HP71B Integral Questions

[Wes Loewer]

Yes, I see what you mean. I get the same values as your tables.

(02-09-2020 01:43 PM)Albert Chan Wrote:  [*] Even though the midpoint numbers does not incorprate previous sample points, it will, when we do the extrapolation.

True, but the fact that for the midpoint method each new row has to start over from scratch is the very thing that makes it not as good as the trapezoid. If the previous points were reusable, then all midpoint values on your Romberg table would shift up by one row, making it the better option. This is the issue that I was missing in my thinking.

Quote:For this example, extrapolated trapezoid numbers seems to converge with half the error.

This makes perfect sense. The error bounds for midpoint and trapezoid methods are: |Em | < k(b-a)^3/(24n^2) and |Et | < k(b-a)^3/(12n^2)
So for a given number of intervals (n) the midpoint method has about half the error of the trapezoid method. But for a given number of evaluation points, the trapezoid will have twice as many intervals, so now trapezoid will have about half the error.

Just for fun, I wanted to see what would happen with Romberg using midpoints with trisected intervals, thereby allowing midpoints to reuse previous points. (If I did my Romberg calculations right, the only difference was changing 4^p-1 in the denominator to 9^p-1.) Here's what I get:

Code:

points  midpoints with trisection
1       0.000000000
3       1.34471664
9       1.01187573
27      1.00152749
81      1.00016874
243     1.00001866
729     1.00000207
2187    1.00000023

This is only slightly better than the trapezoid method for this example. But for the previously mentioned problematic 1/sqrt(1-x^2), it makes a huge difference

Code:

points   midpoints with trisection
1       3.000000000000
3       3.136485386483
9       3.141533702802
27      3.141592493620
81      3.141592652733