02-02-2020, 03:31 PM
Tried on Emu71b, energy integral from thread Sharp EL-506 vs Sharo EL-516T
From the user's manual, about the algoritm, it uses Romberg's method, after a non-linear transform.
>RUN
P=?1E-5
IA( 127 ) = 62563050744 625265.300862 .11 SEC
IB( 255 ) = 62563050665 625640.053047 .3 SEC
>RUN
P=?1E-6
IA( 127 ) = 62563050647 62526.5300862 .11 SEC
IB( 255 ) = 62563050665 62564.0053047 .3 SEC
>RUN
P=?1E-7
IA( 255 ) = 62563050656 6255.39238586 .22 SEC
IB( 255 ) = 62563050665 6256.40053047 .31 SEC
Why does IA(127) have different area, for P=1E-5 vs P=1E-6 ?
Both are using exactly the same 127 sample points.
Does IBOUND (second column) has something to do with different IA(127) results ?
How is IBOUND intepreted ?
At what IBOUND numbers should an integral be splitted ?
From the user's manual, about the algoritm, it uses Romberg's method, after a non-linear transform.
Code:
10 A=6371000 @ B=9.46073047258E+15 @ C=3.98589196E+17 @ A1=A-1
20 DEF FNA(X) @ N=N+1 @ FNA=C/EXP(X) @ END DEF
30 DEF FNB(X) @ N=N+1 @ X=EXP(X) @ FNB=C*X/(X+A1)^2 @ END DEF
40 INPUT "P=?";P
50 T=TIME @ N=0 @ I=INTEGRAL(LN(A),LN(B),P,FNA(IVAR))
60 DISP "IA(";N;") =";IROUND(I),IBOUND,TIME-T;"SEC"
70 T=TIME @ N=0 @ I=INTEGRAL(0,LN(B-A1),P,FNB(IVAR))
80 DISP "IB(";N;") =";IROUND(I),IBOUND,TIME-T;"SEC"
>RUN
P=?1E-5
IA( 127 ) = 62563050744 625265.300862 .11 SEC
IB( 255 ) = 62563050665 625640.053047 .3 SEC
>RUN
P=?1E-6
IA( 127 ) = 62563050647 62526.5300862 .11 SEC
IB( 255 ) = 62563050665 62564.0053047 .3 SEC
>RUN
P=?1E-7
IA( 255 ) = 62563050656 6255.39238586 .22 SEC
IB( 255 ) = 62563050665 6256.40053047 .31 SEC
Why does IA(127) have different area, for P=1E-5 vs P=1E-6 ?
Both are using exactly the same 127 sample points.
Does IBOUND (second column) has something to do with different IA(127) results ?
How is IBOUND intepreted ?
At what IBOUND numbers should an integral be splitted ?