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Math Help - [SOLVED] Complex Taylor, Maclaurin and Laurent series

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Complex Taylor, Maclaurin and Laurent series

    1) Find the Taylor series around the point 2i and specify the radius of convergence:

    (a) f(z)=\frac{1}{z}
    (b) f(z)=log(z)

    (a): f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}} and c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}} where z_0 = 2i

    So the Taylor series is: \sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n

    My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

    2) Find the first three nonvanishing terms for the Maclaurin series of: f(z)=\frac{(1-z)^{1/2}}{1+z^2}

    3) Determine the Laurent series for \frac{1}{z+z^2}, |z+1/2|> 1/2

    I've solved this for different areas, but I'm having a little trouble with this one.
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  2. #2
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    Quote Originally Posted by Spec View Post
    1) Find the Taylor series around the point 2i and specify the radius of convergence:

    (a) f(z)=\frac{1}{z}
    (b) f(z)=log(z)

    (a): f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}} and c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}} where z_0 = 2i

    So the Taylor series is: \sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n =\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n
    Correct as far as there,...

    Quote Originally Posted by Spec View Post
    = \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n
    ... but that step is wrong. In fact, \frac{-1}i = i, so \frac{(-1)^n}{(2i)^n} = \left(\frac{i}{2}\right)^n. The (-1)^{n+1} ought not to be there at all.

    Quote Originally Posted by Spec View Post
    And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?
    The ratio test is always good for finding a radius of convergence. In this case, it's a bit quicker if you notice that this is a geometric series.
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  3. #3
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    Quote Originally Posted by Spec View Post
    1) Find the Taylor series around the point 2i and specify the radius of convergence:

    (a) f(z)=\frac{1}{z}
    (b) f(z)=log(z)

    (a): f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}} and c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}} where z_0 = 2i

    So the Taylor series is: \sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n

    My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

    2) Find the first three nonvanishing terms for the Maclaurin series of: f(z)=\frac{(1-z)^{1/2}}{1+z^2}

    3) Determine the Laurent series for \frac{1}{z+z^2}, |z+1/2|> 1/2

    I've solved this for different areas, but I'm having a little trouble with this one.
    Should not the radius of convergence of a Taylor series simply be the distance to the nearest singularity? In that case, the radius of convergence for (1a) would be 2? I would think the same with (1b) except the series will converge to an analytic sheet of log(z)

    Also, (3) sounds not right. Should it be |z+1/2|<1/2 or no?
    [edit] ok, I understand now. I' wrong. Sorry. . .
    Last edited by shawsend; November 17th 2008 at 02:02 PM. Reason: added same for (1b)
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  4. #4
    Senior Member Spec's Avatar
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    I solved (3) myself by using the substitution w=z+1/2

    Then you have:

    \frac{1}{w^2-1/4}=\frac{1}{w^2}\frac{1}{1-\frac{1/4}{w^2}}=\frac{1}{w^2}\sum_{n=0}^{\infty}\frac{1/4}{w^2}^n = ...


    What about the Taylor series for log z around the point 2i?
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  5. #5
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    Quote Originally Posted by Spec View Post
    What about the Taylor series for \log z around the point 2i?
    \log z = \log(2i + z - 2i) = \log(2i) + \log\bigl(1 + \tfrac{z-2i}{2i}\bigr). Now use the usual series for log(1+w).
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