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5

You can use Trace with TraceDepth option set to 1 to get evaluation steps giving whole expression, and format result as you want it. Function performing this actions can be assigned to $Pre to be automatically used for all inputs. ClearAll[showSetSteps] SetAttributes[showSetSteps, HoldAllComplete] showSetSteps[Set[lhs_, rhs_]] := With[{trace = Replace[...


4

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. For any arbitrary rotation text[txt_String, angle_,size_] := Show[ImageCrop@ Graphics@ Rotate[Text@Style[txt, 80, Bold, ...


3

Original answer The reason for getting extra quotes is that these extra quotes are explicitly present in the box form of the expression generated by such functions as EngineeringForm, NumberForm etc.: ToBoxes@EngineeringForm[6.08717*10^6] TagBox[InterpretationBox[ RowBox[{"\"6.08717\"", "\[Times]", SuperscriptBox["10", "\"6\""]}], 6.08717*10^6, ...


2

I will use Eq. 22.7 of your reference $ v = 2 r + 4 m \ln(|r - 2 m|) + B$ The obliques are, I guess is given by $v=-u$. For scaling I am considering $u=2mr$. m = 1/2; Show[ContourPlot[Evaluate@Table[ v == 2 r + 4 m Log[Abs[r - 2 m]] + n, {n, -5, 5}], {r,0,4 m}, {v,-3m,3m}, ContourStyle -> Blue, AspectRatio -> 3/2], ContourPlot[Evaluate@Table[v == -...


2

Have a look at TeXTableForm.m: Converting Mathematica Lists to LaTeX Tables it is working quite well: t = Table[i, {i, 1, 22}] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, \ 20, 21, 22} TeXTableForm[t, 5, "tab1.tab"] "tab1.tab" Import[".../tab1.tab"]


2

One way to write it is to let $$f(x) = \frac{z \log (1-x)+(1-2 z) \log (x)+(z-1) \log (x z-x+z)}{z (2 z-1)}$$ and write $f^{(-1)}\left(\frac{z \log (1-{a_0})-2 z \log ({a_0})+z \log ({a_0} z-{a_0}+z)-\log ({a_0} z-{a_0}+z)+\log ({a_0})}{z (2 z-1)}-\frac{t}{2}\right)$ for InverseFunction[..][..]. So in regular mathematical notation, the ...


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 ...


2

This changes E**argument to exp(argument): Unprotect[Power]; Power /: Format[Power[E, x_], FortranForm] := exp[x] Protect[Power]; {FortranForm[E^(3*z)], FortranForm[Exp[2*z^3]]} (* Out: {exp(3*z), exp(2*z**3)} *)


2

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=....


1

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 ...


1

I may have found a nice way. However, I think the output is a bit ugly. Here's the working code : NullCurve1[n_] := NDSolve[{ u'[r] == 2/(1 - 1/r), u[0] == 2 n }, {u}, {r, 0, 1}, Method -> Automatic, MaxSteps -> 100000 ] NullCurve2[n_] := NDSolve[{ u'[r] == 2/(1 - 1/r), u[5] == 2 n }, {u}, {r, 1, 10}, Method -&...


1

Using undocumented FrontEnd`ExportPacket command you can get exactly the same formatting as with Copy As ► Plain Text (i.e. without the line breaks and extra spaces): s = First[FrontEndExecute[FrontEnd`ExportPacket[Cell[BoxData[ToBoxes[a]]], "PlainText"]]] {-11 \[Psi]^2 \[Lambda][1]+6 \[Psi] \[Lambda][2],35 \[Psi]^2 \[Lambda][2],11 \[Psi]^2 \[Lambda][1]...



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