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I have this bonus question in an assignment: How is the golden ratio related to the logistic equation in discrete time?

The logistic equation is $x_{n+1}=rx_n(1-x_n)$. The professor suggested looking at the number of 2 point cycles, and 4 point cycles, and 3, 5, 6, 7, point cycles, etc. I started doing this, but it quickly becomes cumbersome to find 4-point cycles and higher.

Any suggestions?

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2  
related : sounding number –  pedja Feb 8 '12 at 7:18
    
I don't see, how is it related to stochastic processes, so let me retag. –  Ilya Feb 8 '12 at 9:08
    
@draks: Transition to chaos actually occurs for $r > 4$, but showing that involves more sophisticated techniques (Schwartzian derivatives); see here: www2.warwick.ac.uk/fac/sci/maths/people/staff/…. This is why I'm a bit confused as to the hint. For instance, in the chaotic case, the map is conjugate to a right shift on the space of 2-symbols, hence periodic points (i.e. $k$-cycles) grow as $2^k$ (i.e. number of $k$-cycles is $2^k$). –  William Feb 12 '12 at 0:35
    
Oh sorry I broke the commenting flow, by deleting mine. They are part of my answer. See below. –  draks ... Feb 16 '12 at 14:56

3 Answers 3

up vote 1 down vote accepted

I was digging on Google and I found your question. Hopefully it's not too late.

The following is an image of some iterations of the logistic map x' = rx(1-x). To get a sense of what's going on, notice that each of the six regions represent x = [0,1] and r = [0,4].

Unfortunately I can't post images directly on this site yet, so I'll have to give a link: http://chaotic-prng.googlecode.com/files/logistic_golden_ratio.png

The bottom-right region shows the values of x ranging from 0 to 1 as a green gradient. The bottom-right region shows the same set of values, but using only 2 colours -- green for values >= 0.5 or black for values < 0.5.

After running each value through the logistic function one time, we arrive at the middle region -- the result of the first iteration.

After running each value through the logistic function one last time, we arrive at the top region -- the result of the second iteration.

You will notice that there is most definitely a branch formed by the values in the top row. It's unmistakable in the 2 colour version. You will also notice from the 2 colour version that the trunk starts at r=2, the branches start at r = sqrt(5) + 1, and the branches effectively end at r=4.

Doing some arithmetic to get the distances involved:

D1 = (sqrt(5) + 1) - 2 = sqrt(5) - 1,

D2 = 4 - (sqrt(5) + 1),

you will notice that D1/D2 (ie. trunk length to branch length ratio) is equal to the golden ratio

phi = (sqrt(5) + 1) / 2.

Also, you will notice that 2/D1 (ie. pretrunk length to trunk length ratio) also equals the golden ratio.

Presumably, this relationship pops up in all similar maps, because...

The paper "Fibonacci order in the period-doubling cascade to chaos" by Linage et al doesn't directly mention this, but does show you from a different point of view how it comes to be.

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Since the correct answer depends on what your professor thinks, and the fact that there are plenty of relations, I'll put my favorite:

EDIT: You might have noticed, that $2$-,$4$-,$3$-,$5$-,$6$- and $7$-point cycles start at increasing values of $r$. See here for a list. So the relation, that your professor bears in mind, could be: Can you show that there is an upper bound for $r$ using the golden ratio. You might have heard already that chaos starts at 4.00 (pm), so you have to choose $r>4$ (and better lock your door at $t<4$).

The logistic map relates to the Golden Ratio, such that a beautiful proof exists, which shows that $x_{n+1}=F(x_n)=rx_n(1-x_n)$ is chaotic, if $r>\phi^3$:

Since $r>4$, the interval $I=[0,1]$ is split into three parts. If $x_0$ lies inside the middle part, it will leave $I$ after one iteration. It is left to show that this will happen for the left ($I_0$) and right ($I_1$) interval too.

Therefore, we require that $|F(x)'|>\lambda>1$ for all $x\in I_0 \bigcup I_1 $. This union set is further called $A$. We have $|F(0)'|=|F(1)'|=r$, so $r>1$. The roots of $F(x)=1$, giving the right/left edges of $I_{0,1}$, are $$x_{-,+}=(1\pm \sqrt{1-4/r})/2.$$ Substituting $x_{-,+}$ in $F(x)'$ gives $$r^2-4r=1, \; \text{ which has roots at } \;\; r_{+,-}=2\pm \sqrt{5}.$$ Since $r_-<0$, you choose $r_+=2+\sqrt{5}=2\phi +1=\phi^2+\phi=\phi^3$.

From here on, I copied what I needed from here, since I'm not at all experienced on this field:

By the chain rule, it follows that $|F_n(x)'|>\lambda^n$ as well. Indeed, if this were so, we could choose two distinct point $x$ and $y$ in $A$ with the closed interval $[x,y]\subset A$. Choose $n$ so that $\lambda^n |y-x|>1$. By the Mean Value Theorem, it then follows that $$|F^n(y)-F^n(x)|>\lambda^n |y-x|>1,$$ which implies that at least one of $F^n(y)$ or $F^n(x)$ lies outside of $I$.


On the very nice linked page, you'll also find that the set $A$ is Cantor Set. (This distracted me quite a time, because I always tried to find the no-middle-third set.)

As WNY points out in his comment, the logistic map becomes chaotic for all $r>4$, but the proofs don't seem share the same kind of beauty.

In total, I think there might be plenty of relations between the logistic map and the golden ratio.

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P.S. Yes this exists in all similar maps. If you look at the second image on the page http://mathworld.wolfram.com/FeigenbaumConstant.html you will see a red-brown line that starts at the top-left and then eventually splits into three lines and then continues on. The ratio of distances covered by the line after and before the split works out to be the golden ratio.

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Hi, I've merged your two accounts. If you register your account (go to your user page by clicking on your name, and select register) the software will be better able to keep track of your postings. That will, in addition to other benefits, allow you to edit your own posts. –  Willie Wong Mar 8 '12 at 11:48

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