limit – Calculus VII

Iterating the logistic map: limsup of nonperiodic orbits

Last time we found that when a sequence with ${x_1\in (0,1)}$ and ${x_{n+1} = 4x_n(1-x_n)}$ does not become periodic, its upper limit ${\limsup x_n}$ must be at least ${\approx 0.925}$ . This time we’ll see that ${\limsup x_n}$ can be as low as ${(2+\sqrt{3})/4\approx 0.933}$ and determine for which ${x_1}$ it is equal to 1.

The quadratic polynomial ${f(x)=4x(1-x)}$ maps the interval ${[0,1]}$ onto itself. Since the linear function ${g(x) = 1-2x}$ maps ${[0,1]}$ onto ${[-1,1]}$ , it follows that the composition ${h=g\circ f\circ g^{-1}}$ maps ${[-1,1]}$ onto ${[-1,1]}$ . This composition is easy to compute: ${h(x) = 2x^2-1 }$ .

We want to know whether the iteration of ${f}$ , starting from ${x_1}$ , produces numbers arbitrarily close to ${1}$ . Since ${f\circ f \circ \cdots \circ f = g^{-1}\circ h \circ h \circ \cdots \circ h\circ g}$ the goal is equivalent to finding whether the iteration of ${h}$ , starting from ${g(x_1)}$ , produces numbers arbitrarily close to ${g(1) = -1}$ . To shorten formulas, let’s write ${h_n}$ for the ${n}$ th iterate of ${h}$ , for example, ${h_3 = h\circ h\circ h}$ .

So far we traded one quadratic polynomial ${f}$ for another, ${h}$ . But ${h}$ satisfies a nice identity: ${h(\cos t)=2\cos^2 t-1 = \cos(2t)}$ , hence ${h_n(\cos t) = \cos (2^n t)}$ for all ${n\in\mathbb N}$ . It’s convenient to introduce ${\alpha = \frac{1}{\pi}\cos^{-1}(1-2x_1)}$ , so that ${ h_n(g(x_1)) = h_n(\cos 2\pi \alpha ) = \cos(2^n\cdot 2\pi \alpha) }$ .

The problem becomes to determine whether the numbers ${2^n\cdot 2\pi \alpha}$ come arbitrarily close to ${\pi}$ , modulo an integer multiple of ${2\pi}$ . Dividing by ${2\pi}$ rephrases this as: does the fractional part of ${2^n \alpha}$ come arbitrarily close to ${1/2}$ ?

A number that is close to ${1/2}$ has the binary expansion beginning either with ${0.01111111\dots}$ or with ${0.10000000\dots}$ . Since the binary expansion of ${2^n\alpha}$ is just the binary expansion of ${\alpha}$ shifted ${n}$ digits to the left, the property ${\limsup x_n=1}$ is equivalent to the following: for every ${k\in\mathbb N}$ the binary expansion of ${\alpha}$ has infinitely many groups of the form “1 followed by k zeros” or “0 followed by k ones”.

A periodic expansion cannot have the above property; this, ${\alpha}$ must be irrational. The property described above can then be simplified to “irrational and has arbitrarily long runs of the same digit”, since a long run of ${0}$ s will be preceded by a ${1}$ , and vice versa.

For example, combining the pairs 01 and 10 in some non-periodic way, we get an irrational number ${\alpha}$ such that the fractional part of ${2^n\alpha}$ does not get any closer to 1/2 than ${0.01\overline{10}_2 = 5/12}$ or ${0.10\overline{01}_2 = 7/12}$ . Hence, ${\cos 2^n 2\pi \alpha \ge -\sqrt{3}/2}$ , which leads to the upper bound ${x_n\le (2+\sqrt{3})/4\approx 0.933}$ for the sequence with the starting value ${x_1=(1-\cos\pi\alpha)/2}$ .

Let us summarize the above observations about ${\limsup x_n}$ .

Theorem: ${\limsup x_n=1}$ if and only if (A) the number ${\alpha = \frac{1}{\pi}\cos^{-1}(1-2x_1)}$ is irrational, and (B) the binary expansion of ${\alpha}$ has arbitrarily long runs of the same digit.

Intuitively, one expects that a number that satisfies (A) will also satisfy (B) unless it was constructed specifically to fail (B). But to verify that (B) holds for a given number is not an easy task.

As a bonus, let’s prove that for every rational number ${y\in (-1,1)}$ , except 0, 1/2 and -1/2, the number ${\alpha = \frac{1}{\pi}\cos^{-1}y}$ is irrational. This will imply, in particular, that ${x_1=1/3}$ yields a non-periodic sequence. The proof follows a post by Robert Israel and requires a lemma (which could be replaced with an appeal to Chebyshev polynomials, but the lemma keeps things self-contained).

Lemma. For every ${n\in \mathbb N}$ there exists a monic polynomial ${P_n}$ with integer coefficients such that ${P_n(2 \cos t) = 2\cos nt }$ for all ${t}$ .

Proof. Induction, the base case ${n=1}$ being ${P_1(x)=x}$ . Assuming the result for integers ${\le n}$ , we have ${2 \cos (n+1)t = e^{i(n+1)t} + e^{-i(n+1)t} }$ ${ = (e^{int} + e^{-int})(e^{it} + e^{-it}) - (e^{i(n-1)t} + e^{-i(n-1)t}) }$ ${ = P_n(2 \cos t) (2\cos t) - P_{n-1}(2\cos t) }$
which is a monic polynomial of ${2\cos t}$ . ${\Box}$

Suppose that there exists ${n}$ such that ${n\alpha \in\mathbb Z}$ . Then ${2\cos(\pi n\alpha)=\pm 2}$ . By the lemma, this implies ${P_n(2\cos(\pi \alpha)) =\pm 2}$ , that is ${P_n(2y)=\pm 2}$ . Since ${2y}$ is a rational root of a monic polynomial with integer coefficients, the Rational Root Theorem implies that it is an integer. ${\Box}$

A limsup exercise: iterating the logistic map

Define the sequence ${\{x_n\}}$ as follows: ${x_1=1/3}$ and ${x_{n+1} = 4x_n(1-x_n)}$ for ${n=1,2,\dots}$ . What can we say about its behavior as ${n\rightarrow\infty}$ ?

The logistic map ${f(x)=4x(1-x)}$ leaves the interval [0,1] invariant (as a set), so ${0\le x_n\le 1}$ for all ${n}$ . There are two fixed points: 0 and 3/4.

Two fixed points of the logistic map: 0 and 3/4

Can ${x_n}$ ever be 0? If ${n}$ is the first index this happens, then ${x_{n-1}}$ must be ${1}$ . Working backwards, we find ${x_{n-2}=1/2}$ , and ${x_{n-3}\in \{1/2 \pm \sqrt{2}/4\}}$ . But this is impossible since all elements of the sequence are rational. Similarly, if ${n}$ is the first index when ${x_n = 3/4}$ , then ${x_{n-1}=1/4}$ and ${x_{n-2}\in \{1/2\pm \sqrt{3}/4\}}$ , a contradiction again. Thus, the sequence never stabilizes.

If ${x_n}$ had a limit, it would have to be one of the two fixed points. But both are repelling: ${f'(x) = 4 - 8x}$ , so ${|f'(0)|=4>1 }$ and ${|f'(3/4)| = 2 > 1}$ . This means that a small nonzero distance to a fixed point will increase under iteration. The only way to converge to a repelling fixed point is to hit it directly, but we already know this does not happen. So the sequence ${\{x_n\}}$ does not converge.

But we can still consider its upper and lower limits. Let’s try to estimate ${S = \limsup x_n}$ from below. Since ${f(x)\ge x}$ for ${x\in [0,3/4]}$ , the sequence ${\{x_n\}}$ increases as long as ${x_n\le 3/4}$ . Since we know it doesn’t have a limit, it must eventually break this pattern, and therefore exceed 3/4. Thus, ${S\ge 3/4}$ .

This can be improved. The second iterate ${f_2(x)=f(f(x))}$ satisfies ${f_2(x)\ge x}$ for ${x}$ between ${3/4}$ and ${a = (5+\sqrt{5})/8 \approx 0.9}$ . So, once ${x_n>3/4}$ (which, by above, happens infinitely often), the subsequence ${x_n, x_{n+2}, x_{n+4},\dots}$ increases until it reaches ${a}$ . Hence ${S\ge a}$ .

The bound ${\limsup x_n\ge a}$ is best possible if the only information about ${x_1}$ is that the sequence ${x_n}$ does not converge. Indeed, ${a}$ is a periodic point of ${f}$ , with the corresponding iteration sequence ${\{(5+ (-1)^n\sqrt{5})/8\}}$ .

Further improvement is possible if we recall that our sequence is rational and hence cannot hit ${a}$ exactly. By doubling the number of iterations (so that the iterate also fixes ${a}$ but also has positive derivative there) we arrive at the fourth iterate ${f_4}$ . Then ${f_4(x)\ge x}$ for ${a\le x\le b}$ , where ${b }$ is the next root of ${f_4(x)-x}$ after ${a}$ , approximately ${0.925}$ . Hence ${S\ge b}$ .

