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3

WDX is not a good format. We are likely to deprecate it or entirely replace it with a different implementation that is not backward-compatible (which is obviously problematic). Dataset will never directly support it. There are candidates for a possible native format for dataset, like Cap'n Proto, HDF5, and a couple others. Or XML. No, just kidding :-)


3

Extended comment: $Version "10.1.0 for Mac OS X x86 (64-bit) (March 24, 2015)" eq = 0.0055356626 + 0.000029 x + 0.000034 y <= 1000.; eq // Simplify x + 1.17241 y <= 3.44826*10^7 Use of assumptions reproduces the problem: Simplify[eq, {x > 0, y > 0}] False However, use of Rationalize avoids the problem with using ...


1

The present state of affairs The kind of manual manipulation demonstrated by kglr should theoretically not be necessary. There may have been some problems in the initial version 10 release but in either 10.0.2 or 10.1.0 many Themes already combine correctly. Emulating his penultimate example: Plot[Evaluate@Table[BesselJ[n, x], {n, 5}], {x, 0, 10}, ...


1

I received an answer from support that it's a Manipulate bug. The following code is stable(it's working on Mathematica 10, Windows 7): co = 2.0*^8; ω = 2 Pi \[ScriptF] 10^6; τ = \[ScriptCapitalT] 10^-9; t = ts*10^-9; sol = Solve[{a + b == 1, (a E^(-I ω τ) + b E^(I ω τ)) == (a E^(-I ω τ) - b E^(I ω τ))*RL/Z0}, {a, b}]; Vi[ts_, x_, ...


0

If you're willing to use a text file instead of XLS, here's the code I use to upload & plot two columns from a text file: CloudDeploy[FormFunction[ "x" -> "Text", ListPlot[ImportString[#x, "Data"]] &, "PNG" ], Permissions -> "Public"]


5

Just to summarize my understanding: First of all, the documentation states that Simplify assumes that variables are real when they occur algebraically in inequalities. Clearly, there are no inequalities in the logical expression y == 0 && x^2 == -1, and therefore x and y should be assumed to be general complex numbers. If they were real, then the ...


16

Preface: For everyone: be aware that this behaviour is very likely to change soon (the name of the hook variable flag, etc). In fact, I did change this for 10.2, where Export would fall back to the standard Export in case when the specialised hook fails. Leonid Shifrin I get the same behaviour using Mathematica version 10.1 on Windows 7/64-bit. ...


2

No problem in version 10: $Version (* Out[3]= "10.1.0 for Microsoft Windows (64-bit) (March 24, 2015)" *) RSolve[{(k + 2) c[k + 2] q^(k + 1) - c[k] == 0}, c[k], k] $$\left\{\left\{c(k)\to \frac{c_1 2^{-\frac{k}{2}} \left(\frac{1}{q^2}\right)^{\frac{1}{8} k (k+2)} q^{k/2}}{\Gamma \left(\frac{k}{2}+1\right)}+\frac{c_2 (-1)^k 2^{-\frac{k}{2}} ...


4

I can confirm the above incorrect results are due to a bug in the Intel MKL library shipping with Mathematica 10.1 which affects FFT convolution. The problem is only known to be triggered on some processors (for example AMD chips, or virtual machine emulated CPUs). The following workaround will use an alternative implementation that does not rely on MKL ...


5

Apparently in version 9 listability of the CDF for TransformedDistribution, at least in this particular case, was broken. As of version 10 this is no longer the case. However, there does appear to be a bug here. Named tests like PearsonChiSquareTest actually call DistributionFitTest under the hood so I will use the former to help point to the problem. The ...


4

I will expand on Szabolcs's answer. Starting with the same triangulation, we can see that the problematic cells consist of "triangle" whose vertices are collinear. These have areas that are either negative (wrong orientation), zero, or nearly zero (below 2.*^-15). They arise no doubt from round-off error. I could not find any means of improving the ...


0

I got a reply from Mathematica indicating that it is a numerical precision issue (the Mathematica implementation of the convex hull seems to work at a lower precision than the data types used). Two work arounds are: Scale the data by a large number Approximate the data by rationals In other words, use PlotPolytope2[n_] := Module[{V = 1000 ...


6

There are a few things to be said here. First, it should not crash and a future version will behave better in this scenario. What I am not quite sure abuot is why the call ToElementMesh[DiscretizeRegion[...]] in the first place. Note that both mesh = ToElementMesh[\[CapitalOmega]]; and DiscretizeRegion[\[CapitalOmega]]; work fine. In your example you ...


1

This is in response to Chris's request in comments, but not an answer to his question. I tried running the code you asked: {ev, if, mesh} = helmholzSolve3D[\[CapitalOmega], 4, MaxCellMeasure -> 0.25] with the definition of Omega that you have in the current question, and the definition of helmholzSolve3D from ...


0

It is a physical phenomenon occurring in a very short time( milliseconds). A = 1; \[Phi] = 0.0257275275; \[Omega] = 151.6*10^3; Cj = 17*10^(-9); Cs = 35*10^(-9); Rt = 223.9; I0 = 5*10^-6; L = 15*10^-3; Cd[t_] := Piecewise[{{Cj/Sqrt[1 - A*Cos[\[Omega]*t]/\[Phi]], t < 0}, {Cs*Exp[A*Cos[\[Omega]*t]/\[Phi]], t >= 0}}]; qn = {V'[t] == (j[t] - ...


1

After removing the superfluous θ[0] == 0 from bc, the key to the solution is a subtle change in the use of Piecewise in the definition of Cd: Cd[x_] := Piecewise[{{Cj/Sqrt[1 - x/ϕ], x < 0}}, Cs*Exp[x/ϕ]] eqn = {V'[t] == (j[t] - Id[V[t]])/Cd[V[t]], j'[t] == (V0[t] - Rt*j[t] - V[t])/L}; bc = {V[0] == A, j[0] == 0 (*, \[Theta][0] == 0*)}; pl = ...


0

Okay, I'm slowly going through the questions involving elliptic integral evaluations. Yet again, none of the software mentioned in this thread have managed to produce a "clean" expression. For the benefit of future readers, here's a tidier closed form for your perusal: N[2 (7 Sqrt[5] + 2 (10 EllipticE[2 π/3, 11/12] + EllipticF[2 π/3, 11/12])/Sqrt[3])/9, 20] ...


1

Even with Rubi's help, the result returned by Mathematica is an ugly looking elliptic integral expression with complex parameter. (Actually, even Maple's result is far from optimal as well.) Using methods similar to what I did here, here is a closed form for the OP's integral: N[(12 Sqrt[7] + 3^(3/4) InverseJacobiCN[(8 Sqrt[3] - 19)/13, (2 + Sqrt[3])/4])/5, ...



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