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Partial Fractions in Laurent Series Expansion

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data
    f = [itex]\frac{1}{z(z-1)(z-2)}[/itex]


    2. Relevant equations
    Partial fraction



    3. The attempt at a solution
    R1 = 0 < z < 1
    R2 = 1 < z < 2
    R3 = z > 2


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

    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;
    [itex]\frac{1}{z-1}[/itex] (dividing 1/z to numerator and denominator at R2,R3)
    and
    [itex]\frac{1}{z-2}[/itex] (dividing 1/z to numerator and denominator at R3)

    so for example,

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

    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?
    ]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 9, 2014 #2

    Zondrina

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    Homework Helper

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

    ##\frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}##
     
  4. Apr 9, 2014 #3
    [itex]\frac{A}{z}[/itex] this part is not necessarily needed for partial fractions because
    it doesn't require any expansions..
     
  5. Apr 9, 2014 #4

    Zondrina

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    Homework Helper

    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.
     
  6. Apr 9, 2014 #5
    Cool, I get it now. You answered my question.

    Thanks Zondrina!
     
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