taylor series

durham2 · #1 $Old$ October 27th, 2009, 03:14 PM

derive taylor series representation for f(z)=1/1-z for |z-i|<squareroot of 2..ie..

1/1-z=summation n=0 to infinity (z-i)^n/(1-i)^n+1

mr fantastic · #2 $Old$ October 27th, 2009, 09:41 PM

Quote:

Originally Posted by durham2 $View Post$

derive taylor series representation for f(z)=1/1-z for |z-i|<squareroot of 2..ie..

1/1-z=summation n=0 to infinity (z-i)^n/(1-i)^n+1

The easy way is to note that:

1. $\frac{1}{1 - z} = \frac{-1}{z - 1} = \frac{-1}{z - i + i - 1} = \frac{-1}{(z - i) - (1 - i)} = \frac{\frac{-1}{1 - i}}{\left( \frac{z - i}{1 - i} \right) - 1}$

$= - \frac{1 + i}{2} \left[ \frac{1}{\left( \frac{z - i}{1 - i} \right) - 1} \right]$ .

2. $\left| \frac{z - i}{1 - i}\right| < 1$ .

Now recall the formula for the sum of an infinite geometric series.