Search Results

Here's the simple approach in case it helps (not a full answer), noting that we can phrase the constraint $u=x$ in terms of $t$ by substituting $x$ in for $u$ in $t = u + \int \frac{dz}{f(u,z)}$: \[Ep …

Here's a fully generic way with ImplicitRegion. First, we might want to simply use scaled, or we might instead want to convert to a BoundaryMeshRegion via bscaled = BoundaryDiscretizeRegion[scaled] …

This is a little tricky, and I'm not sure why BoundaryMesh didn't work in the following. (We have to resort to BoundaryMeshRegion instead.) But here's what we do: Discretize the spline with Discretiz …

I believe you might be looking for InverseFunction; look at the docs and you'll see that Inverse refers to taking the inverse of a square matrix. Also, you'll want to feed InverseFunction a (pure) fun …

It's typically useful in Mathematica to translate mathematical notation of the form $x_i$ into x[i]. If you do that with \[Sigma][i] (which looks much better in a Mathematica notebook!), we could defi …

However, by plotting both sides of the equation in 3D and viewing the result from directly above to see where the intersections lie, it does seem we've got them all in this range. …

You're right that it's a delayed evaluation problem! I traced the problem back to Eigenvectors. Note that it behaves differently when given symbolic arguments and numeric ones: Eigenvectors[hMf[0.01, …

@kglr provided a nice way of calculating it when n is an explicit integer; to enforce that it is explicit, change n to n_Integer: ClearAll[sum]; sum[n_Integer] := Total[Times @@@ Apply[Abs@*W, Tuple …

If you want to mimic MATLAB syntax, you can define quiver as such: quiver[X_, Y_, DX_, DY_, opts___] := Block[{points = Transpose[Flatten /@ {X, Y}]}, ListVectorPlot[Transpose[{points, Transpose[ …

To do this you might have to add extra values to your matrix, essentially to "pin down" the values of ListPlot3D to 0 between your "actual" matrix values. We can do that as follows: mRiffle0[m_?Matrix …

If I'm interpreting your data structure right, then one straightforward way to do this is by the following: datapoints = Flatten[Array[{X[[#1]], Y[[#2]], Z[[#2, #1]]} &, {Length[X], Length[Y]}], 1] A …

Here's something that you could overlay things on with Show (and you could include Axes -> True to show the axes—plus you could style them appropriately)! xmin, xmax, ymin, and ymax are of course adju …

I think your issue might be that $x_1 \leq 0$ and $x_1 \geq 0$ do not properly characterize the parts of the regions you want. It seems that for the bottom one, you want $x_1 \geq -x_2$, and for the t …

I don't think this is actually made with Wolfram graphics primitives, but you can still find the image it uses! Open the application package and navigate to Contents > SystemFiles > FrontEnd > SystemR …

There are a couple ways to do this! One is with ImageSize, and one is with RasterSize (probably recommended), and they have slightly different effects. Specifying ImageSize essentially scales up your …