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.

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.

Sequences and nets

According to the (Banach-)Alaoglu theorem, for any Banach space $X$ the closed unit ball of $X^*$ is compact in the weak* topology (the weakest/coarsest topology that makes all evaluation functionals $f\mapsto f(x)$ continuous).

For example, take $X=\ell_{\infty}$. The dual space $\ell_{\infty}^*$ contains an isometric copy of $\ell_1$ because $\ell_\infty^*=\ell_1^{**}$. The sequence $x_n=n^{-1}(e_1+\dots+e_n)$, where $e_n$ are the standard basis vectors, is contained in the unit sphere of $\ell_1$. Should $(x_n)$ have a weak*-convergent subsequence? Maybe it should, but it does not.

Indeed, take any subsequence $(x_{n_k})$. If necessary, choose a further subsequence so that $n_{k+1}\ge 3n_k$ for all $k$. Define

$\displaystyle y=\sum_{k}(-1)^k\sum_{n_{k-1}< j\le n_k} e_j$

where we set $n_0=0$. Two things to notice here: (1) $y\in \ell_{\infty}$; and (2) at least 2/3 of the coefficients of $e_j$, $1\le j\le n_k$, have the sign $(-1)^k$. Hence, $\langle x_{n_k},y\rangle\le -1/3$ when $k$ is odd and $\langle x_{n_k},y\rangle\ge 1/3$ when $k$ is even. This shows that $(x_{n_k})$ does not converge in the weak* topology.

The above does not contradict the Banach-Alaoglu theorem. Since $\ell_\infty$ is not separable, the weak* topology on the unit ball of its dual is not metrizable. The compactness can be stated in terms of nets instead of sequences: every bounded net in $X^*$ has a convergent subnet. In particular, the sequence $(x_n)$ has a convergent subnet (which is not a sequence). I personally find subnets a recipe for erroneous arguments. So I prefer to say: the infinite set $\{x_n\}$ has a cluster point $x$; namely, every neighborhood of $x$ contains some $x_n$. You can use the reverse inclusion of neighborhoods to define a subnet, but I’d rather not to. Everything we want to know about $x$ can be easily proved from the cluster point definition. For example,

• $\|x\|_{X^*}=1$
• $\langle x, 1_{\infty}\rangle =1$ where $1_{\infty}$ stands for the $\ell_\infty$ vector with all coordinates 1.
• $\langle x, y\rangle = \langle x, Sy\rangle$ where $S$ is the shift operator on $\ell_\infty$