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Notebook[{

Cell[CellGroupData[{
Cell["Solving Equations, Inequalities, and Logical Equations", "Title"],

Cell[CellGroupData[{

Cell["Solving single equations", "Section"],

Cell[TextData[{
  "The most commonly used command for solving equations is the ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  " command.  You can solve a single equation for one variable with ",
  StyleBox["Solve[",
    FontFamily->"Courier"],
  StyleBox["equation",
    FontFamily->"Courier",
    FontSlant->"Italic"],
  StyleBox[", ",
    FontFamily->"Courier"],
  StyleBox["variable",
    FontFamily->"Courier",
    FontSlant->"Italic"],
  StyleBox["]",
    FontFamily->"Courier"],
  ".  You must use two equal signs in the equation, such as ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 + x \[Equal] 2\)]],
  ", to distinguish it from an assignment, such as ",
  StyleBox["x",
    FontSlant->"Italic"],
  " = 5 which assigns the value 5 to the variable ",
  StyleBox["x",
    FontSlant->"Italic"],
  ".  Here is an example of ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  ".  ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s ",
  StyleBox["Log",
    FontFamily->"Courier"],
  " function represents the natural logarithm."
}], "Text"],

Cell[BoxData[{
    \(Clear[a, \ b, \ x]\), "\[IndentingNewLine]", 
    \(Solve[Log[a\ x - 1] \[Equal] b, x]\)}], "Input"],

Cell["\<\
That equation had one solution.  The next one has several \
solutions.\
\>", "Text"],

Cell[BoxData[{
    \(Clear[a, \ b, \ c, \ t]\), "\[IndentingNewLine]", 
    \(Solve[a\ t\^4 + b\ t\^2 + c \[Equal] 2, t]\)}], "Input"],

Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " displays the results as a list of solutions.  Each solution is a list of \
replacement rules for the variables.  In this case, we are only solving for \
one variable, so each inner list only contains one replacement rule.  These \
replacement rules can then be used in subsequent calculations.  For example, \
suppose that you wanted to list the square of each of the solutions above.  \
The first line of output will be a list of the solutions presented as \
replacement rules, and the second line of output will be a list of the \
squares of those solutions.  ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s ",
  StyleBox["ReplaceAll",
    FontFamily->"Courier"],
  " operator, represented by /., means to make the replacements indicated on \
the right side to the expression on the left side."
}], "Text"],

Cell[BoxData[{
    \(Clear[a, b, c, t]\), "\[IndentingNewLine]", 
    \(solutions = 
      Solve[a\ t\^4 + b\ t\^2 + c \[Equal] 2, t]\), "\[IndentingNewLine]", 
    \(t\^2 /. solutions\)}], "Input"],

Cell["We can also check our solutions.", "Text"],

Cell[BoxData[{
    \(Clear[a, b, c, t]\), "\[IndentingNewLine]", 
    \(Simplify[\((a\ t\^4 + b\ t\^2 + c \[Equal] 2)\) /. 
        solutions]\)}], "Input"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Solve[\[ExponentialE]\^\(3  x\) \[Equal] \[ExponentialE]\^\(4  x\) + 2, 
      x]\)}], "Input"],

Cell[TextData[{
  "Sometimes you can get strange results from ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  "."
}], "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Solve[\[ExponentialE]\^\(3  x\) \[Equal] \[ExponentialE]\^\(5  x\) + 2, 
      x]\)}], "Input"],

Cell[TextData[{
  "The solution of this equation requires solving a fifth degree polynomial.  \
If you make the substitution ",
  StyleBox["u",
    FontSlant->"Italic"],
  " = ",
  Cell[BoxData[
      \(TraditionalForm\`\[ExponentialE]\^x\)]],
  ", you will see why.  There are not explicit formulas for solving fifth \
degree polynomials, so ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " only partially solved this equation.  The expression ",
  Cell[BoxData[
      \(Root[2 - #1\^3 + #1\^5 &, 1]\)]],
  " stands for the first root of the function 2 - ",
  Cell[BoxData[
      \(TraditionalForm\`x\^3 + x\^5\)]],
  ".  The first replacement rule in the solution says that if you find the \
first root of this polynomial, let's call it ",
  Cell[BoxData[
      \(TraditionalForm\`\(\(r\_1\)\(,\)\)\)]],
  " then a solution to the original problem is \[ImaginaryI] \[Pi] + Log[",
  Cell[BoxData[
      \(TraditionalForm\`r\_1\)]],
  "].  This is a complex number, with \[ImaginaryI] = ",
  Cell[BoxData[
      \(TraditionalForm\`\@\(-1\)\)]],
  ".  If you change ",
  Cell[BoxData[
      \(TraditionalForm\`\[ExponentialE]\^\(5  x\)\)]],
  " to ",
  Cell[BoxData[
      \(TraditionalForm\`\[ExponentialE]\^\(4  x\)\)]],
  " in the equation above, then the solution will require solving a fourth \
degree polynomial.  This does have an explicit formula, so ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " will give the exact solution."
}], "Text"],

