HP71B Integral Questions

[Wes Loewer]

(02-06-2020 11:16 PM)Albert Chan Wrote:  

(02-06-2020 06:53 PM)Wes Loewer Wrote:  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.

A simple test showed that HP71B INTEGRAL do extrapolations from trapezoids, not mid-points rule.

>10 DISP INTEGRAL(-1, 1, 1E-5, 1/SQRT(1-IVAR^2)), IBOUND
>RUN
3.14156045534              -3.1415397954E-5

this failure to converge (65535 sample points !) is due to missing end points evaluation.

But the standard trapezoid method evaluates the endpoints while the rectangle midpoint method does not. As you said, the endpoints are not evaluated, which points to the midpoint method. The 50g definitely uses Romberg extrapolation of midpoints (see edit below) and gives very similar results as above. Setting the number format to FIX 5, you get:

result = 3.14156045554, IERR = -3.14153979546E-5

I could be wrong, but I would be very surprised if they changed up the algorithm for the 71b.

(02-06-2020 11:16 PM)Albert Chan Wrote:  

Quote:So then the question was "Why make this substitution if the intervals are already nonuniform?"

It does matter, even if gaussian quadrature do not evaluate end points.
u-transformation changes the overall shape of the integrand, affecting all sample points.

Right, that's just what I was trying to explain and demonstrate in the sample graph.

EDIT: Ack! It seems that my statement "The 50g definitely uses Romberg extrapolation of midpoints" was definitely wrong. :-(
Mea culpa