Daily Chaos: Viete's Pi

$\sin(x) = 2\sin \left( \frac{x}{2}\right)\cos \left( \frac{x}{2}\right)$

$\Rightarrow \sin(x) = 2^2\sin \left( \frac{x}{4}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$

$\Rightarrow \sin(x) = 2^3\sin \left( \frac{x}{8}\right)\cos \left( \frac{x}{8}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$

$\ldots$

$\Rightarrow \sin(x) = 2^n\sin \left( \frac{x}{2^n}\right)\cos \left( \frac{x}{2^n}\right)\ldots\cos \left( \frac{x}{8}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{2}\right)$

$\Rightarrow \sin(x) = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)\sin \left( \frac{x}{2^n}\right)2^n$

$\Rightarrow \dfrac{\sin(x)}{x} = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)\dfrac{\sin \left( \frac{x}{2^n}\right)}{\frac{x}{2^n}}$

as $n \to \infty$

$\Rightarrow \dfrac{\sin(x)}{x} = \cos \left( \frac{x}{2}\right)\cos \left( \frac{x}{4}\right)\cos \left( \frac{x}{8}\right)\ldots\cos \left( \frac{x}{2^n}\right)$

$\Rightarrow \dfrac{\sin(x)}{x} =\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{x}{2^k}\right)}$

Plug $x=\frac{\pi}{2}$ we get

$\Rightarrow \dfrac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} =\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{\frac{\pi}{2}}{2^k}\right)}$

$\Rightarrow \dfrac{2}{\pi}=\displaystyle{\Pi_{k=1}^{\infty} \cos \left( \frac{\pi}{2^{k+1}}\right)}$

$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\Pi_{k=2}^{\infty} \cos \left( \frac{\pi}{2^{k}}\right)}$

$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\cos \left( \frac{\pi}{4}\right)\cos \left( \frac{\pi}{8}\right)\cos \left( \frac{\pi}{16}\right)}\ldots$

$\displaystyle{\Rightarrow \dfrac{2}{\pi}=\dfrac{\sqrt{2}}{2}\dfrac{\sqrt{2+\sqrt{2}}}{2}\dfrac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}}\ldots$

Daily Chaos

Tuesday, December 29, 2015

Viete's Pi

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