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cosolver/FM/FMinterval/SampleFMinterval.chk
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SampleFMinterval.chk
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Sample 1. Unstable summation.
1. Do the sum in double precision.
K = 0 S = 1.750000000000000E+01
K = 5 S = -4.955444729786392E+08
K = 10 S = 6.361137619195046E+11
K = 15 S = -6.829259024353563E+12
K = 20 S = 3.022687119371480E+12
K = 25 S = -1.224830662857075E+11
K = 30 S = 7.315904098222299E+08
K = 35 S = -8.834682915557073E+05
K = 40 S = 2.701921995744262E+02
K = 45 S = 2.070038693322906E-02
K = 50 S = 4.546179247215157E-02
K = 55 S = 4.546101674944338E-02
K = 60 S = 4.546101675867138E-02
K = 62 S = 4.546101675867133E-02
2. Use FM with increasing precision.
Setting precision to J digits via CALL FM_SET(J) will actually set the
equivalent number of decimal significant digits slightly higher than J.
For example, if the base used internally in FM is 10**7, then asking for
20 digits with CALL FM_SET(20) gives at least 29 significant digits.
CALL FM_SET(30) gives at least 36 significant digits.
CALL FM_SET(40) gives at least 50 significant digits.
CALL FM_SET(50) gives at least 57 significant digits.
20 digits, 73 terms gave S_FM = .04399094217962562151
This calculation lost about 18 digits.
30 digits, 78 terms gave S_FM = .043990942179625639969699565464
This calculation lost about 18 digits.
40 digits, 88 terms gave S_FM = .0439909421796256399696989706597424719113
This calculation lost about 18 digits.
50 digits, 93 terms gave S_FM = .04399094217962563996969897065974247192700503984511
This calculation lost about 18 digits.
3. Use FM with increasing precision in base 2.
Using 53 bits, 62 terms gave S_FM = .04546101675867133
This calculation lost about 14 decimal digits.
Using 73 bits, 67 terms gave S_FM = .04399094194167266143888
This calculation lost about 14 decimal digits.
Using 93 bits, 72 terms gave S_FM = .04399094217962813948568386511
This calculation lost about 15 decimal digits.
Using 113 bits, 77 terms gave S_FM = .04399094217962563996863023518761958
This calculation lost about 14 decimal digits.
4. Use FM with different rounding modes in base 2.
Using 53 bits, rounding left , 62 terms gave S_FM = .02897151338476893
Using 53 bits, rounding symmetrically, 62 terms gave S_FM = .04546101675867133
Using 53 bits, rounding right , 62 terms gave S_FM = .05194200656637992
These agree to about 0 decimal digits.
Using 113 bits, rounding left , 77 terms gave S_FM = .04399094217962563
Using 113 bits, rounding symmetrically, 77 terms gave S_FM = .04399094217962564
Using 113 bits, rounding right , 77 terms gave S_FM = .04399094217962564
These agree to about 18 decimal digits.
5. Use FM with interval arithmetic in base 2.
Using 53 bits, 62 terms gave S_FM_INTERVAL = [ -.1021766113637601 , .1945740695617867 ]
The two endpoints agree to about 0 decimal digits.
Using 113 bits, 77 terms gave S_FM_INTERVAL = [ .0439909421796257 , .0439909421796257 ]
The two endpoints agree to about 18 decimal digits.
Summary:
For this calculation all four methods for measuring the degree of
instability worked well. When done with double precision carrying
16 significant digits and using the default symmetric rounding,
only 1 digit remained correct at the end of the sum.
Interval arithmetic is probably the strongest of these checks, and
using FM arithmetic with 30 digits and a large base (method 2) is
the fastest of these methods for getting the sum correct to full
double precision accuracy.
Comparing the last value of S in method 1 with the first value of
S_FM in method 3 should show whether this compiler carries extra
digits while evaluating expressions like X / ( (K+1) * FACT**2 ).
If the two values are the same, no extra digits are carried in d.p.
Sample 2. Unstable recurrence.
