11

cp = ContourPlot[EuclideanDistance[{-5, 0}, {x, y}] EuclideanDistance[{5, 0}, {x, y}], {x, -15, 15}, {y, -11, 11}, Contours -> Range[5, 150, 20], Frame -> False, ContourLabels -> (Text[Style[#3, Directive[Blue, 15]], {#1, #2}] &), AspectRatio -> Automatic, ColorFunction -> (If[# < 145, ColorData[{"TemperatureMap",...


9

$Version (* "12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)" *) Clear["Global`*"] Version 12.1.1 also has mismatched FrameTicks on the left and right edges: llp = ListLogPlot[ Table[{x, x^-10}, {x, 1, 5}], GridLines -> Automatic, Frame -> True] Looking at the FrameTicks option values: Cases[llp, HoldPattern[FrameTicks -&...


7

You can use AxisObject to add additional ticks. For example, for the ticks on the horizontal axis: ListLinePlot[ data, PlotRange -> All, PlotStyle -> Black, GridLines -> {(os - 1) 2 + 1, data[[(os - 1) 2 + 1]]}, GridLinesStyle -> Directive[Black], Epilog -> AxisObject[ Line[{{0,0},{34,0}}], {0,34}, ...


6

Simplifying and evaluating the argument will also reduce the redundant labels. $Version (* "13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021)" *) ContourPlot[ Evaluate[ Simplify[ EuclideanDistance[{-5, 0}, {x, y}]* EuclideanDistance[{5, 0}, {x, y}]]], {x, -15, 15}, {y, -11, 11}, Contours -> Range[5, 150, 20], Frame -> False, ...


6

Something like: Clear["Global`*"] dat = RandomReal[{0, 1}, {10, 2}]; t1 = 0.02 Table[{Cos[p], Sin[p]}, {p, 0, 2 Pi, 2 Pi/5}]; t2 = 0.005 Table[{Cos[p], Sin[p]}, {p, 2 Pi/10, 2 Pi, 2 Pi/5}]; star[r_] := Line[(r + #) & /@ Riffle[t1, t2]] Graphics[star /@ dat]


5

plot1 = Plot[{2 x + 4, -6 x - 1}, {x, -2, 4}, PlotLegends -> "Expressions"]; plot2 = Plot[{2 x + 4, -6 x - 1}, {x, -1, 3}, Filling -> {1 -> {2}}]; Show[plot1, plot2]


5

Do not have time to make a full Manipulate, but you can start with Clear["Global`*"] g[t_]:=((1-b) y[t]^2/t^n+1/t^(2+n))^n; ode=y'[t]+3/2 (a-b) y[t]/t+(3/2 a-1)/y[t] t^3==g[t]; pfun=ParametricNDSolveValue[{ode/.{a->3/4,b->5},y[1]==1},y,{t,1,2},{n}] Plot[Evaluate[Table[pfun[n][t],{n,Range[5]}]],{t,1,2}, PlotRange->{Automatic,{-10,10}},...


5

The problem is that you are evaluating Log on all elements of pointsLHTL, not just the x values. The y values are negative so the Log is imaginary. Lm = LinearModelFit[pointsLHTL /. {x_, y_} :> {Log@x, y}, x, x] Lmplot = Plot[Lm[x], {x, -16, -14}, PlotStyle -> Directive[Blue, Dashed], PlotLegends -> {"Linear fit"}] Show[Lmplot, ...


4

Using PolygonMarker: datasets = RandomReal[{0 , 2.5}, {4, 20, 2}]; Graphics[{FaceForm[], EdgeForm[{AbsoluteThickness[1], JoinForm["Miter"]}], EdgeForm[Blue], ResourceFunction["PolygonMarker"]["SixPointedStarSlim", Offset[7], datasets[[1]]], EdgeForm[Red], ResourceFunction["PolygonMarker"]["...


4

Or, simply ListPlot[Prime[Range[10]] -> LabelString[[;; 10]], ImageSize -> 900, Filling -> Axis, PlotTheme -> "Web"] The length of the label list must equal the length of the data list, hence LabelString[[;; 10]]. The option, LabelingFunction, can be used to specify the location of the labels relative to the data points, if ...


4

You need to write"LabelingFunction -> (LabelString[[#[1]]] &)": LabelString = Table[StringForm["Test # ``", LS], {LS, 1, 25}]; ListPlot[Prime[Range[10]], LabelingFunction -> (LabelString[[#[[1]]]] &), Filling -> Axis]


3

Here is some code that should get you going. The positions specify the left of the connection to the elements. Clear["Global`*"] con[pos1_, pos2_] := Line[{pos1, pos2}];(*connections*) switch0[lab_ : ""] := {Line[{{{0, 0}, {0.3, 0}, {0.7, 0.25}}, {{0.7, 0}, {1, 0}}}], Text[lab, {0.5, +0.3}]}; switchh[pos_, lab_ : ""]...


3

Follow the approach by @Bob Hanlon. I think the problem come from the Complex Function EuclideanDistance since it contain Sqrt and Abs. EuclideanDistance[{a, b}, {x, y}] (* Sqrt[Abs[a - x]^2 + Abs[b - y]^2] *) So sometimes I avoid to use EuclideanDistance or Norm. Here we can use #.#& instead. Sqrt[# . # &[{-5, 0} - {x, y}] # . # &[{5, 0} - {x, ...


