Analysis Incarnate 2

16 Aug 2020

This 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 :