\sqrt5+1)/2}. $$ I found this by supposing that $f(x)$ had the form $ax^b$, setting the derivative equal to the inverse function, and solving for $a$ and $b$.

graph of $f(x)$ and $f'(x)$

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    $\begingroup$ Robert Israel beat me by two minutes :) $\endgroup$ – Greg Martin Aug 31 '16 at 6:44
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    $\begingroup$ Nice answer. Now what about the case where $D=\mathbb R$? $\endgroup$ – polfosol Aug 31 '16 at 6:52
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    $\begingroup$ @polfosol: That is impossible, since in order to be injective $f$ must be monotone, so $f'$ must always have the same sign. $\endgroup$ – Eric Wofsey Aug 31 '16 at 7:09
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    $\begingroup$ Is it a coincidence that $f(x)=(\phi-1)^{\phi-1}x^{\phi}$, where $\phi$ is the golden ratio? Probably, but still noteworthy :) $\endgroup$ – Bobson Dugnutt Aug 31 '16 at 10:13
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    $\begingroup$ @Lovsovs: Of course it's not a coindicence. $\phi^2=\phi+1$ or equivalently $\phi^{-1}=\phi-1$. For a function $ax^b$, the derivative has exponent $b-1$ and the inverse has exponent $b^{-1}$, so for them to be equal $b$ must be a solution of $b^{-1}=b-1$, either the normal golden ratio or the other solution. And since the exponent is the golden ratio, and the coefficient is chosen so everything fits, it's natural that it will be based on $\phi$ as well. $\endgroup$ – Meni Rosenfeld Aug 31 '16 at 15:06
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On $(0, \infty)$, take $f(x) = a x^p$ where $p = (\sqrt{5}+1)/2$ (so that $p(p-1) = 1$) and $a = p^{-1/p}$.

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There have already been examples with $f: D \to \mathbb R$, but note that it is not possible with $f:\mathbb R \to \mathbb R$. A simple argument is that for a function $f$ to be injective, necessarily $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$. Thus we can see that for there to be equality between $f'(x)$ and $f^{-1}(x)$, then we must have $f^{-1}(x) \geq 0$ or $f^{-1}(x) \leq 0$ for all $x$.

But this can't happen, because any function defined on $f: \mathbb R \to \mathbb R$ must have its inverse go from positive to negative for some $x$. To confirm this, just look at the fact that the inverse of any horizontal line must cross the x-axis by flipping over the line $y=x$, and then add curves to that line to find that nothing has changed, and it still must cross the x-axis.

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    $\begingroup$ A right arrow symbol for 'tends to' or 'goes to' is not 'dash + greater-than' but rather a LaTeX symbol \to (usually synonym to \rightarrow), which renders as $\to$. $\endgroup$ – CiaPan Aug 31 '16 at 7:54
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    $\begingroup$ @polfosol No, \mapsto is used for denoting a single value mapping. For example, $f: C\to X$ expresses we mean a function $f$ going from a set $C$ to a set $X$, while $q: x\mapsto \sin (x/2)$ expresses we mean a function $q$ which assigns each $x$ argument a value of a sine of a half of $x$. See Wikipedia article 'List of mathematical symbols', section 'Symbols that point left or right' $\endgroup$ – CiaPan Aug 31 '16 at 8:01
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    $\begingroup$ The example given is for $f: R \to R$, so this argument must be flawed. Did you mean $f: \mathbb{R} \to \mathbb{R}$ instead? $\endgroup$ – Peter Taylor Aug 31 '16 at 10:38
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    $\begingroup$ @PeterTaylor Yes, sorry about that. I've changed the respective "R"'s to the symbol for the real numbers instead. Thank you. $\endgroup$ – Striker Aug 31 '16 at 16:30
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    $\begingroup$ An easier way to observe that we can't have $f^{-1}$ always positive or always negative is that it's positive at $f(1)$ and negative at $f(-1)$... $\endgroup$ – Ben Millwood Sep 1 '16 at 8:47
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Since someone mentioned whether the answer given by Robert Israel / Greg Martin might be unique, I thought it is worth noting that the function $$f(x)=-\frac{1}{\phi^\phi}(-x)^{-\frac{1}{\phi}},\quad x<0$$ where $\phi$ is the golden ratio, has the same property on $D=(-\infty,0)$, i.e. $f'(x)\equiv f^{-1}(x)$.

Edit- So if we define: $$f:\mathbb R\to\mathbb R\\x\mapsto a(x/a)^a$$ where $a=\frac{1+\sqrt{5}\text{ sign}(x)}{2}$, we would have a bijection on $\mathbb R$ with that nice property (yay!...).

enter image description here

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    $\begingroup$ And, I know that $f'(x)\ne f^{-1}(x)$, but: $f'(x)\equiv f^{-1}(x)$ :) $\endgroup$ – polfosol Sep 6 '16 at 15:33
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    $\begingroup$ I never imagined the golden ratio would come up in such a strange place! $\endgroup$ – Striker Sep 6 '16 at 17:52

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