Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What kinds of function $f: \mathbb{R} \to \mathbb{R}$ can be written as some function of itself? I.e. $f'(x) = g(f(x))$ for some function $g$?

If $f$ is given, can $g$ be solved in terms of the symbol $f$ (not in terms of specific $f$), if $g$ exists?

My question is related to part 3 of my another question, which asks about when the variance can be represented as a function of mean, both as functions of a distribution parameter, and in particular, when the variance is the derivative of the mean.


share|cite|improve this question
Injective functions and constant ones can but there must be others. – xavierm02 Jun 30 '13 at 14:42
$f(x)=\exp(x)$ and $g(x)=x$. Do you have any additional constraints on $f$ and $g$? – deoxygerbe Jun 30 '13 at 14:43
@deoxygerbe: No additional constraints – Tim Jun 30 '13 at 14:46
If $f$ is given (smooth and injective) then $g=f'\circ f^{-1}$. – Did Jun 30 '13 at 14:58
@deoxygerbe He wasn't asking for examples. – Git Gud Jun 30 '13 at 15:13
up vote 4 down vote accepted

A necessary and sufficient condition is that $[f(x_1)=f(x_2)\implies f'(x_1)=f'(x_2)]$.

When this condition is met, one can define $g$ as follows:

  • If $t$ is not in $f(\mathbb R)$, then $g(t)=0$.
  • If $t$ is in $f(\mathbb R)$, then $g(t)=f'(x)$ for any $x$ such that $t=f(x)$.

The condition above is what is needed for the second part of this definition to be independent of the choice of $x$.

Thus, strictly monotone (smooth) functions $f$ are allright but $f=\cos$ is not.

share|cite|improve this answer

I thought I would somehow find a way to prove $S=L \cup I$ but looking at Did's answer, I start doubting so I'll just post that for the time being and edit if I find something.

Let $S=\left\{f \in \mathcal C^1\left(\Bbb R, \Bbb R\right) \mid \exists g \in \mathcal F\left(\Bbb R,\Bbb R\right), \forall x \in \Bbb R, f'(x) = g(f(x))\right\}$

Let $A=\left\{x \mapsto ax+b \mid a,b \in \Bbb R\right\}$

Let $I=\left\{f\in C^1\left(\Bbb R, \Bbb R\right)\mid \forall x,y \in \Bbb R, f(x)=f(y) \implies x = y\right\}$

Let $f\in A$

$\forall x \in \Bbb R,f'(x)=k=k\Bbb 1(x)$

So we can take $g=k\Bbb 1$

So $f \in S$

$\boxed{A \subset S}$

Let $f\in I$

We can find $h:f\left(\Bbb R\right) \to \Bbb R$ so that

$\forall x \in \Bbb R, h(f(x))=x$

Then, we have that $f'(x)=f'(h(f(x))$ so we can take $g = f'\circ h$

So $f \in S$

$\boxed{I\subset S}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.