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6

To elaborate on the Comment by @Jens, Integrate probably is mishandling this integrand, because ArcSin has branch cuts. The correct solution can be obtained by observing int1 = Integrate[Exp[-x]*DiracDelta[y - Sin[x]], {x, 0, Pi/2}, Assumptions -> {0 < y < 1}] (* 1/(E^ArcSin[y]*Sqrt[1 - y^2]) *) int2 = Integrate[Exp[-x]*DiracDelta[y - Sin[x]], ...


13

Adding up all your kind comments, we can conclude: An enabled Suggestion Bar (standard in Mathematica 10) triggers one WolframKernel process to consume all system memory (possibly up to approximately 4.37 TB, if available). This is a bug in Mathematica 10 (up to and including 10.0.2.0) on multiple operating system platforms. Wolfram has confirmed the ...


5

If you start with the topmost entity and then work your way down step by step you won't miss any data. However, you are correct that some map coordinates appear to be missing: divisions = CountryData["SierraLeone", "AdministrativeDivisions"]; subdivisions = AdministrativeDivisionData[#, "Subdivisions"] & /@ divisions; GeoGraphics[{EdgeForm[Black], ...


0

Maybe you could try this: bc = Plot[{Sqrt[1 - x^2], -Sqrt[1 - x^2]}, {x, -1, 1}, PlotRange -> {-1.2, 1.2}, PlotStyle -> Black, Axes -> False, AspectRatio -> Automatic]; sc = Plot[{(Sqrt[1/4 - x^2]) + .5, (-Sqrt[1/4 - x^2]) + .5}, {x, -1, 1}, PlotRange -> {-1.2, 1.2}, PlotStyle -> Black, AspectRatio -> Automatic, Axes -> ...


6

Slow down the approach, Limit[Log[2 - Sin[x]*Cos[x]] /. x -> x/2, x -> Infinity] or speed it up, Limit[Log[2 - Sin[x]*Cos[x]] /. x -> 2 x, x -> Infinity] -- both yield Interval[{Log[3/2], Log[5/2]}]


4

f[x_] = Log[2 - Sin[x]*Cos[x]]; Simplify[f[x] == f[x + n Pi], Element[n, Integers]] True Since the function is periodic, the limit interval is just the minimum and maximum of the function. Interval@(f[x] /. Solve[{f'[x] == 0, 0 <= x <= 2 Pi}, x, Reals] // FullSimplify // Union) Interval[{Log[3/2], Log[5/2]}] Alternatively, with ...


4

I reported the issue to WRI on November 21. On December 16 they confirmed that it is a known bug. (I forgot to update the question till now.)


5

A limited kind of work-around: expr = Log[2 - Sin[x]*Cos[x]]; TrigReduce //@ expr /. x -> Interval[∞] Interval[{Log[3/2], Log[5/2]}]


4

That's an bug in the 1D case. A somewhat cumbersome way to work around that is to generate the mesh and then generate the mesh a second time where markers are computed and inserted into the mesh elements: Needs["NDSolve`FEM`"] bmesh = ToBoundaryMesh["Coordinates" -> {{0}, {5}, {10}}, "BoundaryElements" -> {PointElement[List /@ Range@3]}]; mesh = ...


1

Assuming you are doing this in the same notebook then: nb = EvaluationNotebook[]; NotebookFind[nb, "tag1", All, CellTags] SelectionEvaluate[nb] If you have several tagged cells that you want to evaluate then (NotebookFind[nb, #, All, CellTags];SelectionEvaluate[nb]) & /@ {"tag1", "tag2"} Edit The above is the way I would normally do it because it ...


2

It seems to be a performance problem (bug?) in version 9.0.1. Version 7.0.1 is also slow, but perhaps is not capable of this calculation. However, version 8.0.4 can produce the correct result quickly, so there seems to be no good reason why 9.0.1 should take so long. I didn't try 9.0.0. The interesting thing, is that 9.0.1 has no problem with the purely ...


3

I don't know what version of Mathematica you're using, but... On V9.0.1, I get the same result as you. On V10.0.1, I get a similar result, but the conditions are embedded in an If statement and the z-integral has not been forgotten; however, the If statement shows that integrand depends on conditions that should have been resolved, given that the z domain ...


