(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 82308, 3236]*) (*NotebookOutlinePosition[ 83264, 3268]*) (* CellTagsIndexPosition[ 83220, 3264]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Algebra", "Title", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "Sean Mauch\nsean@caltech.edu\n", ButtonBox["http://www.its.caltech.edu/~sean", ButtonData:>{ URL[ "http://www.its.caltech.edu/~sean"], None}, ButtonStyle->"Hyperlink"], "\n", "This work is distributed under the GNU FDL. See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Symbolic Calculations"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can do symbolic calculations as you do in algebra. (When you try to \ evaluate the first input in this notebook, ", StyleBox["Mathematica", FontSlant->"Italic"], " will ask you if you want to evaluate the initialization cells. You \ should answer yes.)" }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\^2 - 2\ x + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In the previous input ", Cell[BoxData[ \(TraditionalForm\`x\)]], " is a formal variable. When you enter an expression containing formal \ variables, ", StyleBox["Mathematica", FontSlant->"Italic"], " will automatically perform certain transformations to simplify the \ expression." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(3\ x\^2 + 2\ x - 2\ x\^2 - 4\ x + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "For the above input, ", StyleBox["Mathematica", FontSlant->"Italic"], " combined like terms and sorted the terms in increasing powers of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". ", StyleBox["Mathematica", FontSlant->"Italic"], " can also perform more advanced transformations. " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell["\<\ User-defined constants can also have values that involve formal \ variables. The following input makes a globally defined substitution rule.\ \ \>", "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x = y + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Whenever ", Cell[BoxData[ \(TraditionalForm\`x\)]], " appears in an input the symbolic constant ", Cell[BoxData[ \(TraditionalForm\`y + 1\)]], " is substituted." }], "Text", Evaluatable->False, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\^2 - 2\ x + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Be careful with spaces in your inputs. A space indicates multiplication. \ The following input is ", Cell[BoxData[ \(TraditionalForm\`a\[Times]x\^2\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(a\ x\^2\)], "Input"], Cell[TextData[{ "If you leave out the space between the ", Cell[BoxData[ \(TraditionalForm\`a\)]], " and the ", Cell[BoxData[ \(TraditionalForm\`x\)]], ", then the input is interpretted as the square of the formal variable ", Cell[BoxData[ \(TraditionalForm\`a\[VeryThinSpace]x\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(ax\^2\)], "Input"], Cell[BoxData[ \(Clear[x]\)], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Local Variable Substitution"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "In addition to global variable substitution with user-defined constants, ", StyleBox["Mathematica", FontSlant->"Italic"], " allows you to perform local formal variable substitutions. In the \ following two inputs a numerical and a symbolic constant are locally \ substituted into an algebraic expression. (You can type \[EscapeKey]", Cell[BoxData[ \(TraditionalForm\` -> \)]], "\[EscapeKey] to obtain the \[Rule] symbol, or you can just use ", Cell[BoxData[ \(TraditionalForm\` -> \)]], ".)" }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\^2 + 2\ x + 1 /. x \[Rule] 1\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(x\^2 + 2\ x + 1 /. x \[Rule] y + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Note that the substitution is performed only in that expression and the \ global definition of ", Cell[BoxData[ \(TraditionalForm\`x\)]], " is unchanged." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\)], "Input", AspectRatioFixed->True], Cell[TextData[ "An expression of the form\n\t \[Rule] \nis \ called a rule. The \"/.\" means \"apply the following rule\". You can \ apply multiple rule substitutions to an expression by using a list of \ rules."], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\((x + y)\)\^3 /. {x \[Rule] 1, y \[Rule] 2}\)], "Input", AspectRatioFixed->True], Cell[TextData["User-defined constants can hold rules."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(rules = {x \[Rule] 1, y \[Rule] 2, z \[Rule] 3}\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\((x + y + z)\)\^2 /. rules\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[rules]\)], "Input", AspectRatioFixed->True], Cell["\<\ You can make a list of values of an expression for a sequence of \ values of a formal variable by applying a list of lists of rules. The \ following input makes a list of values of a polynomial.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\^2 + 1 /. {{x \[Rule] 1}, {x \[Rule] 10}, {x \[Rule] 100}}\)], "Input",\ AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Sequences of Expressions"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ You can execute more that one command in a single input cell by \ entering them on separate lines. The following input cell is actually three \ inputs and will produce three outputs.