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Everyone knows that there are at least three functions whose derivative is the function itself, namely $e^x, \ 0$ and $-e^{x}$. ( are there more?)

I was drawing some polynomials and their derivatives and noted that sometimes it was almost like the inverse. This lead me to ask this question: is there a function whose derivative is the inverse of that function?

Well, I figured that at least some kind of answer can be found to be of the form $a x^b$.

Lets solve this:

$f(x) = ax^b, f'(x) = abx^{b-1}$. Then $$f \circ f'(x) = a^{b+1}b^bx^{(b-1)b}=x=a^b b x^{(b-1)b}=f' \circ f (x).$$

Thus $b(b-1) = 1 \iff b^2-b-1=0 \iff b = \phi \vee 1-\phi,\ \phi = \frac{1+\sqrt 5}{2}$

We also see that $ab^{b-1}=1$, because both the multipliers must be one. Thus we get $a= \frac{1}{b^{b-1}}$. If $b=\phi$, we get $a=\phi^{\phi-1}$. If $b=1-\phi, \ a=(1-\phi)^{\phi}$

Thus two functions that satisfy the condition are $\phi^{\phi-1} x^\phi$ and $ (1-\phi)^{\phi}x^{1-\phi}$.

I would like to know if there are more functions like these, and do these functions have any 'interesting' properties, like exponential function, apart from this one condition about inverse being the derivative?

  • 5
    The first assertion is wrong. The functions equal to their derivative are the functions $x\mapsto c\mathrm e^x$, for some $c$.2012-11-18
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    @did: Actually it's not c times e to the power of x. It's c times e to the power of cx2012-11-18
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    @ArmenTsirunyan No.2012-11-18
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    $$(ce^{cx})'=c^2e^{cx}\neq ce^{cx}\Longrightarrow$$ did is right.2012-11-18
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    @ArmenTsirunyan No. $y^{\prime}-y=0\Rightarrow y^{\prime}e^{-x}-ye^{-x}=0\Rightarrow (ye^{-x})^{\prime}=0\Rightarrow ye^{-x}=c\Rightarrow y=ce^{x}$2012-11-18
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    @did: OK, OK! You're right :)2012-11-18
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    Yes, made a stupid mistake there, silly me.2012-11-18
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    The question you are really asking is to determine the functions whose derivative is the inverse of that function, apart from the two you found. This question is interesting.2012-11-18
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    Hmmmm... *(are there more?)*... **YES THERE ARE** (see supra).2012-11-18
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    Ok, i think I got that there are infinitely many of those :)2012-11-18
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    This question appeared some time ago at mathoverflow. There you can find some solutions of the problem which are completely different from the above. See here: http://mathoverflow.net/questions/34052/function-satisfying-f-1-f/34095#340952012-11-18
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    @Christian Blatter Thanks, that place answers OK. I guess the question is more or less answered :) Thanks all.2012-11-18
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    Hey any feedback concerning my answer/idea would be nice...2014-06-03

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