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Question is same as above. Let me write again. Find the satisfying functions such that $f(x^5)=5f(x)$

Clearly seen that $c\ln x$ satisfy the condition. Question is what else?

Thomas Andrews
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Fuat Ray
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    For a first step, let $g(x):=e^{f(x)}$. This satisfies $g(x^5)=g(x)^5$. – Berci Jan 17 '22 at 21:11
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    There are lots of such functions, but few which are continuous. – Thomas Andrews Jan 17 '22 at 21:16
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    You should be specific about the domain of the function. For example, $c\ln x$ is only defined on $x>0.$ – Thomas Andrews Jan 17 '22 at 21:18
  • If $f(0)$, $f(1)$, $f(-1)$, $f(i)$, or $f(-i)$ is defined, then it would have to be $0$, because $f(x)=f(x^5)=5f(x)$ for $x \in {0,1,-1,i,-i}$. – Geoffrey Trang Jan 17 '22 at 21:56
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    Why are people upvoting this terrible question? It shows no work and it is not a complete question. – Thomas Andrews Jan 17 '22 at 23:47
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    In your domain, define an equivalence relation $\sim$ as $x\sim y$ iff $x=y^{5^k}$ for some integer $k.$ Then the equivalence class contain $x$ can be written $[x]={x^{5^n}\mid n\in\mathbb Z}.$ If we pick a unique set of representatives of the equivalence classes, we can define $f$ to be any value at those representatives, and all the other values will be determined. One unique set of representatives in$(1,\infty)$ is $[e,e^5).$ So given any $f_0:[e,e^5)\to\mathbb R,$ we can extend $f_0$ to all of $(1,\infty)$ just by the rule. For example, for $x\in[e^5,e^{25}),$ $f(x)=5f_0(x^{1/5}).$ – Thomas Andrews Jan 18 '22 at 00:12
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    You can get a lot of such functions on $(1,\infty),$ even continuous ones, one for every $f_0:[2,32)\to\mathbb R.$ Similarly on $(0,1)$ and $(1/32,1/2).$ You can define the function piecewise on $(0,1)$ and $(1,\infty)$ with $f(1)=0.$ So there are a lot of such functions, even continuous ones. It will take some work to get analytic ones. – Thomas Andrews Jan 18 '22 at 00:36
  • You need a more complicated equivalence relation on complex number. This is because there are multiple fifth complex roots of any $z.$ On $|z|>1,$ you'd say $z_1\sim z_2$ if $z_1^{5^a}=z_2^{5^b},$ for some natural numbers $a,b.$ Finding a good set of representatives seems difficult. – Thomas Andrews Jan 18 '22 at 01:27

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