Mathematics

I've been solving for solutions to $f^n(x)=x$ where $f(x) = 4x(1-x)$ (the r=4 case of the logistic map, yes) and $f^n(x) = f(f^{n-1}(x))$ is defined recursively. Domain and range are both $[0,1]$.

I think I'm almost there - just need some logic to clear the (apparently) last hurdle.

Since $f^n(x)$ can arise from two different values of $f^{n-1}(x)$ (symmetrical about 0.5) I began to build a sort of reverse tree where I go backwards from $f^n(x)$ to $2^n$ possible values of $x$. This implies that for any $f^n(x) = k, 0 \lt k \lt 1$ there are $2^n$ distinct $x$ available to iterate on. (The $f^n(x) = 1, 0$ cases have some number of repeated roots, but that does not matter, as I am solving for equality to $x$, and 0 is a solution and I leave it at that)

I have this inkling of an idea: since there are infinite such $k$ I should be easily able to find $2^n$ values of $k$ such that $k$ is a part of ${\{x:f^n(x)=x}\}$. But this sort of reasoning seems highly inaccurate to me.

Is this even an acceptable conjecture? Is there any way I can formalize the proof along these lines?

is a duplicate

This question has been asked before and has already been answered. It should be marked as a duplicate.

Please enter the URL of the proposed duplicate in the details field below.

not constructive

This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution for the specific problem of the asker.

Communities

Comments on Formalizing proof of existence of roots of functional equation

Formalizing proof of existence of roots of functional equation

1 comment thread