\ \>", "Text", TextJustification->1], Cell[BoxData[{ \(x = 1\), \(y = 2\), \(x + y\^2\)}], "Input"], Cell["\<\ You can also execute more than one command in a single input cell \ by separating the inputs with semicolons. Note that only the output from the \ last input is returned.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x = 3; \ny = 5; \nx + y\)], "Input"], Cell[BoxData[ \(x = a + b; y = c + d; x\ y\)], "Input", AspectRatioFixed->True], Cell["\<\ Ending an input with a semicolon has the effect of hiding the \ output of an expression. \ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(x\^2\ y\^2; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "There is an output, it's just not shown. You can access this output by \ using the usual ", StyleBox["%", "Input"], " or ", StyleBox["Out[]", "Input"], " operations." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(%\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Clear[x, y]\)], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["User-Defined Functions"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " allows you to define your own functions. The following input defines a \ function, ", Cell[BoxData[ \(TraditionalForm\`f\)]], ", that returns the square of its argument. The underscore following the ", Cell[BoxData[ \(TraditionalForm\`x\)]], " on the left side of the definition identifies ", Cell[BoxData[ \(TraditionalForm\`x\)]], " as a function argument. Note that there is no output when defining a \ function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(f[x_] := x\^2\)], "Input", AspectRatioFixed->True], Cell["\<\ The argument to the function can be a number, a formal variable, ar \ any expression.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(f[5]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[y]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[x + Sin[x]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[f[a]]\)], "Input"], Cell[BoxData[ \(f[{1, 2, 3, 4}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(f[x] /. x -> 2\)], "Input"], Cell[TextData[{ "You can use the function ", Cell[BoxData[ \(TraditionalForm\`f\)]], " within other expressions just as you would built-in functions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sin[f[x + y]]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can use ", StyleBox["Mathematica", FontSlant->"Italic"], " help to get information about user defined functions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(?f\)\)], "Input"], Cell[TextData[{ "The ", StyleBox["Clear[]", FontWeight->"Bold"], " function removes the definition of a user-defined function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f]\)], "Input", AspectRatioFixed->True], Cell["\<\ It is best to have the names of user defined functions begin with a \ lower-case letter so there is no confusion or conflict with built-in \ functions.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 1", "Subsubsection"], Cell[TextData[{ "Define a function, ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ", to be the polynomial ", Cell[BoxData[ \(TraditionalForm\`\((x - 1)\)\^2\)]], ". Make a list of the values of the polynomial when ", Cell[BoxData[ \(TraditionalForm\`x\)]], " is an odd integer between 1 and 9 with the following methods:\n1) Use a \ list as the argument to ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], ".\n2) Apply a list of lists of rules to ", Cell[BoxData[ \(TraditionalForm\`f[x]\)]], "." }], "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Changing the Form of Polynomials"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData["Factoring and Expanding Polynomials"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "When you enter an algebraic expression, ", StyleBox["Mathematica", FontSlant->"Italic"], " will leave it close to its input form. If you enter a product or a sum, \ ", StyleBox["Mathematica", FontSlant->"Italic"], " will output the same kind of form." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\((x + 1)\)\ \((x + 2)\)\ \((x + 3)\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(x\^3 + 6\ x\^2 + 11\ x + 6\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can use the ", StyleBox["Expand[]", FontWeight->"Bold"], " and ", StyleBox["Factor[]", FontWeight->"Bold"], " functions to change the form of an algebraic expression. The ", StyleBox["Expand[]", FontWeight->"Bold"], " function expands products into sums. The ", StyleBox["Factor[]", FontWeight->"Bold"], " function tries to factor a sum into a product." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x + 1)\)\ \((x + 2)\)\ \((x + 3)\)]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x + y)\)\^5]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Factor[x\^3 + 6\ x\^2 + 11\ x + 6]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Factor[x\^3 + x\^2\ y + x\ y\^2 + y\^3]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " factors polynomials over the integers. That is, the factorization will \ involve only integers." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x\^2 + 1\/2)\)\ \((x + 1\/4)\)]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Factor[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Factor[]", "Input"], " function will not factor a polynomial with rational terms into terms that \ involve irrational numbers." