Q: Why is e to the i pi equal to -1?

Physicist: This equation ( $e^{i \pi} = -1$ ) was recently voted one of the most famous equations ever. That isn’t part of the answer, it’s just interesting.

First, you’ll find (by plugging them into a graphing calculator and graphing) that:

1) $Sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots$

2) $Cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots$

3) $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots$

Where $N! = 1 \cdot 2 \cdot 3 \cdots N$ .

These are called “Taylor expansions” of “Sine”, “Cosine”, and “e to the x”. If you were to continue the patterns above forever, then you would find that the equality is exact. There is some very exciting math to back me up on this, but for now just trust.

Know that $i^2 = - 1$ . This is how you define $i$ . Therefore, $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i , \ldots$

Check it!:

$e^{i x} = 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots$ $= 1 + i x + \frac{i^2 x^2}{2!} + \frac{i^3 x^3}{3!} + \frac{i^4 x^4}{4!} + \frac{i^5 x^5}{5!} + \frac{i^6 x^6}{6!} + \frac{i^7 x^7}{7!} + \cdots$ $= 1 + i x - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + i \frac{x^5}{5!} - \frac{x^6}{6!} - i \frac{x^7}{7!} + \cdots$

and grouping terms that have “ $i$ ” in them:

$= ( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots) + i ( x$ $- \frac{x^3}{3!}$ $+ \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots )$