# Analysis Incarnate 2

16 Aug 2020This is a proof of Euler’s Identity using the Taylor Series.

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

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`

.

If we simplify these terms using the powers of `i`

rule, we get :

$\frac{(i\phi)^{0}}{0!}$ 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.

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.

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 :

`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 :