Question

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 :)

Answer 1

How is this done? :)

Answer 2

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.