# Analysis Incarnate 2

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

We can write `e^iphi`

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 :

`(iphi)^0/0!`

is equal to `1/0!`

, as `(iphi)^0 = 1`

, and `0!`

is just itself. On the next term, we simplify to get `iphi/1!`

. On the third term, `i^2`

is just `-1`

, so we get `-phi^2/2!`

. On the fourth term, `i^3`

= `-1*i`

, or `-i`

. The result is multiplied by `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^iphi`

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 :