Results 1 to 5 of 5

Thread: f(x^2+y^2)=g(x)g(y) ...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    f(x^2+y^2)=g(x)g(y) ...

    Hello everybody!

    A problem:

    Find functions $\displaystyle \varphi , \psi $ which are fulfilling:

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

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


    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32
    For example $\displaystyle \phi=\psi =0$. A more interesting example is given by $\displaystyle \phi :x\mapsto e^x$ and $\displaystyle \psi:x\mapsto e^{x^2}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by girdav View Post
    For example $\displaystyle \phi=\psi =0$. A more interesting example is given by $\displaystyle \phi :x\mapsto e^x$ and $\displaystyle \psi:x\mapsto e^{x^2}$.

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

    What about the second part of the question?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    This is how I might attempt to do this problem. I will assume that both $\displaystyle \psi $ and $\displaystyle \phi $ are smooth. Taking the natural log of both sides gives

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

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

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

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

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

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

    Now set $\displaystyle y = 0$ and this gets you the form for $\displaystyle \psi(x) = k e^{ax^2}. $ Then substitute into the original functional equation to determine $\displaystyle k$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Also sprach Zarathustra View Post
    Hello everybody!

    A problem:

    Find functions $\displaystyle \varphi , \psi $ which are fulfilling:

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

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


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

Search Tags


/mathhelpforum @mathhelpforum