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Math Help - Deriving Simpson's Rule

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    Member Mollier's Avatar
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    Deriving Simpson's Rule

    To derive Simpson's Rule and it's error, we can integrate a 2nd degree Lagrange polynomial. To make the process less laborious, we can integrate over the interval [-h,h] using the points (-h,f(-h)), (0,f(0)), (h,f(h)). We end up with,

    \begin{aligned} \int^h_{-h}p_2(x)dx = \frac{h}{3}\bigg(f(-h) + 4f(0) + f(h)\bigg) &+&  \\<br />
                                                     \cdots &+& \int^h_{-h}\frac{f^{(3)}(\xi)}{6}x(x+h)(x-h)dx,<br />
\end{aligned}

    where p_2(x) denotes the second degree Lagrange polynomial. If I now wish to bound the error I can let M=\max_{x\in [a,b]}|f^{(3)}(x)| and call the error E_2(f) such that,

    |E_2(f)| \leq \frac{M}{6}\int^h_{-h}|x(x+h)(x-h)|dx.

    The integral however, is zero and so I end up with a useless error bound. In one of my books they state a similar thing, but instead of using -h,0,h, they use a,\frac{a+b}{2},b to get,

    |E_2(f)| \leq \frac{M}{6}\int^b_{a}|(x-a)(x-\frac{a+b}{2})(x-b)|dx.

    I've tried plugging the integral into wolframalpha and get zero here as well. The book though ends up with,

    |E_2(f)| \leq \frac{(b-a)^4}{196}M.

    Am I doing something wrong here? Are there easier ways of deriving Simpson's along with the error term?Any comments are highly appreciated! Thanks.
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  2. #2
    Member Mollier's Avatar
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    A gentle bumb. I have yet to figure this one out.
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by Mollier View Post

    |E_2(f)| \leq \frac{M}{6}\int^h_{-h}|x(x+h)(x-h)|dx.

    The integral however, is zero and so I end up with a useless error bound. In one of my books they state a similar thing, but instead of using -h,0,h, they use a,\frac{a+b}{2},b to get,

    |E_2(f)| \leq \frac{M}{6}\int^b_{a}|(x-a)(x-\frac{a+b}{2})(x-b)|dx.

    I've tried plugging the integral into wolframalpha and get zero here as well. The book though ends up with,

    |E_2(f)| \leq \frac{(b-a)^4}{196}M.

    Am I doing something wrong here? Are there easier ways of deriving Simpson's along with the error term?Any comments are highly appreciated! Thanks.
    Man oh man, I've totally ignored the absoulte-value sign. The integral is of course not zero!

    |E_2(f)| \leq \frac{M}{6}\int^h_{-h}|x(x+h)(x-h)|dx = \frac{M}{6}\frac{h^4}{2}. .

    Since h = \frac{b-a}{2} then \frac{h^4}{2}= \frac{(b-a)^4}{32}, and so,

    |E_2(f)| \leq \frac{M}{192}h^4 .

    The book gets \frac{1}{196}, but that just might be a typo

    I will stop talking to my self now.
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