# Analysis Incarnate 2

This is a proof of Euler’s Identity using the Taylor Series.

$\dpi{300}&space;\tiny&space;e^{i\pi}+1=0$

We can write $ie^{i\phi}$ as the Taylor Series expansion.

$\dpi{300}&space;\tiny&space;e^{i\phi}=\sum_{n=0}^{\infty}\frac{(i\phi)^{n}}{n!}$

Let’s visualize this sum. It helps to write out all your terms because you know that the powers of i are going to alternate between 1,-1,i,and -i.

$\dpi{300}&space;\tiny&space;=\frac{\left(i\phi&space;\:\right)^0}{0!}+\frac{\left(i\phi&space;\right)^1}{1!}+\frac{\left(i\phi&space;\:\right)^2}{2!}+\frac{\left(i\phi&space;\:\right)^3}{3!}+\frac{\left(i\phi&space;\:\right)^4}{4!}+...$

If we simplify these terms using the powers of i rule, we get :

$\dpi{300}&space;\tiny&space;=\frac{1}{0!}+i\left(\frac{\phi}{1!}\right)-\frac{\phi^2}{2!}-i\left(\frac{\phi^3}{3!}\right)+\frac{\phi^4}{4!}+...$

$\frac{(i\phi)^{0}}{0!}$ is equal to $\frac{1}{0!}$, as $(i\phi)^{0}$ = 1, and $0!$ is just itself. On the next term, we simplify to get $\frac{i\phi}{1!}$. On the third term, $i^{2}$ is just -1, so we get $\frac{-\phi^{2}}{2!}$. On the fourth term, $i^{3}$ = $-1*i$, or $-i$. The result is multiplied by $\frac{\phi^{3}}{3!}$. This process, because the sum is up to infinity, goes on indefinitely. Since we have a complex number, we can factor out the i and compare co-efficients of the front and rear parts.

$\dpi{300}&space;\tiny&space;=\left(\frac{1}{0!}-\frac{\phi&space;^2}{2!}+\frac{\phi&space;^4}{4!}-...\right)+i\left(\frac{\phi&space;^1}{1!}-\frac{\phi&space;^3}{3!}+\frac{\phi&space;^5}{5!}-...\right)$

As you can see, the addition/subtraction operators alternate after every term. The front half of this equation is the Taylor Series expansion of $cos\phi$. The rear half of this equation is the Taylor Series expansion of $sin\phi$. You can take the original sum notation and split it into two partial sums, one with the even part, and one with the odd part.

$\dpi{300}&space;\tiny&space;=cos(\phi)+isin(\phi)$

We can see that $e^{i\phi}$ is equal to $cos\phi$ + $i sin \phi$ (Euler’s formula). If you substitute π into this formula :

$\dpi{300}&space;\tiny&space;e^{i\pi}=&space;cos(\pi)+i&space;sin&space;(\pi)$

$\dpi{300}&space;\tiny&space;=&space;-1+0$

cos(π) is equal to -1 and i sin (π) is just 0 (think of the unit circle!). This leaves us with the most “beautiful” formula in mathematics :

$\dpi{300}&space;\tiny&space;e^{i\pi}+1=&space;0$