# hunting for the closed form of a series

Let $N$ be a positive integer, and $$F(N) = \sum_{n=1}^{N} \frac{1-\cos\left(\frac{(2n-1)\pi}{2N}\right)} {\left[1+\cos\left(\frac{(2n-1)\pi}{2N}\right)\right]\left[5+3\cos\left(\frac{(2n-1)\pi}{2N}\right)\right]^2}$$ I want to konw the closed form of $F(n)$. In order to make an attack on it, I calculate some values, i.e.

$N=16$, $F(n)=117$; $N=32$, $F(n)=490$; $N=64$, $F(n)=2004$; $N=128$, $F(n)=8104$; $N=256$, $F(n)=32592$; $N=512$, $F(n)=130720$, etc.

I can not conclude the general expression of $F(n)$, and oeis does not seem to help a lot.

EDIT: My numerical effort shows that $F(N)=\frac{N(8N-11)}{16}$.

EDIT2:

$N=17$, $F1(N)=132.8125000000000187306828$, $F2(N)=132.8125$;

$N=18$, $F1(N)=149.6250000000000023235341$, $F2(N)=149.625$;

$N=50$, $F1(N)=1215.6250000000000000000000000000000000000000000050$, $F2(N)=1215.625$.

• Let's say that they suggest that $F(2^n) = \frac{2^n(8\cdot 2^n-11)}{16}$. Why are you only calculating $F$ when $N$ is a power of two? What about $N = 17$, $N=3$ and what not? – Ant Jan 20 '16 at 9:35
• Your values and formulas are not correct. I doubt that you will get integer results. The simple case $N=1$ gives $F(1)=1/25$. – gammatester Jan 20 '16 at 9:36
• Your formula may be approximately correct for $N\ge 6$, e.g. $F(6)\approx 13.87508326$ your formula gives $13.875$ or $F(16) \approx 117.0000000000001500253$ instead of $117$ – gammatester Jan 20 '16 at 9:57
• Also, if you don't show the results you get for $N = 17$ or for integers different than powers of 2, then all one can conclude is that the formula seems to hold for power of 2 . But as @gammatester pointed out, your formula is only an approximation :) – Ant Jan 20 '16 at 10:14
• If I didn't make any mistake, we have $$F(N) = \frac{(8N-11)N}{16}+\frac{N((8N+11)9^N + 11)}{8(9^N+1)^2}$$ In particular, the first few values of $F(N)$ are $$\frac{1}{25},\frac{1188}{1681},\frac{327129}{133225},\frac{56551312}{10764961},\frac{316019361}{34869025},\frac{979687020612}{70607649841},\ldots$$ – achille hui Jan 20 '16 at 11:52

Let $\omega = e^{\frac{\pi}{2N}i}$ be the primitive $4N^{th}$ root of unity and $$\Lambda = \bigg\{\; \omega^{2k-1} : -N < k \le N \;\bigg\} = \bigg\{\; \omega^{\pm(2k-1)} : 1 \le k \le N \;\bigg\}$$ be the set of roots for the polynomial $z^{2N} + 1 = 0$.

For any integer $k$, let $c_k = \cos\left(\frac{(2k-1)\pi}{2N}\right) = \frac{\omega^{2k-1} + \omega^{1-2k}}{2}$. Notice $$\frac{1-z}{(1+z)(5+3z)^2} = \frac12\left(\frac{1}{1+z}\right) - \frac12\left(\frac{1}{\frac53 + z}\right) -\frac49\left(\frac{1}{\frac53 + z}\right)^2$$ We can simplify our sum as $$F(N) = \frac12 f(1) - \frac12 f\left(\frac53\right) + \frac49 f'\left(\frac53\right) \quad\text{ where }\quad f(\alpha) = \sum\limits_{k=1}^N \frac{1}{\alpha + c_k}$$ For any $\alpha > 1$, let $\beta = \alpha + \sqrt{\alpha^2 - 1}$, we have $\displaystyle\;\alpha = \frac{\beta + \beta^{-1}}{2}$. Since $c_k = c_{1-k}$, we have

