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Taylor Series Integration

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use taylor series method to compute the integral from 1 to 2 of [sin(x2)] / (x2) with 10 -3 precision

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure where to start. Someone please help me.
     
    Last edited: Dec 2, 2009
  2. jcsd
  3. Dec 2, 2009 #2
    Do you know what a Taylor series is? You need to expand sin(x^2) as a Taylor series and then, given that the series satisfies some very broad conditions, you can integrate each term separately.
     
  4. Dec 2, 2009 #3
    Yes.

    sin(x2) after integration = (x3/3) -(x7)/(7*3!) + (x11)/(11*5!)-...
    right?

    can you help me with the rest of it?
     
  5. Dec 2, 2009 #4
    [tex] \int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx [/tex]
     
    Last edited: Dec 2, 2009
  6. Dec 2, 2009 #5
    Thank you so much! So that stuff to that right of the equal sign, how do I use that solve the integral from 1 to 2 - do I just plug in the upper and lower limits and subtract the lower from the upper value? And the precision of 10 -3 - how do I get that?

    I really appreciate your help.
     
  7. Dec 2, 2009 #6
    The right hand side is easy to integrate now; all you have is a constant times x^(another constant).
     
  8. Dec 2, 2009 #7
    Ok, integrating the right side gives:

    (-1) n+1 * (x 2) 2n) / (2n!*x2)

    right, or wrong?
     
  9. Dec 2, 2009 #8
    Wrong.

    [tex] \int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\int_1^2 dx x^{4n} [/tex]
     
  10. Dec 2, 2009 #9
    I was totally wrong.
    Ok, so now is where I plug into the upper and lower limits and subtract those values?
     
  11. Dec 2, 2009 #10

    ideasrule

    User Avatar
    Homework Helper

    Yes. To get 10^-3 precision, you should keep on calculating terms until they go well below 10^-3.
     
  12. Dec 2, 2009 #11
    Got it.
    Thank you all for your help!
     
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