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Math Help - Functional equation

  1. #1
    Senior Member DeMath's Avatar
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    Functional equation

    Solve the functional equation:

    f(xy)=[f(x)]^{y^\beta}[f(y)]^{x^\beta}, where x>0,~f(x)>0,~\beta\in\mathbb{R}.


    P.S. Good luck!
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  2. #2
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    it seems that f_{\alpha}(x)=x^{\alpha x^{\beta}}, \ \alpha \in \mathbb{R}, are all the solutions ...
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  3. #3
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    let b=beta
    taking the logarithm of the both sides , we get :

    ln(f(xy))= y^bln(f(x))+x^bln(f(y)) , let g(x)=ln(f(x))

    we get :  g(xy)=y^bg(x)+x^bg(y)

    dividing both sides by x^by^b:
    \frac{g(xy)}{x^by^b} = \frac{g(x)}{x^b} + \frac{ g(y)}{y^b}

    let :  h(x)= e^{\frac{g(xy)}{x^by^b}}
    we have :  h(xy)=h(x)+h(y) which is equivalent to : k(xy)=k(x)k(y) , if i remember well , the solution of the Last equation is  x^a for some real a , a difficult result of Erdos !!
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  4. #4
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    you'd need some sort of continuity condition i guess. anyway, just put g(x)=x^{-\beta} \ln f(x) to end up with the simple equation g(xy)=g(x)+g(y).
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    it seems that f_{\alpha}(x)=x^{\alpha x^{\beta}}, \ \alpha \in \mathbb{R}, are all the solutions ...
    Yes, but it would be even more interesting see your full solution, dear NonCommAlg.
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  6. #6
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    Quote Originally Posted by DeMath View Post
    Yes, but it would be even more interesting see your full solution, dear NonCommAlg.
    i've already posted it. see my second response.
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