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all Is there any quadrature rule with error of $O (h^6)$ when using 5 nodal points?

The five nodal points are uniformly distributed. For instance, [-1,0.5,0,0.5,1] or [a, ....,b]

I am trying to find the one which satisfies those condition. If so, could you give an idea or an example?

bluejyellow
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1 Answers1

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A closed Newton-Cotes Formula, with 5 points, with $x_i=x_0+ih$ and $i=0(1)4$ gives:

$$ \int_{x_0}^{x_4}f(x)dx= \frac{2h}{45}\left[ 7f(x_0)+32f(x_1)+12f(x_2)+32f(x_3)+7f(x_4) \right]-\frac{8h^7}{945}f^{(6)}(\xi) $$

for some $\xi\in[x_0,x_4]$, which is $O(h^7)$.

Thales
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  • Hello, Thank you for the reply. However, we are finding the one with O(h^6). Do you have any other idea? – bluejyellow May 30 '17 at 13:44
  • Must it be $O(h^6)$? If $h$ is sufficiently small, then a $O(h^7)$ error is better than a $O(h^6)$ error. All Newton-Cotes formulas with an odd number of nodal points (with $n+1$ points, where $n$ is even) will give an error of $O(h^{n+3})$, so it is not possible to have $O(h^6)$ – Thales May 30 '17 at 13:52