Page 5 - Rhea

$ e $ and Trigonometry

The Taylor series of $ e^x $ is

                $ e^x = \sum^{\infty}_{n=0}{\frac{x^n}{n!}} = 1 + x + \frac{x^2}2 + \frac{x^3}6 + \cdots $

Using this equation, it is possible to relate $ e $ to the seemingly unrelated worlds of trigonometry and the complex numbers by simply plugging in a complex number, $ ix $ for example. This yields:

                $ e^ix = \sum^{\infty}_{n=0}{\frac{(ix)^n}{n!}} = \sum^{\infty}_{n=0}{\frac{i^nx^n}{n!}} = 1 + x + \frac{x^2}2 + \frac{x^3}6 + \cdots $

But by rearranging this, one gets the identity

References:
(Reference 1)
(Reference 2)