1. ## Calc problem

I dug this one up from the internet and thought I'd post it for the next challenge problem. Trouble is I have been unable to find a solution. So I decided to post it here and give what little progress I've made and see if anyone can do any better.

Let f(x) be a function. Solve for f: $\displaystyle f ^{\prime} \left ( f(x) \right ) = x$.

My work: Multiplying both sides by f'(x) and integrating over x gives
$\displaystyle f(x) = \int x f ^{\prime}(x) ~ dx$

and after integration by parts on the right:
$\displaystyle f(x) = x f(x) - \int f(x)~dx$

or
$\displaystyle \int f(x)~dx = (x - 1)f(x)$

All I can say at this point is that if I write f(x) as a power series all the coefficients in the series wind up being 0. I don't know where to take this next.

Note: I saw the problem statement but there was no solution to accompany it. That leaves open the possibility that there is no solution.

-Dan

2. ## Re: Calc problem

Originally Posted by topsquark
I dug this one up from the internet and thought I'd post it for the next challenge problem. Trouble is I have been unable to find a solution. So I decided to post it here and give what little progress I've made and see if anyone can do any better.

Let f(x) be a function. Solve for f: $\displaystyle f ^{\prime} \left ( f(x) \right ) = x$.

My work: Multiplying both sides by f'(x) and integrating over x gives
$\displaystyle f(x) = \int x f ^{\prime}(x) ~ dx$

and after integration by parts on the right:
$\displaystyle f(x) = x f(x) - \int f(x)~dx$

or
$\displaystyle \int f(x)~dx = (x - 1)f(x)$

All I can say at this point is that if I write f(x) as a power series all the coefficients in the series wind up being 0. I don't know where to take this next.

Note: I saw the problem statement but there was no solution to accompany it. That leaves open the possibility that there is no solution.

-Dan
not sure I agree with this

$f^\prime(f(x))=x$

$f^\prime(f(x))f^\prime(x) = x f^\prime(x)$

$\displaystyle{\int}f'(f(x))f^\prime(x)~dx = \displaystyle{\int}x f^\prime(x)~dx$

$u=f(x)$

$du=f^\prime(x)~dx$

$\displaystyle{\int}f^\prime(f(x))f^\prime(x)~dx \Rightarrow \displaystyle{\int}f'(u)~du = f(u) + C = f(f(x))+C$

$f(f(x))+C = \displaystyle{\int}x f'(x)$

ignoring $C$ this isn't

$f(x) = \displaystyle{\int} x f ^{\prime}(x) ~ dx$

as you've written above