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<math> | <math> | ||
| − | \begin{align} e^ix &= \sum^{\infty}_{n=0}{\frac{(ix)^n}{n!}}\\ | + | \begin{align} 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}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots\\ | ||
| + | &= (1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots)\\ | ||
| + | \end{align} | ||
</math> | </math> | ||
Revision as of 12:48, 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:
$ \begin{align} 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}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots\\ &= (1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots)\\ \end{align} $
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:
(Reference 1)
(Reference 2)
