Euler's Formula & Euler's Identity

I remember when I first encountered this problem when I was still studying series and sequences for the first time two years ago. Before I learned about integrating power series, I first encountered this question, and I wrote down the solution, not thinking much of it.

Now that I’ve learned more about how this works, little did I know back then, the math I was learning was far more interesting than I could’ve ever imagined.

Solution

It is known that given the elementary power series:

e^x=\sum_{n=0}^\infty \frac{x^n}{n!}

Substituting ix for x, we obtain:

e^{ix} = \sum_{n=0}^\infty \frac{(ix)^n}{n!}

Writing out the terms of this series,

\begin{aligned} e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots \\ &= 1+ix-\frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \cdots \\[8pt] &= \underbrace{\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)}_{\cos x} +i\underbrace{\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)}_{\sin x} \end{aligned}

Notice that the even powers of x form the Taylor series for \cos x, while the odd powers form the Taylor series form \sin x:

\cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}, \qquad \sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}.

Therefore, we arrive at Euler’s formula:

e^{ix}=\cos x+i\sin x

Additionally, if we plug in \pi, we get Euler’s Identity:

e^{i\pi}=\cos(\pi)+i\sin(\pi) \\ e^{i\pi}=-1 \\ \boxed{e^{i\pi}+1=0}