More fun in 2016, Part 11

I used an approach similar to Comrade Pi Han Goh's, but with some variations. We will show that $S(n)=\sum_{k=0}^{n-1}\tan^2\left(\frac{k\pi}{2n}\right)=\frac{(n-1)(2n-1)}{3}$

The sum we seek is $S(64)-S(32)=\boxed{2016}$ .

To derive the formula for $S(n)$ , consider the equation $\left(e^{\frac{k\pi i}{2n}}\right)^{2n}=(-1)^k$ , expand the LHS in terms of cos and sin, take the imaginary part, and divide by $\cos^2(\frac{k\pi}{2n})$ to find $\sum_{j=0}^{n-1}{2n \choose 2j+1}(i\tan(\frac{k\pi}{2n}))^{2n-2j}=0$ or $\sum_{j=0}^{n-1}{2n \choose 2j+1}(-\tan^2(\frac{k\pi}{2n}))^{n-j}=0$ . Thus we have found a polynomial, $\sum_{j=0}^{n-1}{2n \choose 2j+1}(-1)^{n-j}x^{n-j}$ whose roots are $\tan^2(\frac{k\pi}{2n})$ for $k=0,..n-1$ . By Viete, the sum of these roots is $S(n)=\frac{{2n \choose 3}}{{2n \choose 1}}=\frac{(n-1)(2n-1)}{3}$

Synopsis : We rearrange the terms and notice that the tangent of all $32\times2=64$ angles of are undefined. Then we take advantage of the fact that $\tan(64x)$ is undefined at all these 64 angles. We apply the generalized compound angle formula and Vieta's formula to finish it off.

---

Let $P$ denote the sum in question, we apply one of the trigonometric periodicity identities : $\tan(x) = - \tan(\pi - x) \Rightarrow \tan^2(x) = \tan^2(\pi - x)$ . Then,

$\begin{aligned} P &=& \sum_{k=1}^{32} \tan^2 \left( \pi - \dfrac{(2k-1)\pi}{128} \right) = \sum_{k=1}^{32} \tan^2 \left( \dfrac{(129-2k)\pi}{128} \right) = \sum_{k=33}^{64} \tan^2 \left( \dfrac{(2k-1)\pi}{128} \right) \\ P+P &=& \sum_{k=1}^{32} \tan^2 \left(\dfrac{(2k-1)\pi}{128} \right) + \sum_{k=33}^{64} \tan^2 \left( \dfrac{(2k-1)\pi}{128} \right) \\ 2P &=& \sum_{k=1}^{64} \tan^2 \left( \dfrac{(2k-1)\pi}{128} \right) \end{aligned}$

Note that for $k = 1,2,3,\ldots,64$ , the value of $\tan \left( 64 \cdot \dfrac{(2k-1)\pi}{128} \right)$ is undefined. So we're interested to construct a expression for $\tan(64x)$ such that when it is expressed in its simplest fraction form, the denominator is equal to 0.

Now, for the generalized compound angle formula for the tangent function (currently not in the wiki provided):

$\tan(64x) = \dfrac{e_1 - e_3 + e_5 - \cdots - e_{63}}{1 - e_2 + e_4 -\cdots + e_{64}}$

where $e_m$ denote the $m^\text{th}$ symmetric sum of $\tan \left( \dfrac{(2k-1)\pi}{128} \right)$ , where $k = 1,2,3,\ldots,64$ . Knowing that $e_m= \dbinom {64}{m} (\tan x)^m$ , and we're only interested in the denominator,

$0 = 1 - \dbinom{64}2 \tan^2 x + \dbinom{64}4 \tan^4x - \cdots - \dbinom{64}{62} \tan^{62}x + \dbinom{64}{64} \tan^{64}x.$

By Vieta's formula ,

$\begin{aligned} 2P &=&\sum_{k=1}^{64} \tan^2 \left( \dfrac{(2k-1)\pi}{128} \right) \\ 2P &= & \left[ \sum_{k=1}^{64} \tan \left( \dfrac{(2k-1)\pi}{128} \right) \right]^2 - 2 \left[ \sum_{1\leq j<k\leq64}^{64} \tan\left( \dfrac{(2j-1)\pi}{128} \right) \tan \left( \dfrac{(2k-1)\pi}{128} \right) \right] \\ 2P &=& = (-0)^2 - 2 \cdot \left[ - \dbinom{64}{62} \right] = 64 \cdot 63 \\ P &=& 32 \cdot 63 = \boxed{2016}. \end{aligned}$