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| + | [[Category:Euler's formula]] | ||
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==Euler's Forumla== | ==Euler's Forumla== | ||
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| − | <math> e^{ix} = 1 + ix + | + | <math> e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots </math> |
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| + | Expanding the complex terms yield: | ||
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| + | <math> e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{(ix)^3}{3!} + \frac{x^4}{4!} + \frac{(ix)^5}{5!} - \cdots </math> | ||
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| + | Rearranging the terms into even and odd exponents and factoring out the imaginary number: | ||
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| + | <math> e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots) </math> | ||
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| + | Which is equivalent to: | ||
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| + | <math> e^{ix} = \cos x + i\sin x </math> | ||
Latest revision as of 05:48, 23 September 2011
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 $
Expanding the complex terms yield:
$ e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{(ix)^3}{3!} + \frac{x^4}{4!} + \frac{(ix)^5}{5!} - \cdots $
Rearranging the terms into even and odd exponents and factoring out the imaginary number:
$ e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots) $
Which is equivalent to:
$ e^{ix} = \cos x + i\sin x $
