Find value of $\prod_{k=0}^{2^{1999}}\left(4\sin^2\left(\frac{k\pi}{2^{2000}}\right)-3\right)$

Question

Find value of $$S=\prod_{k=0}^{2^{1999}}\left(4\sin^2\left(\frac{k\pi}{2^{2000}}\right)-3\right)$$

We have for $k=0$ the value as $-3$

and now for $k \ne 0$ $$S_1=\prod_{k=1}^{2^{1999}}\left(\frac{\sin\left(\frac{3k\pi}{2^{2000}}\right)}{\sin\left(\frac{k\pi}{2^{2000}}\right)}\right)$$

Letting $f(k)=\sin\left(\frac{k\pi}{2^{2000}}\right)$ we get:

$$S_1=\prod_{k=1}^{2^{1999}}\frac{f(3k)}{f(k)}$$

Lets consider numerator:

we have the product in Numerator with all arguments multiples of $3$ as:

$$N=f(3)f(6)f(9)\cdots f(2^{1999}-2)f(2^{1999}+1)\cdots f(3.2^{1999})$$

Where as in Denominator we have the product with arguments multiples of $3$ as:

$$D_0=f(3)f(6)f(9)\cdots f(2^{1999}-2) \tag{1}$$

Likewise wit arguments in denominator with reminder $1$ when divided by $3$ as:

$$D_1=f(1)f(4)f(7)\cdots f(2^{1999}-1)\tag{2}$$

Likewise wit arguments in denominator with reminder $2$ when divided by $3$ as:

$$D_2=f(2)f(5)f(8)\cdots f(2^{1999}) \tag{3}$$

So we have:

$$S_1=\frac{N}{D_0D_1D_2}=\frac{f(2^{1999}+1)f(2^{1999}+4)\cdots f(3.2^{1999})}{D_1D_2} \tag{4}$$

Now we know that: $$f(2^{1999}-k)=f(2^{1999}+k)$$

So from backwards we can write $$D_1=f(2^{1999}+1)f(2^{1999}+4)\cdots f(2^{2000}-1)$$

Likewise from backwards we can write $$D_2=f(2^{1999})f(2^{1999}+3)\cdots f(2^{2000}-2)$$

After cancelling terms of $D_1$ from numerator in $(4)$ we get:

$$S_1=\frac{f(2^{2000}+2)f(2^{2000}+5)\cdots f(3.2^{1999})}{f(2^{1999})f(2^{1999}+3)\cdots f(2^{2000}-2)}$$

I am stuck here?

Answer 1

I think this is most easily solved using complex numbers:

Note that $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}.$$ We can thus simplify $$4\sin^2 x-3=-\left(e^{2ix}+e^{-2ix}+1\right).$$ So, using the notation $e(x)=e^{2\pi ix}$ and setting $n=2020$, we have that $$4\sin^2\left(\frac{k\pi}{2^n}\right)-3=\frac{\omega^3-1}{\omega(\omega-1)}$$ where $\omega=e(k/2^n)$. As $k$ ranges from $1$ to $2^{n-1}-1$ (the $k=2^{n-1}$ term gives you a $1$), $$S=-3\prod_{\substack{\omega^{2^n}=1\\\Im(\omega)>0}}-\frac{\omega^3-1}{\omega(\omega-1)}.$$ So, letting $P(x)=\frac{x^{2^n}-1}{x-1}$, $$S\overline{S}=-9\prod_{P(\omega)=0}\frac{\omega^3-1}{\omega(\omega-1)}.$$ As $\omega\to\omega^3$ induces an isomorphism of the roots of $P$, the products of the $\omega^3-1$ and the $\omega-1$ terms cancel, so $$S\overline{S}=-9\prod_{P(\omega)=0}\frac{1}{\omega}=-9e\left(\sum_{k=1}^{2^n-1}\frac{k}{2^n}\right)=-9e\left(\frac{2^n-1}2\right)=9.$$ However, obviously $S$ is real. So it suffices to find the sign of $S$, which can be easily done by counting when each individual term in the product is $<0$.

