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Math Help - error factor of Taylor Series

  1. #1
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    error factor of Taylor Series

    the function is:

     \int^{1}_{0} x cos(x^3) \ dx find the error to 3 decimal places.

    so I expressed cos as a Taylor Series which gives me:

     \int^{1}_{0} x \sum^{\infty}_{n=0} -1^n \frac{x^{6n}}{(2n)!} \ dx

     \int^{1}_{0} \sum^{\infty}_{n=0} -1^n \frac{x^{6n+1}}{(2n)!} \ dx

     \sum^{\infty}_{n=0} -1^n \frac{x^{6n+2}}{(6n+2)(2n)!} \bigg{|}^{1}_{0} = \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!}

    but to find the error to whatever decimal place I get stuck, if someone could show me how to find it, and explain the process behind their approach that would really help me out.
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  2. #2
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    Quote Originally Posted by lllll View Post
    the function is:

     \int^{1}_{0} x cos(x^3) \ dx find the error to 3 decimal places.

    so I expressed cos as a Taylor Series which gives me:

     \int^{1}_{0} x \sum^{\infty}_{n=0} -1^n \frac{x^{6n}}{(2n)!} \ dx

     \int^{1}_{0} \sum^{\infty}_{n=0} -1^n \frac{x^{6n+1}}{(2n)!} \ dx

     \sum^{\infty}_{n=0} -1^n \frac{x^{6n+2}}{(6n+2)(2n)!} \bigg{|}^{1}_{0} = \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!}

    but to find the error to whatever decimal place I get stuck, if someone could show me how to find it, and explain the process behind their approach that would really help me out.
    Reading this thread: http://www.mathhelpforum.com/math-he...ion-error.html

    might help ....?
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  3. #3
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     \sum^{\infty}_{n=0} \frac{-1^n}{(6n+2)(2n)!} <0.0001

    How do you know how many factors to expand out?

    if you expand it to 5 you get it:

     \frac{1}{2}-\frac{1}{16}+\frac{1}{336}-\frac{1}{14400}+\frac{1}{1048320} < 0.0001

    so taking the first 3 terms I get  \frac{37}{84} = 0.4404762 and since I rejected the 4th term will my error be  \leq \left| \frac{-1^3}{(6(3)+2)(2(3))!} \right| = |- 0.000069444| = 0.000069444
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