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# $$\pi$$

$$\pi$$ 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 ${2\over\pi}={\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\cdots$

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