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Power Series

  1. Jun 17, 2010 #1
    [tex]\mbox {Suppose I have: } \sum_{n=1}^\infty (\frac {x} {3})^{2n} [/tex]

    [tex]\mbox{Can I define } y= \frac {x} {3} [/tex]

    a_k(y) = \left\{
    \begin{array}{c l}
    (y)^k, & \mbox{if } k= 2n\\
    (0)^k, & \mbox{otherwise}

    [tex] \mbox {And then use all the cool properties of power series on } \sum_{k=1}^\infty a_k(y)[/tex]

    [I edited my question]
    Last edited: Jun 17, 2010
  2. jcsd
  3. Jun 17, 2010 #2
    Of course, but the convergence properties you know are given in terms of [itex]y[/itex] then. You need to translate them back in terms of [itex]x[/itex]. Essentially, substitute [itex]y = (x/3)^{2}[/itex] everywhere.
    Last edited: Jun 17, 2010
  4. Jun 17, 2010 #3


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    Science Advisor

    The simpler thing to do is write this as
    [tex]\sum_{n=1}^\infty \left(\frac{x^2}{9}\right)^n[/tex]
    so it is a geometric series with "common ratio" of [itex]x^2/9[/itex].
  5. Jun 18, 2010 #4

    Gib Z

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    Homework Helper

    Indeed you can, and that's how you can easily do problems with out getting fooled like

    [tex]\sum_{n=1}^{\infty} n^n z^{n^n}[/tex]
  6. Jun 18, 2010 #5
    Thank you guys!
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