This is a colorful illustration of the estimation process (made with Sage): we are covering the line ${y=x}$ with the iterates of ${f}$ , so that each subsequent one rises above the line the moment the previous one falls. This improves the lower bound on ${S}$ from 0.75 to 0.9 to 0.92.

Although this process can be continued, the gains diminish so rapidly that it seems unlikely one can get to 1 in this way. In fact, one cannot because we are not using any properties of ${x_1}$ other than “the sequence ${x_n}$ is not periodic.” And it’s not true that ${\limsup x_n = 1}$ for every non-periodic orbit of ${f}$ . Let’s return to this later.

Alternating lacunary series and 1-1+1-1+1-1+…

The series ${1-1+1-1+1-1+\cdots}$ diverges. A common way to make it convergent is to replace each ${1}$ with a power of ${x}$ ; the new series will converge when ${|x|<1}$ and maybe its sum will have a limit as ${x\rightarrow 1}$ . And indeed,

$\displaystyle x-x^2+x^3-x^4+x^5-x^6+\cdots = \frac{x}{1+x}$

which tends to ${1/2}$ as ${x}$ approaches ${1}$ from the left.

Things get more interesting if instead of consecutive integers as exponents, we use consecutive powers of ${2}$ :

$\displaystyle f(x) = x-x^2+x^4-x^8+x^{16} -x^{32} +\cdots$

On most of the interval ${(0,1)}$ it behaves just like the previous one:

But there appears to be a little blip near ${1}$ . Let’s zoom in:

And zoom in more:

Still there.

This function was considered by Hardy in 1907 paper Theorems concerning infinite series. On pages 92–93 he shows that it “oscillates between finite limits of indetermination for ${x=1}$ “. There is also a footnote: “The simple proof given above was shown to be by Mr. J. H. Maclagan-Wedderburn. I had originally obtained the result by means of a contour integral.”

Okay, but what are these “finite limits of indetermination”? The Alternating Series Estimation shows ${0<f(x)<1}$ for ${x\in (0,1)}$ , but the above plots suggest that ${f}$ oscillates between much tighter bounds. Let’s call them ${A= \liminf_{x\rightarrow 1-} f(x)}$ and ${B=\limsup_{x\rightarrow 1-} f(x)}$ .

Since ${f(x)+f(x^2)\equiv x^2}$ , it follows that ${f(x)+f(x^2)\rightarrow 1}$ as ${x\rightarrow 1^-}$ . Hence, ${B = \limsup_{x\rightarrow 1-}(1-f(x^2)) = 1-A}$ . In other words, ${A}$ and ${B}$ are symmetric about ${1/2}$ . But what are they?

I don’t have an answer, but here is a simple estimate. Let ${g(x)=x-x^2}$ and observe that

$\displaystyle f(x) = g(x) + g(x^4)+g(x^{16}) + g(x^{64})+\dots \qquad \qquad (1)$

The function ${g}$ is not hard to understand: its graph is a parabola.

Since ${g}$ is positive on ${(0,1)}$ , any of the terms in the sum (1) gives a lower bound for ${f}$ . Each individual term is useless for this purpose, since it vanishes at ${1}$ . But we can pick ${n}$ in ${g(x^{4^n})}$ depending on ${x}$ .

Let ${x_0\in(0,1)}$ be the unique solution of the equation ${x_0^4=1-x_0}$ . It could be written down explicitly, but this is not a pleasant experience; numerically ${x_0\approx 0.7245}$ . For every ${x>x_0}$ there is an integer ${n\ge 1}$ such that ${x^{4^n}\in [1-x_0,x_0]}$ , namely the smallest integer such that ${x^{4^n} \le x_0}$ . Hence,

$\displaystyle f(x)> g(x^{4^n})\ge \min_{[x_0,1-x_0]}g = x_0-x_0^2>0.1996 \qquad \qquad (2)$

which gives a nontrivial lower bound ${A>0.1996}$ and symmetrically ${B<0.8004}$ . Frustratingly, this falls just short of neat ${1/5}$ and ${4/5}$ .

One can do better than (2) by using more terms of the series (1). For example, study the polynomial ${g(t)+g(t^4)}$ and find a suitable interval ${[t_0^4,t_0]}$ on which its minimum is large (such an interval will no longer be symmetric). Or use ${3,4,5...}$ consecutive terms of the series… which quickly gets boring. This approach gives arbitrarily close approximations to ${A}$ and ${B}$ , but does not tell us what these values really are.