# Tag Info

## Hot answers tagged output-formatting

13

Citing this 1996 year MathGroup post by Dave Wagner, A similar function, Continuation[n], is called to format the character at the beginning of continued lines. So, for example, the definition Format[Continuation[n_]] := StringForm["()",n] would place a line number inside of parentheses on the second and subsequent lines of a multi-line ...

7

After a quick Google search I came across this, which provides a description. Translation: Continuation[n] is the result at the beginning of the nth line in a multiline printed Expression. The default value of Continuation is either the string " " or the string ">". The value of Continuation is can be changed with the command Format[Continuation[n_]]:=...

7

\[Continuation] is also the name of the special character used to indicate that input should be interpreted as continuing in the next line, or to break output too long for the current window. I suspect Continuation[] to be the operator form of that infix symbol. On second thought, I suspect that the formatting references you found are linked to special ...

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[... 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 Here is a simple and universal function which formats timings (for real-time monitoring of elapsed time one should replace Round with Floor): formatTiming = StringJoin[{If[# >= 100, ToString@#, IntegerString[#, 10, 2]] &@Floor[#/3600], ":", IntegerString[Floor[Mod[#, 3600]/60], 10, 2], ":", IntegerString[Mod[#, 60], 10, 2]} &@... 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 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 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 == -...

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 FrontEndExportPacket 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[FrontEndExportPacket[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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