HP71B Integral Questions

[Wes Loewer]

(02-05-2020 08:43 AM)J-F Garnier Wrote:  

(02-03-2020 11:15 PM)Albert Chan Wrote:  There was an error in the manual. x = ½(3u-u³) substitution ...

Actually, this description originates from a HP-15C manual (text originally written by W. Kahan) ... and was simplified in the HP71 Math ROM manual.

The 71b and 15c manuals' descriptions both quote from Kahan's article in the August 1980 Hewlett Packard Journal description of the 34c [https://www.hpl.hp.com/hpjournal/pdfs/Is...980-08.pdf]

A couple of notes on the 3/2*u-1/2*u³ substitution.

1) The manuals and article say something like

Quote:Besides suppressing resonance, the substitution has two additional benefits. First, no sample need be taken from either endpoint of the interval of integration...

If I understand correctly (and please correct me if I'm wrong), it's not this substitution that prevents the end points from being evaluated. The reason the endpoints are not evaluated is because the Romberg variation used is based on rectangle midpoint evaluations rather than something like a trapezoid (trapezium) method which does evaluate endpoints. If you were evaluating at the endpoints of -1 and 1, then after this substitution you would still have the endpoints -1 and 1. (This invariance of the endpoints is part of why this substitution is so useful.)


2) Here's an interesting tidbit. I've never seen this in print, but a few years ago I discovered that the ti-89/92/nspire also use this substitution even though the intervals are already nonuniform beforehand.

The ti-83/84 calculators use the standard 15 point Gauss-Kronrod method which uses nonuniform intervals at the following nodes:

Code:

 0.00000000000000
±0.20778495500790
±0.40584515137740
±0.58608723546769
±0.74153118559939
±0.86486442335977
±0.94910791234276
±0.99145537112081

However, the ti-89/92/nspire numeric integration uses the following nodes which didn't match anything I had ever seen.

Code:

 0.00000000000000
±0.30719191764839
±0.57534429140662
±0.77847088404600
±0.90842445832523
±0.97384146143897
±0.99618089849105
±0.99989079590057

I couldn't figure out where these ti-89 values came from until one day on a whim I plugged the Gauss-Kronrod values into 3/2*u-1/2*u³ and, voila!, out came the ti-89 values.

So then the question was "Why make this substitution if the intervals are already nonuniform?" The hp manuals and article go on to say that this substitution allows you to integrate functions whose slope is infinite at an endpoint. While this is not required since the algorithm is not evaluating at the endpoints anyway, experimentation shows that it indeed gives better results when integrating functions like sqrt(1-u) or sqrt(1-u^2) from -1 to 1 where the slope is infinite at one or more endpoints.

By not using the standard Gauss-Kronrod nodes, the algorithm is not as accurate on higher order polynomials (degree > 14), but is more accurate on functions with square roots. Seems like a good trade off.

Why this happens is illustrated by the semi-circular graph of √(1-X^2) before and after the substitution.