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Inverse Binomial Expansion within Laurent Series?

  1. Jan 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the Laurent Series of [itex] f(z) = \frac{1}{z(z-2)^3} [/itex] about the singularities z=0 and z=2 (separately).

    Verify z=0 is a pole of order 1, and z=2 is a pole of order 3.

    Find residue of f(z) at each pole.
    2. Relevant equations

    The solution starts by parentheses in the form (1 - az), where a is some constant.

    [itex] f(z) = \frac{-1}{8z(1-\frac{z}{2})^3} [/itex]

    3. The attempt at a solution

    The solution given expands [itex] \frac{1}{(1-\frac{z}{2})^3} [/itex] as follows, and I just don't understand how it does it! I'm simply trying to recognise what expansion this is, as I can see it's used later too! I understand almost all of the rest of the problem. A general series/hint to what series is used would be amazing!

    [itex] = 1 + (-3)(\frac{-z}{2}) +\frac{(-3)(-4)(\frac{-z}{2})^2}{2!} + \frac{(-3)(-4)(-5)(\frac{-z}{2})^3}{3!} + ... [/itex]

    I think it's a binomial expansion, but how does a binomial expansion work for negative powers?

    Is it a binomial expansion?

    sinkersub
     
  2. jcsd
  3. Jan 12, 2016 #2

    Mark44

    Staff: Mentor

    Yes. See https://en.wikipedia.org/wiki/Binomial_series. The exponent doesn't need to be a positive integer. It can be negative or real, or even complex.
     
  4. Jan 12, 2016 #3
    Thanks for the clarification!
     
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