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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[ 1252203, 18092]*) (*NotebookOutlinePosition[ 1253010, 18120]*) (* CellTagsIndexPosition[ 1252966, 18116]*) (*WindowFrame->Normal*) Notebook[{ Cell["\<\ Cubic Equations-Their Presence, Importance, and Applications, in the Age of \ Technology\ \>", "Title"], Cell["\<\ Robert Kreczner Department of Mathematics and Computing University of Wisconsin Stevens Point, WI 54481 Email: rkreczne@uwsp.edu Telephone: (715) 346-3754 Fax: (715) 346-4260 \ \>", "Author"], Cell[TextData[{ StyleBox["ABSTRACT", FontWeight->"Bold"], ": I in this article we will attempt, with the help of modern mathematical \ technology, to revive an interest and research among teachers and students in \ cubic and higher order equations. The article, written in attractive ", StyleBox["Mathematica's", FontSlant->"Italic"], " notebook format, basically comprises two themes: It uses ", StyleBox["Mathematica ", FontSlant->"Italic"], "3.0 to discover simple and natural methods, algebra and calculus based \ methods, for solving exactly a cubic equation. Once the idea is discover, \ then the computation can be easily carried out by a hand, by anyone with \ standard algebra or calculus skills. In the second part we will show variety \ of different problems whose solution requires solving a non-trivial cubic \ equation. In that part we will take advantage of ", StyleBox["Mathematica", FontSlant->"Italic"], " to do the computation. " }], "Text"], Cell[TextData[{ StyleBox[ "The problem that infected me with such virulence was actually of little \ significance, and even lesser consequence. It concerned solving cubic \ equations, and the answer had been known since Cardano published it in 1545. \ What I did not know was how to drive it.\nThe sages who had designed the \ mathematics curricula for secondary schools had stopped at solving quadratic \ equations...\n \n \ ", FontSlant->"Italic"], "Mark Kac", StyleBox[ "\n \ How I Became a Mathematician\n \ American Scientist, ", FontSlant->"Italic"], "Sep/Oct 1984\n " }], "Text", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "They were very upset when I said the development of the greatest \ importance to mathematics in Europe was the discovery by Tartaglia that you \ can solve a cubic equation: although it is of little use in itself, the \ discovery must have been psychologically wonderful. It therefore helped in \ the Renaissance, which was freeing man from the intimidation of ancients. \ What the Greeks are learning in the school is to be intimidated into thinking \ they have fallen so far bellow their ancestors.\n\n", StyleBox[ "Richard Feynman, physicist,\ncomments about educational attitudes he had \ met during his visit to Greece in 1980.", FontSlant->"Plain"], "\nWhat Do you Care What Other People Think?, ", StyleBox["Unwin, 1988", FontSlant->"Plain"] }], "Text", TextAlignment->Center, FontSlant->"Italic"], Cell[CellGroupData[{ Cell["Discovering Solution of Cubic Equation ", "Section"], Cell[CellGroupData[{ Cell["Taking Advantage of Traditional Notation", "Subsubsection"], Cell[TextData[{ "We will restrict ourselves, as usually in the theory of cubic equations, \ to the equations of the form ", Cell[BoxData[ \(x\^3\)], FontSlant->"Italic"], StyleBox["+px +q=0. ", FontSlant->"Italic"], "For any general cubic equation ", Cell[BoxData[ \(a\ x\^3 + b\ x\^2\ + cx + d = 0\)]], " can be easily converted to the previous one by the means of the \ transformation ", Cell[BoxData[ \(x = y - b\/\(3 a\)\)]], "." }], "Text", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "For example: Consider the equation ", Cell[BoxData[ \(5\ x\^3 + 4\ x\^2 - 2\ x + 3 = 0\)]], " , and apply substitution ", Cell[BoxData[ \(x = y - 4\/\(3*\ 5\)\)]] }], "Example"], Cell[CellGroupData[{ Cell[BoxData[ \(5\ x\^3 + \ 4\ x\^2 - 2\ x + 3 /. x -> y - 4\/\(3*\ 5\)\)], "Input"], Cell[OutputFormData["\<\ 3 - 2*(-4/15 + y) + 4*(-4/15 + y)^2 + 5*(-4/15 + y)^3\ \>", "\<\ 4 4 2 4 3 3 - 2 (-(--) + y) + 4 (-(--) + y) + 5 (-(--) + y) 15 15 15\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell[OutputFormData["\<\ 2513/675 - (46*y)/15 + 5*y^3\ \>", "\<\ 2513 46 y 3 ---- - ---- + 5 y 675 15\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Solve a few equations by using ", FontSize->16, FontWeight->"Bold"], StyleBox["Mathematica'", FontSize->16, FontWeight->"Bold", FontSlant->"Italic"], StyleBox[ "s command Solve, and make an observation about their solution \ structure. To make this observation easier, use the traditional output \ format. For example: ", FontSize->16, FontWeight->"Bold"] }], "Text", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[2513\/675 - \(46\ y\)\/15 + 5\ y\^3 == 0, y]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`{{y \[Rule] \(-\(1\/15\)\)\ \@\(1\/2\ \((2513 - 15\ \@26337)\)\)\%3 + \(-\(46\/15\)\)\ \@\(2\/\(2513 - 15\ \@26337\)\)\%3}, { y \[Rule] 1\/30\ \((1 - \[ImaginaryI]\ \@3)\)\ \@\(1\/2\ \((2513 - 15\ \@26337)\)\)\%3 + 23\/15\ \((\[ImaginaryI]\ \@3 + 1)\)\ \@\(2\/\(2513 - 15\ \@26337\)\)\%3}, { y \[Rule] 1\/30\ \((\[ImaginaryI]\ \@3 + 1)\)\ \@\(1\/2\ \((2513 - 15\ \@26337)\)\)\%3 + 23\/15\ \((1 - \[ImaginaryI]\ \@3)\)\ \@\(2\/\(2513 - 15\ \@26337\)\)\%3}}\)], "Output"] }, Open ]], Cell[BoxData[ \(Solve[x\^3 - 3\ x - 4 == 0, x]\)], "NumberedEquation", TextAlignment->Left, TextJustification->0], Cell[BoxData[ \(TraditionalForm \`{{x \[Rule] \@\(\@3 + 2\)\%3 + 1\/3\ \@\(54 - 27\ \@3\)\%3}, { x \[Rule] \(-\(1\/2\)\)\ \((\((1 - \[ImaginaryI]\ \@3)\)\ \@\(\@3 + 2\)\%3)\) - 1\/6\ \@\(54 - 27\ \@3\)\%3\ \((\[ImaginaryI]\ \@3 + 1)\)}, { x \[Rule] \(-\(1\/2\)\)\ \((\((\[ImaginaryI]\ \@3 + 1)\)\ \@\(\@3 + 2\)\%3)\) - 1\/6\ \@\(54 - 27\ \@3\)\%3\ \((1 - \[ImaginaryI]\ \@3)\)}}\)], "Output"], Cell[TextData[{ StyleBox[ " We can observe that a solution, especially the real one, is of a form\n \ ", FontColor->RGBColor[1, 0, 0]], Cell[BoxData[ \(x = \@u\%3 + \@v\%3\)], FontColor->RGBColor[1, 0, 0]], StyleBox[".", FontColor->RGBColor[1, 0, 0]], "\n We will use this fact, to find a simple and natural way for solving a \ cubic equation. We will illustrate this idea by solving the equation (1). The \ equation (1) we will rewrite as \n", Cell[BoxData[ \(x = \@\(3 x + 4\)\%3\)]], ", \ \nThen set up, as observed, \n", Cell[BoxData[ RowBox[{ StyleBox["x", FontWeight->"Plain"], StyleBox["=", FontWeight->"Bold"], \(\@u\%3 + \@v\%3\)}]], "NumberedEquation", TextAlignment->Left, TextJustification->0], StyleBox[" ", "NumberedEquation"], "\nand\n", Cell[BoxData[ \(\@\(3 x + 4\)\%3\)]], "=", Cell[BoxData[ \(\@u\%3 + \@v\%3\)]], ".\nBy cubing both sides, we get \n", Cell[BoxData[ \(\(\ \ x\ + 4 = u + v + 3 \(\@\( u\ v\)\%3\) \((\@u\%3 + \@v\%3)\)\)\)]], ",\nbut ", Cell[BoxData[ \(3 x + 4\)]], ", by using again the fact that ", Cell[BoxData[ RowBox[{ StyleBox["x", FontWeight->"Plain"], StyleBox["=", FontWeight->"Bold"], \(\@u\%3 + \@v\%3\)}]]], ", can be expressed as \n", Cell[BoxData[ \(3\ x + 4 = 4 + 3 \((\@u\%3 + \@v\%3)\)\)]], ".\nThus,\n", Cell[BoxData[ \(4 + 3 \((\@u\%3 + \@v\%3)\) = u + v + 3 \(\@\( u\ v\)\%3\) \((\@u\%3 + \@v\%3)\)\)]], ",\nand so, the solution of the cubic equation is reduced to a solution of \ this system of equations\n", Cell[BoxData[ \(u + v = 4, \ and\ \ 3 \@\( u\ v\)\%3 = 1\)]], ",\nthat is \n ", Cell[BoxData[ \(u + v = 4, \ and\ \ u\ v = 1\)]], " \n The last system of equations can be reduced to the quadratic equation \ \n ", Cell[BoxData[ \(u\^2 - 4 u + 1 = 0\)]], "\n which has two real solutions, however the expression ", Cell[BoxData[ \(\@u\%3 + \@v\%3\)]], " won't be affected by the choice (verify this). Thus we can take as ", Cell[BoxData[ \(u = 2 - \@3\)]], ", and ", Cell[BoxData[ \(v = 2 + \@3\)]], ", and hence our solution to the equation (1) is \n", Cell[BoxData[ \(x = \@\(2 - \@3\)\%3 + \@\(2 + \@3\)\%3\)]], ", \n which can be verified that is \ the same as the one obtained by ", StyleBox["Mathematica.", FontSlant->"Italic"], " " }], "Example", TextAlignment->Left, TextJustification->0], Cell[TextData[{ "In this example we will consider the equation with all three real roots. \ ", Cell[BoxData[ \(\(\ t\^3 + 6 t^2 + 6 t - 3 = 0\)\)]], ".