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  • $\begingroup$ 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?! $\endgroup$ Commented Jun 19, 2014 at 19:32
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    $\begingroup$ 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). $\endgroup$ Commented Jun 19, 2014 at 23:38
  • $\begingroup$ 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. $\endgroup$ Commented Jun 20, 2014 at 2:00
  • $\begingroup$ 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. $\endgroup$ Commented Jun 20, 2014 at 14:01
  • $\begingroup$ 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 $\endgroup$ Commented Jun 20, 2014 at 15:06