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Math Help - Laurent Series-Complex

  1. #1
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    Laurent Series-Complex

    If anyone could give me a hint that would be great.

    Determine the annulus of convergence of the Laurent series and find the sum fo the series there.

    Σ n(-1)^n 2^ -│n│Z^n
    ( sum of n = neg. infinity to infinity)

    Thanks
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  2. #2
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    Quote Originally Posted by taypez View Post
    If anyone could give me a hint that would be great.

    Determine the annulus of convergence of the Laurent series and find the sum fo the series there.

    Σ n(-1)^n 2^ -│n│Z^n
    ( sum of n = neg. infinity to infinity)

    Thanks
    You have,
    \sum_{n=-\infty}^{\infty} \frac{n(-1)^n}{2^{|n|}} z^n
    Necessay and Sufficient (slip the sums),
    \sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{|n|}}z^n+ \sum_{n=1}^{\infty} \frac{n(-1)^n}{2^{|n|}}z^n
    You can get rid of the absolute value because it takes on positives and negatives,
    \sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{-n}}z^n+ \sum_{n=1}^{\infty} \frac{n(-1)^n}{2^n}z^n
    In the first summand, let u=1/z (if z=0 then it converges, nothing is lost).
    Thus, we can write,
    \sum_{n=0}^{\infty}\frac{n(-1)^{n+1}}{2^n}u^n+\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^n}z^n
    The non-alternating term is,
    \frac{n}{2^n}.
    Use the generalized ratio test,
    \lim_{n\to \infty}\frac{n+1}{2^{n+1}}\cdot \frac{2^n}{n}=\frac{1}{2}.
    Hence the reciprocal of this, which is 2, is the radius of absolute convergence.
    Thus, (in both series)
    |z|<2
    |u|<2
    Take reciprocal of both sides on the second one,
    1/|u|=|z|>1/2
    Thus,
    1/2<|z|<2
    This is an annulus of absolute convergence.
    I still think you need to check the boundary for convergence.

    This is my 38th Post!!!
    Last edited by ThePerfectHacker; December 20th 2006 at 08:12 AM.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    You have,
    \sum_{n=-\infty}^{\infty} \frac{n(-1)^n}{2^{|n|}} z^n
    Necessay and Sufficient (slip the sums),
    \sum_{n=-\infty}^{0}\frac{n(-1)^n}{2^{|n|}}z^n+ \sum_{n=1}^{\infty} \frac{n(-1)^n}{2^{|n|}}z^n
    These two series can be summed as they are the derivatives of geometric series which converge for |z|>1/2 and |z|<2 respectively, and have singularities on the circles |z|=1/2 and |z|=2 respectively, which is sufficient to conclude that the annulus of convergence is 1/2<|z|<2.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    These two series can be summed as they are the derivatives of geometric series which converge for |z|>1/2 and |z|<2 respectively, and have singularities on the circles |z|=1/2 and |z|=2 respectively, which is sufficient to conclude that the annulus of convergence is 1/2<|z|<2.

    RonL
    I was thinking about something else, these are alternating, decreasing sequences to zero. Thus, can we use the Leibniz alternating series test? I was not certain about that because this was a complex analysis type question.
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