# Taylor Series

• Sep 21st 2012, 09:59 AM
linalg123
Taylor Series
Use taylor series to verify $\displaystyle e^{-iz}= cos z - isinz$

i know cos z =$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{2n!}$

and sin z = $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{2n+1!}$

and $\displaystyle e^{-iz} =\sum_{n=0}^{\infty}\frac{(-iz)^n}{n!}$

but i cant get the steps to get there.
• Sep 21st 2012, 12:38 PM
girdav
Re: Taylor Series
The sum for $\displaystyle \cos$ runs over even terms, and the corresponding one for $\displaystyle \sin$ of odd terms. So you just have to check that the coefficients of $\displaystyle e^{-iz}$ for odd/even indexes match.
• Sep 21st 2012, 06:00 PM
johnsomeone
Re: Taylor Series
Start with the $\displaystyle e^{-iz}$ series.
Break it down by what happens to that $\displaystyle (-i)^n$ factor.
That will break the $\displaystyle e^{-iz}$ series into the sum of four series via the values of n mod 4.
Collect real and imaginary terms.
Simpify for the real and imaginary parts (this takes some care - gotta manipulate the series and their indices and such carefully).
That'll do it.