Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$

Question

Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$

My attempt:

$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$

$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \theta}$$

I did not understand how to solve after that. Help required.

A sum of cosines is the Real part of a sum of exponentials... not sure that helps. — Arnaud Mortier, Apr 08 '18 at 17:56
This is strange: the partial sums are produce a linear plot up to $k=90$ where it drops for some time... — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 08 '18 at 18:02
Will be it be $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi k\cdot 3^k}{3^{100}+1}\right]$$ — lab bhattacharjee, Apr 10 '18 at 11:23
@TheSimpliFire Not that strange, considering that $\frac{2\pi k\cdot 3^k}{3^{100}+1}$ is tiny until $k$ gets close to $100$. Consider that $\cos(\epsilon)\geq 1-\frac{\epsilon^2}{2}$ by Taylor's theorem, so most of the terms will be very close to $1$ if $\epsilon$ is small. — Michael L., May 01 '18 at 00:09
@jacky: Just to make sure, you meant $\displaystyle\prod_{k = 1}^{100}\left[1+2\cos\dfrac{2\pi k \cdot 3^k}{3^{100}+1}\right]$ and not $\displaystyle\prod_{k = 1}^{100}\left[1+2\cos\dfrac{2\pi \cdot 3^k}{3^{100}+1}\right]$, i.e. you meant to have the $k$ in the numerator, correct? — JimmyK4542, May 01 '18 at 00:15
Try a Taylor-expansion of that sine-thing you got at the end and look at the values for some k's — Algebear, May 01 '18 at 00:18

2'5 9'2 · Accepted Answer · 2018-05-01T05:47:44.700

It's unclear if you are asking about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$ or about $$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi k\cdot 3^k}{3^{100}+1}\right]$$

I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If it's the former, then using the trig identity you found, you have a telescoping product which then further simplifies nicely: $$ \begin{align} \prod^{100}_{k=1}\frac{\sin\left(\pi\frac{3^{k+1}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{k}}{3^{100}+1}\right)}&=\frac{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\frac{\sin\left(\pi\frac{3^{3}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{2}}{3^{100}+1}\right)}\cdots\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{100}}{3^{100}+1}\right)}\\[1pc] &=\frac{\sin\left(\pi\frac{3^{101}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3^{1}}{3^{100}+1}\right)}\\[1pc] &=\frac{\sin\left(\pi\frac{3\cdot3^{100}}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc] &=\frac{\sin\left(\pi\frac{3\left(\cdot3^{100}+1\right)-3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc] &=\frac{\sin\left(3\pi-\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc] &=\frac{\sin\left(\pi\frac{3}{3^{100}+1}\right)}{\sin\left(\pi\frac{3}{3^{100}+1}\right)}\\[1pc] &=1 \end{align}$$

Note near the end that $\sin(3\pi-X)=\sin(X)$.

score 0 · Answer 2 · answered May 17 '18 at 02:12

Let $$z=\cos\bigg(\frac{2\pi}{3^n+1}\bigg)+i\sin\bigg(\frac{2\pi}{3^n+1}\bigg)$$

Then $z^{3^n+1}=1$ and also $\displaystyle 2\cos \bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)=z^{3^k}+\frac{1}{z^{3^k}}$

Write $$\prod^{n}_{k=1}\bigg[1+2\cos\bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)\bigg]$$ $$=\bigg(1+z^3+\frac{1}{z^3}\bigg)\bigg(1+z^9+\frac{1}{z^9}\bigg)\cdots \cdots \bigg(1+z^{3n}+\frac{1}{z^{3n}}\bigg)$$ $$=\frac{(1+z^3+z^6)(1+z^9+z^{18})\cdots \cdots (1+z^{3^n}+(z^{3^n})^2)}{z^{3+9+\cdots \cdots +3^n}}$$

Multiply both Nr and Dr by $(1-z^3)$

$$=\frac{1-z^{3^{n+1}}}{(1-z^3)\cdot z^{3\frac{(3^n-1)}{2}}}= \frac{1-z^{-3}}{-(1-z^3)\cdot z^{-3}}=1.$$

$\text{Simplification}:\;\; $ From $z^{3n+1}=1\Rightarrow z^{3n}=z^{-1}$

And $$z^{3\frac{(3^n-1)}{2}}=z^{-3}\cdot z^{3\frac{(3^+1)}{2}}=z^{-3}\cdot \bigg(z^{\frac{(3^+1)}{2}}\bigg)^3=-z^3$$

Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$

2 Answers2