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x - \@2)\)\ \((x + \@2)\)]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Factor[%]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x - \@2)\)\^2\ \((x - 2)\)]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Factor[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Notice that in the above calculation, ", StyleBox["Mathematica", FontSlant->"Italic"], " only factored out the term that had rational coefficients." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also factor polynomials over the complex integers, (Gaussian \ integers)." }], "Text", TextJustification->1], Cell[BoxData[ \(Factor[x\^2 + 1]\)], "Input"], Cell[BoxData[ \(Factor[x\^2 + 1, GaussianIntegers -> True]\)], "Input"], Cell[TextData[{ "The ", StyleBox["Factor[]", "Input"], " function was meant to be used only on exact expression. If you use it on \ an approximate expression you may get unexpected results." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Simplifying Polynomials"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Simplify[]", "Input"], " function will try to put a polynomial, (or any other expression), in its \ simplest form. ", StyleBox["Mathematica", FontSlant->"Italic"], " roughly defines the simplest form as the one that has the fewest terms. \ ", StyleBox["Mathematica", FontSlant->"Italic"], " will try a variety of transformations on the expression. Below ", StyleBox["Mathematica", FontSlant->"Italic"], " considers the sum of two terms a simpler form than the input which is the \ product of two terms that each contain two terms." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[\((x + 1)\)\ \((x - 1)\)]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Below ", StyleBox["Mathematica", FontSlant->"Italic"], " considers the square of an expression with two terms to be simpler than \ the sum of three terms." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[x\^2 + 2\ x + 1]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Occasionally ", StyleBox["Mathematica", FontSlant->"Italic"], " may surprise you and yield a really significant simplification." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[ \((x - 1)\)\ \((x + 1)\)\ \((x\^2 + 1)\)\ \((x\^2 - x + 1)\)\ \((x\^2 + x + 1)\)\ \((x\^4 - x\^2 + 1)\)]\)], "Input", AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Changing the Form of Rational Expressions"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Applying ", StyleBox["Expand[]", "Input"], " to a rational polynomial, (a ratio of polynomials), will expand the \ numerator." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Expand[\((x - 1)\)\^2\/\((x + 1)\)\^2]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Apply ", StyleBox["Factor[]", "Input"], " to a rational polynomial will factor both the numerator and denominator." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Factor[\(x\^2 - 2\ x + 1\)\/\(x\^2 + 2\ x + 1\)]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In order to expand both the numerator and the denominator use the ", StyleBox["ExpandAll[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(ExpandAll[\((x - 1)\)\^2\/\((x + 1)\)\^2]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Together[]", "Input"], " puts the terms in a sum over a common denominator and cancels common \ factors." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Together[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Apart[]", "Input"], " function expands an expression in partial fractions, i.e. it expands a \ rational expression into a sum of terms with minimal denominators." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Apart[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can also apply the ", StyleBox["Simplify[]", "Input"], " function to rational expressions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Simplify[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Cancel[]", "Input"], " function cancels common factors in the numerator and denominator of a \ rational expression." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\((x + 1)\)\^3\/\(x\^2 + 2\ x + 1\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Cancel[%]\)], "Input", AspectRatioFixed->True], Cell["\<\ If you want to do a partial fraction expansion where the terms are \ factored over the Gaussian integers you have to factor the expression \ first.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(Apart[1\/\(x\^2 + 1\)]\)], "Input"], Cell[BoxData[ \(Factor[1\/\(x\^2 + 1\), GaussianIntegers -> True]\)], "Input"], Cell[BoxData[ \(Apart[%]\)], "Input"], Cell["\<\ You can factor polynomials that have formal variables as \ coefficients.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(Expand[\((x - a)\) \((x\^2 + b\^2)\) \((x\^2 - c\^2)\)]\)], "Input"], Cell[BoxData[ \(Factor[%]\)], "Input"], Cell[BoxData[ \(Factor[%, GaussianIntegers -> True]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Boolean Expressions"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "You can enter a boolean expression in ", StyleBox["Mathematica.", FontSlant->"Italic"], " If it can trivially evaluate the expression ", StyleBox["Mathematica", FontSlant->"Italic"], " will return True or False. (True and False are built-in constants.) If \ ", StyleBox["Mathematica", FontSlant->"Italic"], " cannot trivially evaluate the expression, it will return it unchanged." