| Line 9: | Line 9: | ||
| | ||
| − | <math> e^ix = \sum^{\infty}_{n=0}{\frac{(ix)^n}{n!}} = \sum^{\infty}_{n=0}{\frac{i^nx^n}{n!}} = 1 + ix - \frac{x^2}2 - i\frac{x^3}6 + \frac{x^4}{24}\cdots </math> | + | <math> e^ix = \sum^{\infty}_{n=0}{\frac{(ix)^n}{n!}} = \sum^{\infty}_{n=0}{\frac{i^nx^n}{n!}} = 1 + ix - \frac{x^2}2 - i\frac{x^3}6 + \frac{x^4}{24} + \cdots </math> |
But by rearranging this, one gets the identity | But by rearranging this, one gets the identity | ||
| − | + | <math> e^ix = \sum^{\infty}_{n=0}{\frac{(-x)^{2n}}{(2n)!}} + i\sum^{\infty}_{n=0}{\frac{(-x)^{2n+1}}{(2n+1)!}} | |
Revision as of 11:41, 2 December 2018
$ 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 + ix - \frac{x^2}2 - i\frac{x^3}6 + \frac{x^4}{24} + \cdots $
But by rearranging this, one gets the identity
$ e^ix = \sum^{\infty}_{n=0}{\frac{(-x)^{2n}}{(2n)!}} + i\sum^{\infty}_{n=0}{\frac{(-x)^{2n+1}}{(2n+1)!}} ''References:'' <br /> (Reference 1) <br /> (Reference 2) [[Category:MA279Fall2018Walther]] $
