Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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? –  Aryabhata Mar 2 '12 at 21:30
    
I wish to know ways to solve such function equations in general. need methods. Thanks in advice –  Mathlover Mar 2 '12 at 21:35
    
The answer to this question is called Maxwell's theorem. See this earlier question about it: math.stackexchange.com/questions/105418/… –  Michael Hardy Mar 3 '12 at 3:15

3 Answers 3

up vote 15 down vote accepted

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.

share|improve this answer
    
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 –  Mathlover Mar 2 '12 at 22:20
    
Nice one.${{}}$ –  Git Gud Sep 1 '13 at 20:18

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

share|improve this answer
    
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? –  Mathlover Mar 2 '12 at 21:47
3  
@Mathlover: Any other solution will have issues with regularity (continuity, differentiability, etc.) and thus will probably not have any global series expansion. –  anon Mar 2 '12 at 21:56

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.