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Math Help - Romberg's Method and Richardson extrapolation

  1. #1
    Newbie russo's Avatar
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    Romberg's Method and Richardson extrapolation

    Hi. I've been having some troubles with a particular integral with Romberg's method plus Richardson's extrapolation.

    The integral is: \int_{0}^{\frac{\pi}{4}}{x^2\ sin(x)\, dx} \approx 0.08875

    And the formulas I have are:

    Romberg: R_{k,1} = \frac{1}{2}\left[R_{k-1,1} + h_{k-1}\left(\sum_{i=1}^{2^{k-2}}f\left(a+\frac{2i-1}{2} \cdot h_{k-1}\right)\right)\right]

    Where:

    a = 0 and b = \frac{\pi}{4}

    R_{1,1} = \frac{b-a}{2} \cdot \left(f(a)+f(b)\right)

    h_{k} = \frac{b-a}{2^{k-1}}

    And we do it from 0 to 3

    Then Richardson to get R_{2,2}, R_{3,2} and R_{3,3}

    R_{i,j} = \frac{4^{j-1}R_{i,j-1} - R_{i-1,j-1}}{4^{j-1} - 1}

    In the end, I get to R_{3,3} = 0.0291 a bit far from the real solution. Could it be an error from the method or something wrong with calculations? I've done it a few times but still I can't find a mistake. I'll keep trying though.

    Thanks in advance.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Romberg's Method and Richardson extrapolation

    Hi russo!

    Looks to me like an error in your calculations.
    Here's what I get:

    Code:
    R(1,1) = 0.17128
    R(2,1) = 0.10881
    R(3,1) = 0.09373
    R(3,2) = 0.08870
    R(2,2) = 0.08799
    R(3,3) = 0.08875
    Thanks from russo
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  3. #3
    Newbie russo's Avatar
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    Re: Romberg's Method and Richardson extrapolation

    Thanks. I could get to those results using the formula shown on wikipedia. So I'm starting to think that I got it wrong here. I'll check on a book later. At least now I have your results as a reference haha.

    Thanks again.
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Romberg's Method and Richardson extrapolation

    Ah well, your formulas are identical to those of wikipedia, except for an offset of 1.
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  5. #5
    Newbie russo's Avatar
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    Re: Romberg's Method and Richardson extrapolation

    Ok so I'll keep checking where is my mistake. I'm sure it's a simple detail.

    EDIT: And it was. I was taking the wrong h
    Last edited by russo; February 9th 2013 at 11:25 AM.
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