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Math Help - Taylor series question

  1. #1
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    Taylor series question

    The question
    For the following function,
    i) Find the largest open disc about the given point z_0 in which f is analytic.
    ii) find the first 3 non-zero terms of the Taylor series of f about z_0
    iii) Find the coefficient a_n of (z - z_0)^n in the Taylor series of f about z_0 either as an explicit function of n, or give a recursive formula for a_n as appropriate.

    f(z) = \frac{1}{z^2 - z - 6}, z_o = 1

    My attempt
    i) I expanded the function as follows:
    \frac{1}{(z - 3)(z + 2)}
    Clearly there's singularities at z = 3, -2
    I draw a quick graph and noticed that the radius of convergence is |z - 1| < 2
    This part I got correct. However, I get stuck for parts ii) and iii)

    ii) I used partial fractions to get:
    \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2}
    I then manipulated each fraction so I could write them in terms of known Macluarin series \frac{1}{z \pm 1}:
    \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n) and
    \frac{-1}{15}(\sum_{n = 0}^{\infty}{(-1)^n(\frac{z - 1}{3})^n)

    Thus I got this for the series:
    \frac{1}{10}(\sum_{n = 0}^{\infty}{(\frac{z - 1}{2})^n) \frac{-1}{15}(\sum_{n = 0}^{\infinity}{(-1)^n(\frac{z - 1}{3})^n)

    Now, upon substituting n = 0, 1, 2 to get the first three terms, I get the following:
    -\frac{1}{5}, \frac{13(z - 1)}{180}, \frac{19(z - 1)^2}{1080}

    These are wrong, and I'm not sure why. :/

    Any assistance would be appreciated.
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  2. #2
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    Re: Taylor series question

    \displaystyle \begin{align*} \frac{\frac{1}{5}}{z - 3} - \frac{\frac{1}{5}}{z + 2} &= -\frac{1}{15}\left(\frac{1}{1 - \frac{z}{3}}\right) - \frac{1}{10}\left[\frac{1}{1 - \left(-\frac{z}{2}\right)}\right] \\ &= -\frac{1}{15}\sum_{r = 0}^{\infty}{\left(\frac{z}{3}\right)^r} - \frac{1}{10}\sum_{s = 0}^{\infty}{\left(-\frac{z}{2}\right)^s}\end{align*}
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  3. #3
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    Re: Taylor series question

    It needs to be around z_0 = 1. Does yours work for that Prove It?
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  4. #4
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    Re: Taylor series question

    OK I worked out what I did wrong with part ii). I thought the series was \frac{1}{z \pm 1}, but it's \frac{1}{1 \pm z}. So my solution is correct if I just add a negative to the start of my series.

    However, I'm still not sure what I'm supposed to do for part iii). Thanks.
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  5. #5
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    Re: Taylor series question

    Quote Originally Posted by Glitch View Post
    It needs to be around z_0 = 1. Does yours work for that Prove It?
    I should have read the question properly :P
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