# Taylor series, complex

• Nov 8th 2011, 06:31 AM
Ulysses
Taylor series, complex
Hi there. I have this exercise, which asks me to develop the Taylor series for the given function, and to determine the convergence ratio r, and the ratio for which the series converges to the function R.

The function is:
$f(z)=\frac{1}{z-1}$ around $z_0=2$

So the series looks like a geometric series to me.
$f(z)=\frac{1}{z-1}=\frac{-1}{1-z}$

Now the thing is that its centered at $z_0=2$, so I'm not sure if this would be right:
$f(z)=\frac{1}{z-1}=\frac{-1}{1-z}=-\sum_0^{\infty}(z-2)^n$

I put (z-2) as the sequence ratio because it was centered at $z_0=2$, but I'm not sure if this is right.
• Nov 8th 2011, 06:42 AM
TheEmptySet
Re: Taylor series, complex
Quote:

Originally Posted by Ulysses
Hi there. I have this exercise, which asks me to develop the Taylor series for the given function, and to determine the convergence ratio r, and the ratio for which the series converges to the function R.

The function is:
$f(z)=\frac{1}{z-1}$ around $z_0=2$

So the series looks like a geometric series to me.
$f(z)=\frac{1}{z-1}=\frac{-1}{1-z}$

Now the thing is that its centered at $z_0=2$, so I'm not sure if this would be right:
$f(z)=\frac{1}{z-1}=\frac{-1}{1-z}=-\sum_0^{\infty}(z-2)^n$

I put (z-2) as the sequence ratio because it was centered at $z_0=2$, but I'm not sure if this is right.

You have a good idea, but this may help clean up the details.

First let $w=z-2 \implies z=w+2$

This gives the function

$f(w)=\frac{1}{w+2-1}=\frac{1}{1+w}=\frac{1}{1-(-w)}=\sum_{n=0}^{\infty}(-1)^nw^n=$
$\sum_{n=0}^{\infty}(-1)^n(z-2)^n$