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Math Help - Laurent series over different domains?

  1. #1
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    Laurent series over different domains?

    Get the Laurent series expansion or f(z) = \frac{1}{z+1} - \frac{4}{(z-2)^2} in the intervals;

    (i) |z-1|<1 and
    (ii) 1<|z-1|< 2

    Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
    For example in (ii) of this problem i would get the Laurent expansions for \frac{1}{z+1}
    for interval 1<|z-1|<2
    and then get the expansions for - \frac{4}{(z-2)^2}
    for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of \frac{1}{z+1} for |z-1|<2 and adds to it the expansion of
    - \frac{4}{(z-2)^2} for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
    r1<|z-zo|< r2. Any help is much appreciated.
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  2. #2
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    Re: Laurent series over different domains?

    use this give you the exact expansion i think is the best posible
    -2 x+\frac{x^2}{4}-\frac{3 x^3}{2}-\frac{5 x^4}{4 (-2+x)^2}+\frac{x^5}{2 (-2+x)^2}+\frac{x^4}{1+x}
    kind regards
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Laurent series over different domains?

    Quote Originally Posted by punkstart View Post
    Get the Laurent series expansion or f(z) = \frac{1}{z+1} - \frac{4}{(z-2)^2} in the intervals;

    (i) |z-1|<1 and
    (ii) 1<|z-1|< 2

    Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
    For example in (ii) of this problem i would get the Laurent expansions for \frac{1}{z+1}
    for interval 1<|z-1|<2
    and then get the expansions for - \frac{4}{(z-2)^2}
    for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of \frac{1}{z+1} for |z-1|<2 and adds to it the expansion of
    - \frac{4}{(z-2)^2} for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
    r1<|z-zo|< r2. Any help is much appreciated.
    (i) If we set z+1=s \implies z=s-1, then we have to find the Laurent expansion around s=0 of...

    f(s)= \frac{1}{s} - \frac{4}{(s-3)^{2}} (1)

    Now we take into account that is...

    \frac{1}{(s-3)^{2}} = -\frac{d}{ds} \frac{1}{s-3} (2)

    ... and that for |s|<3 is...

    - \frac{1}{s-3}= \frac{1}{3}\ \frac{1}{1-\frac{s}{3}} = \frac{1}{3}\ \sum_{n=0}^{\infty} (\frac{s}{3})^{n} (3)

    Deriving (3) and inserting the result in (1) we obtain...

    f(s)= \frac{1}{s} -\frac{4}{3} \sum_{n=1}^{\infty} \frac{n}{3^{n}}\ s^{n-1} (4)

    ... that is the result You are searching for. The (4) converges for 0<|s|<3 and setting s=z+1 You obtain the Laurent expansion of the original f(*) in z...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Laurent series over different domains?

    Quote Originally Posted by punkstart View Post
    Get the Laurent series expansion or f(z) = \frac{1}{z+1} - \frac{4}{(z-2)^2} in the intervals;

    (i) |z-1|<1 and
    (ii) 1<|z-1|< 2

    Now I don't have much trouble finding the Laurent series expansions but am not sure how to do it for specific intervals.
    For example in (ii) of this problem i would get the Laurent expansions for \frac{1}{z+1}
    for interval 1<|z-1|<2
    and then get the expansions for - \frac{4}{(z-2)^2}
    for 1<|z-1|<2,and add them all together. The textbook(correctly, just not sure why yet) just uses the expansion of \frac{1}{z+1} for |z-1|<2 and adds to it the expansion of
    - \frac{4}{(z-2)^2} for 1<|z-1|. I'm not sure as to the procedure or the thinking behind getting Laurent expansions for these types of intervals
    r1<|z-zo|< r2. Any help is much appreciated.
    (ii) Setting z-2=s \implies z=s+2 Tou obtain that for \s|<3 is...

    \frac{1}{z+1}=\frac{1}{s+3}= \frac{1}{3}\ \frac{1}{1+\frac{s}{3}}= \frac{1}{3}\ \sum_{n=0}^{\infty} (-1)^{n} (\frac{s}{3})^{n} (1)

    ... so that is...

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

    Kind regards

    \chi \sigma
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