# My Pi Besides “Wallis’ Pi“, there is another remarkable expression for the number $\pi$ as an infinite product. We derive it as follows:

From the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$,

we have $\sin(x) = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ $= 2\cdot 2\sin(\frac{x}{4})\cos(\frac{x}{4}) \cdot\cos(\frac{x}{2})$ $= 2\cdot 2\cdot 2\sin(\frac{x}{8})\cos(\frac{x}{8})\cdot \cos(\frac{x}{4})\cdot \cos(\frac{x}{2})$ $\ddots$ $= 2^n\sin(\frac{x}{2^n})\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})$.

That is, $\sin(x) = 2^n\sin(\frac{x}{2^n})\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i}).$

Dividing both sides by $x$ yields $\frac{\sin(x)}{x} = \frac{2^n\sin(\frac{x}{2^n})}{x}\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})=\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})$

or, $\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i})=\frac{\sin(x)}{x} /\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}$.

It follows that since $\lim\limits_{n \rightarrow \infty}\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}=1$, $\lim\limits_{n \rightarrow \infty}\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i})=\lim\limits_{n \rightarrow \infty}\frac{\sin(x)}{x}/\lim\limits_{n \rightarrow \infty}\frac{\sin(\frac{x}{2^n})}{\frac{x}{2^n}}=\frac{\sin(x)}{x}$;

i.e., $\frac{\sin(x)}{x}=\lim\limits_{n \rightarrow \infty}\prod\limits_{i=1}^{n} \cos(\frac{x}{2^i}).\quad\quad\quad(1)$

Let $x = \frac{\pi}{2},$

(1) becomes $\frac{2}{\pi} =\lim\limits_{n\rightarrow \infty}\prod\limits_{i=1}^{n}\cos(\frac{\pi}{2\cdot2^i})=\lim\limits_{n\rightarrow \infty}\cos(\frac{\pi}{4})\cos(\frac{\pi}{8})...\cos(\frac{\pi}{2\cdot 2^n}).\quad\quad\quad(2)$

We know $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.$

Applying the half-angle formula $\cos(\frac{x}{2}) = \pm \sqrt{\frac{1+\cos(x)}{2}}$

gives $\cos(\frac{\pi}{8}) = \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} = \frac{\sqrt{2+\sqrt{2}}}{2};$ $\cos(\frac{\pi}{16}) = \sqrt{\frac{1+\cos(\frac{\pi}{8})}{2}} = \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2};$ $\ddots$

Hence, $\frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}...\quad\quad\quad(3)$

We compute the value of $\pi$ according to (3):

Fig. 1

Exercise-1 Compute $\pi$ from (1) by letting $x = \frac{\pi}{6}.$