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Math Help - f(x^2+y^2)=g(x)g(y) ...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    f(x^2+y^2)=g(x)g(y) ...

    Hello everybody!

    A problem:

    Find functions \varphi , \psi  which are fulfilling:

    \varphi(x^2+y^2)=\psi(x) \psi(y) for all x,y.

    Prove that if \varphi , \psi  are fulfilling the above equation then \psi determined by \varphi . How?


    Thank you.
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  2. #2
    Super Member girdav's Avatar
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    For example \phi=\psi =0. A more interesting example is given by \phi :x\mapsto e^x and \psi:x\mapsto e^{x^2}.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by girdav View Post
    For example \phi=\psi =0. A more interesting example is given by \phi :x\mapsto e^x and \psi:x\mapsto e^{x^2}.

    What you wrote and the family of functions: a^x where a>1 ?

    What about the second part of the question?

    Thanks!
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  4. #4
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    This is how I might attempt to do this problem. I will assume that both \psi and  \phi are smooth. Taking the natural log of both sides gives

    \ln \phi(x^2+y^2) = \ln \psi(x) + \ln \psi(y).

    Call the term on the RHS F(x^2+y^2). Differentiating both side wrt x and y gives

    F''(x^2+y^2) = 0.

    Thus, F(x^2+y^2) = a(x^2+y^2) + \ln(b) where a and b are constant.

    So \ln \phi(x^2+y^2) = a(x^2+y^2) + \ln(b)

    so \phi(x^2+y^2) = b e^{a(x^2+y^2)} = \psi(x)\psi(y).

    Now set y = 0 and this gets you the form for \psi(x) = k e^{ax^2}.  Then substitute into the original functional equation to determine k.
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Hello everybody!

    A problem:

    Find functions \varphi , \psi  which are fulfilling:

    \varphi(x^2+y^2)=\psi(x) \psi(y) for all x,y.

    Prove that if \varphi , \psi  are fulfilling the above equation then \psi determined by \varphi . How?


    Thank you.
    Maybe there is something missing from the problem statement. Otherwise
    \phi(x) = e^x and
    \psi(x) = - e^{x^2},
    combined with girdav's solution, provides a counterexample to the second part (uniqueness).
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