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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 …

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 …

Welcome to MMA SE! There are several things preventing this from working. The ; after ContourPlot will suppress the output; simply remove it. Es is not defined as a function, and so the syntax Es[x,y …

You can use the option ScalingFunctions! They're easy to use but a bit tricky to understand. I'll be honest: I don't know what the second function it asks for does, but the documentation says to use t …

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 …

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 …

So, one thing you can do here is note that the equation for a cylinder extending along the z-axis is $x^2+y^2=0$, or $x^2+y^2\leq0$ for a solid cylinder. That is, it's the equation for a circle that i …

Building off of what you have, I think it's nice to put $F(n)$ in the denominator each time, and shift the origin of the axes to {0,1}. Then the one bigger than $F(n)$ is above the axis, and the one s …

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, …

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 …

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 …

One way is to change the construction of Tab1 in the first place, and simply Join together the ranges that you are going to keep. Another is to use Select, Cases, or DeleteCases to modify Tab1. With S …

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[ …

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 …