\nComplete cube\n", Cell[BoxData[ \(\(\ t + 2)\)\^3 - 6 \((t + 2)\) + 1 = 0\)]], ",\nand rewrite it as\n", Cell[BoxData[ \(\((t + 2)\)\^3 = 6 \((t + 2)\) - 1. \)]], "\nSet up \n", Cell[BoxData[ \(\(\ \(t + 2 = \@u\%3 + \@v\%3, \)\)\)]], "\n Then,\n", Cell[BoxData[ \(\(\ 6 \((t + 2)\) - 1 = u + v + 3 \(\@\( u\ v\)\%3\) \((\@u\%3 + \@v\%3)\)\)\)]], ",\n", Cell[BoxData[ \(6 \((t + 2)\) - 1 = \(-1\) + 6 \(\((\@u\%3 + \@v\%3)\) . \)\)]], "\nSolve the system of equations\n ", Cell[BoxData[ \(u + v = \(-1\), 3 \@\( u\ v\)\%3 = 6\)]], "\nIts solution is\n", Cell[BoxData[ \(TraditionalForm \`{{u \[Rule] 1\/2\ \((\(-\[ImaginaryI]\)\ \@31 - 1)\), v \[Rule] 1\/2\ \((\[ImaginaryI]\ \@31 - 1)\)}, { u \[Rule] 1\/2\ \((\[ImaginaryI]\ \@31 - 1)\), v \[Rule] 1\/2\ \((\(-\[ImaginaryI]\)\ \@31 - 1)\)}}\)]], "\nRepresent u and v in trigonometric form.\n", Cell[BoxData[ \(u = 2\ \@2\ \(( cos \((\(tan\^\(-1\)\) \((\@31)\) - \[Pi])\) + \[ImaginaryI]\ \(sin \((\(tan\^\(-1\)\) \((\@31)\) - \[Pi])\)\)) \)\)]], "\n", Cell[BoxData[ RowBox[{"v", "=", FormBox[ RowBox[{"2", " ", \(\@\(\ \)\), " ", RowBox[{"(", RowBox[{\(cos(\[Pi] - \(tan\^\(-1\)\)(\@31))\), "+", RowBox[{ FormBox["\[ImaginaryI]", "TraditionalForm"], "cos", \(sin(\[Pi] - \(tan\^\(-1\)\)(\@31))\)}]}], ")"}]}], "TraditionalForm"]}]]], Cell[BoxData[ \(\n\)]], "\nCompute the first set of cube roots of u and v.\n", Cell[BoxData[ \(\@u\%3 = \(\@\(2\ \@2\)\%3\) \((cos \(( \(tan\^\(-1\)\) \((\(\(\@31)\) - \[Pi]\)\/3)\) + \[ImaginaryI]\ sin \((\(tan\^\(-1\)\) \((\(\(\@31)\) - \[Pi]\)\/3)\)) \)\)\)\)]], "\n", Cell[BoxData[ \(\@v\%3 = \(\@\(2\ \@2\)\%3\) \((cos \((\(\[Pi] - \(tan\^\(-1\)\) \((\@31)\)\)\/3)\) + \[ImaginaryI]\ sin \((\(\[Pi] - \(tan\^\(-1\)\) \((\@31)\)\)\/3)\))\)\)]], "\nSince the above complex numbers are conjugate to each other, their sum \ is real. Thus, the solution of our equation is\n", Cell[BoxData[ \(\(t + 2 = \)\)]], "2 ", Cell[BoxData[ \(\(\@\(2\ \@2\)\%3\) \((cos \((\(\[Pi] - \(tan\^\(-1\)\) \((\@31)\)\)\/3)\)\)\)]], ", or ", Cell[BoxData[ \(\(t = \(-2\) + 2\ \@\(2\ \@2\)\%3)\)\)]], "\nThe other two roots of the cubic equation, we will get by considering \ the other two cube roots of u and v.\n", Cell[BoxData[ \(\(t = \)\)]], Cell[BoxData[ \(\(\(-2\) + \)\)]], "2 ", Cell[BoxData[ \(\(\@\(2\ \@2\)\%3\) \((cos \((\(\[Pi] - \(tan\^\(-1\)\) \((\@31)\) + 2 \[Pi]\)\/3)\)) \)\)]], ", and\n", Cell[BoxData[ \(\(t = \)\)]], Cell[BoxData[ \(\(\(-2\) + \)\)]], "2 ", Cell[BoxData[ \(\(\@\(2\ \@2\)\%3\) \((cos \((\(\[Pi] - \(tan\^\(-1\)\) \((\@31)\) + 4 \[Pi]\)\/3)\)) \)\)]], "." }], "Example", TextAlignment->Left] }, Closed]], Cell[CellGroupData[{ Cell["Solving Cubic Equation in Calculus", "Subsubsection"], Cell[TextData[{ "We will demonstrate now a completely different approach to solution of a \ cubic ", Cell[BoxData[ \(x\^3 + p\ x\ + q = 0\)]], ", by using calculus tools, differentiation, integration, saparation of \ variables. According to [1], this method was already published by John \ Landen in 1775. The q in the cubic equation we will treat as a function of \ x. Thus\n", Cell[BoxData[ \(q = \(\(-x\^3\) - px = \(-x\) \((x\^2 + p)\)\)\)]], " , ", Cell[BoxData[ \(\(q\^'\) = \(-3\) x\^2 - p\)]], ", ", Cell[BoxData[ \(\(q\^\(''\) = \(-6\) x\ \)\)]], ", and\n", Cell[BoxData[ \(q[0] = 0, \ \(q\^'\)[0] = \(-p\), \ \(q\^\(''\)\)[0] = \(0\ . \)\)]], "\nWe can observe that \n", Cell[BoxData[ RowBox[{\(q\^\(''\)\), "=", StyleBox[\(\(\(-18\) q\)\/\(\(q\^'\) - 2 p\)\), FontWeight->"Plain"]}]]], ",\nfrom which we get that \n", Cell[BoxData[ \(\(\(q\^\(''\)\) \(q\^'\) - 2 p\ q = \(-18\)\ q\ \)\)]], ",\nmultiplying by ", Cell[BoxData[ \(\(q\^'\)\)]], ",\n", Cell[BoxData[ \(\(\(q\^\(''\)\) \((\(q\^'\))\)\^2 - 2 p\ q\ \(q\^'\) = \(-18\)\ q\ \(q\^'\)\ \)\)]], ",\nintegrating both sides\n", Cell[BoxData[ \(\((\(q\^'\))\)\^3 - 3\ p\ \((\(q\^'\))\)\^2 = \(-27\)\ \((q)\)\^2 + \(c . \)\)]], "\nBy using the initial conditions, we will find that \n", Cell[BoxData[ \(c = \(-4\) \(p\^3 . \)\)]], "\nThus we have that ", Cell[BoxData[ \(\((\(q\^'\))\)\^3 - 3\ p\ \((\(q\^'\))\)\^2 = \(-27\)\ \((q)\)\^2 - 4 p\^3\)]], ", that is\n", Cell[BoxData[ \(\(\((\(q\^'\))\)\^2\) \((\(q\^'\) - 3\ p)\) = \(-27\)\ \((q)\)\^2 - 4 p\^3\)]], ".