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Computing a geometric series

  1. Apr 13, 2008 #1
    Hi,

    I have a problem with computing this geometric series.

    1. The problem statement, all variables and given/known data
    I have to compute
    [tex]\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}} + \sum_{i=0}^\infty{(\frac{1}{3z})^{2k+1}}[/tex].
    It's for computing the z-transform of
    [tex]f[k]=0[/tex] for [tex]k<0[/tex]
    [tex]f[k]=(\frac{1}{2})^k[/tex] for [tex]k=0,2,4,6,...[/tex]
    [tex]f[k]=(\frac{1}{3})^k[/tex] for [tex]k=1,3,5,...[/tex]

    2. Relevant equations

    3. The attempt at a solution
    It's the [tex]2k[/tex] and [tex]2k+1[/tex] that annoys me in the sum.
    I tried
    [tex]\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}}=\sum_{i=0}^\infty{(\frac{1}{4z^2})^{k}}[/tex]
    but I don't know if that helps?

    Thanks,

    Yoran
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 13, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that helps a great deal! The sum of a geometric series is given by
    [tex]\sum_{i=0}^\infty ar^i= \frac{a}{1- r}[/tex]
    In this case a= 1 and r= 4z2.

    Similarly, for the second sum
    [tex]\sum_{i=0}^\infty{(\frac{1}{3z})^{2k+1}}[/tex]
    You can factor out 1 [itex]1/3z[/itex] and then take the "2" from "2k" 'inside' to get
    [tex]\sum_{i=0}^\infty{\frac{1}{3z}(\frac{1}{9z^2})^k}[/tex]
    Now you have a= 1/3z and r= [itex]1/9z^2[/itex].
     
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