The following has come up in some work I'm doing: If $\frac{f(x+a)}{f(a)}=g(x)$, where $g(x)$ is given and $a \ge 0$ is a constant, what is $f(x)$? We can assume that $g(x)>0 \ \forall x$ . Of course a solution would be great, but I'd appreciate even general information on this equation, such as how it would be referred to (functional equation with translation?), similar equations, etc. Thank you.
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Do you mean $f(x+a)/f(x)$? – mjqxxxx Dec 11 '13 at 21:41
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No, the denominator really is $f(a)$ – Greg Reese Dec 12 '13 at 18:09
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1Then it's trivial; $f(x+a)=g(x)f(a)$, so $f(x)=g(x-a)f(a)$, where $f(a)$ is arbitrary and $g(0)$ must be equal to $1$. – mjqxxxx Dec 13 '13 at 21:18
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$f(x) = e^x$ and $g(x) = e^x$, in fact: $$f(x+a) = e^{x+a} = e^x e^{a} = f(a) g(x)$$
Note that $g(x) = e^x > 0 ~ \forall x$ as for hypotesis.
the_candyman
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$g(x)$ is given, you can't pick it. For example, $g(x)$ may be specified as $x^2$, $1/(1+|x|)$, etc. – Greg Reese Dec 12 '13 at 18:16
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We have $f(x)=f(x-a+a)=g(x-a)f(a)$. Especially for $x=a$ we have $f(a)=g(0)f(a)$ and as $f(a)\neq 0$ we must have $g(0)=1$. This means there exists a solution only if $g(0)=1$ and in that case it is $f(x)=g(x-a)c$ where the constant $f(a)=c$ can be chosen.
flonk
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Thanks! That's what I came up with too but I got stuck on the step $f(x)=g(x-a)f(a)$ and didn't see that $f(a)$ could be arbitrary. – Greg Reese Dec 12 '13 at 18:31
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