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Math Help - Asymptotic error formula for the trapezoidal rule

  1. #1
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    Asymptotic error formula for the trapezoidal rule

    I need to use the asymptotic error formula for the trapezoidal rule to estimate the number n of subdivisions to evaluate $\int_{0}^{2}e^{-x^2}dx$ to the accuracy $\epsilon=10^{-10}$. I also need to find the approximate integral in this case. I would like to know if my attempt is correct. Thanks in advance for any help.


    **My attempt:** $E_n^T(f)\approx -h^2/12[f'(b)-f'(a)]. f(x)=e^{-x^2}, f'(x)=-2xe^{-x^2}.$


    So $E_n^T(f)\approx h^2/(4e^4)$ and since $h=1/n$, we have to find an n that satisfies the inequality $\frac{1}{4e^4n^2}\leq 10^{-10}$. We obtain $n \approx 6767$. The approximate integral is ??
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  2. #2
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    Re: Asymptotic error formula for the trapezoidal rule

    Rewriting this with the "tex" boxes:
    Quote Originally Posted by abscissa View Post
    I need to use the asymptotic error formula for the trapezoidal rule to estimate the number n of subdivisions to evaluate \int_{0}^{2}e^{-x^2}dx to the accuracy \epsilon=10^{-10}. I also need to find the approximate integral in this case. I would like to know if my attempt is correct. Thanks in advance for any help.


    **My attempt:** E_n^T(f)\approx -h^2/12[f'(b)-f'(a)]. f(x)=e^{-x^2}, f'(x)=-2xe^{-x^2}


    So E_n^T(f)\approx h^2/(4e^4) and since h=\frac{1}{n}, we have to find an n that satisfies the inequality \frac{1}{4e^4n^2}\leq 10^{-10}. We obtain n \approx 6767. The approximate integral is ??
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