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Math Help - please help me with this Laurent series :D

  1. #1
    Junior Member
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    Unhappy please help me with this Laurent series :D

    problem says :

    function  \displaystyle f(z) = \frac {z^5}{(z^2-4)^2 } develop in too Laurent series in region  |z|>2


    i really don't have much trouble with series but (perhaps this is more thread for Pre-university math ) i have problem with this one, especially how to do partial fraction of this function ...

    i have perhaps 10 - 15 pages of the work and jet i didn't get the correct answer i'll post here some of the work where I think that if i do this correctly i solve rest myself

    because up (can't be sure how to say it on English ) in fraction have bigger power than down to use partial fractions i need to divide it first so sown should have "z" with higher power (or am I mistaking ? because if not than i really don't know how to do it)

     \dfrac {z^5}{(z^2-4)^2 }

    when dividing z^5 with z^4-8z^2+16 i get

     \dfrac {z^5}{(z^2-4)^2 } = z + \dfrac {8z^3-16z} {z^4-8z^2+16}

    and now i do partial fractions :

    \displaystyle \frac {8z^3-16z} {z^4-8z^2+16}= \frac {A}{z-2} + \frac {B}{(z-2)^2}+\frac {C}{z+2} + \frac {D}{(z+2)^2}

    am I on the good path so far ? if not how can i solve this (even if I am on the good path, is there some more simple way to do this ? )

    sorry if this is wrong sub forum for this type of the question

    thanks in advance
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    A different way to think about it

    You may want to think about it using a geometric series.

    First consider the function

    \displaystyle g(z)=\frac{1}{z^2-4}=\frac{1}{z^2}\cdot \frac{1}{1-\frac{4}{z^2}}=\frac{1}{z^2}\sum_{n=0}^{\infty}\le  ft( \frac{4}{z^2}\right)^n=\sum_{n=0}^{\infty} \frac{4^n}{z^{2n+2}}

    Notice that this geometric series converges when

    \displaystyle \left| \frac{4}{z^2}\right|< 1 \iff |z|> 2

    Now we can take the derivative to get

    \displaystyle g'(z)=\frac{-2z}{(z^2-4)^2}=\sum_{n=0}^{\infty} \frac{(-2n-2)4^n}{z^{2n+3}}=-2\sum_{n=0}^{\infty} \frac{(n+1)4^n}{z^{2n+3}}

    Now notice that

    \displaystyle f(z)=-\frac{z^4}{2}g'(z)=\sum_{n=0}^{\infty} \frac{(n+1)4^n}{z^{2n-1}}
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  3. #3
    Senior Member yeKciM's Avatar
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    i have had similar problem on the exam, and have same issue with decomposing that one perhaps this can help you in any way :









    as far as partial fractions goes I really really like that what TheEmptySet just wrote up there

    (lol they gave us that kind of problem just to see do we remember division, as far as the Laurent series go, it's pretty simple to do when decomposed )


    as far as your partial fractioning goes you should get something like :

     \displaystyle \frac {z^5}{(z^2-4)^2 }=z+ \frac {8z^3-16z} {z^4-8z^2+16}= \frac {A}{z-2} + \frac {B}{z+2}+\frac {C}{(z-2)^2} + \frac {D}{(z+2)^2}

    and get :

     A=2
     B=2
     C=4
     D=4

      \displaystyle \frac {z^5}{(z^2-4)^2 } = z + \frac {2}{z-2} + \frac {2}{z+2}+\frac {4}{(z-2)^2} + \frac {4}{(z+2)^2}


    this shouldn't be a problem to solve from here ?
    Last edited by yeKciM; September 11th 2010 at 01:51 PM.
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  4. #4
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    thank you very very very much !
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