Tag Info

New answers tagged

1

You have two issues - firstly data includes x values and you don't want to convolve those. So use data[[All, 2]] in the convolution. Secondly you need to allow the kernel to overhang the data or you'll just get the single value of the convolution with the kernel aligned to the data. The overhang is determined by the third argument of ListConvolve. See the ...


2

So thanks to @mfvonb I was able to get a start building a preliminary form to create such a dashboard: Mathematica by default needs to load the database functionality into memory (like any header file) in order to function. Needs["DatabaseLink`"]; Creates the connection to the database from which the data can be pulled. conn = ...


0

dataset1 = Table[{x1[n], y1[n]}, {n, 5}]; poly2[x_] = a*x^2 + b*x + c; rms1 = Norm[(poly2 /@ dataset1[[All, 1]]) - dataset1[[All, 2]]]/ Sqrt[Length[dataset1]]; rms2 = RootMeanSquare[(poly2 /@ dataset1[[All, 1]]) - dataset1[[All, 2]]]; (rms1 // ComplexExpand) == rms2 True


1

how about this? lists = Table[ Sin[x*y], {x, 0, 2 \[Pi], \[Pi]/20}, {y, 0, 2 \[Pi], \[Pi]/20}]; ListDensityPlot[lists, Mesh -> None] Or like this lists = DensityPlot[Sin[x*y], {x, 0, 2 \[Pi]}, {y, 0, 2 \[Pi]}][[1, 1]]; ListDensityPlot[{#1, #2, Sin[#1 #2]} & @@@ lists, Mesh -> None]


3

data = Reap[DensityPlot[Sin[x*y], {x, 0, 2 Pi}, {y, 0, 2 Pi}, EvaluationMonitor :> Sow[{x, y, Sin[x*y]}]]][[2, 1]]; ListPlot3D[data] compare to DensityPlot[Sin[x*y], {x, 0, 2 Pi}, {y, 0, 2 Pi}]


3

There is a very quick way to arrive at this. If you are new to Mathematica then maybe this is a starting point. lineData = {allData[[All, {1, 2}]], allData[[All, {1, 3}]], allData[[All, {1, 4}]], allData[[All, {1, 5}]], allData[[All, {1, 6}]], allData[[All, {1, 7}]]} volume = allData[[All, {1, -1}]] p1 = DateListPlot[lineData, Joined -> True, ...


1

Manipulate[ Plot[{Sin@x, Normal@Series[Sin@u, {u, x0, n}] /. u -> x}, {x, -2 Pi, 2 Pi}, PlotRange -> {Automatic, {-2, 2}}, Epilog -> {PointSize[Medium], Point@{x0, Sin@x0}}], {n, 0, 10, 1}, {x0, -Pi, Pi}]


0

Functions are like vectors. Actually you can define a vector space over functions. We can describe any vector in terms of 3 independent vector which may not be orthogonal to each other. Similarly, you can fit a function in terms of other linearly independent functions. The easiest case is fitting to a polynomial of order n. Depending on how well your data ...



Top 50 recent answers are included