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I have a question about period doubling and fixed points in the logistic map. Let's say I have a basic logistic map, $$f(x) = 4\lambda x(1-x).$$ Let me then compare 1,2 and 4 iterations of this map on fixed points. I assume that $\lambda$ is large enough such that two period doublings have occured, and a 4-cycle exists.

Now if $\bar{x}$ represents a fixed point for $f^4(x)$ (4 iterations of the map), then we have by definition, $$f^4(\bar{x}) = \bar{x}.$$ Obviously from this it follows that applying the function $f^2(\bar{x})$ twice should also take one back to point $\bar{x}$, however this is where my confusion comes in. We are in a 4 cycle, so 4 different fixed points relative to $f^4$ exist. Let's call the fixed points closest to $\bar{x}$ (nearest neighbour) $\bar{x}'$.

My lecture notes state that, $$f^2(\bar{x}) = \bar{x}'.$$ In other words, applying the map twice to a fixed point always takes one to the nearest neighbour. It is stated that this can also be extended to the general n-cycle case, where one would have, $$\bar{x}' = f^{2^{n-1}}(\bar{x}).$$

I do not understand why these properties are true. There are 4 different fixed points, and I don't see why the path through these different points has to follow this form, since as I understand it all these points are practically equivalent. Why if I apply $f^2$ can't I go to a different point further away?

I suppose there is some connection between the nearest neighbours $\bar{x}$ and $\bar{x}'$, but I am just not seeing it. Any help would be greatly appreciated!

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    $\begingroup$ If $\bar x$ is a fixed point of $f^4$ then we have $f^4(\bar x) = \bar x$, not necessarily $f(\bar x) = \bar x$. If 4 is the order of the periodic cycle, then we actually have $f(\bar x) \neq \bar x$ because the order of a cycle is the smallest $k$ such that $f^k(\bar x) = \bar x$. $\endgroup$
    – emacs drives me nuts
    Commented Feb 7, 2020 at 15:47
  • $\begingroup$ Yes this I understand, My confusion lies in the application of the second order map, which apparently always takes one to the nearest neighbour. $\endgroup$
    – Jasper
    Commented Feb 7, 2020 at 15:57
  • $\begingroup$ When you have an attractive cycle, then around each element of the cycle you have an area called "immediate basin of attraction". In your example of cycle of order 4 for $f$ the cycle for $f^2$ has order 2. As you are hopping from one immediate basin to the next, the distance to the cycle might change, but as you are cycling, the points in these basins will converge to the cycle. $\endgroup$
    – emacs drives me nuts
    Commented Feb 7, 2020 at 16:20

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I think I understand what you are asking. Apologies if this doesn't address the right question.

The first period doubling splits a single fixed point into two, call them $x_1$ and $x_2$. The next period doubling split each of these into two, call them $x_{1a}$, $x_{1b}$, $x_{2a}$ and $x_{2b}$.

Your question is why is the order something like $x_{1a}$, $x_{2a}$, $x_{1b}$, $x_{2b}$? Why not $x_{1a}$, $x_{1b}$, $x_{2a}$, $x_{2b}$?

Consider f with a $\lambda$ just big enough to split into 4. It is very much like f with a slightly smaller $\lambda$, where the sequence is $x_{1}$, $x_{2}$, $x_{1}$, $x_{2}$.

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  • $\begingroup$ Nice handwaving. ;-) +1 It conveys well the intuition, but do you know if it's really always true or instead model dependent? $\endgroup$
    – stafusa
    Commented Feb 7, 2020 at 17:01
  • $\begingroup$ Ah right I see, thanks a lot! This really helps in understanding, I suppose this also easily rescales to the arbitrary situation, since the operation $$f^{2^{n-1}} \rightarrow f^{2^{n}}$$ is really similar for any n. $\endgroup$
    – Jasper
    Commented Feb 7, 2020 at 17:04
  • $\begingroup$ @stafusa - Thank you. I expect it should be true. Think of the multivalued function that maps $\lambda$ to the fixed points of $f_{\lambda}$. My expectation is that for all reasonable single hump f, the function is continuous and fixed points are separated except at points where the period doubles. But I have no proof. $\endgroup$
    – mmesser314
    Commented Feb 8, 2020 at 1:57
  • $\begingroup$ @mmesser314 Thanks. It makes sense and I expect answer is well known (maybe even by me and I forgot it). It could be interesting to look at not-so-well-behaved maps such as the sawtooth map. $\endgroup$
    – stafusa
    Commented Feb 9, 2020 at 22:53

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