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 :