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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:

  • Radians
  • 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})}

Or in “headache” form

\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 }

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