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}.

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