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anonymous
 one year ago
Explaination wanted > How can e^(i*smallomega*t) be equal to cos(smallomega*t) + i*sin(smallomega*t) ? A big cookie to whomever explains it so I understand the math behind the conversion. I believe it has something to do with Euler's  however I am unable to piece it together myself :)
anonymous
 one year ago
Explaination wanted > How can e^(i*smallomega*t) be equal to cos(smallomega*t) + i*sin(smallomega*t) ? A big cookie to whomever explains it so I understand the math behind the conversion. I believe it has something to do with Euler's  however I am unable to piece it together myself :)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How is this done? :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1using Taylor series: \(e^{x} = \sum^{\infty}_{n=0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\) \(\cos x = \sum^{\infty}_{n=0} \frac{(1)^n}{(2n)!} x^{2n} = 1  \frac{x^2}{2!} + \frac{x^4}{4!}  \cdots\) \(\sin x = \sum^{\infty}_{n=0} \frac{(1)^n}{(2n+1)!} x^{2n+1} = x  \frac{x^3}{3!} + \frac{x^5}{5!}  \cdots\) \(i \, sin x = \sum^{\infty}_{n=0} \frac{(1)^n}{(2n+1)!} x^{2n+1} = i \, x  i\, \frac{x^3}{3!} + i\, \frac{x^5}{5!}  \cdots\) well \(e^{ix} = \sum^{\infty}_{n=0} \frac{x^n}{n!} = 1 + i \,x  \frac{x^2}{2!}  i \, \frac{x^3}{3!} + \cdots\) so \(e^{ix} = \cos x + i \sin x\) that's how Euler found it.....i believe.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444608966526:dw
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