15

Here's a starting point for you: With[{n = 6}, Graphics[MapIndexed[{ColorData[103] @@ #2, #1} &, NestList[MapAt[Composition[ TranslationTransform[AngleVector[2 π/5]/ GoldenRatio], ...


12

1. You can use HorizontalGauge as a control in Manipulate: Manipulate[Plot[Sin[x + phi], {x, 0, 2 Pi}], {phi, 0, 2 Pi, Panel[HorizontalGauge[##, ScaleDivisions -> None, Axes -> {True, False}, ImageSize -> 250, Ticks -> {Transpose[{Subdivide[8], Subdivide[0, 2 Pi, 8]}], None}], #, Right] &}] To allow only discrete values, say ...


10

A starting point: With[{n = 4}, Graphics[MapIndexed[{ColorData[97] @@ #2, #1} &, Append[Riffle[ Table[Rectangle[{1 - 2^(1 - k), 0}, {1 - 2^-k, 2^-k}], {k, n}], Table[Rectangle[{1 - 2^(1 - k), 2^-k}, {1, 2^(1 - k)}], {k, n}]], Rectangle[{1 - 2^-n, 0}...


10

Edition {c, b, a} = SASTriangle[1, 36 Degree, 1][[1]]; α = -1 + GoldenRatio; β = 1 - α; (*p[n_/;n≥4]:=p[n]=α*p[n-1]+β*p[n-3];*) m = {{0, 1, 0}, {0, 0, 1}, {β, 0, α}}; tri[1] = {a, b, c}; tri[n_] := tri[n] = m . tri[n - 1]; mat = Limit[MatrixPower[m, n], n -> Infinity]; center = mat . tri[1] // First; p[n_] := First@tri[n]; θ[1] = ArcTan @@ N@(p[1] - ...


8

There is an option called Paneled to control the outer frame. Setting this option to False should do the trick. Manipulate[Plot[Sin[x (1 + a x)], {x, 0, 6}], {a, 0, 2}, Paneled -> False]


8

Block[{θ = Pi/5., n = 6, c = Cos@θ, s = Sin@θ, f = AffineTransform[{{{c - 1, s}, {-s, c - 1}}, {-c, s}}], pts := NestList[f, {{0, Cot[θ/2]}, {-1, 0}, {1, 0}}, n], vc := Table[Hue[i/(n + 1)], {i, 0, n}, {3}]}, Graphics[Polygon[pts, VertexColors -> vc]] ]


7

Manipulate[ plt = ContourPlot[ Cos[x + y] Sin[k x - 3 y], {x, -2, 2}, {y, -2, 2}, PlotPoints -> 20, PlotRange -> {{-2, 2}, {-2, 2}}, Axes -> True, AxesOrigin -> {0, 0}, AxesStyle -> Directive[GrayLevel[0.5]], AxesLabel -> {Style["X", Bold, 14], Style["Y", Bold, 14]}, ...


7

EDIT I have found something simplier than my first answer: Apply Setting to the Manipulate to obtain a normal graphic. Then to remove all the options of this graphic, one way is to take the first part and reapply Graphic END EDIT OLD ANSWER You can use the Snapshot function in the Manipulate. Then remplace DynamicModule by With (this may not be necessary), ...


7

To remove only the outer panel keep the frame around the content area, you can use Style + DefaultOptions as follows: Style[#, DefaultOptions -> {Panel -> {Background -> White}}] & @ Manipulate[Plot[Sin[n x], {x, 0, 2 Pi}], {{n, 3}, 0, 10, 1}]


7

We need a custom control because TogglerBar does not quite do the job as it shows the value that is being clicked and it allows opened briefcases to be closed again with the next click. With the function togglerGrid below (1) we cannot see the contents of the briefcase until it is opened, and (2) once a briefcase is opened it remains open. ClearAll[...


