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Math Help - Convergent Geometric Series

  1. #1
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    Convergent Geometric Series

    Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?


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    Quote Originally Posted by eg37se View Post
    Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?
    Try x=\frac{-1}{2}~\&~y=\frac{1}{4}.

    There is a general solution. Can you find it?
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    Quote Originally Posted by eg37se View Post
    Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?


    THANKS.
    I would say yes. Assume that the lead term is 1 and that n starts at n = 0 then

    \sum_{n=0}^\infty x^n = \frac{1}{1-x}, \sum_{n=0}^\infty y^n = \frac{1}{1-y} and \sum_{n=0}^\infty x^n y^n = \frac{1}{1-xy}

    so what you're asking is are there soltions to

     <br />
\frac{1}{1-x} \cdot \frac{1}{1-y} = \frac{1}{1-xy} \;\; \text{on}\;\; (-1,1) \times (-1,1)<br />
. I believe there are an infinite number of solutions to this equation on this interval.
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    Quote Originally Posted by Plato View Post
    Try x=\frac{-1}{2}~\&~y=\frac{1}{4}.

    There is a general solution. Can you find it?
    Wouldn't it just be (1-(1/8)+(1/64)-(1/256)+...)
    which is just ∑(-1/8)^n?
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    Quote Originally Posted by eg37se View Post
    Wouldn't it just be (1-(1/8)+(1/64)-(1/256)+...)
    which is just ∑(-1/8)^n?
    If you substitute the x and y values that Plato suggests into the forumula I give

    Quote Originally Posted by Danny View Post
     <br />
\frac{1}{1-x} \cdot \frac{1}{1-y} = \frac{1}{1-xy} \;\; \text{on}\;\; (-1,1) \times (-1,1)<br />
.
    I believe that you'll show they're identical.
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  6. #6
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    Got it. Thank you.
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