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Are there really sufficiently many assignable pixel locations such that you can drag a Controller to such a resolution? Would you be satisfied if the range were small enough that the MinIntervalSize you seek could be rendered? After all, this will work: IntervalSlider[{.003, .007}, {0., .010}, ImageSize -> 600, MinIntervalSize -> .00000001, ...


A way to work with RealDigits, if desired: Drop @@ RealDigits[Round[Sqrt[2], 1*^-7]] // FromDigits 4142136


FractionalPart@N[Sqrt[2], 7] 10^7 // Round (* 4142136*)


If you use inexact arithmetic, you must accept that you will get inexact results, that is, an approximation to the exact result. 2.0 can be represent exactly as a machine float, but 2.0 Pi cannot. Therefore, there is a small error in the argument given to Sin which propagates during evaluation and produces the result you see. High precision numerics is ...


By using 2.0 instead of 2, you are using machine precision instead of symbolic outputs. Try this instead: Sin[2 Pi] That will return exactly 0.


First you must understand the difference between machine precision arithmetic (CPU floating point) and Mathematica's own arbitrary precision arithmetic. You get the first by calling N with one argument and the second by calling N with two arguments, the second being the precision you want to maintain. When using Mathematica's own arbitrary precision ...


I would say that the answer to the first question is, no, if Mathematica does not give any errors when evaluating an expression, then we cannot be absolutely certain that its answer is reliable up to a decent precision. This is shown in the section Examples of Pathological Behavior in the tutorial NIntegrate Integration Rules. Given that the previous ...


In V10 there has been added some symbolic processing of integrands containing an InterpolatingFunction. In particular if the interpolation grid divides the domain of integration into a number of subintervals, the number being at most the value of the option "MaxSubregions", the integrand will automatically be integrated over each subinterval. In V9, this is ...


Because WorkingPrecision determines the minimum number of digits used in the calculations, you can reduce it, if you hope to reduce time by reducing accuracy. However, WorkingPrecision less than MachinePrecision, the default value, is unlikely to have a large effect on timing, as you can see from, Table[First@Timing[Do[NSolve[x^5 - 3 x^4 + 3 x^3 - 4 x^2 + ...

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