If $f : \mathbb{N} \to \mathbb{N}$ and $f(x)=f(2x^2)$ find all possible solutions to the functional equation.
I think it's a constant since it doesn't really fit anything. But how do I show that?
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What do you want to slove? $f$ or $x$? – W. mu May 31 '18 at 03:25
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@W.mu $f$ (updated) – Mathejunior May 31 '18 at 03:26
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4From a fixed x, the orbit is $x,2x^2,2^5x^4,\cdots$. There are more than one orbit, so the functional can be not constant. – W. mu May 31 '18 at 03:30
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@W.mu I meant, the possible solution is $f(x)=a$ $\forall x \in \mathbb{N}$ and for some $a\in \mathbb{N}$. What did you just mean by "not constant"? – Mathejunior May 31 '18 at 03:32
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$f$ can take different values on different orbits. – W. mu May 31 '18 at 03:34
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1To be more specific, $f(1) = f(2) = f(8) = f(128) = f(32768) = ....$ can have one value, whereas $f(3) = f(18) = f(648) = f(839808) = ...$ can have a different value. – Lee Mosher May 31 '18 at 03:36
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You are going to be working a long time as there are continuum many solutions. You can choose $f(1)$ to be anything, then $f(2),f(8),f(128),$ and so on have to have the same value. There is nothing to constrain $f(3)$ so it can be anything you like, then $f(18),f(648)$ and so on have to match that. Every prime starts a new series, as does any product of odd primes. I haven't figured out if the chain that starts with $4$ ever meets the chain that starts with $2$. Yes, $f$ can be constant, but it doesn't need to be.
Ross Millikan
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I feel that $f$ is a constant but I can't really show it. It might not be a constant but still, I am not too sure about that. – Mathejunior May 31 '18 at 03:38
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This answer, and my comment, have given examples where $f$ is not a constant. So, it's not very surprising that you are unable to show that $f$ is a constant. My advice is: never try to prove false statements. – Lee Mosher May 31 '18 at 03:39
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The sequence starting with $x=2$ is $2^{2^j-1}$, the one starting from $4$ is $2^{3 \times 2^{j-1}-1}$. As $2^j \neq 3 \times 2^{k-1}$ the two orbits never meet. – Fabian May 31 '18 at 03:51