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Math Help - Laurent series

  1. #1
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    Laurent series

    This is another past exam question, which we have not been given anwers to, so i was hoping someone could check my answer, so i know if i am on the right track.

    Find the first three non zero terms of the laurent series on an annulus centred at zero.
    f(z)=(1/(z(1-z)))

    Now the sinularities are at 0, and 1, and they are both double poles.
    I did it as follows:

    0<|z|<1

    (1/(z(1+(-z)))=(1/(z))*∑{n=0}to{∞} zⁿ]
    and i used the binomial theorm to get the cooeficients of z.

    This gave me:
    (1/(z))-(2/z)+3-4z

    Does any one know a way that i can check my solution, to make sure i have not stuffed it up.
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  2. #2
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    Quote Originally Posted by thespian View Post
    This is another past exam question, which we have not been given anwers to, so i was hoping someone could check my answer, so i know if i am on the right track.

    Find the first three non zero terms of the laurent series on an annulus centred at zero.
    f(z)=(1/(z(1-z)))

    Now the sinularities are at 0, and 1, and they are both double poles.
    I did it as follows:

    0<|z|<1

    (1/(z(1+(-z)))=(1/(z))*∑{n=0}to{∞} zⁿ]


    I don't understand what you did here: certainly \frac{1}{z^2(1-z)^2}\neq \frac{1}{z^2(1+(-z)^2)}...!

    So you can do \frac{\frac{1}{z^2}}{(1-z)^2}=\frac{1}{z^2}\left(1+z+z^2+...\right)^2 =\frac{1}{z^2}(1+2z+3z^2+4z^3+...)=\frac{1}{z^2}+\  frac{2}{z}+3+4z+...

    Tonio


    and i used the binomial theorm to get the cooeficients of z.

    This gave me:
    (1/(z))-(2/z)+3-4z

    Does any one know a way that i can check my solution, to make sure i have not stuffed it up.
    .
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  3. #3
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    Laurent series still need some help

    I thought that the series would be alternating from positive to negative,
    because the power is -2.
    That is :
    (1/z^2)-(2/z)+3-4z
    Can anyone confirm this because now i am confused.
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  4. #4
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    Quote Originally Posted by thespian View Post
    I thought that the series would be alternating from positive to negative,
    because the power is -2.
    That is :
    (1/z^2)-(2/z)+3-4z
    Can anyone confirm this because now i am confused.

    What power are you talking about? The power series is \frac{1}{1-z}=1+z+z^2+...+z^n+... , as can readily be verified using the sum formula for geometric series.
    This above, squared, is what you have in the denominator multiplying z^2, you don't have anywhere \frac{1}{1-z^2}
    You also have to realize that 1+(-z)^2=1+z^2, so to add that minus sign is pointless.

    Tonio
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  5. #5
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    (1-z)^-2 doesnt the power series for this alternate between positive and negative?
    using the binomial theorm gives me:

    1-2z+3z^2-4z^3

    because the equation is (1/(z))((1/((1-z))

    i am trying to figure out what happens to the exponant of -2 of (1-z)⁻
    because when i expand it using the binomial formula i get alternating sighns.
    but you have only positive.
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  6. #6
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    let

    g(z)=\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n

    This is the definition of the geometric series

    Now taking the derivative of g we get

    g'(z)=\frac{1}{(1-z)^2}=\sum_{n=0}^{\infty}nz^{n-1}=\sum_{n=1}^{\infty}nz^{n-1}

    You are trying to find the series for the function

    f(z)=\frac{1}{z^2(1-z)^2}=\frac{1}{z^2}\cdot \frac{1}{(1-z)^2}=\frac{1}{z^2}\cdot g'(z)=...
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  7. #7
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    Quote Originally Posted by thespian View Post
    (1-z)^-2 doesnt the power series for this alternate between positive and negative?
    using the binomial theorm gives me:

    1-2z+3z^2-4z^3

    because the equation is (1/(z))((1/((1-z))

    i am trying to figure out what happens to the exponant of -2 of (1-z)⁻


    You baffle me: what negative 2 exponent?? It is \frac{1}{(1-z)^2}=(1-z)^{-2} , but the exponent in the denominator is positive, not negative...or at least that's what you wrote already 2-3 times. And you want the positive exponent in the denominator, because if that were a negative one then you'd simply get \frac{1}{(1-z)^{-2}}=(1-z)^2 , and this is a simple polynomial.

    Tonio

    because when i expand it using the binomial formula i get alternating sighns.
    but you have only positive.
    .
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