Answer 2

Let $n=2^{1999}$ so that $2n=2^{2000}$ and we are supposed to find the product $$P=\prod_{k=0}^{n}\left(4\sin^2\frac{k\pi}{2n}-3\right)$$ which equals $$P=-3\prod_{k=1}^{n-1}\left(4\sin^2\frac{k\pi}{2n}-3\right)$$ We will also need the product $$Q=\prod_{k=1}^{n-1}\sin^2\frac{k\pi}{2n}$$ Let's observe that $$-\frac{P} {3\cdot 4^{n-1}Q}=\prod_{k=1}^{n-1}\left(1-\dfrac{\sin^2\dfrac{\pi}{3}}{\sin^2\dfrac{k\pi}{2n}} \right)$$ Let $x=2n\pi/3$ and then the above product on right equals $$\frac{\sin x} {n\sin (x/n)} =-\frac{1}{n}$$ Further one can prove with some effort that $Q=n/4^{n-1}$ so that $P=3$.

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If $n=2^m, m\in\mathbb {N} $ then $$\frac{\sin x} {n\sin (x/n)} =(-1)^{m}\cdot\frac{1}{n}$$ and hence the product in question equals $(-1)^{m-1} \cdot 3$ which sort of confirms the comment (to the current question) by Claude Leibovici.

Answer 3

Well, let's get back to the classic formula, let $ n\in\mathbb{N}^{*} $, we have that : $$ \left(\forall z\in\mathbb{C}\right),\ z^{2n}-1=\left(z^{2}-1\right)\prod_{k=1}^{n-1}{\left(z^{2}-2z\cos{\left(\frac{k\pi}{n}\right)}+1\right)} $$

It can be proved, by looking for the complex roots of the polynomial $ X^{2n}-1 $ and foctoring it out.

Setting $ z=\mathrm{j}=\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}} $, we get : $$ \mathrm{j}^{2n}-1=\left(\,\mathrm{j}^{2}-1\right)\prod_{k=1}^{n-1}{\left(\mathrm{j}^{2}-2\,\mathrm{j}\cos{\left(\frac{k\pi}{n}\right)}+1\right)} $$

But since $ \mathrm{j}^{2}+1=-\,\mathrm{j} $, and $ \frac{\mathrm{j}^{2n}-1}{\mathrm{j}^{2}-1}=\frac{\mathrm{j}^{n}\left(\mathrm{j}^{n}-\mathrm{j}^{-n}\right)}{\mathrm{j}\left(\mathrm{j}-\mathrm{j}^{-1}\right)}=\frac{\mathrm{j}^{n-1}\sin{\left(\frac{2n\pi}{3}\right)}}{\sin{\left(\frac{2\pi}{3}\right)}}=\frac{2\,\mathrm{j}^{n-1}}{\sqrt{3}}\sin{\left(\frac{2n\pi}{3}\right)} $, we have : $$ \frac{2\,\mathrm{j}^{n-1}}{\sqrt{3}}\sin{\left(\frac{2n\pi}{3}\right)}=\prod_{k=1}^{n-1}{\left(-\,\mathrm{j}-2\,\mathrm{j}\cos{\left(\frac{k\pi}{n}\right)}\right)}=\mathrm{j}^{n-1}\prod_{k=1}^{n-1}{\left(-1-2\cos{\left(\frac{k\pi}{n}\right)}\right)} $$

Thus $$ \prod_{k=1}^{n-1}{\left(-1-2\cos{\left(\frac{k\pi}{n}\right)}\right)}=\frac{2}{\sqrt{3}}\sin{\left(\frac{2n\pi}{3}\right)} $$

Now note that : $$ \prod_{k=0}^{n}{\left(4\sin^2\left(\frac{k\pi}{2n}\right)-3\right)}=-3\prod_{k=1}^{n-1}{\left(-1-2\cos{\left(\frac{k\pi}{n}\right)}\right)} $$

Hence $$ \prod_{k=0}^{n}{\left(4\sin^2\left(\frac{k\pi}{2n}\right)-3\right)}=-2\sqrt{3}\sin{\left(\frac{2n\pi}{3}\right)} $$

I Believe setting $ n=2^{1999} $ will give you the result you're looking for.