\nBy substituting explicit formula for ", Cell[BoxData[ \(q\^\('\ \)\)]], " in ", Cell[BoxData[ \(\(q\^'\) - 3\ p\)]], ", on the left side, we get that\n", Cell[BoxData[ \(\(\((\(q\^'\))\)\^2\) \((3 x\^2 + 4 p)\) = 27\ \((q)\)\^2 + 4 p\^3\)]], ".\nIt can be observed now that the above differential equation can be \ solved by separating variables.\n", Cell[BoxData[ \(dx\/\@\(3 x\^2 + 4 p\) = \(-\(dq\/\@\(27 q\^2 + 4 p\^3\)\)\)\)]], "\n\nIf ", Cell[BoxData[ \(p > 0\)]], ", the integration will yield \n", Cell[BoxData[ \(1\/\@3\ ArcSinh\ \((\(\@\(3\/\(4 p\)\ \)\) x)\) = \(-\(1\/\@27\)\) ArcSinh \((\(\@\(27\/\(4\ p\^3\)\)\) q)\)\)]], ",\nfrom which \n", Cell[BoxData[ \(x = \(-\ \@\(\(4 p\)\/3\)\)\ Sinh \((\(1\/3\) ArcSinh \((\@\(27\/\(4 p\^3\)\)\ q)\))\)\)]], ".\nIf ", Cell[BoxData[ \(p < 0\)]], ", the integration will yield , in the above formula ArcSin in place of \ ArcSinh.\n", Cell[BoxData[ \(TraditionalForm \`x = \@\(\(-\(1\/3\)\)\ \((4\ p)\)\)\ \(Sin(1\/3\ \(\((ArcSin(\([\@\(-\(27\/\(4\ p\^3\)\)\)\ q\)))\))\)\)\)]], "\nSince, in the above formula, ArcSin is a multivalued function, by \ selecting an appropriate branch, that is ", StyleBox["n", FontSlant->"Italic"], " in the formula below, we can find all the roots.\nNamely, ", Cell[BoxData[ \(TraditionalForm \`x = \(\[PlusMinus]\@\(\(-\(1\/3\)\)\ \((4\ p)\)\)\)\ \(Sin(1\/3\ \(\((n\ \[Pi] + ArcSin(\([\@\(-\(27\/\(4\ p\^3\)\)\)\ q\)))\)) \)\)\)]], "\nWe would like to write a program that will compute all real roots for a \ real cubic equation by using the formulas with Sin and Sinh.\n First, \ however, we will need to derive a well -known formula for discriminant. The \ discrimminant is \n is defined as ", Cell[BoxData[ \(Discr = \ \(-\(1\/108\)\) \(\((x\_1 - x\_2)\)\^2\) \(\((x\_1 - x\_3)\)\^2\) \((x\_2 - x\_3)\)\^2\)]], "\nIf ", Cell[BoxData[ \(D > 0\)]], ", then the cubic equation has only one real root. If ", Cell[BoxData[ \(D < 0\)]], ", then the cubic equation has three distinct roots. If Discr=0, then the \ cubic equation has all three roots real, but two of them are repeated.\nThe \ constant ", Cell[BoxData[ \(\(\(-1\)\/108\ \)\)]], "is introduced in order to obtain convenient easily remembered final form. \ " }], "Example", TextAlignment->Left, TextJustification->0], Cell[CellGroupData[{ Cell[BoxData[ \(s = x /. Solve[x\^3 + p\ x\ + q == 0, x]; \n Discr[p_, \ q_] = FullSimplify[ \(-\(1\/108\)\) \(\((x\_1 - x\_2)\)\^2\) \(\((x\_1 - x\_3)\)\^2\) \((x\_2 - x\_3)\)\^2 /. {x\_1 -> s[\([1]\)], x\_2 -> s[\([2]\)], x\_3 -> s[\([3]\)]}]\)], "Input"], Cell[OutputFormData["\<\ p^3/27 + q^2/4\ \>", "\<\ 3 2 p q -- + -- 27 4\ \>"], "Output"] }, Open ]], Cell[TextData[{ "Now we are in position to define, in ", StyleBox["Mathematica", FontSlant->"Italic"], "'s syntax, the function ", StyleBox["realroots", FontWeight->"Bold"], " that will compute the real roots of a cubic equation ", Cell[BoxData[ \(x\^3 + p\ x\ + q = 0\)]], ", p and q\[NotEqual]0 , and two auxiliary function", StyleBox[" rootp ", FontWeight->"Bold"], "which will compute the only real root in the case ", Cell[BoxData[ \(p > 0\)]], " since discriminent is positive, and function ", StyleBox["rootn", FontWeight->"Bold"], " which will compote the real roots in the case ", Cell[BoxData[ \(p < 0\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(rootp[p_, q_] := N[\(-\ \@\(\(4 p\)\/3\)\)\ Sinh[\(1\/3\) ArcSinh[\@\(27\/\(4 p\^3\)\)\ q]]]; \n rootn[p_, q_, n_] := \(\((\(-1\))\)\^n\) N[\@\(\(-\(1\/3\)\)\ \((4\ p)\)\)\ Sin[1\/3\ \((\[Pi]\ n + ArcSin[\@\(-\(27\/\(4\ p\^3\)\)\)\ q])\)]]; \n\n \t\t\)\)], "Input"], Cell["\<\ General::spell1: Possible spelling error: new symbol name \"rootn\" is similar to existing symbol \"rootp\".