Being endlessly productive

Geometry Level 3

When $\prod_{k=2}^{\infty}\cos\left(\frac{\pi}{2^k}\right)$

is computed, the result is $a\pi^b$ for integers $a$ and $b.$ What is $a+b?$

The answer is 1.

2 solutions

Brian Charlesworth
Sep 12, 2015

First look at the finite product $P_{n} = \displaystyle\prod_{k=2}^{n} \cos\left(\dfrac{\pi}{2^{k}}\right)$ for $n \gt 2.$

If we then multiply this by $\sin\left(\dfrac{\pi}{2^{n}}\right),$ then since $\sin(x)\cos(x) = \dfrac{\sin(2x)}{2}$ the product will successively "collapse" from the last term to the first, leaving us with the equation

$\sin\left(\dfrac{\pi}{2^{n}}\right)*P_{n} = \dfrac{\sin(\frac{\pi}{2})}{2^{n-1}} = \dfrac{2}{2^{n}} \Longrightarrow P_{n} = \dfrac{2}{2^{n}\sin\left(\dfrac{\pi}{2^{n}}\right)}.$

Now $\lim_{n \rightarrow \infty} 2^{n}\sin\left(\dfrac{\pi}{2^{n}}\right) = \lim_{n \rightarrow \infty} \pi*\dfrac{\sin\left(\dfrac{\pi}{2^{n}}\right)}{\dfrac{\pi}{2^{n}}} = \pi*\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = \pi * 1 = \pi,$

where $x = \dfrac{\pi}{2^{n}} \rightarrow 0$ as $n \rightarrow \infty.$ Thus

$\lim_{n \rightarrow \infty} P_{n} = \dfrac{2}{\pi} = 2\pi^{-1},$ and so $a + b = 2 + (-1) = \boxed{1}.$

Exactly! Very nice! Thank you, Brian.

Otto Bretscher - 5 years, 9 months ago

$\begin{array}{|cc|} \hline \hline\hline \hline\hline \hline\hline \hline\hline \hline \text{Upvoted}\\ \text{Brilliant Solution Sir}\\ \hline \hline \end{array}$

Sir, Can you please elaborate more the fourth line when we multiply sin , because a lot of things are coming and disappearing which caused a little confusion.

Like: $\frac{ sin ( \frac{\pi}{2} )}{ 2^{n-1}}$

And also, $\Longrightarrow P_{ n }=\frac { 2 }{ 2^{ n }\quad sin(\frac { \pi }{ { 2 }^{ n } } ) }$

Syed Baqir - 5 years, 9 months ago

As an example, look at $P_{5} = \cos(\frac{\pi}{4})\cos(\frac{\pi}{8})\cos(\frac{\pi}{16})\cos(\frac{\pi}{32}).$

Then using the identity I've mentioned in my solution, we see that

$\sin(\frac{\pi}{32})*P_{5} = \cos(\frac{\pi}{4})\cos(\frac{\pi}{8})\cos(\frac{\pi}{16})\cos(\frac{\pi}{32})\sin(\frac{\pi}{32}) =$

$\cos(\frac{\pi}{4})\cos(\frac{\pi}{8})\cos(\frac{\pi}{16})*(\frac{1}{2}\sin(\frac{\pi}{16})) =$

$\cos(\frac{\pi}{4})\cos(\frac{\pi}{8})*(\frac{1}{4}\sin(\frac{\pi}{8})) = \cos(\frac{\pi}{4})*(\frac{1}{8}\sin(\frac{\pi}{4})) = \frac{1}{16}\sin(\frac{\pi}{2}) = \frac{1}{16} = \dfrac{2}{2^{5}}.$

So in this case $P_{5} = \dfrac{2}{2^{5}\sin\left(\dfrac{\pi}{2^{5}}\right)},$ and in general $P_{n} = \dfrac{2}{2^{n}\sin\left(\dfrac{\pi}{2^{n}}\right)}.$

Brian Charlesworth - 5 years, 9 months ago

Thank you Sir !

Syed Baqir - 5 years, 9 months ago

Rodrigo Raffa
Oct 25, 2015

Cos (0) = 1 because 1/infinite

Being endlessly productive

The answer is 1.

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