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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 $a \geq 0$ 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. Greg Reese

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  • $\begingroup$ Do you mean $f(x+a)/f(x)$? $\endgroup$ – mjqxxxx Dec 11 '13 at 21:41
  • $\begingroup$ No, the denominator really is $f(a)$ $\endgroup$ – Greg Reese Dec 12 '13 at 18:09
  • $\begingroup$ Then 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 $1$. $\endgroup$ – 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.

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  • $\begingroup$ $g(x)$ is given, you can't pick it. For example, $g(x)$ may be specified as $x^2$, $1/(1+|x|)$, etc. $\endgroup$ – 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.

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  • $\begingroup$ 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. $\endgroup$ – Greg Reese Dec 12 '13 at 18:31
  • $\begingroup$ My pleasure, you're welcome! $\endgroup$ – flonk Dec 12 '13 at 20:19
  • $\begingroup$ @GregReese BTW, you can vote or tick this answer, to close the topic. $\endgroup$ – flonk Dec 17 '13 at 13:42

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