Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
11  
What is the motivation for this question? –  JBL Jul 31 '10 at 20:07
1  
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
4  
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 $f$ –  Chandrasekhar Jul 31 '10 at 21:30
3  
Dear Chandru1, My apologies; I misunderstood the notation (in exactly the way that Robin Chapman suggested). –  Emerton Aug 1 '10 at 1:20
add comment

2 Answers 2

up vote 12 down vote accepted

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.

share|improve this answer
    
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
add comment

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!

share|improve this answer
    
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
add comment

Your Answer

 
discard

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.