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(2 + 3 \[Equal] 5\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(5 < 7\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(3 \[GreaterEqual] 23\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\((x + 1)\)\^2 \[Equal] x\^2 + 2\ x + 1\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " provides the following relational operators: (If you know C these will \ be familiar.)" }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[Cell[BoxData[ FormBox[GridBox[{ {\(Text\ Form\), \(Typeset\ Command\), \(Typeset\ Form\), "Meaning"}, {"==", \(\[EscapeKey] == \[EscapeKey]\), "\[Equal]", "equal"}, {"!=", \(\[EscapeKey] != \[EscapeKey]\), "\[NotEqual]", \(not\ equal\)}, {"<", " ", " ", \(less\ than\)}, {">", " ", " ", \(greater\ than\)}, {"<=", \(\[EscapeKey] <= \[EscapeKey]\), "\[LessEqual]", \(less\ than\ or\ equal\)}, {">=", \(\[EscapeKey] >= \[EscapeKey]\), "\[GreaterEqual]", \(greater\ than\ or\ equal\)} }, ColumnAlignments->{Left}, RowLines->{True, False}, ColumnLines->True], TraditionalForm]]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "In addition, ", StyleBox["Mathematica", FontSlant->"Italic"], " provides the boolean operators:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[Cell[BoxData[ FormBox[GridBox[{ {\(Text\ Form\), \(Typeset\ Command\), \(Typeset\ Form\), "Meaning"}, {"&&", \(\[EscapeKey] && \[EscapeKey]\), "\[And]", "and"}, {"||", \(\[EscapeKey] || \[EscapeKey]\), "\[Or]", "or"}, {"!", \(\(\[EscapeKey]!\) \[EscapeKey]\), "\[Not]", "not"} }, ColumnAlignments->{Left}, RowLines->{True, False}, ColumnLines->True], TraditionalForm]]]], "Text", TextAlignment->Center, TextJustification->0], Cell[BoxData[ \(3 < 5 \[And] 2 \[NotEqual] 3\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The exclamation mark is the negation operator. If its argument is True \ then it returns False and vice versa. Do not begin a boolean expression with \ an excamation mark. Starting an input with an exclamation mark tells ", StyleBox["Mathematica", FontSlant->"Italic"], " to execute a shell command." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(\[InvisibleSpace]\(! \((2 \[NotEqual] 3)\)\)\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(\[InvisibleSpace]\[Not] \((2 \[NotEqual] 3)\)\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(2 < 3 \[Or] 3 < 2\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " will evaluate a wider class of boolean expressions if you ask it to \ simplify the input." }], "Text"], Cell[BoxData[ \(Sin[x + y] == Cos[y]\ Sin[x] + Cos[x]\ Sin[y]\)], "Input"], Cell[BoxData[ \(Sin[x + y] == Cos[y]\ Sin[x] + Cos[x]\ Sin[y] // Simplify\)], "Input"], Cell[BoxData[ \(Sin[x]\^5 == 1\/16\ \((10\ Sin[x] - 5\ Sin[3\ x] + Sin[5\ x])\) // Simplify\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Solving Equations of a Single Variable"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "An equallity relation that involves formal variables is an equation in ", StyleBox["Mathematica", FontSlant->"Italic"], ". The following are equations:" }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`x\^2 + x + 1 \[Equal] 2\)]], " \n", Cell[BoxData[ \(TraditionalForm\`Sin[x]\ + \ Tan[y]\ \[Equal] \ 3\)]] }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell["The following are not equations:", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`Cos[x]\ + \ 1\ < \ 3/2\)]], "\n", Cell[BoxData[ \(TraditionalForm \`x\^2\ \[Equal] \ 5\ \ \[And] \ \ y\^3\ + \ y\ \[Equal] \ 3\)]] }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Roots[]", "Input"], " function can find the roots of many types of equations. The following \ example finds the roots of a quadratic polynomial." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Roots[x\^2 + 5\ x + 6 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The result of ", StyleBox["Roots[]", "Input"], " is a boolean expression describing when the equation is satisfied. From \ the above result we see that ", Cell[BoxData[ \(TraditionalForm\`x\^2\ + \ 5\ x\ + \ 6\)]], " is zero when ", Cell[BoxData[ \(TraditionalForm\`x = \(-3\)\)]], " or ", Cell[BoxData[ \(TraditionalForm\`x = \(-2\)\)]], ". You can change the boolean expression above into a set of substitution \ rules with the ", StyleBox["ToRules[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \({ToRules[%]}\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "To get a list of the roots, apply the rules to the variable ", Cell[BoxData[ \(TraditionalForm\`x\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x /. {{x \[Rule] \(-3\)}, {x \[Rule] \(-2\)}}\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can evaluate an arbitrary expression in ", Cell[BoxData[ \(TraditionalForm\`x\)]], " at a root of the quadratic. The next input evaluates ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " at the second solution." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(x\^3 + x\^2 /. {x \[Rule] \(-2\)}\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Solve[]", "Input"], " function combines the operations of ", StyleBox["Roots[]", "Input"], " and ", StyleBox["ToRules[]", "Input"], ". ", StyleBox["Solve[]", "Input"], " is equivalent to ", StyleBox["{ToRules[Roots[]]}", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[x\^2 + 5\ x + 6 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The list of substitution rules is usually a more convenient form than the \ boolean expression given by ", StyleBox["Roots[]", "Input"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "If an algebraic equation has a double, triple, etc. root, that root will \ be given multiple times in the ouput of ", StyleBox["Solve[]", "Input"], ". The following cubic equation has a double root at 0." