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I would like to evaluate the number $c$ given by $$ c = \lim_{m\to\infty} \frac{1}{\log m}\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)} $$ where $\tau = (1+\sqrt{5})/2$.

My attempt: my guess was this sum would be dominated by the terms for which $n$ is a Fibonacci number. I considered the sum of this sub series using the relation $F_a \tau = F_{a+1}-(-\tau)^{-a}$, the small angle approximation, and that $F_a\tau^{-a} = 1/\sqrt{5}+\mathrm{O(\tau^{-2a})}$. This yielded $$ \sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)} \approx \sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{1}{F_a^2 \sin^2(\pi \tau^{-a})} \approx \sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{5}{\pi^2} = \frac{5}{\pi^2} \frac{\log m \sqrt{5}}{\log \tau} $$ this would imply $c = 5/(\pi^2 \log \tau)$, however this seems about a factor $1.2$ smaller than the correct value, implying I have missed some important terms.

Is it possible to obtain an exact expression? Or at least a better a tighter lower (and upper) bound?

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  • $\begingroup$ I would also be extremely interested in approaches that could be extended to generic algebraic or irrational values of $\tau$! $\endgroup$ Commented Aug 25, 2017 at 22:17
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    $\begingroup$ I suppose it's off by (at least) a constant factor due to multiples of Fibonacci numbers such as $2F_a$. Those should contribute $1/16$ of the original sum, so it isn't enough to explain a factor of $1.2$, even if we allow other fixed multiples. We can ignore any values of $n$ for which $n\tau$ is more than $1/\sqrt{\log n}$ away from any integer, but this still leaves a lot of room between the logarithmically-sparse sequence of Fibonacci numbers and the modestly-good approximations of $(\tau^{-1})\mathbb N$ which, while $o(m)$ in quantity, are still quite numerous. $\endgroup$
    – Erick Wong
    Commented Aug 31, 2017 at 20:57
  • $\begingroup$ Ah yes. I've found it very difficult to account for the cases $n F_a$ in general though without accidentally over counting due to the solutions of $n F_a = m F_b$. Yes I take your point in the second part, summing only this sub-series could not have worked. $\endgroup$ Commented Sep 1, 2017 at 2:50
  • $\begingroup$ I think the other important contributions come from other series with the same recursion relation as the Fibonacci numbers, but different initial conditions. The ratio of consecutive terms in these series also converge on $\tau$. But I cannot see for the moment a way to sum over these different series without over counting. Multiples $n F_a$ are one example of such a series, but so are eg the Lucas numbers. $\endgroup$ Commented Sep 1, 2017 at 3:26
  • $\begingroup$ That's an interesting idea. Maybe something can come out of enumerating the values by the number of 1s in the Zeckendorff representation (or perhaps the distance between the first and last 1s). Then Fibonacci numbers would be 1-bit values and Lucas numbers 2-bits. At least that gives a way of indexing them without duplications. $\endgroup$
    – Erick Wong
    Commented Sep 1, 2017 at 3:45

2 Answers 2

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$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cI{\mathcal{I}}$Amazingly, your sum can actually be computed in closed form. The answer is $$c=\frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$

Notation: Let $\tau = \tfrac{1+\sqrt{5}}{2}$ and $\bar{\tau} = \tfrac{1-\sqrt{5}}{2}$, so $\tau+\bar{\tau}=1$ and $\tau \bar{\tau} = -1$. Let $R$ be the ring $\ZZ[\tau]$. The ring $R$ is known to be a PID with unit group $\pm \tau^k$. Let $\cI$ be the set of nonzero ideals of $R$. For $m+n \tau \in R$, set $N(m+n \tau) = m^2+mn-n^2 = (m+n \tau) (m+n \bar{\tau})$; for an ideal $I \subseteq R$ set $N(I) = |R/I|$. The relation between these notations is that $|N(m+n \tau)| = N(\langle m+n \tau \rangle)$.


