HP71B Integral Questions

[J-F Garnier]

I ran the same tests on the HP75 Math ROM, here is the program adapted from Albert's code for the 75C:

10 A=6371000 @ B=9.46073047258E15 @ C=3.98589196E17 @ A1=A-1
20 DEF FNA(X)
22 N=N+1 @ FNA=C/EXP(X)
25 END DEF
30 DEF FNB(X)
32 N=N+1 @ X=EXP(X) @ FNB=C*X/(X+A1)^2
35 END DEF
40 INPUT "P=?";P
50 T=TIME @ N=0 @ I=INTEGRAL(LOG(A),LOG(B),P,FNA(X))
60 DISP "IA(";N;") =";ROUND(I,0),IBOUND,TIME-T;"SEC"
70 ! T=TIME @ N=0 @ I=INTEGRAL(0,LOG(B-A1),P,FNB(X))
80 ! DISP "IB(";N;") =";ROUND(I,0),IBOUND,TIME-T;"SEC"

The results are a bit different from the HP71 outputs:

P=?2e-2
IA( 63 ) = 62563132323 422758135.629 .11 SEC

P=?1e-3
IA( 127 ) = 62563050559 21501195.6588 .22 SEC

P=?1e-4
IA( 255 ) = 62563050656 2072473.90909 .495 SEC

P=?1e-5
IA( 255 ) = 62563050656 207247.390909 .494 SEC

P=?1e-6
IA( 511 ) = 62563050656 20718.6780627 .989 SEC

For the same target accuracy P, the HP75 uses more samples:
with P=1E-6, the HP71 uses only 127 samples, and the HP75 511 samples.

Comparing with the Excel simulations from Albert for the first two results:

Code:

Intervals       Richardson Extrapolations
64  62401515156    62564066925 62562986272 62563074044 62563117315 62563129267 62563132322
128 62522713769    62563113307 62563049732 62563050740 62563050648 62563050583 62563050564 62563050559

it seems that the HP75 uses the last extrapolated value. In a way it could make sense since each extrapolated value only requires a few operations.

So it turns out that the HP71 and HP75 INTEGRAL are based on the same algorithm but use different criteria for the choice of the number of samples and returned value.

The IBOUND value is still not clear for me, both on the HP71 and HP75. With same number of samples and same extrapolation, why is the IBOUND value different and depends on the user-supplied target accuracy P? For instance in the cases above, IBOUND is exactly divided by 10 when P is changed from 1E-4 to 1E-5 - same samples, same extrapolation, same INTEGRAL result.

J-F