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Let n be even. Show how the composite Simpson rule with 2n equally spaced nodes can be computed from the case of n equally spaced nodes with a minimum amount of additional work.

I've been working on this problem for almost 2 hours now and can't seem to get anywhere. I feel it has something to do with the initial conditions on the [a,b] interval ($x_i=a+ih \text{ and } h=\frac{b-a}{2}$) and working the problem so there are the same number of terms to calculate in 2n as there are in n. There's always replacing any instance of n with 2n and halving h, but I don't think that's what they're looking for.

Apologies of I don't make any sense, any help would be much appreciated, I just cannot seem to wrap my head around this.

James Snyder
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  • You need to define your $n$. The basic Simpson's rule uses function values at three points, the two ends of the interval plus the center. Cutting the interval in half would use five points with separation of $\frac 14$ the interval. Is that two nodes? Note that the three points you started with are reused in this calculation, which will continue to be true as you cut the intervals in half. – Ross Millikan Nov 21 '13 at 21:13
  • I understand that part and I wrote out the terms for n=2,4,8 on [0,1] and I noticed that the points that were in previous iterations were multiplied by 2 and the newly added points at 2n were multiplied by 4(or the odd $x_i$ are the new points and the even points were in the previous iterations), but I'm still not too sure where to go from here. – James Snyder Nov 22 '13 at 02:43

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The starting Simpson's rule on $[0,1]$ is $I_1 \approx \frac 16 f(0)+ \frac 46 f(\frac 12) + \frac 16 f(1)$ When you subdivide the interval it is $I_2\approx \frac 1{12}f(0)+\frac 4{12}f(\frac 14)+\frac 2{12}f(\frac 12)+\frac 4{12}f(\frac 34)+\frac 1{12}f(1)=\frac 12 I_1+\frac 4{12}(f(\frac 14)+f(\frac 34))$ Then when you double again, you divide $I_2$ in half, then add the contributions for the new points. You are being asked to show that this continues, so all you need to do is evaluate $f$ at the new points, sum the values, multiply by an appropriate constant, and add half the previous approximation to get the new approximation.

Ross Millikan
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