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Homework Help: Partial Sum with in 10^-10

  1. Mar 10, 2010 #1
    determine the smallest positive integer N such that the partial sum is within 10^-10 of the series sum.

    {Sigma} 2(-1/4)^(n-1) < 10^-10

    ok divide the 2 out

    (-1/4)^(n-1) < 5^-10


    I know i have to take the natural log like so:
    (n-1) ln|-1/4| < nl|5^-10|

    But theres the negative 1/4, how do i get it to be positive? i get the right answer when its positive but im not sure how to do it.
     
  2. jcsd
  3. Mar 10, 2010 #2

    Dick

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    It's an alternating series sum{a_n}. You have an error estimate for the partial sum up to n in terms of |a_(n+1)|. Note the absolute value.
     
  4. Mar 11, 2010 #3

    HallsofIvy

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    Is [itex]\sum 2 (-1/4)^{n-1}[/itex] the sum? You have stated this wrong- you don't want the partial sum, which is what you have here, to be less than [itex]10^{-10}[/itex], you want its difference from the actual sum to be less than [itex]10^{-10}[/itex] but have then done that.

    One way to do that is to find the actual sum. That is not difficult because this is a "geometric" series and there is a formula for the sum.

    But, better, as Dick says, because this is alternating + and -, the "error" is always less than the next term. You just want to find n so that the absolute value of the next term is less than [itex]10^{-10}[/itex]

    [tex]\left|\left(-\frac{1}{4}\right)^{n-1}\right|= \left(\frac{1}{4}\right)^{n-1}< 5^{-10}[/tex]
     
  5. Mar 11, 2010 #4
    so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much
     
  6. Mar 11, 2010 #5

    Dick

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    Pretty much, yes. Provided you understand why that works.
     
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