1

The issue in V10.0.1 is that LogLogPlot uses x = 0 as a test point, instead of a value inside the specified plot domain. The issue is resolved in V10.0.2. Confirmed by WRI.


7

Let k^2 = x^2 + y^2 + z^2 + 1. Then substitute and note that the integral of F[x,y,z] over all {dx, dy, dz} space is the integral over G[k] 4 Pi k^2 dk from 0 to Infinity, bring out the constant factors, change the integration limits, and then your integral is: 4 Pi Exp[-2] Integrate[Exp[2/k^2]/k^8 k Sqrt[k^2-1], {k, 1, Infinity}] which yields (\[Pi]^2 ...


4

This error should probably be reported to WRI as a bug, most likely in Experimental`NumericalFunction; you should not be seeing this come back up to the top level. I see no obvious reason why memory allocation should fail, as this is not really a large or difficult problem, despite the apparent complexity of the expression. However, we do not really need ...


0

This is the best I can get: qfB = Uncompress["You expression here"]; myfun = qfB /. {\[Lambda] -> 1., \[Gamma] -> 2., a -> 0.5162124455872646`} // N; (* so that the expression only involves t *) NMaximize[{myfun, t > 0}, t] {0.0175048, {t -> 0.570513}} Hope this helps.


9

The function you wish to plot happens to be the InverseSurvivalFunction of a MixtureDistribution with component distributions NormalDistribution[1, 0.3] and NormalDistribution[3, 0.3], and weights .6 and .4, respectively. Using the built-in functions MixtureDistribution and InverseSurvivalFunction we get the desired result without an issue: dist = ...


7

We get a little insight to the "bug" by writing the CDF in terms of Erfc: InverseFunction[ 1 - 0.6*CDF[NormalDistribution[1, 0.3], #] - 0.4*CDF[NormalDistribution[3, 0.3], #] &][.3] InverseFunction[ 1 - 0.6 1/2 Erfc[2.3570226039551585` (1 - #)] - 0.4 1/2 Erfc[2.3570226039551585` (3 - #)] &][.3] 1.47655 + ...


6

f[x_] = 1 - 0.6*CDF[NormalDistribution[1, 0.3], x] - 0.4*CDF[NormalDistribution[3, 0.3], x]; ParametricPlot[{f[x], x}, {x, -1, 3}, AspectRatio -> 1/GoldenRatio]


5

To start with, this should not be a problem for Mathematica. The non-inverse function is reasonably well-behaved: f[u_] := 1-0.6 CDF[NormalDistribution[1, 0.3], u] - 0.4 CDF[NormalDistribution[3, 0.3], u]; Plot[f[u], {u, -1, 5}] (and one can verify, plotting or otherwise, that f' remains $<0$ so that f is 1-to-1). All I can suggest is to increase ...


7

You have complex numbers as a result in these ranges. Check this: Table[{i, InverseFunction[ 1 - 0.6*CDF[NormalDistribution[1, 0.3], #] - 0.4*CDF[NormalDistribution[3, 0.3], #] &][i]}, {i, 0, 1, 0.01}] The problem seems to be generated internally because of the real number in the function and also because of the fact that the plot uses a ...


0

Not a solution but worth noting. ParallelTable[randSphere[], {250}] Does seem to work. I dont know about its limit, but eventually, it crashed (as expected for "big" data).


2

As an alternative, you can use the following function to generate pseudo-random points on the sphere: PointOnTwoSphere[] := Module[ {z, phi, rho, x, y}, z = RandomReal[{-1, 1}]; phi = RandomReal[{0, 2*Pi}]; rho = Sqrt[1 - z^2]; x = rho * Cos[phi]; y = rho * Sin[phi]; {x, y, z} ] This algorithm, due to Marsaglia, has the virtue that it does ...


6

There appears to be a bug in the type inference mechanism here. If you evaluate <<Dataset` <<TypeSystem` and then res = drugNames[byPrefix]; res // GetType (* Assoc[Atom[String], Atom[String], AnyLength] *) which is bogus. Basically that says it thinks the result is an Association with keys and values that are both String. We can compare ...


22

I can reproduce this on OS X in M10.0.2 and M9.0.1, so it looks like a bug. Please report it to Wolfram support. Table will automatically try to compile its argument above a table length threshold. This threshold is 250 by default and can be set to a different value using SetSystemOptions["CompileOptions" -> "TableCompileLength" -> ...]. It seems ...



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