i

pi\(\pi\) is defined as the ratio between the diameter and the circumference of a circle.

It has been proven to be transcendental and therefore irrational, but we still don't know whether it is normal.

Although it is defined as a geometric constant, it turned out to be useful in other braches of mathematics. One of the main bridges is the unit circle; we choose the period of \(\sin(x)\) and \(\cos(x)\) to be \(2\pi\) (since the diameter of the unit circle is 2) and the nice thing of choosing this, is that their derivatives become exactly \(\cos(x)\) and \(-\sin(x)\) respectively. Then creating their Taylor series (an infinite polynomial) allows us to prove the following:\[\sum_{k\in\mathbb{N}}{1\over k^2}={\pi^2\over6}\] The same value has been proven to be the chance that two randomly chosen numbers are relatively prime.

We can also do the following. We know \(\sin(\alpha)=2\sin({\alpha\over 2})\cos({\alpha\over 2})\) (it's the duplication formula), and substituting this into itself over and over gives us \[\sin(\alpha)=\lim_{k\rightarrow\infty}2^k\sin(2^{-k}\alpha)\prod_{i=1}^{k}{\cos(2^{-i}\alpha)}\]We can rewrite \(2^k\sin(2^{-k}\alpha)\). We'll do this by taking the derivative of \(\sin(\alpha x)\). We see that \({d\over dx}\sin(\alpha x)={\sin(\alpha x+\alpha dx)-\sin(\alpha x)\over dx}\). Setting \(x=0\) and using the fact that the derivative is actually also \(\alpha\cos(\alpha x)\), gives us \(\alpha={\sin(\alpha dx)\over dx}\). Since \(dx\) goes to 0, we may substitute it by \(2^{-k}\) as \(k\) approaches infinity. This gives us \[\sin(\alpha)=\alpha\prod_{i=1}^{\infty}{\cos(2^{-i}\alpha)}\]We also know that \(\cos^2(x)+\sin^2(x)=1\) so with the duplication formula for \(\cos(x)\) we can get \(\cos(2x)=\cos^2(x)-\sin^2(x)\) which is \(2\cos^2(x)-1\). Rewriting this gives us \(\cos({\alpha\over 2})={\sqrt{2\cos(\alpha)+2}\over 2}\). Since \(\cos({\pi\over 4})={\sqrt{2}\over 2}\), we can get a nice result by setting \(\alpha={\pi\over 2}\)in our original equation, and using the recurrence relation we just derived, we get the expression \[{2\over\pi}={\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\cdots\]

1

2

3

4

5

6

7

8

9

0