(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 156238, 4783]*) (*NotebookOutlinePosition[ 171150, 5326]*) (* CellTagsIndexPosition[ 171106, 5322]*) (*WindowFrame->Normal*) Notebook[{ Cell["Mathematica Primer (the first one.)", "Title", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "This notebook has been written in ", StyleBox["Mathematica ", FontSlant->"Italic"], "by \n\nMark J. McCready\nProfessor and Chair of Chemical Engineering\n\ University of Notre Dame\nNotre Dame IN 46556\nUSA\n\n", ButtonBox["mjm@nd.edu", ButtonData:>{ URL[ "mailto:mjm@nd.edu"], None}, ButtonStyle->"Hyperlink"], "\n", ButtonBox["http://www.nd.edu/~mjm/", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/"], None}, ButtonStyle->"Hyperlink"], "\n\n\nIt is copyrighted to the extent allowed by whatever laws pertain to \ the World Wide Web and the Internet.\n\nI would hope that as a professional \ courtesy that this notice remain visible to other users. \nThere is no charge \ for copying and dissemination \n\nVersion: 8/20/00\nMore recent versions \ of this notebook should be available at the web site:\n", ButtonBox["http://www.nd.edu/~mjm/", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None}, ButtonStyle->"Hyperlink"], StyleBox[ButtonBox["Mathematicaprimer.1", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None}, ButtonStyle->"Hyperlink"], FontSlant->"Italic"], ButtonBox[".nb", ButtonData:>{ URL[ "http://www.nd.edu/~mjm/dimensionless.nb"], None}, ButtonStyle->"Hyperlink"] }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Basic ", StyleBox["Mathematica", FontSlant->"Italic"], " commands" }], "Subtitle"], Cell[CellGroupData[{ Cell["Arithmetic", "Subsubsection"], Cell["\<\ A system of such colossal power probably can do simple \ mathematics\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2 + 2\)], "Input"], Cell[BoxData[ \(TraditionalForm\`4\)], "Output"] }, Open ]], Cell["\<\ Note that for multiplication, there is no need for a sign. Just \ use a space.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(3\ 4\)], "Input"], Cell[BoxData[ \(TraditionalForm\`12\)], "Output"] }, Open ]], Cell["Don't forget the space.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(34\)], "Input"], Cell[BoxData[ \(TraditionalForm\`34\)], "Output"] }, Open ]], Cell["You can use a * if you wish.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(3*4\)], "Input"], Cell[BoxData[ \(TraditionalForm\`12\)], "Output"] }, Open ]], Cell["This one is fun", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(35/17\)], "Input"], Cell[BoxData[ \(TraditionalForm\`35\/17\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " does not switch to floating point, which is inherently less precise, \ unless you want it to.\n\nWe could do N[%], which says make the previous \ result floating point" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(N[%]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`2.0588235294117645`\)], "Output"] }, Open ]], Cell["Or", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(N[35/17]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`2.0588235294117645`\)], "Output"] }, Open ]], Cell["It gives", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2^10\)], "Input"], Cell[BoxData[ \(TraditionalForm\`1024\)], "Output"] }, Open ]], Cell["or", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(3^19\)], "Input"], Cell[BoxData[ \(TraditionalForm\`1162261467\)], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Vectors and Matrices", "Subsubsection"]], \ "Subsubsection"], Cell[CellGroupData[{ Cell["A vector is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({a, b, c}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{a, b, c}\)], "Output"] }, Open ]], Cell["Another is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({d, e, f}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{d, e, f}\)], "Output"] }, Open ]], Cell["Take the \"dot\" product of these.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({a, b, c}\ \ . {d, e, f}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`a\ d + b\ e + c\ f\)], "Output"] }, Open ]], Cell["\<\ This looks good for the scalar multiplication of vectors. We can \ also do:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Dot[{a, b, c}, {d, e, f}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`a\ d + b\ e + c\ f\)], "Output"] }, Open ]], Cell["One kind of vector product is:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Cross[{a, b, c}, {d, e, f}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{b\ f - c\ e, c\ d - a\ f, a\ e - b\ d}\)], "Output"] }, Open ]], Cell["\<\ The normal cross product can also be obtained from the determinant \ of a matrix as you will see below.