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

**Also sprach Zarathustra** · May 26th 2011, 04:18 AM

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.

**girdav** · May 26th 2011, 04:26 AM

For example $\phi=\psi =0$ . A more interesting example is given by $\phi :x\mapsto e^x$ and $\psi:x\mapsto e^{x^2}$ .

**Also sprach Zarathustra** · May 26th 2011, 04:59 AM

Originally Posted by girdav

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!

**Danny** · May 27th 2011, 07:45 AM

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 $k$ .

**awkward** · May 29th 2011, 02:05 PM

Originally Posted by Also sprach Zarathustra

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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