# If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$

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 $a_{2n}$

• 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

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.

• 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 $g(u+v)=g(u)g(v)$.

• 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
• @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

Let' assume you want solution to $$f:\mathbb{R}\mapsto\mathbb{R}$$ that satisfies this equation $$\forall x,y\in\mathbb{R},f(x)f(y)=f\left(\sqrt{x^2+y^2}\right)\tag{1}$$

Let's plug in $$(x,y)=(0,0)$$ in the equation $$(1)$$, $$f(0)^2=f(0)\Rightarrow f(0)\in\{0,1\}$$

Case-1 : $$f(0)=0$$

For $$x>0, f(x)=f\left(\sqrt{x^2+0^2}\right)=f(x)f(0)=0$$.

For $$x<0, f(x)^2=f(x)f(x)=f\left(\sqrt{x^2+x^2}\right)=f\left(\sqrt{2}|x|\right)=0\Rightarrow f(x)=0$$.

So, if $$f(0)=0$$, then $$\forall x\in\mathbb{R}, f(x)=0$$

Case-2 : $$f(0)=1$$

In this case, we claim that this function must be even.

Proof :

$$f(x)=f(x)f(0)=f\left(\sqrt{x^2+0^2}\right)=f\left(\sqrt{(-x)^2+0^2}\right)=f(-x)f(0)=f(-x)\tag{2}$$

Now, we will consider two assumptions under this case.

Assumption-A :

$$\exists a>0, f(a)=0$$.

Then, $$\forall x\geq a$$, $$f(x)=f\left(\sqrt{a^2+\left(\sqrt{x^2-a^2}\right)^2}\right)=f(a)f\left(\sqrt{x^2-a^2}\right)=0$$

Assumption-B :

$$\exists b>0, f(b)>0$$

Then, $$f\left(\sqrt2b\right)=f\left(\sqrt{b^2+b^2}\right)=f(b)^2>0$$ Proceeding this way, we can find arbitrarily large real number $$x$$ for which $$f(x)>0$$. Which contradicts with Assumption-A that says after reaching a bound, we can't find and real number $$x$$ for which $$f(x)>0$$.

So, we conclude that $$a$$ and $$b$$ can't exist simultaneously. If we assume the existence of $$a$$, We get the solution : $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$ We can check and indeed, this is a valid solution.

Now we assume the existence of $$b$$, that is, $$\exists b>0, f(b)>0$$. Since this assumption excludes the possibility of the existence of real number $$a$$ for which $$f(a)=0$$, we only have two options : either $$f(x)>0$$ or $$f(x)<0$$ for any real number $$x$$. We claim that $$\forall x, f(x)>0$$.

Proof : $$f(x)=f\left(|x|\right)=f\left(\sqrt{\left(\frac{x}{\sqrt2}\right)^2+\left(\frac{x}{\sqrt2}\right)^2}\right)=\left(f\left(\frac{x}{\sqrt2}\right)\right)^2>0$$ So, $$\forall x, f(x)>0$$.

Now, we will plug in $$(\sqrt{x},\sqrt{y})$$ where $$x,y\geq0$$ in equation $$(1)$$, $$f\left(\sqrt{x}\right)f\left(\sqrt{y}\right)=f\left(\sqrt{x+y}\right)\\\Leftrightarrow \ln\left(f\left(\sqrt{x}\right)\right)+\ln\left(f\left(\sqrt{y}\right)\right)=\ln\left(f\left(\sqrt{x+y}\right)\right)$$

Let's define $$\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$$ such that $$\phi(x)=\ln\left(f\left(\sqrt{x}\right)\right)$$. Note that $$\phi(x)+\phi(y)=\phi(x+y)$$ which satisfies the Cauchy's functional equation. So, $$\phi$$ is a solution to Cauchy's functional equation. The most trivial solution is $$\forall x\geq0,\phi(x)=kx$$ where $$k\in\mathbb{R}$$. There are non trivial solutions which are highly pathological functions.

So, analyzing all the cases we got this solutions :

1. $$\forall x, f(x)=0$$
2. $$f(x) =\left\{\begin{array}{10}1 & \mbox{if } x=0\\0 & \mbox{otherwise}\end{array}\right.$$
3. $$\forall x, f(x)=e^{\phi(x^2)}$$ where $$\phi:\mathbb{R}_{\geq0}\mapsto\mathbb{R}$$ and $$\phi$$ satisfies Cauchy's functional equation.

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