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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.

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    $\begingroup$ Do you want all or just one non-trivial one? $e^{kx^2}$ seems to be a set of non-trivial solutions. Are there any assumptions about $f$? like continuous/differentiable, non-negative etc? $\endgroup$ – Aryabhata Mar 2 '12 at 21:30
  • $\begingroup$ I wish to know ways to solve such function equations in general. need methods. Thanks in advice $\endgroup$ – Mathlover Mar 2 '12 at 21:35
  • $\begingroup$ The answer to this question is called Maxwell's theorem. See this earlier question about it: math.stackexchange.com/questions/105418/… $\endgroup$ – Michael Hardy Mar 3 '12 at 3:15
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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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  • $\begingroup$ It is interesting that such function equations are directly related to $h(x)+h(y)=h(x+y)$. I will focus on that function equation now.thanks $\endgroup$ – Mathlover Mar 2 '12 at 22:20
  • $\begingroup$ Nice one.${{}}$ $\endgroup$ – Git Gud Sep 1 '13 at 20:18
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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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  • $\begingroup$ Thanks for answer. If so, Aryabhata is right that one of solution is $e^{kx^2}$. Is it possible to find other solutions in my power series method? $\endgroup$ – Mathlover Mar 2 '12 at 21:47
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    $\begingroup$ @Mathlover: Any other solution will have issues with regularity (continuity, differentiability, etc.) and thus will probably not have any global series expansion. $\endgroup$ – anon Mar 2 '12 at 21:56
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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

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