# Function satisfying $f^{-1} =f'$

How many functions are there which are differentiable on $(0,\infty)$ and they satisfy the relation $f^{-1}=f'$.

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What is the motivation for this question? –  JBL Jul 31 '10 at 20:07
Unless I am missing something, this is an elementary differential equations question, and so might be more appropriately asked at math.stackexchange.com . –  Emerton Jul 31 '10 at 20:34
If $f^{-1}$ means $1/f$, then yes, it is an easy differential equations question. On the other hand if $f^{-1}$ is the functional inverse of $f$, then it looks pretty hard. –  Robin Chapman Jul 31 '10 at 21:18
$f^{-1}$ means inverse of $f$ –  S.C. Jul 31 '10 at 21:30
Dear Chandru1, My apologies; I misunderstood the notation (in exactly the way that Robin Chapman suggested). –  Emerton Aug 1 '10 at 1:20

Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum_{k \ge 1} c_k t^k$ we obtain from $(*)$ a recursion formula for the $c_k$, and one can show that $0< c_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.

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I know little about analysis, so my apologies if this is a silly question. Are you solving a version of the problem where the function f is not necessarily defined on (0,∞) but is defined on some interval (0,b) (which is also an interesting problem)? –  Tsuyoshi Ito Aug 1 '10 at 14:16
The function $f$ constructed here is a priori only defined in a neighborhood of the point $a$. –  Christian Blatter Aug 1 '10 at 14:57
Thanks for the clarification. –  Tsuyoshi Ito Aug 1 '10 at 14:59

Wow. I remember that I thought exactly the same problem out of curiosity as a high school student but did not reach an answer. In fact, I was thinking about posting this problem on MathOverflow!

At least it is easy to construct one solution: f(x)=xφφ−1, where φ=(1+√5)/2 is the golden ratio.

Edit: Corrected the calculation. Thanks to Aaron Meyerowitz for spotting the error!

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Which does not answer the question, as far as I can tell. –  Did Nov 18 '12 at 12:53
@Didier Piau: It clearly does not. In case anyone is wondering, the asker posted the answer shortly after Christian Blatter posted his related analysis, and deleted it after I asked him if he had posted the question knowing the answer. –  Tsuyoshi Ito Nov 20 '12 at 1:04