Partial Fractions in Laurent Series Expansion

[tadf2]

Homework Statement

f = \frac{1}{z(z-1)(z-2)}

Homework Equations

Partial fraction

The Attempt at a Solution

R1 = 0 < z < 1
R2 = 1 < z < 2
R3 = z > 2


f = \frac{1}{z(z-1)(z-2)} = \frac{1}{z} * (\frac{A}{z-1} + \frac{B}{z-2})
Where A = -1 , B = 1.
f = \frac{1}{z} * (\frac{1}{z-2} + \frac{1}{1-z})

My question is, why do you have to use partial fractions?

Can't you just leave the initial multiplication of three terms and expand them individually;
\frac{1}{z-1} (dividing 1/z to numerator and denominator at R2,R3)
and
\frac{1}{z-2} (dividing 1/z to numerator and denominator at R3)

so for example,

for R2,
f = \frac{1}{z} * \frac{1}{z-2} * (-\frac{1}{z}*\frac{1}{1-z^{-1}})

You can expand 1/(1-z) and 1/(z-2) using taylor series.

Should you always use partial fractions?
Or does expanding without using partial fractions still work?
]

 

[STEMucator]

I believe your partial fraction expansion is incorrect, you should check it again.

$\frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}$

 

[tadf2]

\frac{A}{z} this part is not necessarily needed for partial fractions because
it doesn't require any expansions..

 

[STEMucator]

\frac{A}{z} this part is not necessarily needed for partial fractions because
it doesn't require any expansions..

You don't really even need partial fractions at all. You could write it like this:

$(\frac{1}{z})(\frac{1}{z-1})(\frac{1}{z-2})$

Then write the terms as their respective series, for example:

$\frac{1}{z} = \sum_{n=0}^{∞} (-1)^n (-1+z)^n$ where $|1-z|<1$

I believe that's what you were asking anyway.

 

[tadf2]

Cool, I get it now. You answered my question.

Thanks Zondrina!