\ \>", "Text"], Cell["You can also do", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({a, b, c}\ {d, e, f}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{a\ d, b\ e, c\ f}\)], "Output"] }, Open ]], Cell["\<\ But this multiplication does not arise in any physical problems \ that we will be considering this semester. It could be useful for numerical \ solution techniques. \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "We would like to make a ", StyleBox["matrix", FontWeight->"Bold", FontSlant->"Italic"], " that ", StyleBox["Mathematica", FontSlant->"Italic"], " can understand. Use an opening {{ then close each row with } and then \ open the next with { ....... after the last row you obviously need a closing \ }. The one that is defined here has the name \"a\"" }], "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(a = {{1, 2, 3}, {7, 5, 6}, {1, 6, 9}}\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "2", "3"}, {"7", "5", "6"}, {"1", "6", "9"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(a + a\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"2", "4", "6"}, {"14", "10", "12"}, {"2", "12", "18"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "The matrix ", StyleBox["a", FontSlant->"Italic"], " is a 3 by 3 square matrix. \n\nA matrix could correspond to many \ different physical entities. For example a 3 by 3 could be the elements in a \ second order tensor which corresponds to the stresses in a fluid at a point \ and some time. Of course, if this were the case, the matrix would be \ symmetric for all simple fluids. (Yes, this course will be complicated at \ times!)\n\nHow can we get a symmetric matrix from ", StyleBox["a", FontSlant->"Italic"], " and what is \"symmetric\". \n\nFirst define transpose, which is a \ switching of a[i,j] --> a[j,i]" }], "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(Transpose[a]\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "7", "1"}, {"2", "5", "6"}, {"3", "6", "9"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ You see that we left the diagonal elements unchanged and \ \"reflected\" the off diagonal elements across the diagonal. For example \ elements a[i,j] , where i is row and j is column were transformed to a[ \ j,i]\ \>", "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(a\[LeftDoubleBracket]1, 2\[RightDoubleBracket]\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ \(TraditionalForm\`2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(a\[LeftDoubleBracket]2, 1\[RightDoubleBracket]\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ \(TraditionalForm\`7\)], "Output"] }, Open ]], Cell[TextData[{ "Now we can easily get the symmetric part of ", StyleBox["a", FontSlant->"Italic"], " (such that a[i,j]=a[j,i] and (of course) its antisymmetric piece." }], "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(asymm = 1\/2\ \((a + Transpose[a])\)\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", \(9\/2\), "2"}, {\(9\/2\), "5", "6"}, {"2", "6", "9"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(aanti = 1\/2\ \((a - Transpose[a])\)\)], "Input", CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"0", \(-\(5\/2\)\), "1"}, {\(5\/2\), "0", "0"}, {\(-1\), "0", "0"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ We see that the antisymmetric part does not have any diagonal \ elements because a[i,i] cannot be equal to -a[i,i]. Thus our definition of \ an anti symmetric matrix is that a[i,j]=-a[j,i]. \ \>", "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell["\<\ Symmetric and antisymmetric matrixes are useful, for example, in \ defining the basic elements of fluid motion in a study of \"kinematics\". It \ is convenient that\ \>", "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[BoxData[ \(asymm + aanti\)], "Input", CellLabelAutoDelete->True], Cell[BoxData[ FormBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "2", "3"}, {"7", "5", "6"}, {"1", "6", "9"} }, ColumnAlignments->{Decimal}], "\[NoBreak]", ")"}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "thus we have recovered", StyleBox[" a", FontSlant->"Italic"], " again." }], "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True], Cell[TextData[{ "We need to clear the value of ", StyleBox["a", FontSlant->"Italic"], " because we use it again below." }], "Text"], Cell[BoxData[ \(Clear[a]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(a\)], "Input"], Cell[BoxData[ \(TraditionalForm\`a\)], "Output"] }, Open ]], Cell["The determinant is a standard operation. You recall", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Det[{{a, b}, {c, d}}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`a\ d - b\ c\)], "Output"] }, Open ]], Cell["\<\ As mentioned above, the cross product can be defined with the \ determinant\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Det[{{ex, ey, ez}, {a, b, c}, {d, e, f}}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(-c\)\ e\ ex + b\ f\ ex + c\ d\ ey - b\ d\ ez + a\ e\ ez - a\ ey\ f\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \({Coefficient[%, ex], Coefficient[%, ey], Coefficient[%, ez]}\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{b\ f - c\ e, c\ d - a\ f, a\ e - b\ d}\)], "Output"] }, Open ]], Cell["\<\ Where we had to work a little too hard with the {ex,ey,ez} being a \ unit vector and the Coefficient being used to recover the components of each \ direction. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Cross[{a, b, c}, {d, e, f}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`{b\ f - c\ e, c\ d - a\ f, a\ e - b\ d}\)], "Output"] }, Open ]], Cell["\<\ From these operations we see that addition and subtraction of \ matrices is done by simple addition and subtraction of corresponding \ elements. \ \>", "Text", Evaluatable->False, CellLabelAutoDelete->True, AspectRatioFixed->True] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Making plots, special functions and Series expansion", "Subsubsection"], Cell["\<\ It knows all of the functions, including yours and my favorite \ orthogonal polynomials.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Plot[ChebyshevT[3, y], {y, \(-1\), 1}];\)\)], "Input"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: .61803 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations 0.5 0.476191 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Cell["\<\ You have already figured out that if the first number in the \ argument, is odd, the polynomial is an odd function. If it is even, the \ polynomial is an even function. \ \>", "Text"], Cell["From math you may already recall:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_\(-1\)\%1\( ChebyshevT[2, y]\ ChebyshevT[1, y]\ 1\/\@\(1 - y\^2\)\) \[DifferentialD]y\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(2/\[Pi]\ Integrate[ ChebyshevT[8, y]\ ChebyshevT[3, y]\ 1\/\@\(1 - y\^2\), {y, \(-1\), 1}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0\)], "Output"] }, Open ]], Cell["Was this magic?", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2/\[Pi]\ Integrate[ ChebyshevT[4, y]\ ChebyshevT[7, y]\ 1\/\@\(1 - y\^2\), {y, \(-1\), 1}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`0\)], "Output"] }, Open ]], Cell["This is also magic.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(2/\[Pi]\ Integrate[ ChebyshevT[4, y]\ ChebyshevT[4, y]\ 1\/\@\(1 - y\^2\), {y, \(-1\), 1}]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`1\)], "Output"] }, Open ]], Cell["One of the most fun things to do is expand things.", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Series[BesselJ[0, x], {x, 0, 5}]\)], "Input"], Cell[BoxData[ FormBox[ InterpretationBox[ RowBox[{"1", "-", \(x\^2\/4\), "+", \(x\^4\/64\), "+", InterpretationBox[\(O(x\^6)\), SeriesData[ x, 0, {}, 0, 6, 1]]}], SeriesData[ x, 0, {1, 0, Rational[ -1, 4], 0, Rational[ 1, 64]}, 0, 6, 1]], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ A secret command that you always need with \"Series\" is \ \"Normal\". Normal turns the expression with the error term into just a \ regular expression.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(b5 = Normal[Series[BesselJ[0, x], {x, 0, 5}]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`x\^4\/64 - x\^2\/4 + 1\)], "Output"] }, Open ]], Cell["Try some other lengths", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(b3 = Normal[Series[BesselJ[0, x], {x, 0, 3}]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`1 - x\^2\/4\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(b9 = Normal[Series[BesselJ[0, x], {x, 0, 9}]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`x\^8\/147456 - x\^6\/2304 + x\^4\/64 - x\^2\/4 + 1\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(b13 = Normal[Series[BesselJ[0, x], {x, 0, 13}]]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`x\^12\/2123366400 - x\^10\/14745600 + x\^8\/147456 - x\^6\/2304 + x\^4\/64 - x\^2\/4 + 1\)], "Output"] }, Open ]], Cell["Do you see the pattern?", "Text"], Cell[CellGroupData[{ 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An equal sign in ", StyleBox["Mathematica", FontSlant->"Italic"], " means the thing on the left is replaced by the thing on the right -- when \ ever you type the thing on the left. \n\nFor example" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(b = 6\)], "Input"], Cell[BoxData[ \(TraditionalForm\`6\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(b\)], "Input"], Cell[BoxData[ \(TraditionalForm\`6\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(b^2\)], "Input"], Cell[BoxData[ \(TraditionalForm\`36\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Sqrt[b]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\@6\)], "Output"] }, Open ]], Cell[TextData[{ "Since we might use the symbol b elsewhere here or in another open ", StyleBox["Mathematica", FontSlant->"Italic"], " notebook, I need to unassign the value of 6 to b. 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1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3)\) + 1\/2\ \@\(41\/3 - 2951\/\(6\ \@\(\(-571157\) + 300\ \@3910209\ \)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\)\ \[Sqrt]\((82\/3 + 2951\/\(6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\) - 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3 - 50\/\@\(41\/3 - 2951\/\(6\ \@\(\(-571157\) + 300\ \ \@3910209\)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\))\) + 1\/4\ \((\(-\(82\/3\)\) - 2951\/\(6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3 + 50\/\@\(41\/3 - 2951\/\(6\ \@\(\(-571157\) + 300\ \ \@3910209\)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\))\))\), y \[Rule] 1\/2\ \@\(41\/3 - 2951\/\(6\ \@\(\(-571157\) + 300\ \ \@3910209\)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\) - 1\/2\ \[Sqrt]\((82\/3 + 2951\/\(6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\) - 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3 - 50\/\@\(41\/3 - 2951\/\(6\ \@\(\(-571157\) + 300\ \ \@3910209\)\%3\) + 1\/6\ \@\(\(-571157\) + 300\ \@3910209\)\%3\))\)}, {x \ \[Rule] 2, z \[Rule] 1\/5\ \((\(-4\) + 1\/4\ \((\(-3\) - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\))\) - 1\/2\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\)\ \ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\ \%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) - \ 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\))\) + 1\/4\ \((\(-6\) + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) + 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\))\), y \[Rule] \(-\(1\/2\)\)\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\ \/\(2\ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\) - 1\/2\ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\)}, {x \[Rule] 2, z \[Rule] 1\/5\ \((\(-4\) + 1\/4\ \((\(-3\) - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\))\) + 1\/2\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\)\ \ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\ \%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) - \ 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\))\) + 1\/4\ \((\(-6\) + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) + 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\))\), y \[Rule] \(-\(1\/2\)\)\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\ \/\(2\ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\) + 1\/2\ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\)}, {x \[Rule] 2, z \[Rule] 1\/5\ \((\(-4\) + 1\/4\ \((\(-3\) - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\))\) + 1\/4\ \((\(-6\) + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\) - 1\/2\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\)\ \ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\ \%3\) + 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\) - \ 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\))\))\), y \[Rule] 1\/2\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ 3\^\(2/3\)\ \) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\)\) + 1\/2\ \[Sqrt]\((6 - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) + 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\)}, {x \[Rule] 2, z \[Rule] 1\/5\ \((\(-4\) + 1\/4\ \((\(-3\) - \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) + 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\))\) + 1\/4\ \((\(-6\) + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \@1085073)\)\)\%3\) - 50\/\@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 1517\/\(2\ \@\(3\ \((\(-19431\) + 100\ \ \@1085073)\)\)\%3\)\))\) + 1\/2\ \@\(3 + \@\(\(-19431\) + 100\ \@1085073\)\%3\/\(2\ \ 3\^\(2/3\)\) - 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