Numerical integration methods - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: HP Calculators (and very old HP Computers) ( /forum-3.html)+--- Forum: HP Prime ( /forum-5.html)+--- Thread: Numerical integration methods ( /thread-18566.html)Pages: 1 2 |

RE: Numerical integration methods - lrdheat - 07-25-2022 10:36 PM
That is what is going on! When, on the Classpad 400, I write integral from 1 to 100 of x^(2(e^((1/x)^3)-1)), I get the 99.5996+ answer! RE: Numerical integration methods - Nigel (UK) - 07-26-2022 05:18 PM
I never considered that as a possible interpretation of the original expression. Well done for figuring out what was going on! I'm glad that's sorted. Nigel (UK) RE: Numerical integration methods - parisse - 07-31-2022 04:59 PM
(07-24-2022 08:55 PM)Wes Loewer Wrote: (Note: It's possible that I am mistranslating "ordre". What the paper called "ordre 30" seems to be "degree 29" (ie, 30 terms), but I could be wrong about this. Francophones, please correct me if I am mistaken.)"order" may mean max degree of polynomial where the method is exact, or the exponent in the step of the error, up to the author choice... Quote:Since the 15-node Gauss-Kronrod is exact up to degree 22 and the Prime's method is exact up to degree 29, it would seem that the Prime's method is a bit better with only a small added overhead. A slightly more accurate calculation per iteration could occasionally reduce the need for further recursion. It would be interesting to do some comparisons of these two methods with some well-behaved functions as well as more "temperamental" ones.Yes, it is important not to re-evaluate the function for efficiency reasons. The idea is to estimate the error of the order 30 method in h^30 by err1*(err1/err2)^2 where err1=abs(i30-i14) is in h^14 and err2=abs(i30-i6) in h^6 RE: Numerical integration methods - Wes Loewer - 08-01-2022 03:43 AM
(07-31-2022 04:59 PM)parisse Wrote: "order" may mean max degree of polynomial where the method is exact, or the exponent in the step of the error, up to the author choice... I see. The order of the error, not the order of the polynomial, in this case. Thank you. (07-31-2022 04:59 PM)parisse Wrote: Yes, it is important not to re-evaluate the function for efficiency reasons. The idea is to estimate the error of the order 30 method in h^30 by This method works beautifully, but I am fuzzy about some of the details. For instance, why *(err1/err2)^2 ? Why not *(err1/err2) or *(err1/err2)^3 ? Was ^2 derived mathematically, or experimentally? I'm guessing it was experimentally determined to give a reasonable error approximation. And why does the 3rd calculation use 6 nodes? Why not 5 or 7? I realize you want the 3rd calculation to be different enough (but not too different) from the 2nd calculation to determine the error. Was 6 determined experimentally to produce an optimal error estimate? Or is there something mathematical that makes 6 the best choice? Using 6 leaves the entire middle third of the interval unevaluated. Seems like including the center node would have been advantageous. RE: Numerical integration methods - Albert Chan - 08-01-2022 03:01 PM
(08-01-2022 03:43 AM)Wes Loewer Wrote: For instance, why *(err1/err2)^2 ? Why not *(err1/err2) or *(err1/err2)^3 ? Was ^2 derived mathematically, or experimentally? I'm guessing it was experimentally determined to give a reasonable error approximation. This is my guess ... err1 = k1 * h^15 err2 = k2 * h^7 integral error estimate = err1*(err1/err2)^2 = k1*(k1/k2)^2 * h^(15+8*2) = k * h^31 If (k, k1, k2) are similar in size, constant term matched too. Exponents cannot be picked in random; it had to produce O(h^31) on the right. If we use 5 or 7 nodes, instead of square, it would need some non-integer exponents. RE: Numerical integration methods - Wes Loewer - 08-01-2022 03:46 PM
(08-01-2022 03:01 PM)Albert Chan Wrote: This is my guess ... Okay, I must be missing something. Why does it have to produce O(h^31) on the right? What if the algorithm turned out to be O(h^33) or some other order? How would using a different number of nodes produce non-integer exponents? Even if it did produce non-integer exponents, what would be wrong with that? Sorry if I'm missing something obvious. RE: Numerical integration methods - parisse - 08-01-2022 07:01 PM
The exponent of the error estimate must match the error majoration of the best quadrature. RE: Numerical integration methods - Wes Loewer - 08-01-2022 07:24 PM
(08-01-2022 07:01 PM)parisse Wrote: The exponent of the error estimate must match the error majoration of the best quadrature. Okay, that's the piece I was missing. Everything else falls into place now. Thank you. RE: Numerical integration methods - jonmoore - 08-04-2022 12:54 PM
(07-19-2022 05:16 PM)parisse Wrote: int uses an adaptive method, as described here in French (Hairer: https://www.unige.ch/~hairer/poly/chap1.pdf). Bernard, out of interest, is a version of the Risch Algorthm still used for symbolic integration in XCAS/HP Prime? RE: Numerical integration methods - parisse - 08-05-2022 09:45 AM
Yes, the rational version is implemented. |