7
$\begingroup$

This is a follow up to this question. Any help would be very much appreciated.

Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ or some other $N>ak^2$.

Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.

Here there are five questions of varying intricacy. An answer to four is what I am hoping to achieve myself but an answer to an earlier part should go a long way towards helping and obviously an answer to part 5. would be amazing.

I have given a fairly trivial bound below which is good for my needs. If I don't get a better answer by the end of the bounty period I will accept my own (CW) answer and grant charMD the bounty.

  1. Simplify, where $v\in\{1,2,\dots,(k-1)/2\}$, $$1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}.$$

  2. Upper bound, where $v\in\{1,2,\dots,(k-1)/2\}$, $$\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq f_2(v,k,N).$$

  3. Simplify, where $v\in\{1,2,\dots,(k-1)/2\}$, $$\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right).$$

  4. Upper bound $$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq f_4(k,N).$$

  5. Sum $$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2 \left(\frac{2\pi\,v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right).$$

$\endgroup$
  • 2
    $\begingroup$ I haven't made the computations myself, but checking for some values of $k$, it seems to me that your very first formula should rather be $\frac{1}{4^{2N-1}}\sum \limits_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)=2 \cdot\frac{k^2-1}{4^{2N}}$ (with a $2$ instead of $\frac{1}{2}$) $\endgroup$ – charmd Apr 26 '16 at 20:39
  • $\begingroup$ @charMD thanks - I had it written up elsewhere and wasn't careful with the constant. It can be shown using the secant to tangent identity and if you search this site there is a sum of tangent-squared identity. $\endgroup$ – JP McCarthy Apr 26 '16 at 20:57
3
+500
$\begingroup$

The calculations were too awful and I haven't managed to find a simple form yet, but along with your first equality ($\sum \limits_{v=1}^{\frac{k-1}{2}} \sec^2 (\frac{2\pi v}{k} ) = \frac{1}{2} (k^2 - 1)$ ), you have also

$\sum \limits_{v=1}^{\frac{k-1}{2}} \frac{1}{\cos \big( \frac{2 \pi v}{k}\big)} = (-1)^\frac{k-1}{2} 2 \lfloor \frac{k-1}{4} \rfloor$.

And you can prove that $\sum \limits_{v=0}^{k-1} \cos ^j \big( \frac{2\pi v}{k} \big) = \frac{k}{2^j} \sum \limits_{\substack{0 \leqslant p \leqslant j\\ k \mid 2p - j}} \binom{j}{p}$.

Thus $$\sum \limits_{v=1}^{\frac{k-1}{2}} \cos ^j \big( \frac{2\pi v}{k} \big) = \frac{1}{2} \Big( \frac{k}{2^j} \sum \limits_{\substack{0 \leqslant p \leqslant j\\ k \mid 2p - j}} \binom{j}{p} - 1\Big)$$

And you can write your sum as a sum of sums of these types, which you can rewrite with the previous formula (but as I said, for now I haven't been able to do some real simplifications)

$\endgroup$
  • $\begingroup$ That looks like something I can work with but I can't see how I can write my sum as a sum of those terms... $\endgroup$ – JP McCarthy May 1 '16 at 9:02
  • 2
    $\begingroup$ I will try to obtain a more simplified form and complete my answer $\endgroup$ – charmd May 1 '16 at 17:29
  • $\begingroup$ I awarded you the bounty because nobody else had a stab at it. My answer below is sharp enough though... there is another term in the calculation that dominates the upper bound below so this works fine --- wasn't that hard to get this crude bound in the end. $\endgroup$ – JP McCarthy May 3 '16 at 8:19
0
$\begingroup$

Consider first $$\begin{align} 1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}&\leq 1+\sin^2\left(\frac{2\pi v}{k}\right)\left|\Re\left((\alpha_v\zeta)^N\right)\right|+|\alpha_v|^{2N} \\&\leq 1+\sin^2\left(\frac{2\pi v}{k}\right)\left|(\alpha_v\zeta)^N\right|+|\alpha_v|^{2N} \\&\leq 1+\sin^2\left(\frac{2\pi v}{k}\right)\left|\alpha_v\right|^N+|\alpha_v|^{2N} \\&\leq 1+\sin^2\left(\frac{2\pi v}{k}\right)3^N+3^{2N} \end{align}$$ In terms of efficiency, I am interested in $k$ large but $N=\mathcal{O}(k^2)$ and for $N\approx \frac{k}{2}\mod k$, $$-\Re\left((\alpha_v\zeta)^N\right)\approx +\Re\left((\alpha_v)^N\right).$$ The largest problem is that $$\alpha_v=2\cos\left(\frac{2\pi v}{k}\right)+\zeta^{-1}$$ has a large real part for $k$ large and $v$ small but as $v\rightarrow \frac{k-1}{2}$ $$\alpha_v\approx -1,$$ rather than $\alpha_v\approx 3$ as is the case for $v$ small. Hopefully this doesn't make the bound unusable (should find out soon).

Therefore

$$ \begin{align} &\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\\ &\leq \frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1+\sin^2\left(\frac{2\pi v}{k}\right)3^N+3^{2N}\right) \\&=\frac{1}{4^{2N-1}}\left((1+3^{2N})\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)+3^{N}\sum_{v=1}^{\frac{k-1}{2}}\tan^2\left(\frac{2\pi v}{k}\right)\right). \end{align}$$ Now using $\sec^2A=1+\tan^2A$ and this answer of Joriki this is equal to \begin{align*} \frac{1}{4^{2N-1}}\left((1+3^{2N})\left(\frac{k-1}{2}+\frac{k(k-1)}{2}+3^{N}\frac{k(k-1)}{2}\right)\right)&=2\frac{k-1}{4^{2N}}\left[(k+1)3^{2N}+k\cdot 3^N+k+1\right] \end{align*}

Any sharpening would be most welcome in an answer. An ideal answer would be a good bound of the form:

$$\frac{1}{4^{2N-1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(1-\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)+|\alpha_v|^{2N}\right)\leq f(N,k)e^{-\pi^2(2N-1)/k^2},$$

with the 'smaller' $f(N,k)$ the better. This answer here leads to an $$f(N,k)=6(k^2-1)\left(\frac{3}{4}e^{\pi^2/k^2}\right)^{2N}$$

$\endgroup$

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.