# 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. ${\Box}$

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