I've been reading through the implementation of MidpointIntegrator, and I've discovered an issue with the algorithm as implemented.
The midpoint method generates an estimate for the definite integral of f between a and b by
- subdividing (a, b) into n intervals
- estimating the area of each interval as a rectangle (b-a)/n wide and as tall as the midpoint of the interval - ie, ((b-a)/n) * f((a + b) / 2)
- adding up all estimates.
The MidpointIntegrator implementation here sticks to n = powers of 2 for this reason, stated in the comments:
The interval is divided equally into 2^n sections rather than an arbitrary m sections because this configuration can best utilize the already computed values.
Here is the problem: the midpoint method can't reuse values if you split an interval in half. It can only reuse previous values if you split the interval into 3; ie, use 3^n sections, not 2^n.
What's actually implemented in `MidpointIntegrator` is a left Riemann sum without the leftmost point. Or, identically, a right Riemann sum without the rightmost point: https://en.wikipedia.org/wiki/Riemann_sum
This matters because the error of a (left, right) Riemann sum is proportional to 1/n, while the error of the midpoint method is proportional to 1/n^2... a big difference.
The idea behind the midpoint method's point reuse is that if you triple the number of integral slices, one of the midpoints with n slices will overlap with the n/3 slice evaluation:
You can incrementally update the n=1 estimate by
- sampling the points at (1/6) and (5/6) of the way across the n=1 interval
- adding them to the n=1 estimate
- scaling this sum down by 3, to scale down the widths of the rectangles
For values of n != 1, the same trick applies... just sample the 1/6, 5/6 point for every slice in the n/3 evaluation.
What happens when you try and scale from n => 2n? The old function evaluations all fall on the boundaries between the new cells, and can't be reused:
The implementation of "stage" in MidpointIntegrator is combining them like this:
which is actually:
- tripling the number of integration slices at each step, not doubling, and
- moving the function evaluation points out of the midpoint.
In fact, what "stage" is actually doing is calculating a left Riemann sum, just without the left-most point.
Here are the evaluation points for a left Riemann sum for comparison:
Here's the code from "stage" implementing this:
To keep the exact same results, I think the only solution is to remove the incorrect incremental "stage"; or convert it to the algorithm that implements the correct incremental increase by 3, and then don't call it by default.
Everything does work wonderfully if you expand n by a factor of 3 each time. Press discusses this trick in Numerical Recipes, section 4.4 (p136): http://phys.uri.edu/nigh/NumRec/bookfpdf/f4-4.pdf and notes that the savings you get still make it more efficient than NOT reusing function evaluations and implementing a correct scale-by-2 each time.
Happy to provide more information if I can be helpful!