Consider $I = \int_a^b f(x) dx = 267.25$ and the approximations $\overline{I}_8 = 295.27$, $\overline{I}_{16} = 274.15$, $\overline{I}_{32} = 268.97$ and $\overline{I}_{64} = 267.68$ of $I$ given by the composite Simpson's rule with $9,17,33$ and $65$ nodes. Give an $a$, $b$ and $f(x)$ for the above scenario.
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Can you provide some context? What have you done, and what is giving you trouble? – Ron Gordon Apr 29 '13 at 21:57
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All problems like these can be solved by considering what points are used in $I_{2^{n+1}}$ that are not used in $I_{2^n}$. Since the numbers are decreasing in $n$, that means that the new points must be lower in value then the old points; since the integral itself is lower than the approximations, the average value off the points must be lower than the points.
So just consider functions that have peaks at the sample points, with smaller and smaller peaks as $n$ increases.
Brian Rushton
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Hint: note that $\bar I_{16}=\frac 12 \bar I_8 + \frac 1{16}\left(f(a+\frac 1{16}(b-a))+f(a+\frac 3{16}(b-a))+\ldots + f(a+\frac {15}{16}(b-a))\right)$ so you have the sum of the function values at those points. Similarly the next two values give you sums of the function values at the nodes that get added at each step. I would take $a=0,b=1,f(x)=267.25$ and then perturb $f(x)$ specifically at a few nodes to make the approximations come out.
Ross Millikan
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