Post History
#2: Post edited
by
zayn
·
2024-07-02T18:23:15Z (6 months ago)
- Hey, codidact!
- 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 there any way I can formalize the proof along these lines? Or is this idea unworkable with?- I'd appreciate help with tags.
- Hey, codidact!
- 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?
- I'd appreciate help with tags.
#1: Initial revision
by
zayn
·
2024-07-02T18:22:00Z (6 months ago)
Formalizing proof of existence of roots of functional equation
Hey, codidact!
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 there any way I can formalize the proof along these lines? Or is this idea unworkable with?
I'd appreciate help with tags.