You want to evaluate $$\lim_{K \to \infty} \frac{1}{\log K} \sum_{n=1}^K \frac{1}{n^2 \sin^2 (\pi n \tau)}.$$ We recall the identity $$\frac{\pi^2}{\sin^2 (\pi x)} = \sum_{m=-\infty}^{\infty} \frac{1}{(m-x)^2}$$ to rewrite this as $$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \sum_{m=- \infty}^{\infty} \frac{1}{n^2 (m-n \tau)^2}. \quad (1)$$

All the terms are positive, so we may rearrange the sum at will; we group together terms $m-n \tau$ which generate the same ideal $I$ in $R$, giving $$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{I \in \cI} \sum_{\begin{matrix} \langle m-n \tau \rangle=I \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2} \quad (2).$$

I assume it is legitimate to exchange the limit and the outer sum in (2) (should be easy, but I haven't checked). So we want to consider $$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{\begin{matrix} \langle m-n \tau \rangle=I \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2}. \quad (3)$$

If $\gamma$ is a generator of $I$, then the list of all generators is the numbers of the form $\pm \tau^k \gamma$. All of these points lie on the hyperbolas $m^2-mn-n^2 = \pm N(I)$, with asymptotes $m = \bar{\tau} n$ and $m = \tau n$.

As $k \to - \infty$, we approach the asymptote $m=\bar{\tau} n$. Both $n$ and $m-n \tau$ grow exponentially, so the contribution from those summands is bounded and is wiped out by the $\log K$ term.

As $k \to \infty$, we approach the $m = \tau n$ asymptote. The number of terms is $\tfrac{\log K + O(1)}{\log \tau}$ and each of those terms is $$\frac{1}{n^2 (m-n \tau)^2} = \frac{(m- \bar{\tau} n)^2}{n^2 N(I)^2} = \frac{(m/n- \bar{\tau})^2}{N(I)^2}.$$ Since we are approaching the asymptote $m/n = \tau$, the numerator approaches $(\tau - \bar{\tau})^2 = 5$. We have a sum of $\tfrac{\log K+O(1)}{\log \tau}$ terms which approach $\tfrac{5}{N(I)^2}$, so $(3)$ is $$\frac{5}{\pi^2 (\log \tau) N(I)^2}.$$

Plugging into $(2)$, $$c=\frac{5}{\pi^2 \log \tau} \sum_{I \in \cI} \frac{1}{N(I)^2} .$$ That last sum is your "about 1.2"; you only computed the contribution from the ideal $\langle 1 \rangle$. (To see the connection, note that $\tau^k = F_{k} \tau + F_{k-1}$.)


I expected this to be the end of the line, but it turns out this sum can actually be evaluated! Recall that the $\zeta$ function of $R$ is defined to be $$Z(s) := \sum_{I \in \cI} \frac{1}{N(I)^s}.$$ So we want to evaluate $Z(2)$.

We know $Z(s)$ factors as $$Z(s) = \zeta(s) L(s)$$ where $\zeta$ is the Riemann $\zeta$ function and $$L(s) = \sum_{n=1}^{\infty} \frac{\left( \tfrac{5}{n} \right)}{n^s}.$$ We know that $\zeta(2) = \tfrac{\pi^2}{6}$, so we are left to evaluate $L(2)$.

Using quadratic reciprocity, $$\left( \frac{5}{n} \right) = \begin{cases} 0 & n \equiv 0 \bmod 5 \\ 1 & n \equiv \pm 1 \bmod 5 \\ -1 & n \equiv \pm 2 \bmod 5 \end{cases}$$ from which we deduce $$\left( \frac{5}{n} \right) = \frac{2}{\sqrt{5}} \left( \cos \tfrac{2 \pi n}{5} - \cos \tfrac{4 \pi n}{5} \right).$$ So $$L(2) = \frac{2}{\sqrt{5}} \left( \sum_{n=1}^{\infty} \frac{\cos \tfrac{2 \pi n}{5} }{n^2} - \sum_{n=1}^{\infty} \frac{\cos \tfrac{4 \pi n}{5} }{n^2} \right).$$ We now recall that, for $0 \leq x \leq 2 \pi$, we have $$\sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{(\pi-x)^2}{4} - \frac{\pi^2}{12}.$$ Plugging in $x=2 \pi/5$ and $4 \pi/5$ we get $$L(2) = \frac{4 \pi^2}{25 \sqrt{5}}.$$ Turning quadratic reciprocity symbols into linear combinations of trigonometric functions, and then recognizing the Fourier series that results, is a standard way to evaluate $L$-functions.

