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I am currently trying to understand the analysis of a logistic-like map $$f_\mu (x) = 1-\mu x^2$$

after section 2.2 in "Renormalization Methods" by A. Lesne.

As I understand it, the physical situation is that $f_\mu$ has exactly $2^n$ attractors $x_{i\dots 2^n}$ in a certain region $\mu\in [0,\mu_c)$. Since one can order them such that $f_\mu(x_i) = x_{i+1}$ and the number of attractors grows in some way always by a factor of two, it is called "double-periodic scenario".

However, at $\mu_c\approx1.401$, this behavior cannot be found any more and one might interprete this as critical point since the system appears to be chaotic from there, or, in some sense, the number of iterations it takes until a point $x$ is reached again becomes infinity.

Here is a scetch of the situation: Logistic-Like-Map

At certain points $\mu_i$, $i=[i,\infty]$, the number of attractors doubles until $\mu_c$ is reached where this periodicity cannot be found. $N$ has been chosen to be sufficiently large.

To understand the transition from the periodic point of view, Lesne conjectures that some delta exists such that $$\lim_{i\rightarrow \infty}\delta^i (\mu_c - \mu_i)=A\neq0\ .$$ Then, it is stated that $\delta \approx 4.67$ is somehow universal and can be derived using a renormalization approach of the form $R\left[ f\right] \propto f^k$ with $R$ beeing an operator that contains in the end all the informations about the system.

Two things are unclear to me:

How can one use the renormalization operator $R$ to analyze the critical behaviour, thus $\delta$?


Is there a systematic way to find an appropriate $R$ better than just looking at self-similarities of some graphs?

Thank you in advance

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The book by Sethna has a very good exercise on understanding the logistic map with renormalisation; it walks you through the steps but leaves the calculations up to you. Using renormalisation in practise needs practice, so I think the best way forward here is to actually do it. – genneth Oct 1 '11 at 13:35
Link to the exercise:… – genneth Oct 1 '11 at 13:40
@genneth: Thank you for the link, this seems to be exactly what I need. As you say, understanding renormalization needs practice and this is what I need :) If you want, you can kind of sum up the main points of the exercise as an answer and I would be glad to accept it. Greets – Robert Filter Oct 2 '11 at 11:50
I think you should take Ron's answer --- it's correct and concise, and contains a great reference. – genneth Oct 2 '11 at 13:47
@genneth: So shall it be. – Robert Filter Oct 3 '11 at 18:28
up vote 5 down vote accepted

The best reference for this is Feigenbaum's original article, reprinted in "Universality in Chaos" by Cvitanovic. The point is that when you iterate a map, every time you period double, you fold up the function one more time. The behavior is dominated by the solution to the following equation:

$$ \alpha g(g(x/\alpha)) = g(x)$$

Which says that g iterated with itself and rescaled (both in the domain and range) looks just like g. The function $g$ is shifted relative to f, so that it's maximum is at 0, not at some point between 0 and 1, which means you don't have to follow the critical point under iteration. You can solve this condition more easily by imposing the symmetry $g(x)=g(-x)$ and using a Taylor expansion, and this gives g and $\alpha$, and $\alpha$ is the scale exponent. Everything about the critical behavior is determined by g, and this is described best in the original article.

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Good answer, though I think you should make it clear that your $g$ is a shifted version of the $f$ used by the questioner, such that the hump is centred at $x=0$. Also, I'm not sure the condition $g(x) = g(-x)$ is necessary --- I was under the impression that as long as you shift the window appropriately it will renormalise to the same universal function anyway. – genneth Oct 1 '11 at 16:25
@genneth: the condition is not necessary, but it halves the work of solving the equation, so it is useful. – Ron Maimon Oct 1 '11 at 16:40
@Ron: Thank you for your nice answer, I am sure that it is conceptually right. However, I will not accept it right now since someone might come up with something more detailed. Greets – Robert Filter Oct 2 '11 at 11:55

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