7

It's a very broad question (too many controls, options, etc.), but one can find out what the output is equivalent to like this: Manipulate[x + y, {x, 0, 9, 1}, {y, 0, 90, 10}] // MakeBoxes[#, StandardForm] & // First // ToExpression // InputForm (* DynamicModule[{ x = 0, y = 0, Typeset`show = True, Typeset`bookmarkList = {}, Typeset`...


6

Try this: Manipulate[ SomeFunction[SensorSize], {SensorSize, {A122 -> "A122", A122B -> "A122B", A322 -> "A322"}, ControlType -> SetterBar}] with the effect Have fun!


6

As suggested in the comments, the current version of Mathematica (12.1.1) import functionality of VTK files seems restricted to legacy format files, and it appears to ignore field data. Sample VTK file Here's a simple example of a cube with scaler nodal data taken from here. SetDirectory[NotebookDirectory[]] writeScript[filename_, txt_] := Module[{file}, ...


6

Using ArrayPad + Fold + Nest + MatrixPlot ClearAll[padMat] padMat = Fold[ArrayPad[#, RotateLeft[{{0}, {Length @ #, 0}}, #2 - 1], 1 + Max @ #] &, #, {1, 2}] &; Examples: With[{n = 6}, MatrixPlot[Nest[padMat, {{1}}, n] /. x_Integer :> ColorData[97][x], ImageSize -> 600, Frame -> False, Mesh -> All]] Grid @ Partition[#, 3] & @ ...


6

The most general solution is to do the whole layout yourself. You can actually use Control and put the controls themselves in the Grid along with your output. So you will have only ONE grid, which now includes everything in it. Both the controls and the final output of your Manipulate code. This gives you full control of what you want to do. For your example,...


5

Use your function to define a parametric curve (say, curve) and use it with ParametricPlot and add the moving circle asEpilog. The normal to the curve at point t is Cross[Normalize[curve'[t]]] so we can easily find the center of moving circle: ClearAll[curve] curve[x_] := {x, x} Manipulate[ParametricPlot[curve[x], {x, -Pi, 3 Pi}, Epilog -> {Red, ...


5

In Manipulate >> Details and Options allowed forms of the second argument are listed: So HorizontalGauge[Dynamic[x], {0,10}] (or, for that matter, Slider[Dynamic[x], {0, 10}] or Checkbox[Dynamic[z]] or InputField[Dynamic[w]] etc.) is not allowed in the second (or later) argument. The highlighted row of the list is elaborated on as So using Row @ {...


5

You can try ParametricPlot: ParametricPlot[{x, a x^2}, {x, 0, 10}, {a, 1, 3}, AspectRatio -> 1] You can highlight lines corresponding to specific values of a using the options MeshFunctions and Mesh: alist = {1, 2, 3}; colors = Opacity[1, #] & /@ {Red, Green, Purple}; mesh = Thread[{alist, Directive[Thick, #] & /@ colors}]; ParametricPlot[{x, a ...


5

With your revised initial and boundary conditions $Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global`*"] kc = 0; eqns = { D[y[t, x], t] == D[D[y[t, x], x], x] + y[t, x]^3 - z[t, x], D[z[t, x], t] == D[D[z[t, x], x], x] + y[t, x] - k*z[t, x] + kc, y[0, x] == 0.3, z[0, x] == 0.1, y[t, -2] == 0.3, ...


5

Manipulate[Plot[Evaluate@ OutputResponse[TransferFunctionModel[(5/(s^2 + 2 s)*K1)/(1 + 5/(s^2 + 2 s)*(K2*s + K1)), s], UnitStep[t], {t, 0, 10}], {t, 0, 4}, PlotRange -> {0, 3}], {{K1, 1}, 0, 3}, {{K2, 2}, 0, 3}]


5

Clear["Global`*"] y[a_, k_, l_, d_] := a*k^d*l^(1 - d); mpl[a_, k_, l_, d_] := (1 - d)*a*k^d*l^(-d); lsupply[b_, m_, l_] := b + m*l; l should not be an argument to intersectlabor. Use NSolve intersectlabor[a_?NumericQ, k_?NumericQ, d_?NumericQ, b_?NumericQ, m_?NumericQ] := {l, mpl[a, k, l, d]} /. NSolve[{mpl[a, k, l, d] == lsupply[b, m, l], ...


