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Comparison Test Problem & Estimitaing it's error

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    The question in the book is: "Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

    My problem is estimating the error I'm looking for. I just need help with finding the integral.

    2. Relevant equations

    ((sin n)^2)/(n^3)

    3. The attempt at a solution

    ∞∫n ((sin n)^2)/(n^3)

    u = ((sin n)^2) du = 2 ((sin n)^2)cos n dv = n^3 dx v = 1/4 x^4

    This is as far as I got.
     
  2. jcsd
  3. Nov 13, 2011 #2

    Ray Vickson

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    What integral? Aren't you just given an infinite series, and are asked to estimate the error when you stop at 10 terms? Anyway, if you did have to integrate the function you would need to use the non-elementary function Ci(x). Maple 11 gets:
    f:=sin(x)^2/x^3;
    > Jt:=Int(f,x=1..t);
    [tex] \int_1^\infty \frac{\sin^2(x)}{x^3} \, dx [/tex]
    > value(Jt) assuming t>1;
    [tex]
    \frac{1}{4} \frac{t-t^2 \cos(2)+2t^2 \sin(2)-2\mbox{Ci}(2) t^2 - 2 + 2\cos^2(t)
    - 4t \cos(t) \sin(t) + 4\mbox{Ci}(2t) t^2}{t^2} [/tex]

    RGV
     
    Last edited: Nov 13, 2011
  4. Nov 13, 2011 #3
    Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.
     
  5. Nov 13, 2011 #4

    Dick

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    Your error is the sum of (sin(k))^2/k^3 over all of the terms you didn't include. Don't sweat the Ci function. All you need is an estimate to bound the error. (sin(k))^2/k^3<=1/k^3. Estimate that instead.
     
  6. Nov 14, 2011 #5

    Ray Vickson

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    You still have not answered my question: what integral? Are you not just dealing with a sum? (The Ci stuff was just to show you that the integral is not 'doable', so you should not waste time on it!)

    RGV
     
  7. Nov 14, 2011 #6
    Alright guys. Thanks for the tip. I think get it now.
     
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