3

Alternatively, you could use Callout ListPlot[ Callout[Prime[#], StringForm["Test # ``", #], Above] & /@ Range[10], ImageSize -> 900, Filling -> Axis, PlotTheme -> "Web"]


2

Let d be your data. Then d1 = Cases[d, {x_, y_} /; Im[y] > -5] d2 = Cases[d, {x_, y_} /; Im[y] < -5] ListLinePlot[{Re[d1], Re[d2]}] It happens that the Im[y] > -5 criterion works just fine for the data in the OP. In more complicated situations, a more complicated criterion will be needed, but the syntax remains.


2

plt = Plot[x, {x, 0, 25} , Ticks -> { Automatic, {#, DecimalForm[1.0 #, {6, 2}]} & /@ Range[0, 25, 5] } ]


2

$Version "13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021)" Clear["Global`*"] plt = Plot[x, {x, 0, 25}]; Modify the existing Ticks, Show[plt, Ticks -> {Automatic, (Ticks /. AbsoluteOptions[plt, Ticks])[[2]] /. {y_, lbl_String?(StringLength[#] > 0 &), rest_} :> {y, NumberForm[y, {5, 2}], rest}}]


2

We can divide data into two list as follows up = Select[data, #[[2]] >= data[[2, 2]] &]; dw = Select[data, #[[2]] <= data[[1, 2]] &]; Now we can plot {ListPlot[{up, dw}, PlotStyle -> Blue], ListLinePlot[{up, dw}, Mesh -> All, MeshStyle -> Red, PlotRange -> {{1.88, 2.2}, {0.39, 0.42}}, PlotStyle -> Blue]} Update 1. Second ...


2

I figured out a solution: I needed to set the Mesh option to be higher. In this case, Mesh->100 worked fine.


1

Clear[F, G]; F = -((3 Sin[3 x])/(2 x)) - Sqrt[-1 + (Cos[3 x] + (2 Sin[3 x])/ x + (Cos[2 x] - Cos[y]) Csc[2 x] Sin[3 x])^2]; G = Cos[3 x] + ((1 + 2 x Cot[2 x] - x Cos[y] Csc[x] Sec[x]) Sin[ 3 x])/(2 x); Reduce[{F == 0, G == 1, 0 < x <= 3, -1 <= y <= 1}, {x, y}] Reduce[{F == 0, G == -1, 0 < x <= 3, -1 <= y <= 1}, {...


1

pl1 = Plot[Exp[x], {x, 0, 10}, PlotRange -> All] pl2 = Plot[Exp[x], {x, 0, 5} , GridLines -> Automatic , PlotRange -> 50 , Background -> Nest[Lighter, Yellow, 3]] pl3 = Plot[Exp[x] , {x, 0, 10} , PlotRange -> All , GridLines -> Automatic , Epilog -> {Inset[pl2, {2, 10000}, {0, 0}]} ] Here (0,0) of the inset plot is ...


1

You are nearly there. But you misplaced the "Evaluate" command: NumericalParametricPlot1 = Show[ParametricPlot[ Evaluate[ Callout[{q[\[Phi]], q'[\[Phi]]} /. NumericalDiffequaWith, StringForm["Test # ``", i], Above]], {\[Phi], 0, 100}, PlotStyle -> ColorList, PlotLegends -> Placed[LegendWith, {Top, Left}], ...


1

use Rotate on the text directly angles2 = {{0., 0.05922829252191935`}}; labels = {"AAAAAA"}; names = Text[ Style[Rotate[labels[[1]], Mean[angles2[[1]]] - \[Pi]/2], Red, 20], 1.1*{1.0995176865382061`, 0.032570799647401755`}, {0, 0}]; gg = Graphics[{names, Opacity[0.5], Red, Circle[], Rectangle[]}, ImageSize -> 200] Export["xxx.pdf&...


1

Note that your data contains 2 y values for every x value. However, there is a bug, the exception is the 69'th point that does not have a pair. therefore delete point 69: data = d0 = ToExpression /@ Import["d:/Downloads/testdata.csv"]; data = Delete[data, 69]; Now we separate the 2 curves using "Partition" and "Transpose": data ...


1

Straight replacing the way you have done does not honor derivatives. You can do it this way: sl[x_] = -D[p[x]*D[y[x], x], x] - q[x]*y[x] res[x_] = sl[x] /. {p -> (1 &), q[x] -> 0} /. {y -> (Sin[#] &)} (* Sin[x] *) And you will get the plot you expect. If you don't want to use pure functions you can do it this way: sl[x] /. {p -> ...


1

Clear["Global`*"] sl[p_, q_, y_, xv_, var_ : x] := (-D[p*D[y, var], var] - q*y) /. var -> xv p[x_] = 1; q[x] = 0; y[x] = Sin[x]; res[xv_] := sl[p[x], q[x], y[x], xv] Plot[Evaluate@res[x], {x, 0, Pi}] Another example, p[x_] = x^2; q[x] = Sqrt[x]; y[x] = x*Sin[x]; res[xv_] := sl[p[x], q[x], y[x], xv] Note the equivalent use of the option ...


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