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Math Help - Calc problem

  1. #1
    Forum Admin topsquark's Avatar
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    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: f ^{\prime} \left ( f(x) \right ) = x.

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

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

    or
    \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
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  2. #2
    MHF Contributor
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    Re: Calc problem

    Quote Originally Posted by topsquark View Post
    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: f ^{\prime} \left ( f(x) \right ) = x.

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

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

    or
    \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
    Thanks from topsquark
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