\begin{align} f(\alpha) &= \sum_{k=1}^{N} \frac{1}{\alpha + c_k} = \frac12 \sum_{k=-N+1}^N \frac{1}{\alpha + c_k} = \frac12\sum_{\lambda \in \Lambda}\frac{1}{\frac{\beta + \beta^{-1}}{2} + \frac{\lambda + \lambda^{-1}}{2}}\\ &= \sum_{\lambda \in \Lambda}\frac{\lambda}{(\lambda+\beta)(\lambda+\beta^{-1})} = \frac{1}{\beta-\beta^{-1}}\sum_{\lambda\in\Lambda}\left(\frac{\beta}{\lambda+\beta} - \frac{\beta^{-1}}{\lambda+\beta^{-1}}\right) \end{align} Notice $\lambda \in \Lambda \iff -\lambda \in \Lambda$ and taking logarithm derivatives, we have $$\prod_{\lambda\in\Lambda} (z + \lambda) = \prod_{\lambda\in\lambda}(z - \lambda) = z^{2N} + 1 \quad\implies\quad \sum_{\lambda\in\Lambda} \frac{z}{\lambda + z} = \frac{2N z^{2N}}{z^{2N}+1}$$ This leads to $$f(\alpha) = \frac{2N}{\beta-\beta^{-1}}\left(\frac{\beta^{2N} - 1}{\beta^{2N} + 1}\right)$$ Notice $f(\alpha)$ is continuous at $\alpha = 1$. We find

$$f(1) = \lim_{\alpha\to 1}f(\alpha) = \lim_{\beta\to 1}f(\alpha) = \frac{2N}{2}\left(\frac{2N}{2}\right) = N^2$$

This expression for $f'(\alpha)$ is pretty messy. If I didn't make any mistake, it is \begin{align} f'(\alpha) &= \frac{d\beta}{d\alpha}\frac{df(\alpha)}{d\beta} = \frac{\beta}{\sqrt{\alpha^2-1}} \frac{d\log f(\alpha)}{d\beta} f(\alpha)\\ &= \frac{4N}{(\beta-\beta^{-1})^2}\left(-\frac{\beta+\beta^{-1}}{\beta - \beta^{-1}} + \frac{4N\beta^{2N}}{\beta^{4N}-1}\right)\left(\frac{\beta^{2N} - 1}{\beta^{2N} + 1}\right)\end{align}

For the problem at hand, we need $f(\alpha)$ and $f'(\alpha)$ at $\alpha = \frac53$ which is equivalent to $\beta = 3$. Throwing the function $f(\alpha)$ and $f'(\alpha)$ to a CAS and ask it to simplify the mess at $\beta = 3$, we get

\begin{align} F(N) &= \frac12 N^2 - \frac{3N}{8}\left(\frac{9^N-1}{9^N+1}\right) -\frac{N \left( 5\cdot 9^{2N}-16N\cdot 9^N-5\right)}{16\left( 9^N+1\right)^2}\\ &= \frac{(8N-11)N}{16}+\frac{N((8N+11)9^N + 11)}{8(9^N+1)^2} \end{align} In particular, the first few values of $F(N)$ are $$\frac{1}{25},\frac{1188}{1681},\frac{327129}{133225},\frac{56551312}{0764961}‌​,\frac{316019361}{34869025},\frac{979687020612}{70607649841},\ldots$$

As a double check, I have computed the values of these $F(N)$ numerically using original expression and they match up to $50$ decimal places.

• Very brilliant thought! Thank you very very much. – Roger209 Jan 20 '16 at 15:18
• @achillehui: Instructive answer! (+1) – Markus Scheuer Jan 20 '16 at 15:55
• A minor error: $\omega=e^{\frac{\pi}{\color{red}{2N}}i}$ in the first line. – Roger209 Jan 21 '16 at 1:25
• @Roger209 thanks, fixed. – achille hui Jan 21 '16 at 1:28