Question

having trouble with a series expansion.
\[f(z)=\frac{z+1}{z-1}\]I need series expansion in both |z|<1 and 1<|z|<∞

Answer 1

for |z|<1 I rewrite as
\[-(z+1)\frac{1}{1-z}=\frac{-z}{1-z}+\frac{1}{1+z}\]

Answer 2

oops missing a negative sign there. anyways I get\[=-\sum^\infty_{n=0}z^{n+1}-\sum^\infty_{n=0}z^n\]

Answer 3

for |z|<1. Now Apparently I need to replace n by n-1 in the first series. Can someone explain to me why I do this?

Answer 4

There's more than one possible answer, but here's what you could also do for \(|z|<1\):
\[\frac{z+1}{z-1}=\frac{z-1+2}{z-1}=1-2\sum_{n=0}^\infty z^n\]
What you did:
\[\frac{z+1}{z-1}=-\frac{z}{1-z}-\frac{1}{1-z}=-\sum_{n=0}^\infty z^{n+1}-\sum_{n=0}^\infty z^n\]
Changing \(n\) to \(n-1\) requires changing the starting index so that you have
\[\sum_{n=0}^\infty z^{n}=1+z+z^2+z^3+\cdots\\
\sum_{n=1}^\infty z^{n-1}=1+z+z^2+z^3+\cdots\]
As for why? I don't see a good reason for it. I like to consolidate and combine the series whenever I can, but changing the index is counterproductive in that regard.

Answer 5

ah, I missed that my text changed the starting index. thanks.

Answer 6

but are you not missing a negative sign? wouldn't it be \[-1-2\sum_{n=0}^\infty z^n\]

Answer 7

oh wait nvm. you also started at n = 0. ok thanks.