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[x\^3 + x\^2 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If ", StyleBox["Mathematica", FontSlant->"Italic"], " cannot find a closed form solution to an algebraic equation it will \ return a list of rules that can be evaluated numerically to give approximate \ solutions to the equation." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[x\^6 + x\^5 - x\^4 + x\^3 - x\^2 + x - 1 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(N[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Trigonometric equations often have an infinite number of solutions. In \ this case ", StyleBox["Mathematica", FontSlant->"Italic"], " will warn you that it may not have found all the solutions." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[Sin[x] == 1\/2, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Many transcendental functions do not have closed-form solutions. In this \ case ", StyleBox["Mathematica", FontSlant->"Italic"], " will return the input." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[Tan[x] == x, x]\)], "Input", AspectRatioFixed->True], Cell["\<\ Sometimes the solutions to an equation can be quite complicated and \ a numerical solution is more useful.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[x\^3 + 2 x\^2 + 3 x + 4 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In this case you can apply the numerical function to the output or you can \ use the ", StyleBox["NSolve[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(N[%]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(NSolve[x\^3 + 2 x\^2 + 3 x + 4 == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In equations containing parameters, the ", StyleBox["Solve[]", "Input"], " function finds the general solution and does not take into account \ special values of the parameters. Consider the following input. If ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is nonzero, then the only solution is ", Cell[BoxData[ \(TraditionalForm\`x\ == \ 0\)]], ". If ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is zero then any value of ", Cell[BoxData[ \(TraditionalForm\`x\)]], " is a solution to the equation. The ", StyleBox["Solve[]", "Input"], " function only finds the solution for general values of ", Cell[BoxData[ \(TraditionalForm\`a\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[a\ x == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If you want the full solution to an equation use the ", StyleBox["Reduce[]", "Input"], " function. " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Reduce[a\ x == 0, x]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Rememeber that you can convert this boolean expression into a list of \ rules with the ", StyleBox["ToRules[]", FontWeight->"Bold"], " function." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \({ToRules[%]}\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 2", "Subsubsection"], Cell[TextData[{ "Find the general solutions to ", Cell[BoxData[ \(TraditionalForm\`a\ x\^2 + b\ x + c = 0\)]], ". Now find all of the solutions and explain then in a paragraph. Recall \ that you can write text in a notebook by first clicking to make a horizontal \ line where you want the text cell to appear and then selecting ", StyleBox["Style\[Rule]CellStyle\[Rule]Text", FontWeight->"Bold"], " from the menu bar." }], "Text", TextJustification->1] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Solving Systems of Equations"], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[x, y]; \n plane1 = Plot3D[{2 - x - y, Hue[0]}, {x, 0, 2}, {y, 0, 2}, DisplayFunction \[Rule] Identity]; \n plane2 = Plot3D[{\(-1\) - x + 2\ y, Hue[1\/3]}, {x, 0, 2}, {y, 0, 2}, DisplayFunction \[Rule] Identity]; \n plane3 = Plot3D[{1\/2\ \((\(-4\) + x + 3\ y)\), Hue[2\/3]}, {x, 0, 2}, { y, 0, 2}, DisplayFunction \[Rule] Identity]; \n plane4 = Plot3D[{1\/2 - x + y\/2, Hue[2\/3]}, {x, 0, 2}, {y, 0, 2}, DisplayFunction \[Rule] Identity]; \n plane5 = Plot3D[{2 - x - y, Hue[0]}, {x, 0, 2}, {y, 0, 2}, DisplayFunction \[Rule] Identity]; \n plane6 = Plot3D[{\(-x\) - y, Hue[1\/3]}, {x, 0, 2}, {y, 0, 2}, DisplayFunction \[Rule] Identity]; \n par1 = Plot3D[x\^2 + 2\ y\^2, {x, \(-1\), 1}, {y, \(-1\), 1}, DisplayFunction \[Rule] Identity]; \n sphere1 = ParametricPlot3D[{Cos[t]\ Cos[u], Sin[t]\ Cos[u], Sin[u]}, {t, 0, 2\ \[Pi]}, {u, \(-\(\[Pi]\/2\)\), \[Pi]\/2}, DisplayFunction \[Rule] Identity]; \n temp1 = ParametricPlot3D[{x, \(-\(\@\(\(-1\) - 4\ x\^2 + \@\(17 - 8\ x\^2\)\)\/\(2\ \@2\)\)\), 1\/4\ \((\(-1\) + \@\(17 - 8\ x\^2\))\)}, {x, \(-\(\@\(\@5 - 1\)\/\@2\)\), \@\(\@5 - 1\)\/\@2}, DisplayFunction \[Rule] Identity]; \n temp2 = ParametricPlot3D[{x, \@\(\(-1\) - 4\ x\^2 + \@\(17 - 8\ x\^2\)\)\/\(2\ \@2\), 1\/4\ \((\(-1\) + \@\(17 - 8\ x\^2\))\)}, {x, \(-\(\@\(\@5 - 1\)\/\@2\)\), \@\(\@5 - 1\)\/\@2}, DisplayFunction \[Rule] Identity]; \nintersect1 = Show[temp1, temp2]; \)], "Input", Editable->False, CellOpen->False, InitializationCell->True, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Plotting Functions", "Subsection"], Cell[TextData[{ "It will be useful to introduce plotting here. We'll learn much more about \ it later. The ", StyleBox["Plot[]", FontWeight->"Bold"], " function graphs a given function over a specified range." }], "Text", TextJustification->1], Cell[BoxData[ \(\(?Plot\)\)], "Input"], Cell[BoxData[ \(\(Plot[x\^3\/2 - x + 1, {x, \(-2\), 2}]; \)\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " will make an educated guess for the y-axis range that you would like to \ see. It may not show you the full range of the function." }], "Text", TextJustification->1] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Linear Equations"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "You can solve a system of equations by giving the ", StyleBox["Solve[]", "Input"], " function a list of equations and a list of variables for which to solve." