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Have you any reason to think that there exist nontrivial solution to your equations? This plot may suggest that the only solution may be `x==0`, except for periodicity: eqs = Rationalize[{Sin[x]*Cos[2.5*y]*Cosh[2.5*z] + ...
You can do a lot this way: sol = NDSolveValue[{D[r[t], t] == (r[t].r[t])^-1 RotationMatrix[Pi/2, {0, 0, 1}].r[t], r[0] == {1, 1, 1}}, r, {t, 0, 1}] sol[1]
The solution set is more complicated than that: Reduce[(js + ps) 6 == w && js (pd + 5) == w && ps pd == w, Reals] // LogicalExpand I don't know the exact reason why `Solve` gives an empty set in that special case, but with slight...
The code pp/.Line[x_]:>{Arrowheads[Table[.04, {4}]], Arrow[x]} targets the hidden internal structure of the output of `Plot` and `ParametricPlot`. Try Plot[x,{x,0,1}][[1]] you will see that it contains a Line primitive, which...
Each of the points has integer distance from the others. This finds a new point with integer distances from the other four: pts = {{0, 0}, {20, 0}, {138/5, (24 Sqrt[6])/5}, {25, 10 Sqrt[6]}}; dists = MapIndexed[({x, y} - #) . ({x, y} - #)...
Have you tried with parentheses: Cases[{"u1", "p1"}, _?(StringStartsQ[#, "u"] &)] Cases[{"u1", "p1"}, _?(StringStartsQ["u"])]
Maybe the wireframe version is clearer: Plot3D[2 x^2 y/(x^4 + y^2), {x, -1, 1}, {y, -1, 1}, PlotPoints -> 200, Exclusions -> Automatic, MeshFunctions -> {#2/#1 &}, Mesh -> {Tan[Pi/2 Range[-19, 19]/20]}, PlotStyle -> None]
Try plotting with `PlotRange -> All`.
The solution uses complex numbers, but its values are real and correct: sol1 = DSolveValue[{eq, i[0] == i0}, i, t] FullSimplify[{eq, i[0] == i0} /. i -> sol1] Block[{\[Gamma] = 1, \[Mu] = 1, \[Lambda] = 1, i0 = 1}, ...
I would use `Piecewise`: f[x_] := Piecewise[{{40 - 6 x, 0 Axis] Plot[f[x], {x, 0, 10}, Exclusions -> Automatic, ExclusionsStyle -> Directive[Red, Dashed]] Show[%, %%] Curiously, I could not combine `Exclusions` with...