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Math Help - Power series

  1. #1
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    Power series

    Starting with the geometric series 1+x+x^2+x^3+... (-1<x<1), find the power series for x/[(1-x)^2] in powers of x.
    Where is the series valid?
    Find the sum ∑(1,∞) n/(2^n).
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  2. #2
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    Quote Originally Posted by schnek View Post
    Starting with the geometric series 1+x+x^2+x^3+... (-1<x<1), find the power series for x/[(1-x)^2] in powers of x.
    Where is the series valid?
    Find the sum ∑(1,∞) n/(2^n).
    What have you tried already?

    You are missing the fact that the geometric series 1+x+x^2+x^3...=\sum_{k=0}^{\infty}x^k is \frac{1}{1-x} when |x|<1

    If \frac{d}{dr} \sum_{k=0}^{\infty}x^k=\sum_{k=0}^{\infty}kx^{k-1} differentiate the left hand side from above to get the same thing in a different form. With me so far?
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  3. #3
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    yes? I understand what you did, but I don't quite see how I need to apply that to the question.
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  4. #4
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    Quote Originally Posted by schnek View Post
    yes? I understand what you did, but I don't quite see how I need to apply that to the question.
    \frac{d}{dx} \frac{1}{1-x}=\frac{1}{(1-x)^2}

    That is equal to the derivative of the power series I showed previously. So you have a power series for the above, now all you need to do is account for an x in the numerator and you will have an answer.

    From my last post, the above is equal to \sum_{k=0}^{\infty}kx^{k-1}. Now multiply this by x. x represents a constant thus can be included inside the summation. Simplification leads to the power series of the expression from your problem.
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  5. #5
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    Quote Originally Posted by schnek View Post
    Find the sum ∑(1,∞) n/(2^n).
    here's a way on doing this without the above result:

    \sum\limits_{j=1}^{\infty }{\frac{j}{2^{j}}}=\sum\limits_{j=1}^{\infty }{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}}=\sum\lim  its_{k=1}^{\infty }{\sum\limits_{j=k}^{\infty }{\frac{1}{2^{j}}}}=\sum\limits_{k=1}^{\infty }{\sum\limits_{j=0}^{\infty }{\frac{1}{2^{j+k}}}}=\left( \sum\limits_{k=1}^{\infty }{\frac{1}{2^{k}}} \right)\left( \sum\limits_{j=0}^{\infty }{\frac{1}{2^{j}}} \right),

    hence your sum equals

    \frac{\frac{1}{2}}{1-\frac{1}{2}}\cdot \frac{1}{1-\frac{1}{2}}=2.
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