5

Actually, g[a,T] have analytic expression. Integrate[f[v, T], {v, a, Infinity}, Assumptions -> T > 0] $$1.\, -\frac{1. \sqrt{a^2} \text{erf}\left(0.0438624 \sqrt{\frac{a^2}{T}}\right)}{a}+\frac{0.0494935 a e^{-\frac{0.00192391 a^2}{T}}}{\sqrt{T}}$$ Integrate[f[v, T], {v, 0, Infinity}, Assumptions -> T > 0] 1 kb = 1.381*10^-23; Nav = 6.022*...


5

It's a simple method: Graphics[Map[HalfLine, Partition[ AnglePath[ Prepend[ Table[{(2 Sin[18 Degree])^n, -108 Degree}, {n, 1, 10}], {1, -72 Degree}]], 2, 1]]] But this method has a lot of extra half-lines. The following code works better: Graphics[{Opacity[ 0.6], {Table[FaceForm[RandomColor[]], 10], Map[Triangle, ...


5

We can get the coordinates of successive triangles by finding the "AngleBisectingCevianEndpoint" using TriangleConstruct: nextTriangle[{a_, b_, c_} ]:= {TriangleConstruct[{a, b, c}, "AngleBisectingCevianEndpoint"][[1]], a, b} Use NestList starting with N @ SASTriangle[1, 36 Degree, 1] to get n triangles: triangleList[n_]:= NestList[...


5

If you want the control variable to only take exact values you need to specify all values including the step value as exact values, e.g., Manipulate[var, {{var, Pi}, 0, Pi, Pi/16, Appearance -> "Labeled"}]


5

Use the slider to control the number of rows and click on a cell in MatrixPlot to toggle between "+" and "-": Manipulate[Row[{EventHandler[ Dynamic @ MatrixPlot[m[[;; nr]], Mesh -> All, ColorRules -> {"+" -> Red, "-" -> Yellow}, PlotRangePadding -> 0, FrameTicksStyle -> 18, ...


4

This is the well-functioning code due to changes made according to commentators. thank you for your time. Clear["Global`*"] kc = 0; func[k_] := NDSolve[{ D[ y[t, x], t] == D[D[ y[t, x], x], x] + y[t, x]^3 - z[t, x], D[ z[t, x], t] == D[D[ z[t, x], x], x] + y[t, x] - k*z[t, x] + kc, y[0, x] == 0.3, z[0, x] == 0.1, y[t, -2] == 0.3, z[t,...


4

Dynamic[HorizontalGauge[Dynamic[x], {0, range}], TrackedSymbols :> {range}] Slider[Dynamic[range], {5, 50}]


4

You can use the options MeshFunctions, Mesh and MeshStyle as follows: Manipulate[Plot[x^2 - 2*(a - 2)*x + a - 2, {x, -20, 20}, ClippingStyle -> False, MeshFunctions -> {# &}, Mesh -> {{a - 2}}, MeshStyle -> Directive[PointSize[Large], Red], PlotRange -> {-200, 200}], {a, -10, 10}] Alternatively, you can use Epilog Manipulate[...


4

We can use ParametricPlot instead of PolarPlot to construct the region. Here we change the definition of domain from π/2 to 2 π + π/2 in order to make the boundary of region fine. ρ[a_, θ_] := a (1 - Sin[θ]); region = ParametricPlot[ρ[a, θ]*{Cos[θ], Sin[θ]}, {θ, π/2, 2 π + π/2}, {a, 0, 2}, Axes -> False, BoundaryStyle -> {Thick, Brown}, ...


Only top voted, non community-wiki answers of a minimum length are eligible