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Thread: limiting sum of a series??

  1. #1
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    limiting sum of a series??

    the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x
    how do you do this??
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  2. #2
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    Quote Originally Posted by deej813 View Post
    the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x
    how do you do this??
    $\displaystyle 3+x+x^2+...=18$

    $\displaystyle 3+\frac{x}{1-x}=18$

    $\displaystyle \frac{x}{1-x}=15$

    $\displaystyle x=15-15x$

    $\displaystyle 16x=15$

    $\displaystyle x=\frac{15}{16}$
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  3. #3
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    how does that work??
    where does the x/1-x come from??
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  4. #4
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    Everything after the 3 is an infinite geometric series.

    For any infinite geometric series of the form

    $\displaystyle S_n = a + ar + ar^2 + ar^3 + \dots$

    converges to $\displaystyle \frac{a}{1 - r}$, provided $\displaystyle |r| < 1$.


    In your case, the infinite geometric series is

    $\displaystyle S_n = x + x^2 + x^3 + x^4 + \dots$.


    Can you see what $\displaystyle a$ and $\displaystyle r$ have to be?

    Can you see what it converges to if $\displaystyle |x| < 1$?
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  5. #5
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    wouldn't it be 3/1-x cos isn't a=3??
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  6. #6
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    No.

    I said the terms AFTER the 3.


    In other words, your equation is

    $\displaystyle 3 + \left(x + x^2 + x^3 + \dots\right) = 18$.


    We are only dealing with the stuff inside the brackets.


    Edit: It might help if you subtract the 3 to the other side.

    So the equation becomes

    $\displaystyle x + x^2 + x^3 + \dots = 15$.
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  7. #7
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    oh ok i get it
    thanks
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