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As we know, for the $f(x)f(y)=f(x+y)$ $f(x)=\mathrm e^{\alpha x}$ is a solution.

What about $f(x)f(y)=f(\sqrt{x^2+y^2})$? Does anybody know about the solution of the function equation?

I tried to find $f(x)$. See my attempts below to find $f(x)$.

$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$

$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$

$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$

$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$

$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$

if we use binom expansion for $(1+(x/y)^2)^{m}$

$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$

Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$

$ \begin{align} & f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\ \\ & \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\ \\ & + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\ \\ & = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots \end{align} $

If we equal for all $x^n$ terms in both sides

we can see $a_{2n-1}=0$, but to find $a_{2n}$ seems hard for me. Any idea to find $a_{2n}$

Thanks in advice.

  • 0
    The answer to this question is called Maxwell's theorem. See this earlier question about it: http://math.stackexchange.com/questions/105418/very-elementary-proof-of-maxwells-theorem2012-03-03

3 Answers 3

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If you set $g(x) := f(\sqrt{x})$ for $x \in [0, \infty)$ then you get $g(x)g(y) = f(\sqrt{x})f(\sqrt{y}) = f(\sqrt{x+y}) = g(x+y)$

You see that $g(x) \geq 0$, and if $g(x) = 0$ for some $x > 0$ then $g \equiv 0$. Thus, you can look at $h(x) := \log(g(x))$ It satisfies $h(x) + h(y) = \log(g(x)g(y)) = \log(g(x+y)) = h(x+y)$ If you impose any reqularity condition on $f$ you can think of, you will get $h(x) = \alpha x$, and consequently $f(x) = \exp(\alpha x^2)$ for $x > 0$. You can generalise this result to $x < 0$ using the fact that from the initial equation it follows that $f$ is even.

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    Nice one.${{}}$2013-09-01
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Change variable, $g(u) = f(\sqrt{u})$. You need to decide what you want for negative $u$. Then this functional equation becomes $g(u+v)=g(u)g(v)$.

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    @Mathlover: Any other solution will have issues with regularity (continuity, differentiability, etc.) and thus will probably not have any global series expansion.2012-03-02
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The answer to this question is a well known result called Maxwell's theorem, after James Clerk Maxwell. This earlier question deals with it:

very elementary proof of Maxwell's theorem