Why do I insert 2015 anywhere I like? #2

Let $m=2n+1$ be an odd positive integer, and consider $w=e^{2\pi{i}/m}.$

Now $z^m-1=(z-1)\prod_{k=1}^{m-1}(z-w^k)$ . Substituting $z=-1$ and simplifying a bit, we find that $\prod_{k=1}^{m-1}(1+w^k)=1.$

Now $\prod_{k=1}^{m-1}\cos\left(\frac{2k\pi}{m}\right)= \prod_{k=1}^{m-1}\frac{1}{2}(w^k+w^{-k})=\frac{1}{2^{m-1}} \prod_{k=1}^{m-1}(w^k+w^{-k})$ $= \frac{1}{2^{m-1}} \prod_{k=1}^{m-1}(w^{2k}+1)= \frac{1}{2^{m-1}} \prod_{k=1}^{m-1}(w^{k}+1)=\frac{1}{2^{m-1}}$ .

The penultimate step follows from the fact that $w^2$ is a primitive $m$ th root of unity. The third step from the back follows from the fact that the product of the $m$ th roots of unity is 1, by Viete.

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:

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

Thus, in the problem, we are asked the value of:

$\log_2 \left[ \displaystyle \prod_{k=0}^{2014} \cos \left( \dfrac{2k\pi}{2015} \right) \right] = \log_2 (2^{-2014} ) = \boxed{-2014}$

Lemma 1 : $\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1} = \dfrac{1}{2^n}$

Proof : Let, $\displaystyle\text{C}=\prod_{r=1}^{n}\cos{\left(\dfrac{r\pi}{2n+1}\right)}$

and

$\displaystyle\text{S}=\prod_{r=1}^{n}\sin{\left(\dfrac{r\pi}{2n+1}\right)}$

Now,

$\displaystyle \text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{2n+1}} \cdot \cos{\dfrac{\pi}{2n+1}}\right) \cdot \left(\sin{\dfrac{2\pi}{2n+1}}\cdot\cos{\dfrac{2\pi}{2n+1}}\right)\cdot \ldots \cdot\left(\sin{\dfrac{n\pi}{2n+1}} \cdot \cos{\dfrac{n\pi}{2n+1}}\right)$

$\displaystyle \implies \text{C}\cdot\text{S}= \dfrac{1}{2^n} \left(2\sin{\dfrac{\pi}{2n+1}} \cdot \cos{\dfrac{\pi}{2n+1}}\right) \cdot \left(2\sin{\dfrac{2\pi}{2n+1}}\cdot\cos{\dfrac{2\pi}{2n+1}}\right)\cdot \ldots \cdot\left(2\sin{\dfrac{n\pi}{2n+1}} \cdot \cos{\dfrac{n\pi}{2n+1}}\right)$

$\displaystyle \implies \text{C}\cdot\text{S}= \dfrac{1}{2^n} \ \sin{\dfrac{2\pi}{2n+1}}\cdot\sin{\dfrac{4\pi}{2n+1}}\cdot \ldots \cdot\sin{\dfrac{2n\pi}{2n+1}}$

$\displaystyle \{\because \sin(2x) = 2\sin (x) \cos (x) \}$

$\displaystyle \implies \text{C}\cdot\text{S}= \dfrac{1}{2^n} \ \sin{\dfrac{\pi}{2n+1}}\cdot\sin{\dfrac{2\pi}{2n+1}} \cdot \ldots \cdot \sin{\dfrac{n\pi}{2n+1}} \\\\ \{\because \sin(\pi-x)=\sin(x)\}$

$\displaystyle \implies \text{C}\cdot\text{S}= \dfrac{1}{2^n} \cdot \text{S}$

since $\text{S} \neq 0$ ,

$\therefore {\text{C}=\dfrac{1}{2^n}}$

Lemma 2 : $\prod_{r=1}^{2n} \cos \dfrac{2r\pi}{2n+1} = \left(\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1}\right)^2$

Proof : $\displaystyle \prod_{r=1}^{2n} \cos \dfrac{2r\pi}{2n+1} = \left( \prod_{r=1}^{n} \cos \dfrac{2r\pi}{2n+1}\right) \cdot \left(\prod_{r=n+1}^{2n} \cos \dfrac{2r\pi}{2n+1}\right)$

$\displaystyle = (-1)^n\left( \prod_{r=1}^{n} \cos \dfrac{2r\pi}{2n+1}\right) \cdot \left(\prod_{r=1}^{n} \cos \dfrac{(2r-1)\pi}{2n+1}\right) \ \{ \because \cos(\pi+x)=-\cos(x) \}$

$\displaystyle = (-1)^n\prod_{r=1}^{2n}\cos \dfrac{r\pi}{2n+1}$

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

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

$\displaystyle =(-1)^n\left(\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1}\right)\cdot\left(\prod_{r=1}^{n} \cos \dfrac{(2n+1-r)\pi}{2n+1}\right) \ \left( \because \prod_{r=1}^{n} f(r) = \prod_{r=1}^{n}f(n+1-r)\right)$

$\displaystyle =\left(\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1}\right)\cdot\left(\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1}\right) = \left(\prod_{r=1}^{n} \cos \dfrac{r\pi}{2n+1}\right)^2 \ \{\because \cos(\pi-x) = \cos x \}$

Combining the above two Lemmas, we have,

$\displaystyle \prod_{r=0}^{2n} \cos \dfrac{2r\pi}{2n+1} = \dfrac{1}{2^{2n}}$

$\therefore \log_2 \left[ \displaystyle \prod_{r=0}^{2014} \cos \left( \dfrac{2r\pi}{2015} \right) \right] = \log_2 (2^{-2014} ) = \boxed{-2014}$

Let $w=e^{4\pi i/n}$ where $n$ is odd, and let $S=\displaystyle \prod_{k=1}^{n-1} \cos\left(\dfrac{2\pi k}{n}\right)$

Since $\cos(\theta)=\cos(2\pi-\theta)$ , we know that $S$ is always positive.

Now, with the identity $|1+w^k|=2\left | \cos\left(\dfrac{2 \pi k}{n}\right)\right|$ that becomes to:

$S=\displaystyle \prod_{k=1}^{n-1} \dfrac{|1+w^{k}|}{2}=\dfrac{|1+w||1+w^2|\cdots|1+w^{(n-1)}|}{2^{n-1}}$

But also $P(x)=\dfrac{x^n-1}{x-1}=(x-w)(x-w^2)\cdots(x-w^{n-1})$ , that implies that $x^{n-1}+x^{n-2}+\cdots+x+1=(x-w)(x-w^2)\cdots(x-w^{n-1})$ . If we let $x=-1$ we obtain: $1=(-1-w)(-1-w^2)\cdots(-1-w^{n-1})$ , or $1=(1+w)(1+w^2)\cdots(1+w^{n-1})$ .

Finally, apply absolute value to both sides:

$1=|1+w||1+w^2|\cdots|1+w^{n-1}|$

So, the product is: $S=\dfrac{1}{2^{n-1}}$

In this case $n=2015$ , so $\log_2(2^{-2014})=\boxed{-2014}$ .