The Most Beautiful Equation

Euler’s formula is stated as the following.

e^(ix) = cos(x) + i.sin(x)

where i = sqrt(-1)

Meanwhile, Euler’s identity is stated as the following.

e^(i.pi) + 1 = 0

where i = sqrt(-1)

In this post, we’re going to prove the above beautiful equations.

The main approach here is by leveraging the Maclaurin series for e^x which is formed as 1 + (x)/(1!) + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + . . ..

Simply by replacing x with i.x we’ll get a new Maclaurin series as shown below.

e^(i.x) = 1 + [(i.x) / (1!)] + [(i^2.x^2) / (2!)] + [(i^3.x^3) / (3!)] + [(i^4.x^4) / (4!)] + . . .

Simplifying the i part of the above equation yield the following.

e^(i.x) = 1 + [(i.x) / (1!)] + [(-1.x^2) / (2!)] + [(-i.x^3) / (3!)] + [(1.x^4) / (4!)] + . . .

e^(i.x) = 1 + [(i.x) / (1!)] - [(x^2) / (2!)] - [(i.x^3) / (3!)] + [(x^4) / (4!)] + . . .

Notice that the above series seems to have a similar pattern to the Maclaurin series of certain functions. Those functions are the sine and cosine. As a refresher, let’s take a look at the Maclaurin series for both of these functions as well as the above e^(ix).

Maclaurin series for Sine

sin(x) = x - [(x^3) / (3!)] + [(x^5) / (5!)] - [(x^7) / (7!)] + . . .

Maclaurin series for Cosine

cos(x) = 1 - [(x^2) / (2!)] + [(x^4) / (4!)] - [(x^6) / (6!)] + . . .

Maclaurin series for e^(i.x)

e^(i.x) = 1 + [(i.x) / (1!)] - [(x^2) / (2!)] - [(i.x^3) / (3!)] + [(x^4) / (4!)] + [(i.x^5) / (5!)] - . . .

e^(i.x) = {1 - [(x^2) / (2!)] + [(x^4) / (4!)] - . . .} + i {[(x) / (1!)] - [(x^3) / (3!)] + [(x^5) / (5!)] - . . .}

Take a look at the above Maclaurin series for e^(i.x). The group of terms without i is the Maclaurin series for cos(x), while the ones with i is the Maclaurin series for sin(x).

Therefore, we can rewrite the Maclaurin series of e^(i.x) to the following.

e^(i.x) = cos(x) + i.sin(x)

Well, we’ve just proved the Euler’s formula!

Now, how about the Euler’s identity?

Recall that the Euler’s identity is stated as the following.

e^(i.pi) + 1 = 0

where i = sqrt(-1)

Well, what’s the result of e^(i.pi) then? Seems that working with x = pi in the Euler’s formula might be the easiest way to prove this identity.

e^(i.pi) = cos(pi) + i.sin(pi)

e^(i.pi) = -1

e^(i.pi) + 1 = 0

Q.E.D