\ \>", "Message"] }, Open ]], Cell[BoxData[ \(realroots[x_\^3 + p_\ \ x_\ + q_] := \n\t Which[\n\t\tp > 0, rootp[p, q], \n\t\t \(p < 0 && q < 0\) && Discr[p, q] > 0, rootn[\ p, q, 2], \n\t\t p < 0 && q > 0 && Discr[p, q] > 0, rootn[\ p, q, 1], \n\ \ \ \ \ p < 0 && Discr[p, q] <= 0, {rootn[p, q, 0], rootn[p, q, 1], rootn[p, q, 2]}, \np == 0, \(-\@q\%3\)] // Chop\)], "Input"], Cell[TextData[{ "Now we will examine how ", StyleBox["realroots ", FontWeight->"Bold"], "function works in practice, pointing out possible problems with their \ solutions. " }], "Text"], Cell[TextData[{ "Example: One real root, ", Cell[BoxData[ \(p > 0\)]], ", ", Cell[BoxData[ \(x\^3 + 2 x + 4 == 0\)]] }], "Example"], Cell[CellGroupData[{ Cell[BoxData[ \(realroots[x\^3 + 2 x + 4]\)], "Input"], Cell[OutputFormData["\<\ -1.179509024602916\ \>", "\<\ -1.17951\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[x\^3 + 2 x + 4 == 0, x]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -1.179509024602916}, {x -> 0.5897545123014583 - 1.744543250922657*I}, {x -> 0.5897545123014583 + 1.744543250922657*I}}\ \>", "\<\ {{x -> -1.17951}, {x -> 0.589755 - 1.74454 I}, {x -> 0.589755 + 1.74454 I}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Comments: ", FontWeight->"Plain"], "realroots", StyleBox[" finds correctly the only one solution", FontWeight->"Plain"] }], "Text", FontWeight->"Bold"], Cell[TextData[{ "Example: One real root,", Cell[BoxData[ \(\(\ p < 0\)\)]], ", ", Cell[BoxData[ \(x\^3 - 2 x + 3 == 0\)]], ". " }], "Example"], Cell[CellGroupData[{ Cell[BoxData[ \(realroots[x\^3 - 2 x + 3]\)], "Input"], Cell[OutputFormData["\<\ -1.893289196304497\ \>", "\<\ -1.89329\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[x^3 - 2 x + 3 == 0, x]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -1.893289196304497}, {x -> 0.9466445981522488 - 0.8297035528624054*I}, {x -> 0.9466445981522488 + 0.8297035528624054*I}}\ \>", "\<\ {{x -> -1.89329}, {x -> 0.946645 - 0.829704 I}, {x -> 0.946645 + 0.829704 I}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Comments: ", FontWeight->"Plain"], "realroots", StyleBox[" finds correctly the only one solution", FontWeight->"Plain"] }], "Text", FontWeight->"Bold"], Cell[TextData[{ "Example: Three real roots,", Cell[BoxData[ \(\(\ p < 0\)\)]], ", ", Cell[BoxData[ \(x\^3 - 3 x - 1 == 0\)]] }], "Example"], Cell[CellGroupData[{ Cell[BoxData[ \(realroots[x\^3 - 3 x - 1]\)], "Input"], Cell[OutputFormData["\<\ {-0.3472963553338606, -1.532088886237955, 1.879385241571816}\ \>", "\<\ {-0.347296, -1.53209, 1.87939}\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[x\^3 - 3 x - 1 == 0, x]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -1.532088886237956}, {x -> -0.3472963553338606}, {x -> 1.879385241571816}}\ \>", "\<\ {{x -> -1.53209}, {x -> -0.347296}, {x -> 1.87939}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Comments: ", FontWeight->"Plain"], "realroots", StyleBox[" finds correctly all the solutions", FontWeight->"Plain"] }], "Text", FontWeight->"Bold"], Cell[TextData[{ "Example: Repeated roots, ", Cell[BoxData[ \(x\^3 - 3 \(\@ 0.25\%3\) x - 1 == 0\)]] }], "Example"], Cell[CellGroupData[{ Cell[BoxData[ \(realroots[x\^3 - 3 \(\@ 0.25\%3\) x - 1]\)], "Input"], Cell[OutputFormData["\<\ 1.587401051968199\ \>", "\<\ 1.5874\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[x\^3 - 3 \(\@ 0.25\%3\) x - 1 == 0, x]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -0.7937005283975395}, {x -> -0.7937005235706599}, {x -> 1.587401051968199}}\ \>", "\<\ {{x -> -0.793701}, {x -> -0.793701}, {x -> 1.5874}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Comments: ", FontWeight->"Plain"], "realroots", StyleBox[" failed to find all the roots. Reason, the ", FontWeight->"Plain"], "Discr", StyleBox[ " funtion was not evaluated to zero because the coefficients were entered \ numerically.", FontWeight->"Plain"] }], "Text", FontWeight->"Bold"], Cell[CellGroupData[{ Cell[BoxData[ \(realroots[x\^3 - 3 \(\@\( 1\/4\)\%3\) x - 1]\)], "Input"], Cell[OutputFormData["\<\ {-0.7937005259840997, -0.7937005259840997, 1.587401051968199}\ \>", "\<\ {-0.793701, -0.793701, 1.5874}\ \>"], "Output"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Solution of Cubic Equation by Mark Kac", "Subsubsection"], Cell[TextData[{ "In the book ", StyleBox["Mathematics and Logic,", FontSlant->"Italic"], " Mark Kac provides yet another method to solve a cubic equation. A method \ discovered by himself in his youth. He made an observation, using alternating \ groups and symmetric polynomials, that the expressions ", Cell[BoxData[ \(TraditionalForm\`f(x, y, z) := \((z\ w\^2 + y\ w + x)\)\^3\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\(\ g(x, y, z) := \((y\ w\^2 + z\ w + x)\)\^3\)\)]], ", where w is the cubic root of unity and x,y,z are the roots of a cubic \ equation, can be solely expressed by the coefficients of a cubic equations. \ In this section we will assume that a cubic equation is of the form ", Cell[BoxData[ \(x\^3 + px\ + q = 0\)]], ". Then by using exact formulas for the roots, obtained by applying ", StyleBox["Solve ", FontWeight->"Bold"], "command , we compute f and g at these roots and simplify the result by \ using the command ", StyleBox["FullSimplify. ", FontWeight->"Bold"] }], "Text"], Cell[BoxData[ \(Clear[w, f, g, a, b]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`w = \(-\(1\/2\)\) + \(\@3\ \[ImaginaryI]\)\/2; Expand[w\^3]\)], "Input"], Cell[OutputFormData["\<\ 1\ \>", "\<\ 1\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`f(x_, y_, z_) := \((z\ w\^2 + y\ w + x)\)\^3; roots = x /. Solve(x\^3 + p\ x + q == 0, x); \nFullSimplify(f@@roots)\)], "Input"], Cell["\<\ General::spell: Possible spelling error: new symbol name \"roots\" is similar to existing symbols {rootn, rootp, Roots}.\ \>", "Message"], Cell[OutputFormData["\<\ (-3*(9*q + Sqrt[3]*Sqrt[4*p^3 + 27*q^2]))/2\ \>", "\<\ 3 2 -3 (9 q + Sqrt[3] Sqrt[4 p + 27 q ]) ------------------------------------- 2\ \>"], "Output"] }, Open ]], Cell[TextData[{ "Since ", StyleBox["Mathematica", FontSlant->"Italic"], " renders a cube root of a negative real number as a complex number, we \ would like it to be a real number, we defined our own cube root function." }], "Text"], Cell[BoxData[ \(cuberoot[x_] := If[Re[x] < 0, w \@ x\%3, \@x\%3]\)], "Input"], Cell[BoxData[ \(a[p_, q_] := cuberoot[\(-\(3\/2\)\)\ \((9\ q + \@3\ \@\(4\ p\^3 + 27\ q\^2\))\)]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`g(x_, y_, z_) := \((z\ w + y\ w\^2 + x)\)\^3; roots = x /. Solve(x\^3 + p\ x + q == 0, x); \nFullSimplify(g@@roots)\)], "Input"], Cell[OutputFormData["\<\ (-27*q + Sqrt[108*p^3 + 729*q^2])/2\ \>", "\<\ 3 2 -27 q + Sqrt[108 p + 729 q ] ----------------------------- 2\ \>"], "Output"] }, Open ]], Cell[BoxData[ \(\(b[p_, q_] := cuberoot[1\/2\ \((\@\(108\ p\^3 + 729\ q\^2\) - 27\ q)\)]\n\)\)], "Input"], Cell[TextData[{ " By applying the fact that ", Cell[BoxData[ \(x + y + z = 0\)]], ", we can set up a system of linear equations." }], "Text"], Cell[BoxData[ \(kacway[p_, q_] := NSolve[{x + y + z == 0, z\ w\^2 + y\ w + x == a[p, q], y\ w\^2 + z\ w + x == b[p, q]}, {x, y, z}]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(kacway[\(-2\), 4] // Chop\)], "Input"], Cell[OutputFormData["\<\ {{x -> -2., y -> 1. + 0.9999999999999989*I, z -> 1. - 0.9999999999999992*I}}\ \>", "\<\ {{x -> -2., y -> 1. + 1. I, z -> 1. - 1. I}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], "'s solution." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`NSolve(x\^3 - 2\ x + 4 == 0)\)], "Input"], Cell[OutputFormData["\<\ {{x -> -2.}, {x -> 1. - 1.*I}, {x -> 1. + 1.*I}}\ \>", "\<\ {{x -> -2.