Euler's Gem

Problem

First prove the formula ${e}^{ix} = \cos(x) + i\sin(x)$. Then discover the beautiful identity ${e}^{i\pi} + 1 = 0$.

Hint: Use the Maclaurin series of exponential and sinusoidal functions.

Solution

Begin with the Maclaurin series of $e^x, \sin(x), \cos(x)$.

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum\limits_{n=0}^\infty \frac{x^n}{n!}$

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... = \sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... = \sum\limits_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}$

Replace $x$ with $ix$, where $i$ is the imaginary unit.

$e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} +... = \sum\limits_{n=0}^\infty \frac{(ix)^n}{n!}$

The powers of $i$ follow a pattern:

${i}^{4n} = 1$, for $n = 0, 1, 2,...$

${i}^{4n+1} = i$, for $n = 0, 1, 2,...$

${i}^{4n+2} = -1$, for $n = 0, 1, 2,...$

${i}^{4n+3} = -i$, for $n = 0, 1, 2,...$

Thus,

$e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} +... = \sum\limits_{n=0}^\infty \frac{(ix)^n}{n!}$

$e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -...\right) + \left(ix - \frac{ix^3}{3!} + \frac{ix^5}{5!} + ...\right)$

$e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -...\right) + i \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...\right)$

This shows that ${e}^{ix} = \cos(x) + i\sin(x)$.

Setting $x = \pi$ yields

${e}^{i\pi} = \cos(\pi) + i\sin(\pi)$

${e}^{i\pi} = -1$.

Therefore,

${e}^{i\pi} + 1 = 0$.