## Very fractional geometric progressions with an integer ratio

The geometric progression 1/3, 2/3, 4/3, 8/3, 16/3,… is notable for being dyadic (ratio 2) and staying away from integers as much as possible (distance 1/3 between this progression and the set of integers; no other dyadic progression stays further away from integers). This property is occasionally useful: by taking the union of the dyadic partition of an interval with its shift by 1/3, one gets a system of intervals that comfortably covers every point: for every point x and every (small) radius r there is an interval of size r, in which x is near the middle.

It’s easy to see that for any real number x, the distance between the progression {x, 2x, 4x, 8x, …} and the set of integers cannot be greater than 1/3. Indeed, since the integer part of x does not matter, it suffices to consider x between 0 and 1. The values between 0 and 1/3 lose immediately; the values between 1/3 and 1/2 lose after being multiplied by 2. And since x and 1-x yield the same distance, we are done.

Let’s find the most fractional progressions with other integer ratios r. When r is odd, the solution is obvious: starting with 1/2 keeps all the terms at half-integers, so the distance 1/2 is achieved. When r is even, say r = 2k, the best starting value is x = k/(2k+1), which achieves the distance x since rx = k-x. The values between 0 and k/(2k+1) are obviously worse, and those between k/(2k+1) and 1/2 become worse after being multiplied by r: they are mapped to the interval between k-x and k.

The problem is solved! But what if… x is irrational?

Returning to ratio r=2, it is clear that 1/3 is no longer attainable. The base-2 expansion of x cannot be 010101… as that would be periodic. So it must contain either 00 or 11 somewhere. Either of those will bring a dyadic multiple of x within distance less than 0.001111… (base 2) of an integer, that is distance 1/4.

The goal is to construct x so that its binary expansion is as balanced between 0 and 1 as possible, but without being periodic. The Thue-Morse constant does exactly this. It’s constructed by starting with 0 and then adding the complement of the sequence constructed so far: x = .0 1 10 1001 10010110 … which is approximately 0.412. The closest the dyadic geometric progression starting with x comes to an integer is 2x, which has distance about 0.175. The Wikipedia article links to the survey The ubiquitous Prouhet-Thue-Morse sequence by Allouche and Shallit, in which Corollary 2 implies that no other irrational number has a dyadic progression with a greater distance from integers, provided that this distance is attained. I have not been able to sort out the case in which the distance from a progression to the integers is not attained, but it seems very likely that Thue-Morse remains on top.

What about other ratios? When the ratio r is even, the situation is essentially the same as for r=2, for the following reason. In base r there are two digits nearest (r-1)/2, for example 4 and 5 in base 10. Using these digits in the Thue-Morse sequence, we get a strong candidate for the most fractional progression with ratio r: for example, 0.455454455445… in base 10, with the distance of about 0.445. Using any other digit loses the game at once: for example, having 3 in the decimal expansion implies that some multiple of 10 is within less than 0.39999… =  0.4 of an integer.

When the ratio is odd, there are three digits that could conceivably be used in the extremal x: namely, (r-1)/2 and its two neighbors. If the central digit (r-1)/2 is never used, we are back to the Thue-Morse pattern, such as  x = 0.0220200220020220… in base 3 (an element of the standard Cantor set, by the way). But this is an unspectacular achievement, with the distance of about 0.0852. One can do better by starting with 1/2 = 0.1111111… and sprinkling this ternary expansions with 0s or 2s in some aperiodic way, doing so very infrequently. By making the runs of 1s extremely long, we get the distance arbitrarily close to 1 – 0.2111111… base 3, which is simply 1/2 – 1/3 = 1/6.

So it seems that for irrational geometric progressions with an odd ratio r, the distance to integers can be arbitrarily close to the number 1/2 – 1/r, but there is no progression achieving this value.

## 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. ${\Box}$

## Wild power pie

Many people are aware of ${\pi}$ being a number between 3 and 4, and some also know that ${e}$ is between 2 and 3. Although the difference ${\pi-e}$ is less than 1/2, it’s enough to place the two constants in separate buckets on the number line, separated by an integer.

When dealing with powers of ${e}$, using ${e>2}$ is frequently wasteful, so it helps to know that ${e^2>7}$. Similarly, ${\pi^2<10}$ is way more precise than ${\pi<4}$. To summarize: ${e^2}$ is between 7 and 8, while ${\pi^2}$ is between 9 and 10.

Do any two powers of ${\pi}$ and ${e}$ have the same integer part? That is, does the equation ${\lfloor \pi^n \rfloor = \lfloor e^m \rfloor}$ have a solution in positive integers ${m,n}$?

Probably not. Chances are that the only pairs ${(m,n)}$ for which ${|\pi^n - e^m|<10}$ are ${m,n\in \{1,2\}}$, the smallest difference attained by ${m=n=1}$.

Indeed, having ${|\pi^n - e^m|<1}$ implies that ${|n\log \pi - m|, or put differently, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n \,\pi^n}}$. This would be an extraordinary rational approximation… for example, with ${n=100}$ it would mean that ${\log \pi = 1.14\ldots}$ with the following ${50}$ digits all being ${0}$. This isn’t happening.

Looking at the continued fraction expansion of ${\log \pi}$ shows the denominators of modest size ${[1; 6, 1, 10, 24, \dots]}$, indicating the lack of extraordinarily nice rational approximations. Of course, can use them to get good approximations, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n^2}}$, which leads to ${\pi^n\approx e^m}$ with small relative error. For example, dropping ${24}$ and subsequent terms yields the convergent ${87/76}$, and one can check that ${\pi^{76} = 6.0728... \cdot 10^{37}}$ while ${e^{87} = 6.0760...\cdot 10^{37}}$.

Trying a few not-too-obscure constants with the help of mpmath library, the best coincidence of integer parts that I found is the following: the 13th power of the golden ratio ${\varphi = (\sqrt{5}+1)/2}$ and the 34th power of Apèry’s constant ${\zeta(3) = 1^{-3}+2^{-3}+3^{-3}+4^{-4}+\dots}$ both have integer part 521.