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5

It's a bug in TeXForm. It shows up when there are exponents in the argument for Abs[], suggesting that it adds \left and \right when it thinks it needs to to make the absolute value bars bigger for the exponent, not realizing that Abs[] already generated \left| and \right|. This applies the equation generation, the TeXForm, and a workaround for the bug to ...


5

Borrowing some of the display-related code for TransformationFunction[], I came up with this: MakeBoxes[PartitionedMatrixForm[M_, {p_Integer, q_Integer}], fmt_] ^:= Module[{mat}, mat = If[Unevaluated[M] =!= {{}} && MatrixQ[Unevaluated[M]] && And @@ MapThread[0 <= Abs[#2] <= #1 - 1 &, {Dimensions[M], {p,...


5

You can convert your Text to Graphics and then use any size Show[ImageCrop@Graphics@Text@Style["Text Text", 80, Bold, FontFamily -> Times], ImageSize -> 200] You can change the font size to get a better resolution. With a little modification you can use it inside a canvas in the way you want. insert[txt_String, col_, pos_List, angle_, ...


4

You can convert text to FilledCurve graphics primitives by converting via PDF. See here for example. In the example below I get the bounding box of the resulting graphics using PlotRange, then use Scale and Translate to size and position the text as required. t = Style["Text Text", FontFamily -> "Times New Roman", 60]; g = ImportString[ExportString[t, "...


4

I think you need the SignPadding option: NumberForm[-60., {3, 1}, NumberPadding -> {" ", "0"}, NumberSigns -> {{"(", ")"}, ""}, SignPadding -> True] (60.0)


3

Use Show to combine a ListPlot of the data and Plot of the fit functions. If the data is listed in the same order as the fit functions then the data will match the fit color. dataA = {{0, 21}, {4, 18}, {9, 10}}; dataB = {{0, 45}, {5, 44}, {10, 38}}; dataC = {{0, 32}, {6, 22}, {11, 26}}; fitA = LinearModelFit[dataA, x, x]; fitB = LinearModelFit[dataB, x, x];...


3

The following also works. It doesn't require any string replacements in the output. All I did is to re-define the set of variables u such that it contains the second power of x as a String: v={5,-13,5,6,-20,17}; u={"\!\(\*SuperscriptBox[\(x\), \(2\)]\)",x,1}; Abs[{v[[1]],v[[2]],v[[3]]}.u]=={v[[4]],v[[5]],v[[6]]}.u//TeXForm \left| 5 x^2-13 x+5\right| =6 ...


3

NumberForm[f, {∞, n}] with n being the number of digits beyond the decimal separator


3

You can manually apply a style to each heading: GM = TableForm[GraphM, TableHeadings -> {{Style["v=1", FontSize -> 18], Style["v=.5", FontSize -> 18], Style["v=.25", FontSize -> 18], Style["v=.125", FontSize -> 18]}, {Style["k=.4", FontSize -> 18], Style["k=.2", FontSize -> 18], Style["k=.1", FontSize -> 18], Style["k=....


2

I've edited my answer in the linked thread so that it can now be used without modification. Previously, you had to define the functions uv that you wish to differentiate in a more general way, replacing the explicit 0 in their argument with y. The reason is that my earlier answer assumed differentiations are performed on the given function, not on the new ...


2

Based on the findings from this and this answers, I can suggest the following approach. Let us make a histogram: SeedRandom[10]; hist = Histogram[RandomVariate[NormalDistribution[0, 1], 1000]] It has AspectRatio -> 1/GoldenRatio (the default): Options[hist, AspectRatio] {AspectRatio -> 1/GoldenRatio} Now we can place this histogram as ...


1

Your code contains numerous syntax errors. The first formula has no sense in Mma. I tried to rewrite its part that seems reasonable: expr = 10* Log[10, Re[(1/Subscript[R, p] + 1/(2 \[Nu]*\[Pi]*L*I) + 2 \[Pi]*\[Nu]*Subscript[C, p]*I)^-1]] + Subscript[P, 0]; Then you may do the following: TraditionalForm[expr] There are also multiple other ways ...


1

Let,s say this is your input list = {{-1, 1}, {-1, 2}, {0, 3}, {{1, -1}, {20, 12}}, {2, 4}, "0.01*Round[{}+{-1,-1}]"} First replace the String list = list /. x__ /; StringQ[x] -> {0, 0} Join[{list[[1]]}, Table[s = list[[i]]; If[Length[Flatten@s] > 2, d = Min[{Norm[s[[#]] - list[[i - 1]]], Norm[s[[#]] - list[[i + 1]]]}]...


1

First, I'll have to admit that those data should be modified a bit: wd2="56个7"; dat={{#1[[1]], -#1[[2]]}, #2} & @@@ Map[Reverse, **Those data**, {2}]; Then, just following @Sumit's answer, we'll have to use Inset to do this job, but we need to further specify the alignment by using Inset's Third options. Then, we also need to use the size option(the ...



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