Question

expand

Answer 1

http://math.stackexchange.com/questions/172235/expanding-fractions-as-powers-of-z

Answer 2

When I add I don't get the series -3+9/z+...

Answer 3

@SithsAndGiggles

Answer 4

\[\sum_{n=0}^{\infty}((-z)^n +\frac{ 1 }{ 2 }(\frac{ z }{ 2})^n)\]

Answer 5

\[\sum_{n=0}^\infty \left((-z)^n+\frac{1}{2}\left(\frac{z}{2}\right)^n\right)=\sum_{n=0}^\infty\underbrace{\left((-1)^n+\frac{1}{2^{n+1}}\right)}_{a_n}z^n=\frac{3}{2}-\frac{3}{4}z+\cdots\]

Answer 6

yes, but f(z)= -3+9z/2-15z^2/4+....

Answer 7

how to expand in powers of 1/z?

Answer 8

may be we need to add both z and 1/z to get f(z)

Answer 9

That's not necessary. The series is multiplied by \(\dfrac{4}{z}\), and so distributing that among the terms you have
\[\frac{3\times4}{2z}-\frac{3\times4}{4z}z+\frac{9\times4}{8z}z^2-\frac{15\times4}{16z}z^3+\cdots\\
\quad\quad=\frac{6}{z}-3+\frac{9}{2}z-\frac{15}{4}z^2+\cdots\]

Answer 10

oh, they started from n=1

Answer 11

I did the 6/z and thought I was wrong and stopped the process. Thanks. But second part asks for 1/z expansion. Is there a formual for it?

Answer 12

Sorry, what's the second part? Just find the Laurent expansion?

Answer 13

expand in powers of 1/z

Answer 14

Okay, easy enough. Consider just one series as an example:
\[\frac{1}{1-z}=\sum_{n=0}^\infty z^n\]is valid for \(|z|<1\), but we want a series valid for \(|z|>1\). We can get what we want as follows:
\[\frac{1}{1-z}=\frac{1}{z}\times\frac{1}{\frac{1}{z}-1}=-\frac{1}{z}\times\frac{1}{1-\frac{1}{z}}=-\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n\]

Answer 15

\[\sum_{n=0}^{\infty} (\frac{ 1 }{ z^{n+1} }-\frac{ 2^n }{ z^{n+1} })\]

Answer 16

when I add two

Answer 17

Almost right, the first term should be an alternating power of negative \(1\), since
\[\frac{1}{1+z}=\frac{1}{z}\times\frac{1}{1-\left(-\frac{1}{z}\right)}=\frac{1}{z}\sum_{n=0}^\infty \left(-\frac{1}{z}\right)^n=\sum_{n=0}^\infty \frac{(-1)^n}{z^{n+1}}\]

Answer 18

Got you!

Answer 19

Adding 1 and 2

Answer 20

\[\sum_{n=0}^{\infty} ((-1)^n+\frac{ 1 }{ 2^{n+1}})z^n+\frac{ (-1)^n }{ z^{n+1}}-\frac{ 2^n }{ z^{n+1}}\]