1. Use non-interval FM arithmetic with 30 digits.
After 5 terms with 30 digit accuracy, the result is 4.91084749908279
After 10 terms with 30 digit accuracy, the result is 4.99277028806207
After 15 terms with 30 digit accuracy, the result is 4.99943593714684
After 20 terms with 30 digit accuracy, the result is 4.99995612709047
After 25 terms with 30 digit accuracy, the result is 5.00009039073834
After 30 terms with 30 digit accuracy, the result is 77.16163142845431
After 35 terms with 30 digit accuracy, the result is 99.99999060423245
After 40 terms with 30 digit accuracy, the result is 99.99999999999706
After 45 terms with 30 digit accuracy, the result is 100.00000000000000
After 50 terms with 30 digit accuracy, the result is 100.00000000000000
After 55 terms with 30 digit accuracy, the result is 100.00000000000000
After 60 terms with 30 digit accuracy, the result is 100.00000000000000
2. Use interval arithmetic with 30 digit accuracy.
After 1 terms the result is [ 4.4705882352941176 , 4.4705882352941176 ]
After 2 terms the result is [ 4.6447368421052632 , 4.6447368421052632 ]
After 3 terms the result is [ 4.7705382436260623 , 4.7705382436260623 ]
After 4 terms the result is [ 4.8557007125890736 , 4.8557007125890736 ]
After 5 terms the result is [ 4.9108474990827932 , 4.9108474990827932 ]
After 6 terms the result is [ 4.9455374041239167 , 4.9455374041239167 ]
After 7 terms the result is [ 4.9669625817627006 , 4.9669625817627006 ]
After 8 terms the result is [ 4.9800457013556312 , 4.9800457013556312 ]
After 9 terms the result is [ 4.9879794484783923 , 4.9879794484783923 ]
After 10 terms the result is [ 4.9927702880620681 , 4.9927702880620681 ]
After 11 terms the result is [ 4.9956558915066340 , 4.9956558915066340 ]
After 12 terms the result is [ 4.9973912683813441 , 4.9973912683813441 ]
After 13 terms the result is [ 4.9984339439448169 , 4.9984339439448169 ]
After 14 terms the result is [ 4.9990600719708939 , 4.9990600719708939 ]
After 15 terms the result is [ 4.9994359371468391 , 4.9994359371468392 ]
After 16 terms the result is [ 4.9996615241037670 , 4.9996615241037680 ]
After 17 terms the result is [ 4.9997969007134058 , 4.9997969007134286 ]
After 18 terms the result is [ 4.9998781354776765 , 4.9998781354781580 ]
After 19 terms the result is [ 4.9999268794992229 , 4.9999268795094152 ]
After 20 terms the result is [ 4.9999561269476688 , 4.9999561271634113 ]
After 21 terms the result is [ 4.9999736736100334 , 4.9999736781766820 ]
After 22 terms the result is [ 4.9999841549419453 , 4.9999842516043860 ]
After 23 terms the result is [ 4.9999894553132943 , 4.9999915013665818 ]
After 24 terms the result is [ 4.9999717607369067 , 5.0000150694764154 ]
After 25 terms the result is [ 4.9995202522450967 , 5.0004369664045127 ]
After 26 terms the result is [ 4.9899353973829161 , 5.0093395647887771 ]
After 27 terms the result is [ 4.7869972899223478 , 5.1977831184897810 ]
After 28 terms the result is [ .3000582860996429 , 9.0357119352120193 ]
After 29 terms the result is [ -.1646377251287M+4 , 52.4812789961537590 ]
After 30 terms the result is [ - OVERFLOW , + OVERFLOW ]
Summary:
The general solution of this recurrence has a term that causes
the values to converge to 100, but for these initial conditions
4, 4.25, the specific solution has a coefficient of zero on that
term, so this sequence converges to 5 mathematically.
When rounding errors occur in later x-values, they introduce a
very small but nonzero amount of the term that causes convergence
to 100, and that term grows rapidly and soon swamps the rest of
the solution.
Interval arithmetic tracks the growing uncertainty in the x-values,
and when an interval gets big enough to include zero, dividing by
that interval is undefined and the result is
[ -overflow , +overflow ].
Interval arithmetic works better than a sequence of increasing
precision FM results here, since comparing FM results at 30, 40, 50
digits gives 100 each time.
Sample 3. Unstable sum.
For x = -25.00 The sum gave [ 1.3887943864963056M-11 , 1.3887943864964986M-11 ]
correct = 1.3887943864964021M-11
For x = -30.00 The sum gave [ 9.3576229646761189M-14 , 9.3576229730042186M-14 ]
correct = 9.3576229688401746M-14
For x = -35.00 The sum gave [ 6.3046409205537791M-16 , 6.3055928170842012M-16 ]
correct = 6.3051167601469894M-16
Summary:
As the input x becomes more negative, the instability increases.
This is shown as the left and right endpoints of the interval
result agree to fewer digits.
The mathematically correct value of the sum lies within the
interval in each case.
All results were ok.
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