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Find the power series

  1. Jan 31, 2012 #1
    z is a complex number.

    [tex]f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}[/tex]

    [tex]\frac{1}{1 + z} = \frac{1}{1 - (-z)} = 1 + (-z) + (-z)^2 + \cdots = \sum_{n = 0}^{\infty}(-z)^n[/tex]

    [tex]\frac{1}{(z + 2)^2} = -\frac{d}{dz} \ \frac{1}{1 - (-z - 1)} = -\frac{d}{dz}\sum_{n = 0}^{\infty}(-z - 1)^n = \sum_{n = 0}^{\infty}-n(-z-1)^{n-1} [/tex]

    [tex]f(z) = \sum_{n = 0}^{\infty}-(4 + 3z) n (-z)^n (-z - 1)^{n - 1}[/tex]

    Radius of convergence by the ratio test:

    [tex]\lim_{n\to\infty}\left|\frac{-(4+3z)(n+1)(-z)^{n+1}(-z-1)^n}{-(4+3z)n(-z)^n(-z-1)^{n-1}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)(z)(z+1)}{n}\right|[/tex]

    [tex]|z(z+1)|<1[/tex]

    Is the above correct?

    If so, how do I find the Radius of Convergence from that?
     
  2. jcsd
  3. Feb 1, 2012 #2

    Char. Limit

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    Gold Member

    No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.
     
  4. Feb 1, 2012 #3
    Then how can I find the power series of that expression?
     
  5. Feb 1, 2012 #4

    Char. Limit

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    Gold Member

    I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.
     
  6. Feb 1, 2012 #5
    Could I instead use the rule [itex]R\geq\min{R,S} = 1[/itex]?

    Isn't there a way to do this without the use of partial fractions?
     
  7. Feb 4, 2012 #6
    By partial fractions, we can re-write [itex]f(z)[/itex] as
    [tex]
    \frac{A}{z + 1} + \frac{B}{z + 2} + \frac{C}{(z + 2)^2} = \frac{4 + 3z}{(z + 1)(z + 2)^2}.
    [/tex]
    By direct computations, [itex]A(z + 2)^2 + B(z + 2)(z + 1) + C(z + 1)[/itex] [itex]= Az^2 + 4Az + 4A + Bz^2 + 3Bz + 2B + Cz + C.[/itex]
    To find the coefficients, we solve the system of linear equations.
    [tex]
    \begin{pmatrix}1 & 1 & 0 & 0\\
    4 & 3 & 1 & 3\\
    4 & 2 & 1 & 4\end{pmatrix}\Rightarrow
    \begin{pmatrix}1 & 0 & 0 & 1\\
    0 & 1 & 0 & -1\\
    0 & 0 & 1 & 2\end{pmatrix}
    [/tex]
    So the coefficients are [itex]A = 1[/itex], [itex]B = -1[/itex], and [itex]C = 2[/itex].
    Therefore, [itex]\displaystyle\frac{4 + 3z}{(z + 1)(z + 2)^2} = \frac{1}{z + 1} - \frac{1}{z + 2} + \frac{2}{(z + 2)^2}[/itex] [itex]\displaystyle\Rightarrow\frac{1}{1 - (-z)} - \frac{1}{2}\cdot\frac{1}{1 - \left(-\frac{z}{2}\right)} - \frac{d}{dz}\frac{1}{1 - \left(-\frac{z}{2}\right)}[/itex].
    The rational expressions can be represented as the following power series
    [tex]
    \sum_{n = 0}^{\infty}(-z)^n - \frac{1}{2}\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^n - \sum_{n = 0}^{\infty}n\left(-\frac{z}{2}\right)^{n - 1}.
    [/tex]

    Now how do I find the radius of convergence
     
  8. Feb 4, 2012 #7

    Char. Limit

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    Gold Member

    First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

    [tex]\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}[/tex]

    Which, in a simpler form, is:

    [tex]\sum_{n=0}^{\infty} \left( (-1)^n \left(1 - \frac{1}{2^{n+1}} - \frac{n}{2^{n-1}} \right) \right) z^n[/tex]

    Now how would you find the radius of convergence of a single sum?
     
  9. Feb 4, 2012 #8
    Ratio test

    So we end up getting

    [tex]|z| < 1[/tex]

    and R = 1

    Correct?
     
    Last edited: Feb 4, 2012
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