(Proof)
(Proof)
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<math> e^{ix} = 1 + ix + /frac{(ix)^2}{2!} + /frac{(ix)^3}{3!} + /frac{(ix)^4}{4!} + /frac{(ix)^5}{5!} + \cdots </math>
+
<math> e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots </math>

Revision as of 04:53, 5 September 2008

Euler's Forumla

$ e^{ix} = \cos x + i * \sin x $

Proof

Using Taylor Series:

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

$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $

$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots $


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

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood