Timeline for NSolve for high degree univariate polynomials
Current License: CC BY-SA 3.0
9 events
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| Jun 20, 2014 at 19:41 | comment | added | John | My question is, does the tighter tolerance 10^(-100) puts 'closer' to the original floating point problem than the tolerance 10^(-10) ? | |
| Jun 20, 2014 at 15:15 | comment | added | user15996 |
lis = Tally[ ParallelTable[ Length[IntegerDigits[ Denominator[Rationalize[RandomReal[{-1, 1}], 0]]]], {10^7}]] // Union with output (* {{5, 1}, {6, 363}, {7, 33752}, {8, 3608865}, {9, 6064836}, {10, 286763}, {11, 5332}, {12, 87}, {13, 1}} *), which didn't find any 16-digit denominators, but they would be there in a large sample. ListLogPlot of output is nice.
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| Jun 20, 2014 at 15:06 | comment | added | user15996 |
You are correct. I was only making the point that numbers with only 16 digits of precision are not going to get turned into rationals with 100-digit denominators by Rationalize. To see what Rationalize does, I ran
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| Jun 20, 2014 at 14:01 | comment | added | Daniel Lichtblau |
I don't understand the comment about Rationalize not being able to get within 10^(-100). The numbers in question had around 16 decimal places so these can typically be approximated by rationals with 10 or fewer digits in numerator and denominator (worst case: 16 digits). But that doesn't mean we fell outside the 10^(-100) bound in terms of how close we are.
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| Jun 20, 2014 at 2:00 | comment | added | user15996 |
In principle, the tolerance on the coefficients could be set tighter, to 10^-100, as Daniel suggests, but Mathematica's Rationalize function isn't up to that task. It can't do much better than 10^-10, as indicated in the P.S. to my original answer.
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| Jun 20, 2014 at 1:49 | history | edited | user15996 | CC BY-SA 3.0 |
added 1501 characters in body
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| Jun 19, 2014 at 23:38 | comment | added | Daniel Lichtblau | Could rationalize it to tighter tolerance, say 10^(-100). I'm not clear on what is meant by "solving a different problem" though. It's certainly different from what's there if no rationalization is done-- any rationalizing will move it to a "nearby' problem. But that was true of the original formulation, when rationalizing actually takes place (which it does, on some coefficients). | |
| Jun 19, 2014 at 19:32 | comment | added | John | Thanks for your answer. But by rationalizing the coefficient by approximating it with 0.0001 precision, aren't you solving a different problem than the original one?! | |
| Jun 18, 2014 at 22:55 | history | answered | user15996 | CC BY-SA 3.0 |