HP71B Integral Questions

[J-F Garnier]

(02-03-2020 11:15 PM)Albert Chan Wrote:  

(02-02-2020 03:31 PM)Albert Chan Wrote:  From the user's manual, about the algoritm, it uses Romberg's method, after a non-linear transform.

There was an error in the manual. x = ½(3u-u³) substitution *required* integrand limit -1 to 1, not a to b

Thanks, Albert.
Actually, this description originates from a HP-15C manual (text originally written by W. Kahan) ... and was simplified in the HP71 Math ROM manual.
In the HP-15C Advanced Functions Handbook p.41, the integration limits are -1 and 1, and the exact sentence is: "Their spacing can be demonstrated by substituting, say" (emphasis added) to indicate it is an example for illustration only, not an exact detailed description, The important word "say" was removed in the HP75 and HP71 manuals...

Quote:Trivia:
For 2 intervals, h = 1 → 2^16 intervals, h = 1/2^15 = 0.00003 05175 78125 (exact)
With 15 digits precision, and |u| ≤ 1, this implied INTEGRAL internally calculated u's are all exact.

Interesting. The older HP-34C and HP-15C (and HP-41C Advantage ROM) used 13 digits internally, maybe were they more limited in the number of intervals.

J-F