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Quick help in Geometric series question

  1. Dec 11, 2007 #1

    rock.freak667

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    Homework Helper

    1. The problem statement, all variables and given/known data
    The common ratio,ratio,r, of a geometric series is given by:
    [tex]r=\frac{5x}{4+x^2}[/tex]

    Find all the values of x for which the series converges


    2. Relevant equations



    3. The attempt at a solution

    For the series to converge |r|<1
    so that
    [tex]|\frac{5x}{4+x^2}|<1[/tex]


    this means that:

    [tex]-1<\frac{5x}{4+x^2}<1[/tex]

    For

    [tex]-1<\frac{5x}{4+x^2}[/tex]
    I get

    {x:x<-4}U{x:x>-1}

    and for

    [tex]\frac{5x}{4+x^2}<1[/tex]

    I get
    {x:x<1}U{x:x>-1}

    The answer I want would be the region that satisfies both {x:x<1}U{x:x>-1} AND {x:x<-4}U{x:x>-1} right?

    So I get

    {x:x<1}U{x:x>-1} which is basically {x:-1<x<1}...(*answer I got)
    is that correct? because I see that the answer is {x:-1<x<1}U{x:x<-4}U{x:x>4}
    I am not sure if that is the same as what I have because I think that the answer I got includes the rest of {x:x<-4}U{x:x>4}
     
  2. jcsd
  3. Dec 12, 2007 #2
    no. You are incorrect.

    The answer you got includes all reals: x > -1 in union with x < 1 includes all numbers less than one and greater than -1 or all reals. And all reals is not correct (take 3 as a counter example).

    The answer in the book is correct: {x:-1<x<1}U{x:x<-4}U{x:x>4}
     
  4. Dec 12, 2007 #3

    dynamicsolo

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    I'm finding in this case (x+1)(x+4) > 0 , so either

    x>-1 and x>-4 implies x>-1
    or
    x<-1 and x<-4 implies x<-4

    So we are in agreement here.


    In this case, I find (x-1)(x-4) > 0 , so either

    x>1 and x>4 implies x>4
    or
    x<1 and x<4 implies x<1

    Perhaps you "miswrote" yourself here?

    Combining the two cases gives

    x < -4 , -1 < x < 1 , x > 4 .

    As a check, you'll find that values of x with absolute value between 1 and 4 don't work in the original inequality.
     
    Last edited: Dec 12, 2007
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