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x + y == 1, x - 3\ y == 4}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " is able to solve any set of linear equations and many polynomial and \ trigonometric equations. For linear equations there are three possibilities \ for solutions. First, the set of equations may have a unique solution." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x + y + z == 2, x - 2\ y + z == \(-1\), x + 3\ y - 2\ z == 4}, { x, y, z}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If we plot the three planes described by the above equations we see that \ they intersect at one unique point. (Don't worry about the ", StyleBox["Show[]", FontWeight->"Bold"], " function yet. We'll cover that when we learn about plotting.)" }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(Show[plane1, plane2, plane3, DisplayFunction \[Rule] $DisplayFunction]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If the equations are not linearly independent, then there will be an \ infinite nmber of solutions. Here the solution is a line in ", Cell[BoxData[ \(TraditionalForm\`x\[VeryThinSpace]y\[VeryThinSpace]z\)]], " space." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x + y + z == 2, x - 2\ y + z == \(-1\), x - y\/2 + z == 1\/2}, { x, y, z}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(Show[plane1, plane2, plane4, DisplayFunction \[Rule] $DisplayFunction]; \)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If the equations are inconsistent, then there will be no solutions. ", StyleBox["Mathematica", FontSlant->"Italic"], " represents this as ", StyleBox["{}", "Input"], ". In the plot below, we see that two of the planes are parallel." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x + y + z == 2, x + y + z == 0, x + 3\ y - 2\ z == 4}, {x, y, z}] \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(Show[plane5, plane6, plane3, DisplayFunction \[Rule] $DisplayFunction]; \)\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 3", "Subsubsection"], Cell[TextData[{ "Using the ", StyleBox["Solve[]", FontWeight->"Bold"], " function, find a cubic polynomial that passes through the points: ", Cell[BoxData[ \(TraditionalForm\`\((1, 3)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\((2, 1)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\((3, 4)\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\((4, 1)\)\)]], ". Plot this function." }], "Text", TextJustification->1] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 4", "Subsubsection"], Cell[TextData[{ "Using the ", StyleBox["Solve[]", FontWeight->"Bold"], " function, solve the set of equations:\n\t", Cell[BoxData[ \(TraditionalForm\`x + y = 1\)]], ",\n\t", Cell[BoxData[ \(TraditionalForm\`x + a\ y = 2\)]], ".\nFor what values of the parameter, ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", is the solution valid? Now use the ", StyleBox["Reduce[]", FontWeight->"Bold"], " function to solve the same set of equations. Next do the same for the \ set of equations:\n\t", Cell[BoxData[ \(TraditionalForm\`x + y = 1\)]], ",\n\t", Cell[BoxData[ \(TraditionalForm\`x + a\ y = 1\)]], "." }], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData["Nonlinear Equations"], "Subsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Suppose we want to find the points where the parabola\n\t", Cell[BoxData[ \(TraditionalForm\`y = 4 x\^2\)]], "\nand the ellipse\n\t", Cell[BoxData[ \(TraditionalForm\`x\^2 + 2 y\^2 = 3\)]], "\nintersect. Here is a plot of the two functions. 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Cell[TextData[{ StyleBox["Solve[]", "Input"], " has returned four solutions instead of the two we expected. Why is this? \ Notice that in the first two solutions ", Cell[BoxData[ \(TraditionalForm\`x\)]], " is complex-valued. This is more clear if we find a numerical \ approximation to the solutions" }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(N[%] // TableForm\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Always remember that ", StyleBox["Mathematica", FontSlant->"Italic"], " finds both real and complex-valued solutions. Often when we visualize a \ problem we only think of the real-valued solutions. " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "Now suppose we want to find the intersection of the unit sphere and the \ parabloid ", Cell[BoxData[ \(TraditionalForm\`z\ = \ x\^2\ + \ 2\ y\^2\)]], ". Below the surfaces are shown from two different angles. We could look \ for a parametric representation of the intersection curve in terms of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". We see that there will be two solution curves, (the upper and lower \ halves of the curve). " }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Show[par1, sphere1, BoxRatios \[Rule] {1, 1, 2}, DisplayFunction \[Rule] $DisplayFunction]; \n Show[par1, sphere1, ViewPoint \[Rule] {2, \(-2\), 4}, BoxRatios \[Rule] {1, 1, 1}, DisplayFunction \[Rule] $DisplayFunction]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We use the ", StyleBox["Mathematica", FontSlant->"Italic"], " ", StyleBox["Solve[]", "Input"], " function to find ", Cell[BoxData[ \(TraditionalForm\`y\)]], " and ", Cell[BoxData[ \(TraditionalForm\`z\)]], " as functions of ", Cell[BoxData[ \(TraditionalForm\`x\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x\^2 + y\^2 + z\^2 == 1, z == x\^2 + 2\ y\^2}, {y, z}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Solve[]", "Input"], " returns four solutions. For the first two solutions, ", Cell[BoxData[ \(TraditionalForm\`y\)]], " is complex-valued for all values of ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". For the last two solutions, ", Cell[BoxData[ \(TraditionalForm\`y\)]], " and ", Cell[BoxData[ \(TraditionalForm\`z\)]], " will both be real-valued for certain values of ", Cell[BoxData[ \(TraditionalForm\`x\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "Now to find where the above two solutions are real-valued. ", Cell[BoxData[ \(TraditionalForm\`y\)]], " is real-valued when the argument of the outer square root function is \ real and positive. If ", Cell[BoxData[ \(TraditionalForm\`y\)]], " is real-valued, then ", Cell[BoxData[ \(TraditionalForm\`z\)]], " will also be real-valued. Hence we look for the range where ", Cell[BoxData[ \(TraditionalForm\`y\)]], " is real. We do this by finding where the argument of the square root \ vanishes." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[\(-1\) - 4\ x\^2 + \@\(17 - 8\ x\^2\) == 0, x]\)], "Input", AspectRatioFixed->True], Cell["\<\ Thus a parametric representation of the intersection curve is given \ by the two real-valued solutions in the interval defined above. Below the \ curve is plotted alone and on the surface of the unit spere.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Show[intersect1, DisplayFunction \[Rule] $DisplayFunction]; \n Show[intersect1, sphere1, Boxed \[Rule] False, DisplayFunction \[Rule] $DisplayFunction]; \)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 5", "Subsubsection"], Cell[TextData[{ "Using the Solve[] function, find the points where the parabolas ", Cell[BoxData[ \(TraditionalForm\`y = x\^2\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y = 2 x\^2 + a\ x + b\)]], " intersect in the real ", Cell[BoxData[ \(TraditionalForm\`x\[VeryThinSpace]y\)]], " plane. Consider all possible real values of the parameters ", Cell[BoxData[ \(TraditionalForm\`a\)]], " and ", Cell[BoxData[ \(TraditionalForm\`b\)]], ". There will be three distinct cases. Plot an example of each." }], "Text", TextJustification->1] }, Open ]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["*The Potential Evils of Subscripts.", "Section"], Cell[TextData[{ "The superscript has a mathematical meaning: raise the argument to a power. \ The subscript does not have a mathematica meaning in ", StyleBox["Mathematica", FontSlant->"Italic"], " and should not be used in inputs. Doing so will yield strange and \ misleading results." }], "Text", TextJustification->1], Cell[TextData[{ "Below we make some definitions and ask ", StyleBox["Mathematica", FontSlant->"Italic"], " what it knows about the variables. " }], "Text", TextJustification->1], Cell[BoxData[ \(Clear[a, b, c, d]\)], "Input"], Cell[BoxData[ \(a = 1\n?a\)], "Input"], Cell[BoxData[ \(b[1] = 1\n?b\)], "Input"], Cell[TextData[{ "Here we see that ", Cell[BoxData[ \(TraditionalForm\`c\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\_1\)]], " are not independent variables in ", StyleBox["Mathematica", FontSlant->"Italic"], ". " }], "Text"], Cell[BoxData[{ \(c = 1\), \(c\_1\)}], "Input"], Cell[TextData[{ "Assigning a value to ", Cell[BoxData[ \(TraditionalForm\`d\_1\)]], " is asking for trouble. ", StyleBox["Mathematica", FontSlant->"Italic"], " will not give you information about the variable." }], "Text", TextJustification->1], Cell[BoxData[ \(d\_1 = 1\n?d\_1\)], "Input"], Cell[TextData[{ "Asking about the variable ", Cell[BoxData[ \(TraditionalForm\`d\)]], " is no help either." }], "Text", TextJustification->1], Cell[BoxData[ \(\(?d\)\)], "Input"], Cell[BoxData[ \(d\)], "Input"], Cell[TextData[{ "However, you can see that ", StyleBox["Mathematica", FontSlant->"Italic"], " has stored the value of ", Cell[BoxData[ \(TraditionalForm\`d\_1\)]], "." }], "Text"], Cell[BoxData[ \(d\_1\)], "Input"], Cell[TextData[{ "Strangly, clearing the definition of ", Cell[BoxData[ \(TraditionalForm\`d\_1\)]], " or ", Cell[BoxData[ \(TraditionalForm\`d\)]], " does not get rid of this definition." }], "Text", TextJustification->1], Cell[BoxData[ \(Clear[d\_1]\)], "Input"], Cell[BoxData[ \(Clear[d]\)], "Input"], Cell[BoxData[ \(d\_1\)], "Input"], Cell[TextData[{ "You have to remove the variable ", Cell[BoxData[ \(TraditionalForm\`d\)]], " to clear the definition of ", Cell[BoxData[ \(TraditionalForm\`d\_1\)]], "." }], "Text", TextJustification->1], Cell[BoxData[ \(Remove[d]\)], "Input"], Cell[BoxData[ \(d\_1\)], "Input"], Cell[BoxData[ \(Clear[a, b, c, d]\)], "Input"], Cell["\<\ Sometimes you can get away with using subscripts in inputs. But it \ is risky and error-prone.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(Solve[\(a\_2\) x\^2 + \(a\_1\) x + a\_0 == 0, x]\)], "Input"], Cell["Use one of the following forms instead.", "Text", TextJustification->1], Cell[BoxData[ \(Solve[a2\ x\^2 + a1\ x + a0 == 0, x]\)], "Input"], Cell[BoxData[ \(Solve[a[2]\ x\^2 + a[1]\ x + a[0] == 0, x]\)], "Input"], Cell[TextData[{ "If you really want to see constants with subscripts in the output, use ", Cell[BoxData[ \(TraditionalForm\`C[i]\)]], "'s for the constants and display the result in ", StyleBox["TraditionalForm", FontWeight->"Bold"], "." }], "Text", TextJustification->1], Cell[BoxData[ \(Solve[C[2]\ x\^2 + C[1]\ x + C[0] == 0, x]\)], "Input"], Cell[BoxData[ \(% // TraditionalForm\)], "Input"], Cell["\<\ So in summary: Only use subscripts for expressions that are meant \ to be read, they are not suitable for input.