The basic example of such a function is $f(x) = e^{ax^2}$ for any constant $a$. Are these the only functions with this property, or are there others?
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Please clarify your question. Do you mean that if $$x^2+y^2 = z^2 + t^2$$ then $$f(x)f(y)=f(z)f(t)$$? – Crostul Aug 23 '22 at 18:49
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If you are looking for continuously differentiable functions which satisfy this relation, then yes, only non-trivial functions would be the Gaussians, per Mariano's answer. – Doug Aug 23 '22 at 19:02
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What do you mean by probably? :-| – Mariano Suárez-Álvarez Aug 23 '22 at 19:02
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Well, the zero function is a trivial solution, right? – Doug Aug 23 '22 at 19:03
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$\def\RR{\mathbb{R}}$Suppose that $f$, $g:\RR\to\RR$ is smooth functions such that $f(x)f(y)=g(x^2+y^2)$ for all choices of $x$ and $y$ in $\RR$, and to avoid trivialities that there exists $y_0$ in $\RR$ such that $y_0f(y_0)\neq0$.
Taking derivatives with respect to $x$ and $y$ we see that $f'(x)f(y)=2xg'(x^2+y^2)$ and $f(x)f'(y)=2yg'(x^2+y^2)$ identically, so that $yf'(x)f(y)-xf(x)f'(y)=0$.
In particular, $y_0f'(x)f(y_0)-xf(x)f'(y_0)=0$ for all $x$, and the choice of $y_0$ implies that $$ f'(x) = \frac{f'(y_0)}{y_0f(y_0)}xf(x)$$.
You can now solve this differential equation.
Mariano Suárez-Álvarez
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1The function $g(x^2+y^2)$ has rotational symmetry, so the function $f(x)f(y)$ also does. Now a function $h(x,y)$ has rotational symmetry if and only if $yh_x-xh_y=0$. This is how you come up with this. – Mariano Suárez-Álvarez Aug 23 '22 at 19:10
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Can you find the functions $f$ such that there is a $g$ with $f(x)f(y)=g(xy)$? – Mariano Suárez-Álvarez Aug 23 '22 at 19:15