Chebyshev polynomial practice: medium hard product

Let $z = e^{i\theta} = x+iy$ . Then:

$\cos((2n+1)\theta) = Re(z^{2n+1}) = \displaystyle \sum_{k=0}^n \binom{2n+1}{2k} x^{2(n-k)+1} (iy)^{2k}$ $= \displaystyle \sum_{k=0}^n (-1)^k \binom{2n+1}{2k} x^{2(n-k)+1} (1-x^2)^k$

which we denote as $f(x)$ . It's easy to see that:

$f(x) = a_{2n+1}x^{2n+1} + \ldots + a_1x^1 \quad ...(1)$

where

$a_1 = (-1)^n \binom{2n+1}{2n} = (-1)^n (2n+1) \quad ...(2)$

On the other hand, since $\cos((2n+1)\theta) = \dfrac12 \left( z^{2n+1} + \dfrac{1}{z^{2n+1}} \right) = 1 + \dfrac12 \left( z^{2n+1} + \dfrac{1}{z^{2n+1}} -2 \right)$

$= 1 + \dfrac{1}{2z^{2n+1}}(z^{2n+1} - 1)^2$ and

$(z^{2n+1} - 1)^2 = \displaystyle \prod_{k=0}^{2n} \left( z-e^{\frac{2k\pi i}{2n+1}} \right)^2 = \prod_{k=0}^{2n} \left(z-e^{\frac{2k\pi i}{2n+1}} \right) \left(z-e^{\frac{-2k\pi i}{2n+1}} \right)$

$= \displaystyle \prod_{k=0}^{2n} \left( z^2 + 1 - 2z \cos \left( \dfrac{2k\pi}{2n+1} \right) \right)$

We also have:

$f(x) = 1 + \dfrac12 \prod_{k=0}^{2n} \left( z + \dfrac1z - 2 \cos \left( \dfrac{2k\pi}{2n+1} \right) \right)$

$= 1 + 2^{2n} \prod_{k=0}^{2n} \left( x - 2 \cos \left( \dfrac{2k\pi}{2n+1} \right) \right) \quad ...(3)$

Now we compare the expressions $(1)$ and $(3)$ for $f(x)$ . We see from $(1)$ that the constant term in $(3)$ is zero. Therefore:

$2^{2n} \prod_{k=0}^{2n} \cos \left( \dfrac{2k\pi}{2n+1} \right) = 1 \quad \Longrightarrow \prod_{k=0}^{2n} \sec \left( \dfrac{2k\pi}{2n+1} \right) = 2^{2n}$

In the problem, we are given $n=1008$ , so the answer to the problem would be $2^{2 \cdot 1008} = 2^{2016}$ . Thus $a = 2, b = 2016, a+b= \boxed{2018}$ .

Let $T_n$ be the nth Chebyshev Polynomial where $T_n(\cos(x))=\cos(nx)$

We are looking at a product of 2017 terms from

$\large \displaystyle \prod_{n=0}^{2016} \sec\left(\dfrac{n\pi}{2017}\right)$

We can make each term a root of the polynomial $T_{2017}(n)=1$

Now, the trick here is to realize that not every term of $cos(n\pi)$ is equal to 1. 1008 are odd n and are equal to -1. To accommodate this, we shift every odd term $\pi$ to the right (we add \pi). This won't change the value of the equation so it won't change the polynomial value either.

Now, for the product of the roots, we need the constant term divided by the lead coefficient. The constant term will be -1 of course and the lead coefficient $2^{n-1}=2^{2016}$

$T_{2017}(x)-1=2^{2016}x^{2017}+\text{stuff}-1$

Product of roots

$\dfrac{1}{2^{2016}}$

So $\sec(x)$ is

$2^{2016}$

$\large \displaystyle \prod_{n=1}^{2016} \sec\left(\dfrac{n\pi}{2017}\right)=\frac{1}{\displaystyle \prod \limits^{2016}_{n=1}\cos \left( \frac{n\pi }{2017} \right)}$ $x = \displaystyle \prod \limits^{2016}_{n=1}\cos \left( \frac{n\pi }{2017} \right)$ $y = \displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right)$ $x. y =\displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right) \cos \left( \frac{n\pi }{2017} \right)$ $x. y = \displaystyle\frac{1}{2}\prod_{n=1}^{2016}\sin\left( \frac{2n\pi}{2017} \right) = \bigg(\frac{1}{2}\bigg)^{2016}[\sin \left( \frac{2\pi }{2017} \right) \sin \left( \frac{4\pi }{2017} \right) \cdots \sin \left( \frac{14\pi }{2017} \right) \sin \left( \frac{16\pi }{2017} \right) \sin \left( \frac{18\pi }{2017} \right) \cdots \sin \left( \frac{4032\pi }{2017} \right)]$ $\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{\theta+\alpha=2\pi \rightarrow \sin(\theta)=-\sin(2\pi-\alpha)}}}}$ $x. y = \displaystyle\frac{1}{2}\prod_{n=1}^{2016}\sin\left( \frac{2n\pi}{2017} \right) = \bigg(\frac{1}{2}\bigg)^{2016}(-1)^{1008}[\displaystyle\prod \limits^{2016}_{n=1}\sin\left( \frac{n\pi }{2017} \right) ]$ $x. y =\left( \frac{1}{2} \right) ^{2016}\left( -1\right) ^{1008}. y$ $x =\left( \frac{1}{2} \right) ^{2016}$ $\frac{1}{x} =\left({2}\right) ^{2016}=a^b$ $\Large\color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{A+B=2018}}}}$

Just to get something posted, I used the identity here to get an answer of

$2^{2017 - 1} = 2^{2016} \Longrightarrow a + b = 2 + 2016 = \boxed{2018}.$

A variety of proof techniques are provided in the link, including a Chebyshev-inspired one.