Putting it all together, we deduce $$c=\frac{5}{\pi^2 \log \tau} \frac{\pi^2}{6} \frac{4 \pi^2}{25 \sqrt{5}} = \frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$

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  • $\begingroup$ Very nice answer! Could this be extended to find asymptotics of sums with other powers on the $n$ and $\sin(\pi n \tau)$ term (like the Flint-Hills series, where $\tau = 1/\pi$ and the exponent of $n$ is $3$)? $\endgroup$ Commented Sep 6, 2017 at 3:47
  • $\begingroup$ (+1) What a thrilling conclusion to an interesting and well-researched question! $\endgroup$
    – Erick Wong
    Commented Sep 6, 2017 at 3:48
  • $\begingroup$ Really great! I have been mucking around with a complex-analysis result but this is so eloquent. $\endgroup$
    – Ron Gordon
    Commented Sep 6, 2017 at 6:51
  • $\begingroup$ Consistently performed considerations in a greater/longer context : great (+1) $\endgroup$
    – user90369
    Commented Sep 6, 2017 at 8:37
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    $\begingroup$ @CarlSchildkraut I think replacing $\tfrac{1}{\sin^2 \pi x}$ with $\tfrac{1}{\pi^2} \sum \tfrac{1}(m-x)^2$, and analogous formulas for other powers of $\sin$, will always be a good first step. I doubt you can prove anything unconditionally about the Flint-Hill series, because we know so little about Diophantine approximation of $\pi$. This method should work for $\tau$ a quadratic algebraic integer. I also expect that someone who understood the Gauss-Kuzmin statistics mathworld.wolfram.com/Gauss-KuzminDistribution.html well enough could prove something about almost all $\tau$. $\endgroup$ Commented Sep 6, 2017 at 14:34
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Responding to questions in comments about the behavior of $$\lim_{K \to \infty} \frac{1}{\log K} \sum_{n=1}^K \frac{1}{n^2 \sin^2 (n \theta)}$$ for other $\theta$: We can transform this to $$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \sum_{m=- \infty}^{\infty} \frac{1}{n^2 (m-n \theta)^2}$$ as before. For arbitrary $\theta$, I doubt we can say anything. But for almost all $\theta$ (in other words, on the complement of a set of measure $0$), we should be able to.

Let $R(x)$ be the closest integer to $x$ ($R$ stands for "round") and let $\langle x \rangle = x-R(x)$. The inner sum is $$\frac{1}{n^2 \langle n \theta \rangle^2} + \frac{1}{n^2}\sum_{k=1}^{\infty} \left( \frac{1}{(\langle n \theta \rangle +k)^2}+\frac{1}{(\langle n \theta \rangle -k)^2} \right). \quad (4)$$ The sum in (4) is at most $2 \sum_{k=1}^{\infty} \tfrac{1}{(k-1/2)^2}$, which is less than some constant $C$, and then summming $C/n^2$ converges, so this is wiped out by the $\log K$ factor. So what we really care about is $$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \frac{1}{n^2 \langle n \theta \rangle^2}. \ (5)$$

I learned from this answer that the number of $n \leq K$ for which $\langle n \theta \rangle \in (x/n, (x+dx)/n)$ is roughly $dx \log K$. So I would predict that $(5)$ behaves like $$\int_{x=-\infty}^{\infty} \frac{dx}{x^2}.$$ This, of course, diverges, so I would guess the limit does too. Perhaps I'll think about the actual rate of growth later tonight.

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  • $\begingroup$ Using Roth's theorem we know $|n\left<n\theta\right>| > c n^{-\epsilon}$ this bounds $x$ away from $x=0$ at finite $K$. A simple correction to account for this could be to remove the region $[-c K^{-\epsilon},c K^{-\epsilon}]$ from the interval of integration. Doing this and performing the integral suggests the sum scales as $\sum_{n=1}^K \frac{1}{n^2\left< n \theta \right>^2} \sim K^\epsilon \log K$. This suggests $\lim_{K \to \infty} \frac{1}{\log K}\sum_{n=1}^K \frac{1}{n^2\left< n \theta \right>^2}$ converges only for badly approximable numbers (for which $\epsilon=0$). Would you agree? $\endgroup$ Commented Dec 4, 2017 at 4:05
  • $\begingroup$ Roth's theorem of course holds for algebraic numbers, but I understand the same bound holds for almost all irrational numbers. $\endgroup$ Commented Dec 8, 2017 at 20:01

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