3

I'm trying to prove the following approximate identity for $p$ integer: $$ \sum_{l=1}^m\frac{\sin^2\left(\frac{\pi l}{p}\right)}{\sin^2\left(\frac{\pi l}{mp}\right)}\sim \frac{m^2(p-1)}{2}+O(m) $$

Things I have tried:

  • Convert to an integral through a Riemann sum, however, the function $1/\sin^2(x)$ and it's derivatives are unbounded for small $x$

  • I tried to relate it to this problem, but I found it impossible to eliminate the numerator, simply averaging over one period of $p$ doesn't work.

Any help would be much appreciated.

Community
  • 1
rmatthys
  • 33

1 Answers1

2

For large $m$, the quantity $\pi \ell/(m p)$ is small except where $\ell \approx m$. Even then, the argument of the sine is small for even moderate values of $p$, so to first order we can replace the sine by its argument. Thus the ratio looks like, approximately,

$$\left (\frac{m p}{\pi} \right )^2 \sum_{\ell=1}^m \frac{\sin^2{\frac{\pi \ell}{p}}}{\ell^2}$$

Now, you can evaluate the sum as $m \to \infty$ using the following relation, derived here:

$$\sum_{\ell=-\infty}^{\infty} \frac{\sin^2{a \ell}}{\ell^2} = \pi a$$

when $a \in [0,\pi)$. Here, $a=\pi/p$, so we are OK. Thus we have that the fraction is approximately

$$\frac12 \left (\frac{m p}{\pi} \right )^2 \left (\frac{\pi^2}{p}-\frac{\pi^2}{p^2} \right ) = \frac12 m^2 \left (p-1 \right )$$

ADDENDUM

You can also show that the next order term is $O(m)$ by considering the next term in the expansion of the sine in the denominator. That is, we can easily show that, when the argument is "small":

$$\frac1{\sin^2{\frac{\pi \ell}{m p}}} \sim \left (\frac{m p}{\pi \ell} \right )^2 + \frac13$$

The first term gives the above result. The second term may be evaluated exactly:

$$\frac13 \sum_{\ell=1}^m \sin^2{\frac{\pi \ell}{p}} = \frac1{12} \left [2 m+1 - \frac{\sin{\left ( (2 m+1) \frac{\pi}{p}\right )}}{\sin{\frac{\pi}{p}}} \right ] \sim \frac{m}{6}$$

for large $m$. Thus the next order term is $O(m)$ as expected.

Community
  • 1
Ron Gordon
  • 138,521
  • But doesn't that need a bit more arguing, as $\ell$ can take values up to $m$ — in which case the quantity $\frac{\ell}{m}$ is clearly not small enough to replace $\sin$ by its Taylor series? – Clement C. Apr 14 '14 at 13:53
  • @ClementC.: True. But then the worst-case scenario is $p=2$ ($p=1$ gives an identically zero sum), in which we can say that it's just a numerically mediocre approximation which gets better for large $p$. – Ron Gordon Apr 14 '14 at 13:59
  • I agree --- just wanted to point out that this needed saying :) As a matter of fact, for $p\geq 2$ (since $p=1$ is trival as you mentioned), one can use the loose inequality $$x\leq \sin x \leq \frac{2}{\pi}x,\qquad x\in\left[0,\frac{\pi}{2}\right]$$ to derive the result (up to a constant multiplicative factor), as one can apply the inequality to the denominator as long as $0\leq \ell\leq \frac{p}{2}m$, which is always true. – Clement C. Apr 14 '14 at 14:06
  • @ClementC.: I agree with that too. Thank you for the edifying comments. – Ron Gordon Apr 14 '14 at 14:08
  • @ClementC.: Out of curiosity: the approximation seems to hold for arbitrary $p\geq 1$, with no integer restrictions. The case $p\geq 2$ follow from your answer. However, for $p$ close to 1 the denominator gets small for both $l\approx 1$ and $l\approx m$, with the Taylor expansion only small for $l\approx 1$. Any ideas on how to prove the approximation for all $p\geq 1$? – rmatthys Apr 14 '14 at 14:37
  • PS: the inequalities above were reversed (LateX typo): it should read $\frac{2}{\pi} x \leq \sin x \leq x$, where the upper bound holds no matter where $x$ lies and the lower bound only for $0\leq x \leq \frac{\pi}{2}$. – Clement C. Apr 14 '14 at 15:13
  • I suppose one could try to prove something of the form $$ \alpha \frac{\sin\frac{\pi\ell}{x}}{\sin\frac{\pi\ell}{m x}} \leq \beta$$ for any fixed $x\in[1,2]$ and $\ell,m$; ($\alpha,\beta$ not depending on $\ell,m$). Or, equivalently, $$ \alpha \frac{\sin\frac{\pi m r}{x}}{\sin\frac{\pi r}{x}} \leq \beta$$ for any fixed $x\in[1,2]$ and $r\in[\frac{1}{m}{1}],m\in\mathbb{N}$. This seems to hold, in particular since intuitively the hard case should be the lower bound around $x=1$, and the limit exists. – Clement C. Apr 14 '14 at 15:24