Timeline for how to get single, and quad Precision out of Mathematica which match Fortran result
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2018 at 14:39 | vote | accept | Nasser | ||
| Jan 25, 2018 at 2:31 | answer | added | Michael E2 | timeline score: 4 | |
| Jan 24, 2018 at 22:52 | history | tweeted | twitter.com/StackMma/status/956298745506942979 | ||
| Jan 24, 2018 at 19:48 | answer | added | Bob Hanlon | timeline score: 2 | |
| Jan 24, 2018 at 18:50 | comment | added | george2079 | right, for anything other than machine precision mathematica is not doing IEEE arithmetic and so you will never get the same results. | |
| Jan 24, 2018 at 18:45 | comment | added | Nasser |
@george2079 I do not think this gives same as Fortran. For example, $x1=0.00001\`7$ then $sum=0$ results in $0.9999999999999972025\`6$ while Fortran gives $1.00099015$ I think setting sum=0 instead of sum=0.0 makes it do arbitray precision somewhere internally?
|
|
| Jan 24, 2018 at 18:38 | comment | added | george2079 |
One thing going on here is 0.0`32 is for some reason MachinePrecision. You can actually initialize sum1=0; and get your desired precision in the calculation. (It does not match the fortran results though)
|
|
| Jan 24, 2018 at 18:26 | history | asked | Nasser | CC BY-SA 3.0 |