Let $a<0$ and $b>0$ be real numbers such that $a<-2b$. Let $n>1$ be a positive integer and consider the following partial sum: $$ f(n) = \frac{1}{(n+1)^2}\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j}\frac{(1-\cos(2x_i))(1-\cos(2x_j))}{-2a-2b(\cos(x_i)+\cos(x_j))}, \ \ \text{with }\ x_k:=\frac{k\pi}{n+1}. $$

Problem. Numerical simulations seem to suggest that $$\lim_{n\to\infty}\frac{f(n)}{c^{n}}= \infty$$ for small enough $c>0$. Is there a way to prove this claim? If so, is it possible to characterize the values of $c$ that satisfy the above condition?

I would like to stress that this is not an homework question, even if it may look so at a first glance. I encountered this fact in my research, and, after several unsuccessful proof attempts, I decided to post the problem here. If this is a trivial fact (it doesn't look so to me), please feel free to close this OP.

  • $\begingroup$ Something must be missing, as by trivial estimate the absolute value of f(n) is bounded above by 4/(-2a-4b). $\endgroup$ – Cherng-tiao Perng Oct 30 '18 at 3:34
  • $\begingroup$ @Cherng-tiaoPerng: There was a typo in my question ("for large enough $c>0$"). I've just fixed it ("for small enough $c>0$"). Does it make sense to you now? $\endgroup$ – Ludwig Oct 30 '18 at 4:02
  • $\begingroup$ Right, I saw it. That makes the question more interesting. $\endgroup$ – Cherng-tiao Perng Oct 30 '18 at 4:04

The desired inequality should be true iff $$ c < c_0 := (r - \sqrt{r^2-1})^2 \quad\ \text{where} \quad\ r = \frac{|a|}{2b} $$ (NB the hypotheses $b>0$ and $a < -2b$ imply $r>1$, so $0 < c_0 < 1$). Numerical computation suggests that $f(n) \sim A c_0^n \left/ \sqrt{n} \right.$ for some $A>0$. It should be possible to prove this by completing the following analysis.

It will be convenient to set $N = n+1$, and sum over $j,k$ rather than $i,j$ because we'll need $i = \sqrt{-1}$. For $x,y \in ({\bf R}/{\bf Z})^2$ define $$ F(x,y) = \frac1{2b}\frac {(1 - \cos(4\pi x)) (1 - \cos(4\pi y))} {2r - (\cos 2\pi x + \cos 2 \pi y)}. $$ Then $$ f(n) = \frac1{N^2} \sum_{j=1}^{N-1} \sum_{k=1}^{N-1} (-1)^{j+k} F\Bigl(\frac{j}{2N}, \frac{k}{2N}\Bigr) \, . $$ Now $F(x,y) = F(-x,y) = F(x,-y)$, and $$ F(0,y) = F(1/2,y) = F(x,0) = F(x,1/2) = 0 $$ for all $x,y$ thanks to the factors $1 - \cos(4\pi x)$ and $1 - \cos(4\pi y)$ in the numerator of $F$. So we can write $f(n)$ as an alternating sum over $\frac1{2N}$-lattice points: $$ f(n) = \frac1{(2N)^2} \sum_{j=1}^{2N} \sum_{k=1}^{2N} (-1)^{j+k} F\Bigl(\frac{j}{2N}, \frac{k}{2N}\Bigr) $$ (each term in the original sum for $f$ appears four times here, and the added terms with $j,k \in \{N, 2N\}$ all vanish).

But such a sum can be expressed in terms of the Fourier expansion $$ F(x,y) = \sum_{r\in\bf Z}\sum_{s\in\bf Z} \phi(r,s) \exp 2\pi i (rx+sy) $$ of $F$, which converges absolutely because $F$ is smooth. The alternating sum of $\exp 2 \pi i (rx+sy)$ is $(2N)^2$ if $(r,s) \equiv (N,N) \bmod 2N$, and zero otherwise. Therefore $f(n) = \sum\!\sum_{r,s} \phi(r,s)$ with the sum extending over all $(r,s)$ such that $r \equiv s \equiv N \bmod 2N$. The simplest terms in this sum are the four coefficients $\phi(\pm N, \pm N)$, which are all equal (since they satisfy $\phi(r,s) = \phi(-r,s) = \phi(r,-s)$ for all $r,s$). Thus we expect that $$ f(n) \sim 4 \phi(-N,-N) = 4 \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \exp (2\pi i N (x+y)) \, F(x,y) \, dx \, dy $$ unless $\phi(N,N)$ is unusually small.

We can now explain and quantify the observed exponential cancellation in the alternating sum that defines $f$. For large $r,s$ the Fourier coefficients $\phi(r,s)$ decay exponentially but no faster, because $F$ is not just smooth but analytic for $x,y \in ({\bf C}/{\bf Z})^2$ in some neighborhood of $({\bf R}/{\bf Z})^2$, but does have singularities for some complex $(x,y)$. We can shift the integral in the complex direction: $$ \phi(N,N) = e^{-4\pi N w} \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \exp (2\pi i N (x+y)) \, F(x+iw,y+iw) \, dx \, dy $$ as long as $F$ is analytic for $\left|\mathop{\rm Im}(x)\right|, \left|\mathop{\rm Im}(y)\right| \leq w$. This is the case for all $$ w < w_0 := \frac1{2\pi} \cosh^{-1} r = \frac1{2\pi} \log \bigl(r + \sqrt{r^2-1}\bigr) = -\frac1{4\pi} \log c_0. $$ In fact we can take $w = w_0$: the integrand $F(x+iw_0,y+iw_0)$ blows up at $(x,y)=(0,0)$, but the integral still converges absolutely because if $\left| F(x+iw_0,y+iw_0) \right| > M$ then $|x+y| \ll M^{-1}$ and $|x-y| \ll M^{-1/2}$. Thus $$ \phi(N,N) = c_0^N \int_{-1/2}^{1/2} \int_{-1/2}^{1/2} \exp (2\pi i N (x+y)) \, F(x+iw_0,y+iw_0) \, dx \, dy. $$ This shows that $f(n) \ll c_0^N$, and thus also $f(n) \ll c_0^n$; some more analysis of the double integral near the singularity at $(0,0)$ should show that $f(n) / c^n \to \infty$ for all $c<c_0$.

  • $\begingroup$ Thanks, and good questions. Answers: (1) The fancy answer is that it's the sum of the values of a character over an abelian group (product of two cyclic groups of order 2N), which is zero unless the character is trivial in which case it's the group size. The simple answer is to write the sum as the product of two geometric series. (2) Yes, sorry. They were all c(r,s) until I noticed that this clashed with the notation in the first display, and then I missed one of the c's when changing each to $\phi$. Will fix soon. (3) The integral's absolute value is bounded, being at most $\iint |F|$. $\endgroup$ – Noam D. Elkies Nov 2 '18 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.