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Math Help - Finding a Laurent series

  1. #1
    MHF Contributor arbolis's Avatar
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    Finding a Laurent series

    The exercise states : Find the first 4 terms different from 0 of the Laurent series of g(z)=e^z \cos z, around z_0=0.

    My attempt : g(z)=\left ( 1+z+\frac{z^2}{2}+ \frac{z^3}{3!}+ \frac{z^4}{4!}+...  \right ) \left (  1- \frac{z^2}{2}+\frac{z^4}{4!}-\frac{z^6}{6!}             +... \right )=1+... ( I don't really know the other terms but I think they are all 0 when z=0.)

    So what can I answer? The only term seems to be 1, and if there are others they are all worth 0 in z=0. ( I realize it could not be so, but I don't think so).
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  2. #2
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    Quote Originally Posted by arbolis View Post
    The exercise states : Find the first 4 terms different from 0 of the Laurent series of g(z)=e^z \cos z, around z_0=0.

    My attempt : g(z)=\left ( 1+z+\frac{z^2}{2}+ \frac{z^3}{3!}+ \frac{z^4}{4!}+...  \right ) \left (  1- \frac{z^2}{2}+\frac{z^4}{4!}-\frac{z^6}{6!}             +... \right )=1+... ( I don't really know the other terms but I think they are all 0 when z=0.)

    So what can I answer? The only term seems to be 1, and if there are others they are all worth 0 in z=0. ( I realize it could not be so, but I don't think so).
    I think you're confusing the value of the function at z with the point at which the power series is centered. Since both the series for e^z and \cos z are centered at 0, by using the product of the power series you're already expanding around 0. Also notice that the Laurent series is just the Taylor expansion since it's a product of entire functions.

    Now using the Cauchy product formula we have:

    Let e^z= \sum_{n=0} ^{ \infty } \frac{z^n}{n!} = \sum_{n=0} ^{ \infty} a_nz^n and \cos z = \sum_{n=0} ^{ \infty } \frac{(-1)^nz^n}{(2n+1)!} = \sum_{n=0} ^{ \infty} b_nz^n then e^z\cos z = \sum_{n=0} ^{ \infty} c_nz^n where c_n = \sum_{k=0} ^{n} a_kb_{n-k}

    So to compute the second term in the expansion we have a_0b_2+a_1b_1+a_2b_0= \frac{1}{5!} + \frac{-1}{3!} + \frac{1}{2!} = \frac{41}{5!} which would be the second coefficient in the series (if I made no mistake), now just compute two more which are non-zero.
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  3. #3
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    Sorry, I messed up the power series for \cos z, it should be \sum_{n=0} ^{ \infty} \frac{(-1)^nz^{2n}}{2n!} =\sum_{n=0} ^{ \infty} b_nz^{2n}. The rest is the same although with the exception that c_n= \sum_{k+2l=n} a_kb_l so c_2 should be 0.
    Last edited by Jose27; October 25th 2009 at 03:11 PM.
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