\ \>", "Text", TextJustification->1] }, Closed]], Cell[CellGroupData[{ Cell["*Immediate and Delayed Assignment", "Section"], Cell[TextData[{ "Above I indicated that one should use the assignment operator, ", Cell[BoxData[ \(TraditionalForm\` = \)]], ", to assign values to constants and the delayed assignment operator, ", Cell[BoxData[ \(TraditionalForm\` := \)]], ", to define functions. This is a rule of thumb that usually gives the \ desired result. 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Everytime \ rand is encountered on input, a new random number is generated.\ \>", "Text", TextJustification->1], Cell[BoxData[ \(rand := Random[Integer, {0, 100}]\)], "Input"], Cell[BoxData[{ \(rand\), \(rand\), \(rand\)}], "Input"], Cell[BoxData[ \(Clear[rand]\)], "Input"], Cell["The following commands illustrate a subtle mistake.", "Text", TextJustification->1], Cell[BoxData[ \(Solve[3 x + 5 y == 1, y]\)], "Input"], Cell[BoxData[ \(y[x_] := y /. First[%]\)], "Input"], Cell[BoxData[ \(\(Plot[y[x], {x, \(-1\), 1}]; \)\)], "Input"], Cell[TextData[{ "The solution of the linear equation is not stored in ", Cell[BoxData[ \(TraditionalForm\`y[x]\)]], ". Instead the command \"apply the first part of the previous expression \ to ", Cell[BoxData[ \(TraditionalForm\`y\)]], "\" is stored. 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If ", Cell[BoxData[ \(TraditionalForm\`a\)]], " is zero and ", Cell[BoxData[ \(TraditionalForm\`b\)]], " is nonzero then there is one solution, ", Cell[BoxData[ \(TraditionalForm\`x\ \[Equal] \ \(\(-c\)\ \)\/b\)]], ". If ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", ", Cell[BoxData[ \(TraditionalForm\`b\)]], " and ", Cell[BoxData[ \(TraditionalForm\`c\)]], " are all zero then all values of ", Cell[BoxData[ \(TraditionalForm\`x\)]], " are a solution to the equation." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Solution 3", "Subsubsection"], Cell["\<\ First we define a function to be the general cubic polynomial.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\(cubic[x_] := a\ x\^3 + b\ x\^2 + c\ x + d; \)\)], "Input", AspectRatioFixed->True], Cell["Now we solve a list of equations for the coefficients.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{cubic[1] == 3, cubic[2] == 1, cubic[3] == 4, cubic[4] == 1}, {a, b, c, d}]\)], "Input", AspectRatioFixed->True], Cell["The polynomial solution is:", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(soln[x_] = cubic[x] /. {a \[Rule] \(-\(11\/6\)\), b \[Rule] 27\/2, c \[Rule] \(-\(89\/3\)\), d \[Rule] 21}\)], "Input", AspectRatioFixed->True], Cell["Here is a plot of the solution.", "Text"], Cell[BoxData[ \(\(Plot[soln[x], {x, 1, 4}]; \)\)], "Input"], Cell[BoxData[ \(Clear[cubic, soln]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Solution 4", "Subsubsection"], Cell[BoxData[ \(Solve[{x + y == 1, x + a\ y == 2}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Due to the ", Cell[BoxData[ \(TraditionalForm\`\((a - 1)\)\)]], " in the denominator, the solution is valid when ", Cell[BoxData[ \(TraditionalForm\`a \[NotEqual] 1\)]], ". When ", Cell[BoxData[ \(TraditionalForm\`a = 1\)]], " the equations are inconsistent." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Reduce[{x + y == 1, x + a\ y == 2}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using the ", StyleBox["Reduce[]", "Input"], " function tells us explicitly that there are only solutions if ", Cell[BoxData[ \(TraditionalForm\`a \[NotEqual] 1\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{x + y == 1, x + a\ y == 1}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The above solution is valid for all values of ", Cell[BoxData[ \(TraditionalForm\`a\)]], ", but does not give the full set of solutions when ", Cell[BoxData[ \(TraditionalForm\`a = 1\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Reduce[{x + y == 1, x + a\ y == 1}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Using ", StyleBox["Reduce[]", "Input"], " shows us that if ", Cell[BoxData[ \(TraditionalForm\`a = 1\)]], " then there are an infinite number of solutions of the form ", Cell[BoxData[ \(TraditionalForm\`x = 1 - y\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Solution 5", "Subsubsection"], Cell["First solve the set of equations for general parameters.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Solve[{y == x\^2, y == 2\ x\^2 + a\ x + b}, {x, y}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "These solutions are real if the argument of the square root is \ non-negative. If ", Cell[BoxData[ \(TraditionalForm\`a\^2 > 4 b\)]], " then there are two distinct real solutions. The curves intersect at two \ points given by the above set of rules. If ", Cell[BoxData[ \(TraditionalForm\`a\^2 = 4 b\)]], " then the solutions are no longer distinct. The curves intersect at a \ single point given by either of the above rules. If ", Cell[BoxData[ \(TraditionalForm\`a\^2 < 4 b\)]], " then the solutions are complex-valued. 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*******************************************************************)