most recent 30 from http://mathoverflow.net 2013-05-25T14:41:09Z http://mathoverflow.net/feeds/question/34052 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34052/function-satisfying-f-1-f Chandrasekhar 2010-07-31T19:57:06Z 2010-08-02T09:05:39Z

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

http://mathoverflow.net/questions/34052/function-satisfying-f-1-f/34061#34061 Tsuyoshi Ito 2010-07-31T22:02:18Z 2010-07-31T22:25:42Z

<p>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!</p> <p>At least it is easy to construct one solution: f(x)=x^φ/φ^φ−1, where φ=(1+√5)/2 is the golden ratio.</p> <p>Edit: Corrected the calculation. Thanks to Aaron Meyerowitz for spotting the error!</p>

http://mathoverflow.net/questions/34052/function-satisfying-f-1-f/34095#34095 Christian Blatter 2010-08-01T09:48:32Z 2010-08-01T12:45:58Z

<p>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 $0&lt; c_r&lt;1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|&lt; p$, 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.</p>