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Math Help - Another Power Series

  1. #1
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    Another Power Series

    This one seems a lot harder...

    Find the power series about the origin of:

    z^2/(4-z)^2
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  2. #2
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    Quote Originally Posted by jzellt View Post
    This one seems a lot harder...

    Find the power series about the origin of:

    z^2/(4-z)^2
    You only need to bother calculating the series for 1/(4 - z)^2 (why?).

    Did you bother to read the link I gave in my reply here: http://www.mathhelpforum.com/math-he...er-series.html

    Now look for the pattern in f^{n}(0) where n = 0, 1, 2, 3, ....

    If you need more help, please say where you are stuck.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by jzellt View Post
    This one seems a lot harder...

    Find the power series about the origin of:

    z^2/(4-z)^2
    If you know the power series of \frac{1}{(1-z)^2} you are essentially done. Consider:
    \frac{1}{(1-z)^2}=\frac{d}{dz}\frac{1}{1-z}=\frac{d}{dz}\sum_{n=0}^\infty z^n=\sum_{n=1}^\infty n z^{n-1}=\sum_{n=0}^\infty (n+1)z^n

    To know this is perhaps generally useful, but now let's solve your original problem:

    \frac{z^2}{(4-z)^2}=\left(\frac{z}{4}\right)^2\cdot\frac{1}{\lef  t(1-\left(\frac{z}{4}\right)\right)^2}=\left(\frac{z}{  4}\right)^2\cdot\sum_{n=0}^\infty(n+1)\left(\frac{  z}{4}\right)^n=\ldots
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