anonymous
  • anonymous
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 :)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
How is this done? :)
1 Attachment
IrishBoy123
  • IrishBoy123
using 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
  • IrishBoy123
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