## 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 :)

1. anonymous

How is this done? :)

2. 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.

3. IrishBoy123

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