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Observe In: Precision[6613] Out: Infinity In: Precision[6613.3] Out: MachinePrecision The first number is an integer and has infinite precision the second number is a numerical value that is represented with a finite precision.


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Perhaps someone who knows how NSolve works will comment on my speculations or even answer. But since this question has languished, I'll share how it appears to me. It appears to me that increasing the precision of the approximations of the roots is a minor expense relative to finding the roots. It is perhaps so minor that a PrecisionGoal option seemed ...


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Give this a try: eq1 = x^2 + y^3 + 10 y^2 - 40 x == 1; eq2 = 2 x + 3 y - 20 y^2 == -10; Do[a1 = NSolve[{eq1, eq2}, {x, y}, Reals], {100}] // Timing eq1 = SetPrecision[eq1, 4]; eq2 = SetPrecision[eq2, 4]; Do[a2 = NSolve[{eq1, eq2}, {x, y}, Reals], {100}] // Timing {Sort@a1,Sort@a2} (* {1.326008, Null} {0.826805, Null} {{{x -> 0.0712826, y -> ...



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