# Euler's Formula, or a Study in Cool

## 25 November 2013

Let $i=\sqrt{-1}$, and let
$e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.^{1} If follows that

Utilizing series expansion, this can be equivalently expressed this as

$\begin{array}{l} =1+ix+\frac{\left(ix\right)^{2}}{2!}+\frac{\left(ix\right)^{3}}{3!}+\frac{\left(ix\right)^{4}}{4!}+\frac{\left(ix\right)^{5}}{5!}+\frac{\left(ix\right)^{6}}{6!}+\frac{\left(ix\right)^{7}}{7!}+\frac{\left(ix\right)^{8}}{8!}+\cdots\\ =1+ix-\frac{x^{2}}{2!}-\frac{ix^{3}}{3!}+\frac{x^{4}}{4!}+\frac{ix^{5}}{5!}-\frac{x^{6}}{6!}-\frac{ix^{7}}{7!}+\frac{x^{8}}{8!}+\cdots \end{array}.$Because the series is absolutely convergent, the terms can be rearranged so to be expressed as

$\begin{array}{l} =\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)\\ =\sum_{k=0}^{\infty}\frac{\left(-1\right)^k x^{2k}}{\left(2k\right)!}+i\sum_{k=0}^{\infty}\frac{\left(-1\right)^k x^{2k+1}}{\left(2k+1\right)!} \end{array}.$By recognizing the Maclaurin series for $\cos x$ and $\sin x$,^{2} we can collapse the series to

This is called Euler’s Formula,^{3} and perhaps one of the coolest things I’ve seen in mathematics. It establishes a relationship between exponentiation ($e^x$) and trigonometric functions - even if it is just imaginary.