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.
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.