}, {x -> 1. - 1. I}, {x -> 1. + 1. I}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ "Unfortunately,", StyleBox[" kacway", FontWeight->"Bold"], " does not always produces a correct solution. This problem is caused by \ multiple-valued root functions in complex domain. One way to solve this \ problem, probably not the best, is to multiply the right side of the system \ of equation by", Cell[BoxData[ \(\(\ w\ \)\)]], "or ", Cell[BoxData[ \(w\^2\)]], ". In other words, consider different branch For example:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[x^3 - 4 x + 1 == 0]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -2.114907541476755}, {x -> 0.2541016883650524}, {x -> 1.860805853111703}}\ \>", "\<\ {{x -> -2.11491}, {x -> 0.254102}, {x -> 1.86081}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ "But,", StyleBox[" kacway ", FontWeight->"Bold"], "will do incorrectly." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\nkacway[\(-4\), 1] // Chop\)\)], "Input"], Cell[OutputFormData["\<\ {{x -> -0.9304029265558517 + 1.611505140305509*I, y -> 1.057453770738377 - 1.831563657574161*I, z -> -0.1270508441825259 + 0.2200585172686523*I}}\ \>", "\<\ {{x -> -0.930403 + 1.61151 I, y -> 1.05745 - 1.83156 I, z -> -0.127051 + 0.220059 I}}\ \>"], "Output"] }, Open ]], Cell["To correct this problem, consider", "Text"], Cell[BoxData[ \(kacway2[p_, q_] := NSolve[{x + y + z == 0, z\ w\^2 + y\ w + x == w\^2\ *a[p, q], y\ w\^2 + z\ w + x == w\^2*\ b[p, q]}, {x, y, z}]\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(kacway2[\(-4\), 1] // Chop\)], "Input"], Cell[OutputFormData["\<\ {{x -> 1.860805853111702, y -> -2.114907541476755, z -> 0.2541016883650526}}\ \>", "\<\ {{x -> 1.86081, y -> -2.11491, z -> 0.254102}}\ \>"], "Output"] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Problems with Cubic Eqauation", "Section"], Cell[CellGroupData[{ Cell["Making Box", "Subsubsection"], Cell[TextData[{ "From a rectangular piece of cardboard 20 by 15 cm, by cutting the same \ square of a side length x cm from each corner, an open-top box of volume 250 \ ", Cell[BoxData[ \(cm\^3\)]], " has to be made. 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Substituting 10000 ", Cell[BoxData[ \(cm\^3\)]], " for V, we will find r in cm." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(TraditionalForm\`V = 10000.00; \ \ \ r = \@\(\(3\ V\)\/\(2\ \[Pi]\)\)\%3\)], "Input"], Cell[OutputFormData["\<\ 16.83890300960629\ \>", "\<\ 16.8389\ \>"], "Output"] }, Open ]], Cell[BoxData[ \(TraditionalForm \`The\ empty\ space\ in\ the\ tank\ equals\ 2\ liters = 2\ 000\ cm\^3 . \ On\ the\ other\ hand, \ it\ can\ be\ computed\ by\ \ using\ \ methods\ of\ \(disks . \)\)], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(volume = \[Integral]\_0\%x \[Pi] \((r\^2 - s\^2)\) \[DifferentialD]s\)], "Input"], Cell[OutputFormData["\<\ 890.7943701227308*x - 1.047197551196597*x^3\ \>", "\<\ 3 890.794 x - 1.0472 x\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(volume == 2000\)], "Input"], Cell[OutputFormData["\<\ 890.7943701227308*x - 1.047197551196597*x^3 == 2000\ \>", "\<\ 3 890.794 x - 1.0472 x == 2000\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[volume == 2000, x]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -30.22953123846837}, {x -> 2.258734173736224}, {x -> 27.97079706473215}}\ \>", "\<\ {{x -> -30.2295}, {x -> 2.25873}, {x -> 27.9708}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Answer: ", FontWeight->"Bold"], "The distanse x equals 2.258 cm " }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["The Smallest Distance from a Parabola", "Subsubsection"], Cell[TextData[{ "Find the smallest distance from the point ", Cell[BoxData[ \(\((1, \ 3)\)\)]], "to the parabola ", Cell[BoxData[ \(\(y = x\^2\ \)\)]], "." }], "Example"], Cell[TextData[{ StyleBox["Solution", FontWeight->"Bold"], ": Square of a distance between a point on parabola ", Cell[BoxData[ \(\((x, x\^2)\)\)]], "and the point ", Cell[BoxData[ \(\((1, 3)\)\)]], " is given by ", Cell[BoxData[ \(\((1 - x)\)\^2 + \((3 - x\^2)\)\^2\)]], ". By equating the derivative of this expression to 0, we will find the \ desired x." