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I need to show that one extrapolation of the trapezoid rule leads to Simpson's rule.

I've looked through the other posts on ME, specifically the post with the same title, and this for help, but I still don't have a clear idea of how to answer this.

I understand that the extrapolation from $n$ subintervals to $2n$ subintervals reduces the error term, and helps the formula converge faster.

Can anyone help direct me toward a constructed argument? I'm not seeing how to directly make the inference.

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Neurax
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  • Which part of the post is confusing you? – qqo May 01 '14 at 20:46
  • I'm having trouble understanding the actual process of starting with the trapezoid rule, performing an explicit extrapolation step, and having the result be something resembling Simpson's rule. I understand that they're in the same "family", but what is the direct correlating step that shows this to be the case. – Neurax May 01 '14 at 22:02

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Indeed, Simpson's method can be obtained by applying Richardson's extrapolation to the trapezoidal method. We begin with the trapezoidal method: $$\int_a^b f(x)\,dx \approx \frac{h}{2} \left\{ f(x_0)+ 2f(x_1)+2f(x_2)+2f(x_3)+\dots \\ \dots +2f(x_{n-1}) + f(x_n) \right\}\tag{1}$$ where $h=(b-a)/n$ and $x_i=a+ih$. The error of trapezoidal method is known to be of order $h^2$. Therefore, if we double $n$, thus halving $h$, the error should go down by about the factor of $4$. We can then subtract $1/4$ of the original formula (1) from the new one, thus canceling out the error term with $h^2$. The result is Simpson's formula, and this is the derivation you are likely to see in a textbook.

But I find it notationally easier to proceed in the opposite direction: using $h$ and $2h$, not $h$ and $h/2$. To this end: let $n$ in the formula (1) be even. Next, write down the trapezoidal rule with step $2h$. It will involve only $x_0,x_2,x_4,\dots,x_n$:
$$\int_a^b f(x)\,dx \approx h \left\{ f(x_0)+ 2f(x_2)+2f(x_4)+2f(x_6)+ \dots \\ \dots +2f(x_{n-2}) + f(x_n) \right\}\tag{2}$$ The error of formula (2) is about $4$ times the error in formula (1), since we doubled the step size. So we divide both sides of (2) by $4$ and subtract the result from (1), hoping to cancel out the error.
$$\frac34 \int_a^b f(x)\,dx \approx \frac{h}{2} \left\{ f(x_0)+ 2f(x_1)+2f(x_2)+2f(x_3)+\dots \\ \dots +2f(x_{n-1}) + f(x_n) \right\} \\ - \frac{h}{4}\left\{ f(x_0)+ 2f(x_2)+2f(x_4)+2f(x_6)+\dots \\ \dots +2f(x_{n-2}) + f(x_n) \right\} $$ Collect like terms:
$$\frac34 \int_a^b f(x)\,dx \approx \frac{h}{4} \left\{ f(x_0)+ 4f(x_1)+2f(x_2)+4f(x_3)+\dots \\ \dots +4f(x_{n-1}) + f(x_n) \right\} $$ Finally, multiply both sides by $\frac43$ and you'll get Simpson's formula.