# Tag Info

3

This has been fixed as of version 10.0.2. – ilian May 6 at 20:30 Just to take this off the unanswered list.

2

I think you've found a bug. It seems to me that the Encrypt/Decrypt functionality introduced in 10.1 needs more work. But, I have found that the following work around may help. If you explicitly pass the EncryptedObject properties into an EncryptedObject via its Association parameter, they will be correctly interpreted with no errors. Try this: encryptedObj ...

1

Wolfram has confirmed that this is a bug.

1

In version 10, I think that much of the behavior is simply a difference between how the year 0 is handled by astronomers (and therefore the page you were getting Julian Dates from) and everybody else. For example, take this calculation: jd[{-3000, 11, 24, 12, 0, 0}] - 625660 I believe that you have actually pulled the wrong number from the website due to ...

4

The reason this is happening might be because Mathematica is internally producing AppellF1 functions. If you replace Sqrt[b^2 - c^2] with Sqrt[k] and integrate, then afterwards put back the b^2 - c^2 you get this mess ... tmp = Integrate[1/(-Sqrt[k] + b*Cosh[x] + c*Sinh[x])^(1/2), x]; tmp //. k -> b^2 - c^2 (* Result *) (1/(Sqrt[1 - b^2/c^2] c))2 ...

15

In addition to the error messages quoted in the question the line returns: GeneralUtilitiesBenchmarkingPackagePrivateplot[ IndexBy[{{{16, 9.37132*10^-6}, . . . IndexBy was removed from 10.1.0: Note that IndexBy will be removed in a future version of Mathematica. It was something that was considered for 10.0.0 but didn't make the cut. – Stefan R ...

3

Update 2015-05-14 I contacted WRI support and they confirmed that this is a known issue with Export (support case: 3206586). Summary @ChenStatsYu seems to have found an unexplained behavior of the Export function for PDF files that looks like a bug. Detailed results 1) I generated two graphics similar to those in the OP's original question, then ...

0

Plot[{PDF[NormalDistribution[], x], PDF[NormalDistribution[-4, 1], x], PDF[NormalDistribution[2, 1], x]}, {x, -8, 6}, AxesLabel -> {None, None}, Ticks -> {Automatic, None}, Axes -> {True, False}, Background -> None, PlotLegends -> Placed[{"N(0,1)", "N(-4,1)", "N(2,1)"}, {1, .5}]]

5

I wonder if this is a bug that appears when using regions and {"MonteCarlo"} as a method. It hangs my machine. This might be a possible workaround: NIntegrate[1, {x, y} \[Element] Triangle[{{0,0},{1,2},{2,1}}], Method-> "MonteCarlo", MaxPoints -> 10^5}] There isn't mention of the new arbitrary region functionality with the "MaxPoints" option or ...

2

I wass going to post an answer about scoping issues but it seems that this problem can be narrowed down. As MarcoB has noticed, there appears to be a bug in Manipulator. var = 75; Manipulator[Dynamic[var], {50, 100, 0.2}] when you open it, var is reset to 50... If you start with "Open" it is ok. However, if you put Manipulator to dialog, each option ...

3

The behavior seems to be a bug of Mathematica. Here is an excerpt from an email I got from Wolfram after asking them about the problem: It does seem that the answer of FullSimplify is incorrect especially since the exponential function is not identical to zero (or a very-close-to-zero constant). Therefore, I filed a report with our development team ...

2

According to a comment by @ilian, this has been fixed as of version 10.0.0. It certainly works in version 10.1, as we can see below: $Version (* 10.1.0 for Linux x86 (64-bit) (March 24, 2015) *) fC[randomdata] == (fC /@ randomdata) (* True *) totalC /@ randomdata totalC[randomdata] 1 Token That this menu item is doing is: FrontEndExecute[FrontEndToken[InputNotebook[],"GenerateNotebook"]] Keep in mind you have to add there FrontEnd` context. You can use it in other palette after previously setting focus on different one. Or you can put it in joker.m from 68871 Or use as a manual function for any NotebookObject in place of ... 6 This is a bug in Mathematica 10.1.0 on OSX, and seems to primarily, but not exclusively, affect older MacBookPro machines with Core 2 Duo processors. We have identified the cause and devised a fix. Unfortunately, the fix is not simple, and requires replacement of a few key components. We have created a patched version of Mathematica 10.1.0 which is ... 1 It is not a good idea to feed approximate numbers like 0.1 to symbolic methods. Have a look at the indefinite integral Integrate[g[Cos[\[Theta]], Sin[\[Theta]], x2, y2], \[Theta]] to see what's going on. Mathematica has to go very far into complex analysis to solve this integral symbolically. I suspect the several terms with branch points at \[Theta] == ... 1 To record an answer officially: As noted by Daniel Lichtblau in a comment, it is a bug. Szabolcs comments that it affects versions 10.0.1, 10.0.2, and 10.1.0, but not versions before 10 (and unclear about 10.0.0). As of the time of writing, version 10.1.0 is the most recently released. However, Daniel further adds that it has already been fixed in the ... 0 As Albert Retey notes in a comment, this seems to be a bug specific to version 7 (or perhaps just 7.0.1), given that this behavior is not observed in versions 6.0.2, 8.0.4, 9.0.1 or 10.0.2. Incidentally, you are absolutely right that the file is not exported correctly. If you use a different version to create it as it should be, version 7 will then be able ... 11 I think the problem might be related to a bug in FullForm when applied to a ByteArray object: ByteArray["aV+jpGtfd3BHhoSvOthJpQ=="] // FullForm (* List[105,95,163,164,107,95,119,112,71,134,132,175,58,216,73,165] *) The full form has lost information regarding the structure of the ByteArray. The box-form of the button is using this list form but the ... 2 This appears to be a bug in V10.0.x which was fixed in V10.1.0.$Version "10.1.0 for Mac OS X x86 (64-bit) (March 24, 2015)" Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi}, Assumptions -> {x > 0, y > 0, x > y}] -((π y)/(4 x^2))

4

\$Version "10.0 for Mac OS X x86 (64-bit) (September 10, 2014)" As entered Mathematica returns the wrong result. Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi}, Assumptions -> {x > 0, y > 0, x > y}] Pi*(1/(8*y) - (3*y)/(8*x^2)) However, a workaround is to convert the trig functions to exponentials ...

6

OK, now I have another suggestion: With[{x = ToString[encryptedObj, InputForm]}, Print[x]; Button["Try with", foo = Decrypt["pass", ToExpression[x]]]] The button generates no error and foo is set to "TestCase".

4

It seems to me that Integrate can do some strange things with your function g. From plotting g, we can see the integral should clearly be zero. g[x1_, y1_, x2_, y2_] = -Log[Sqrt[(x2 - x1)^2 + (y2 - y1)^2]] Plot[g[Cos[θ], Sin[θ], 1/10, 1/10], {θ, 0, 2 π}] Further, Integrate[g[Cos[θ], Sin[θ], 1/10, 1/10], {θ, 0, 2 π}] gives zero as expected, but ...

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