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[\[PartialD]\_x\((\((1 - x)\)\^2 + \((3 - x\^2)\)\^2)\)]\)], "Input"], Cell[OutputFormData["\<\ -2 - 10*x + 4*x^3\ \>", "\<\ 3 -2 - 10 x + 4 x\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(NSolve[\(-2\) - 10\ x + 4\ x\^3 == 0]\)], "Input"], Cell[OutputFormData["\<\ {{x -> -1.469617434058037}, {x -> -0.203364213796905}, {x -> 1.672981647854942}}\ \>", "\<\ {{x -> -1.46962}, {x -> -0.203364}, {x -> 1.67298}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Answer:", FontWeight->"Bold"], " The smallest distance will occur for the point on the parabola with the \ abscissa equal 1.67298." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Pumping Water out of Tank", "Subsubsection"], Cell["\<\ Suppose that a pump of 60% efficiency, after using 0.5 kWh, broke down while \ pumping the water out of the tank full of water. 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:> $DisplayFunction, ColorOutput -> Automatic, Axes -> Automatic, AxesOrigin -> Automatic, PlotLabel -> None, AxesLabel -> {h, work}, Ticks -> Automatic, GridLines -> Automatic, Prolog -> {}, Epilog -> {}, AxesStyle -> Automatic, Background -> Automatic, DefaultColor -> Automatic, DefaultFont :> $DefaultFont, RotateLabel -> True, Frame -> True, FrameStyle -> Automatic, FrameTicks -> Automatic, FrameLabel -> None, PlotRegion -> Automatic, ImageSize -> Automatic, TextStyle :> $TextStyle, FormatType :> $FormatType}]\ \>", "\<\ -Graphics-\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Answer:", FontWeight->"Bold"], " h\[Rule]2.4253" }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Equation of State for Real Gases", "Subsubsection"], Cell[TextData[{ StyleBox[ "Van der Waal equation , a generalization of the Ideal Gas Law, is the \ equation which describes the relationship between three basic parameters for \ a real gas; P, T, and V; pressure, temperature and volume.\n \ ", "Example"], Cell[BoxData[ \(\((P + \(a\ n\^2\)\/V\^2)\) \((V - n\ b)\) = n\ R\ T\)]], ",\n ", StyleBox[ "where a and b are the constants peculiar to each gas, and are determined \ experimentally. R is the Universal Gas Constant and it equals 0.082 l \ Atm/mole K\n Find the volume of the hydrogent if ", "Example"], Cell[BoxData[ \(P = 3\ Atm, T = 400\ K, \ R = \(0.082\ \((l\ Atm)\)\)\/\(mole\ K\), \ n = 1\ mole, \ \na = \(0.24\ \((l\^2\ Atm)\)\)\/mole\^2, \ b = \(0.0226\ l\)\/\(mole\^2 . \)\)]] }], "Text"], Cell[TextData[StyleBox["Solution:", FontWeight->"Bold"]], "Text"], Cell[BoxData[ \(TraditionalForm\`P = 3\ ; T = 400\ ; R = 0.082\ ; \ n = 1\ ; \ a = 0.24\ ; b = 0.0226\ ; \)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(equation = Together[\((P + \(a\ n\^2\)\/V\^2)\) \((V - n\ b)\) - n\ R\ T]\)], "Input"], Cell[OutputFormData["\<\ (-0.005423999999999998 + 0.2399999999999999*V - 32.8678*V^2 + 3*V^3)/ V^2\ \>", "\<\ 2 3 -0.005424 + 0.24 V - 32.8678 V + 3 V -------------------------------------- 2 V\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Numerator[equation]\)], "Input"], Cell[OutputFormData["\<\ -0.005423999999999998 + 0.2399999999999999*V - 32.8678*V^2 + 3*V^3\ \>", "\<\ 2 3 -0.005424 + 0.24 V - 32.8678 V + 3 V\ \>"], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Solve[\n\tNumerator[equation] == 0, V]\)], "Input"], Cell[OutputFormData["\<\ {{V -> 0.003645879939670853 - 0.01232242674398953*I}, {V -> 0.003645879939670853 + 0.01232242674398953*I}, {V -> 10.94864157345399}}\ \>", "\<\ {{V -> 0.00364588 - 0.0123224 I}, {V -> 0.00364588 + 0.0123224 I}, {V -> 10.9486}}\ \>"], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Answer:", FontWeight->"Bold"], " The volume of hydregen is 10.9486\n", StyleBox["Remark:", FontWeight->"Bold"], " Since the solution has only one real root, we do not have any difficulty \ to identify the good one." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Electrical Resistance", "Subsubsection"], Cell["\<\ Three electrical resistors are connected in parallel, first resistor is of R \ ohms, second 2 more ohms, and third 3 more ohms. 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