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se.mathworks.com/help/matlab/ref/ezplot.html: ezplot(fun) plots the expression fun(x) over the default domain -2π < x < 2π, where fun(x) is an explicit function of only x. (... and pretty much t …

Combine Plots with Show, and remember to use PlotRange -> All: Show[ Plot[x, {x, 0, 1}, PlotStyle -> Blue], Plot[x^2, {x, 1, 2}, PlotStyle -> Red], PlotRange -> All] Another option, as Karsten …

I'm not quite following your question, but are you possibly looking for this? Pp = TransformedProcess[{g[t] E^(j[t]), j[t]}, {g \[Distributed] GeometricBrownianMotionProcess[v, \[Sigma], 1], …

Fresh v10.2 introduces ListStepPlot which provides just the right facility for this task: pts = {{1.81515, 1.46912}, {4.99276, 1.87137}, {9.12065, 6.83238}, {9.9288, 5.30858}, {5.6387, 4.30558}, …

Use Reduce to get the region as inequalities: f[x_] := Sin[4 x] + x; Reduce[-1 <= f[x] <= 1, x, Reals] (* Root[{1 + Sin[4 #1] + #1 &, -1.45320652256538767403}] <= x <= Root[{1 + Sin[4 #1] + #1 &, …

I'm not entirely certain if this is a particularly useful approach to the problem hidden behind your question, but this should help while maintaining the form: RevolutionPlot3D[Sign[x] Sqrt[4 - x^2], …

Sometimes I would want to Plot simple functions on unconventional projection, but without specifying all the styling myself. If one looks at InputForm of a normal plot, all this would appear to be dee …

Looks like a bug to me. Nonetheless, you can add padding to it: PolarPlot[1, {t, -Pi, Pi}, PolarAxes -> {True, False}, PolarTicks -> {"Degrees", Automatic}, PolarGridLines -> {True, False}, PlotRa …

A quick hack: n = 10; f[x_] := Sin[x]; rectangles[f_, a_, b_, n_] := {Opacity[0.0], Blue, EdgeForm[Black], N@Table[Rectangle[{a + k (b - a)/n, 0}, {a + (k + 1) (b - a)/n, f[a + k (b - a)/n …

EDIT: With help from @cvgmt, both fixing a wrong term in the equation(!) and adding constraints for finding lines: With[{expr = 1 + 54 x y z - 9 (x + y + z) + 126 (x y + x z + y z) - 9 (x^2 + …

Add all the items you want in the Epilog to it as a list: data3 = Table[{x, x^2 + 2}, {x, 0, 5}]; Show[Plot[x^2, {x, 0, 5}, PlotRange -> All], ListPlot[data3, PlotStyle -> Red], ImageSize -> Large, …

Partially based on Chris Degnen's answer, and all that predates it: F[k_, x_, y_, z_, w_] = (-1)^(z + w - 1) ((2 x + 1) (2 y + 1) (2 w + 1) (2 k + 1))^(0.5) ThreeJSymbol[ {x, 1}, {y, -1}, {k, …

It is easiest to use ParametricPlot and RotationTransform. ParametricPlot[ RotationTransform[a][{1, 0}] + RotationTransform[4 a][{1/4, 0}], {a, 0, 2 Pi}, Evaluated -> True] Evaluated -> T …

Consider this: ParametricPlot3D[ RotationTransform[a, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}], {a, 0, 2 Pi}, Evaluated -> True] Now rotate this around a circle, while rotating it at the same ti …

Considering your GPS-inspired goal, here's an approach which visualizes a least-squares solution to distances from sphere shells, and the corrected standard deviation as size of the red sphere: Modul …