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Derivative of Geometric Series

  1. Jun 10, 2010 #1
    1. The problem statement, all variables and given/known data
    I am having trouble following what is going on in this solution. We are looking to find the expectation value of:

    [tex]f(x,y)=\frac{1}{4^{x+y}}\cdot\frac{9}{16}[/tex]

    I have gotten it down to:

    [tex]E(X) = \frac{3}{4}\sum_{x=0}^\infty x\cdot\left(\frac{1}{4}\right)^x\qquad(1)[/tex]

    We know that for a geometric series with an initial value of 1 we can write for 0 < r < 1:

    [tex]\sum_{x=0}^\infty r^x = \frac{1}{1-r}[/tex]

    taking the derivative of both sides wrt 'r' yields:

    [tex]\sum_{x=1}^\infty r^{x-1} = \frac{1}{(1-r)^2}\qquad(2)[/tex]


    Here is where I get confused:

    I thought it was a simple matter of plugging in:

    [tex]\frac{3}{4}\cdot\frac{1}{(1-1/4)^2} = 4/3[/tex]

    However, the solution gives:


    [tex]\frac{3}{4}\cdot\underbrace{\frac{1}{4}}\cdot\frac{1}{(1-1/4)^2} = 1/3[/tex]

    I am a little confused as to where the factor of 1/4 is coming from. I am having a feeling that it has something to do with the fact that (1) runs from 0 to infinite and (2) runs from 1 to infinite.

    Any thoughts?
     
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2
    Ah, you forgot to divide out a (1/4) in order to get the summand into the form [tex] x r^{x-1} [/tex]. In particular:

    [tex]\sum_{x=0}^\infty x r^{x}[/tex]

    Divide out an r, dropping the power in the summand down by 1:

    [tex] r \sum_{x=0}^\infty x r^{x-1} [/tex]

    Recognize that this is the derivative of the series with respect to r:

    [tex] = r \sum_{x=0}^\infty \frac{d}{dr} \left(r^x\right) [/tex]

    Take the derivative outside of the sum and apply your knowledge about the geometric series:

    [tex] = r \frac{d}{dr} \left( \sum_{x=0}^\infty r^x \right) = r \frac{d}{dr} \left( \frac{1}{1-r} \right) = r \frac{1}{(1-r)^2} [/tex]

    So that r out in front is where the extra (1/4) factor comes from.
     
  4. Jun 10, 2010 #3
    Wow, great explanation hbweb500! Thanks for the help :smile:
     
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