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Mathematica Help Browser from Mathematica 5 in new versions of Mathematica

Tag : development , By : mylonov
Date : March 29 2020, 07:55 AM
I wish did fix the issue. At this moment I have found two ways to invoke the legacy Help Browser by clicking a button:
``````DisplayForm@ButtonBox["Preface",
Active -> True,
ButtonData :> {"PCT Preface", None}]
``````
``````DisplayForm@
ButtonBox[" »", BaseStyle -> "Link", Evaluator -> Automatic,
ButtonFunction :> (PacletManager`Package`helpBrowserLookup[#] &),
ButtonData -> "Help Browser"]
``````

List Manipulation in Mathematica pertaining to Lagrange Interpolation Polynomials in Mathematica

Tag : math , By : Michael Gunderson
Date : March 29 2020, 07:55 AM
This might help you Given your last question, I'm guessing that you mean a Lagrange Polynomial, so
``````LagrangePoly[pts_?MatrixQ, var_: x] /; MatchQ[Dimensions[pts], {_, 2}] :=
With[{k = Length[pts]}, Sum[pts[[j, 2]] Product[
If[j != m, (var - pts[[m, 1]])/(pts[[j, 1]] - pts[[m, 1]]), 1],
{m, 1, k}], {j, 1, k}]]
``````
``````In[2]:= points = Table[{x, Tan[x]}, {x, -1.2, 1.2, .2}]
Out[2]= {{-1.2, -2.57215}, {-1., -1.55741}, {-0.8, -1.02964},
{-0.6, -0.684137}, {-0.4, -0.422793}, {-0.2, -0.20271},
{0., 0.}, {0.2, 0.20271}, {0.4, 0.422793},
{0.6, 0.684137}, {0.8, 1.02964}, {1., 1.55741}, {1.2, 2.57215}}

In[3]:= Plot[Evaluate[Expand[LagrangePoly[points, x]]], {x, -1.2, 1.2},
Epilog -> Point[points]]
``````
``````In[4]:= FindMaximum[{Abs[Tan[x] - LagrangePoly[points, x]], -1.2<x<1.2}, x]
Out[4]= {0.000184412, {x -> 0.936711}}
``````
``````In[5]:= InterpolatingPolynomial[points, x]-LagrangePoly[points, x]//Expand//Chop
Out[5]= 0
``````

Photo Mosaic in Mathematica: an example from 2008 doesn't work in Mathematica 9

Tag : development , By : fukas78
Date : March 29 2020, 07:55 AM
it fixes the issue As the comments note, your problem is because the images you're using for the imagePool are not all the same number of channels, and that's upsetting the Nearest function. Probably the easy way to fix this is:
``````imagePool = Map[With[{i = Import[#]}, {i,
N@Mean[Flatten[ImageData[RemoveAlphaChannel[i]], 1]]}] &,
FileNames["*.png", "/tmp"]]
``````

Is that possible that Mathematica can locate the maximum of a scalar function whereas Mathematica Alpha cannot?

Tag : development , By : user183275
Date : March 29 2020, 07:55 AM
will help you It is certainly possible that Mathematica will solve problems that Alpha cannot - or will not because of the time limits. The desktop product does not have any such limits.
From the Wolfram|Alpha FAQs:

Trying to use Edge Count Mathematica command on a list of subgraphs but Mathematica doesn't recognise the objects as gra

Tag : development , By : Helpful Dude
Date : March 29 2020, 07:55 AM
Hope this helps There is a minor detail of defining Subcount but then using Subcount2
Subcount returns a Table of {graph} and thus you are trying to do Edgecount[{graph}] instead of Edgecount[graph]