## Blog

There are many interesting ways of representing ${\pi}$. Here is one of the more approachable and fun representations.

$\pi=\frac{2}{\sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})} \sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}} \sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}})} \ldots }$

We will prove the above formula using little more than High School mathematical concepts and techniques.

Some of the notation may look a bit strange and the level of conceptual difficulty is quite high. The main concepts that we will be using here are:

• Trigonometric Identities
• Limits
• Recursion Formulas
• Summation/Product terminology

The last one will look very strange.

We will be using $\prod$ to show an infinite product of terms.

#### Groundwork

We will start with a fundamental concept about the function $f(x)=\sin{x}$.

It can be proved that if $x$ is very small, then  $\sin{x}\approx x$.

Another way of writing this is to say that $\lim_{x \to 0}\frac{\sin{x}}{x}=1$

We can also write this as $\lim_{x \to 0}\sin{x}=x$

Next up come the three main Trigonometric Identities that we will be using.

These are the two double angle formulas and the fundamental link between $\sin{x}$ and $\cos{x}$.

• $\cos{2 x}=\cos^{2}{x}-\sin^{2}{x}$
• $\sin{2 x}=2 \sin{x}\cos{x}$
• $\cos^{2}{x}+\sin^{2}{x}=1$

With a bit of rewriting and redefining we get the following:

$\sin{x}=2 \sin{\frac{x}{2}}\cos{\frac{x}{2}}$

$\cos{x}=2 \cos^{2}{\frac{x}{2}}-1$

or,

$\cos{\frac{x}{2}}=\sqrt{\frac{1}{2}(1+\cos{x})}$

#### Begin!

We are now ready to begin:

$\sin{\frac{\pi}{2}}=2 \sin{\frac{\pi}{4}}\cos{\frac{\pi}{4}}\newline \newline =2 (2 \sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}) \cos{\frac{\pi}{4}} \newline \newline =4 \sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}\cos{\frac{\pi}{4}}\newline \newline =4 (2\sin{\frac{\pi}{16}}\cos{\frac{\pi}{16}})\cos{\frac{\pi}{8}}\cos{\frac{\pi}{4}}\newline \newline =8\sin{\frac{\pi}{16}}\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}\cos{\frac{\pi}{4}}\newline \newline =8 (2\sin{\frac{\pi}{32}}\cos{\frac{\pi}{32}})\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}\cos{\frac{\pi}{4}}\newline \newline =16\sin{\frac{\pi}{32}}\cos{\frac{\pi}{32}}\cos{\frac{\pi}{16}}\cos{\frac{\pi}{8}}\cos{\frac{\pi}{4}}$

Now, with a bit of abstraction, we get the following:

$\sin{\frac{\pi}{2}}=2^{n} \sin{\frac{\pi}{2^{n+1}}}\prod_{i=1}^{n}\cos{\frac{\pi}{2^{i+1}}}$

or

$1=2^{n} \sin{\frac{\pi}{2^{n+1}}}\prod_{i=1}^{n}\cos{\frac{\pi}{2^{i+1}}}$

#### Keep going!

Now we consider the Product terms.

First off, note that $\cos{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$

Note that as above,

$\cos{\frac{x}{2}}=\sqrt{\frac{1}{2}(1+\cos{x})} \newline \newline \Leftrightarrow \cos{\frac{\pi}{8}}=\sqrt{\frac{1}{2}(1+\cos{\frac{\pi}{4}})}\newline \newline \Leftrightarrow \cos{\frac{\pi}{8}}=\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}$

Also,

$\cos{\frac{\pi}{16}}=\sqrt{\frac{1}{2}(1+\cos{\frac{\pi}{8}})}\newline \newline = \sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}}$

And,

$\cos{\frac{\pi}{32}}=\sqrt{\frac{1}{2}(1+\cos{\frac{\pi}{16}})}\newline \newline =\sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}})}$

Thus we get a recursive formula for all of the $\cos$ terms.

This recursive formula can be written as

$x_{n+1}=\sqrt{\frac{1}{2}(1+x_{n})}$ where

$x_{1}=\cos{\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$

Now, we can rewrite our Formula as

$1=2^{n} \sin{\frac{\pi}{2^{n+1}}}\prod_{i=1}^{n}x_{i}$ ,  $x_{n+1}=\sqrt{\frac{1}{2}(1+x_{n})}$$x_{1}=\frac{1}{\sqrt{2}}$.

As we let $n$ get bigger and bigger we see that the  $\sin{\frac{\pi}{2^{n+1}}}$ gets smaller and smaller and if we let  $n \to \infty$ we get:

$1=\lim_{n \to \infty}(2^{n} \sin{\frac{\pi}{2^{n+1}}}\prod_{i=1}^{n}x_{i})$ ,  $x_{n+1}=\sqrt{\frac{1}{2}(1+x_{n})}$$x_{1}=\frac{1}{\sqrt{2}}$.

which becomes

$1=2^{n} {\frac{\pi}{2^{n+1}}}\prod_{i=1}^{n}x_{i}$ ,  $x_{n+1}=\sqrt{\frac{1}{2}(1+x_{n})}$$x_{1}=\frac{1}{\sqrt{2}}$ as $\sin{x} \to x$ when $x$ is small.

So,

$1= \frac{\pi}{2}\prod_{i=1}^{\infty}x_{i}$

#### And Finally…

Or, finally and succinctly,

$\pi=\frac{2}{\prod_{i=1}^{\infty}x_{i}} , x_{1}=\frac{1}{\sqrt{2}} , x_{n+1}=\sqrt{\frac{1}{2}(1+x_{n})}$

$\pi=\frac{2}{\sqrt{\frac{1}{2}} \sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})} \sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}} \sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\sqrt{\frac{1}{2}(1+\frac{1}{\sqrt{2}})}})} \ldots }$