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Having some problems with this functional equation:

$\frac{g(x)}{g(-x)} = r^{2x}$

Given from the assignment is that $x \in\mathbb{R}$ and $r > 0$ ($r \in\mathbb{R}$).

We are rather confident that $g(x) = r^x p(x)$, where $p(x)$ is an even function, is a solution. But, we lack a solid agrument for this, and have no good way of proving that it's the only solution (or that thare are more..).

Orre Snorre
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  • Why don't you try to define $p(x)$ so, that $g(x)=r^xp(x)$ -- it's one-to-one for $r>0$ and substitute it back into the equation? – Alexey Burdin Nov 19 '19 at 16:05

1 Answers1

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Define $p(x)=g(x)r^{-x}$. Then $p(-x)=g(-x)r^x=g(x)r^{-2x}r^x=g(x)r^{-x}=p(x)$. So indeed, $g(x)$ must be $r^{x}$ times some even function.

ajotatxe
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