Cell[TextData[{
  "Although ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " cannot find the exact solution to ",
  Cell[BoxData[
      \(\[ExponentialE]\^\(3  x\) \[Equal] \[ExponentialE]\^\(5  x\) + 2\)]],
  ", you can ask for an approximate solution, using ",
  StyleBox["NSolve",
    FontFamily->"Courier"],
  " or ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  ".  ",
  StyleBox["NSolve",
    FontFamily->"Courier"],
  " has the same syntax as ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  "."
}], "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(NSolve[\[ExponentialE]\^\(3  x\) \[Equal] \[ExponentialE]\^\(5  x\) + 
          2, x]\)}], "Input"],

Cell["That's better!", "Text"],

Cell[TextData[{
  StyleBox["Exercise 2.1:",
    FontWeight->"Bold",
    FontColor->RGBColor[0, 0.500008, 0.250004]],
  " Use the second derivative test to find the relative maxima and mimima of \
",
  StyleBox["f",
    FontSlant->"Italic"],
  "(",
  StyleBox["x",
    FontSlant->"Italic"],
  ") = 3",
  Cell[BoxData[
      \(TraditionalForm\`x\^5 - 4  x\^3 + 2  x - 1\)]],
  ".  Try to do this as elegantly and consicely as possible, using the \
techniques above."
}], "Text",
  Background->GrayLevel[0.750011]],

Cell[TextData[{
  "Sometimes, ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  " does not give you all of the solutions.  For example, it ignores special \
cases that may arise for specific values of parameters.  It only give general \
solutions.  We get the familiar quadratic formula when we ask for a solution \
of the quadratic equation."
}], "Text"],

Cell[BoxData[{
    \(Clear[a, b, c, x]\), "\[IndentingNewLine]", 
    \(Solve[a\ x\^2 + b\ x + c \[Equal] 0, x]\)}], "Input"],

Cell["\<\
However, there is a specific value for one of the parameters, a, b, \
or c, that gives a different solution.  What is that specific solution?  \
Hint:  Look to see which values of the parameters are not allowed in the \
general solution.\
\>", "Text"],

Cell[TextData[{
  StyleBox["Reduce",
    FontFamily->"Courier"],
  " gives you a more complete solution.  It presents the solution as an \
equivalent logical expression.  The symbols ",
  StyleBox["||",
    FontFamily->"Courier"],
  " means \"or\", ",
  StyleBox["&&",
    FontFamily->"Courier"],
  " means \"and\".  This notation is borrowed from most computer programming \
languages."
}], "Text"],

Cell[BoxData[{
    \(Clear[a, b, c, x]\), "\[IndentingNewLine]", 
    \(Reduce[a\ x\^2 + b\ x + c \[Equal] 0, x]\)}], "Input"],

Cell[TextData[{
  "Solve can also miss some solutions because it uses the standard inverse \
functions, which are based on restricted domains.  For example, ",
  StyleBox["ArcSin",
    FontFamily->"Courier"],
  " is the inverse of the ",
  StyleBox["Sin",
    FontFamily->"Courier"],
  " function with restricted domain ",
  Cell[BoxData[
      \(TraditionalForm\`\(-\(\[Pi]\/2\)\) \[LessEqual] \[Theta] \[LessEqual] \
\[Pi]\/2\)]],
  ".  So, when Solve uses ",
  StyleBox["ArcSin",
    FontFamily->"Courier"],
  " to solve the following equation, it only gives one solution instead of an \
infinite number of them."
}], "Text"],

Cell[BoxData[{
    \(Clear[\[Theta]]\), "\[IndentingNewLine]", 
    \(Solve[2  Sin[2  \[Theta] + 1] == 1\/2, \[Theta]]\)}], "Input"],

Cell[TextData[{
  "At least ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " was kind enough to give us a warning that it may have missed some \
solutions.  Try using ",
  StyleBox["Reduce",
    FontFamily->"Courier"],
  " instead and see what happens.  You can use ",
  StyleBox["N[ ]",
    FontFamily->"Courier"],
  ", too, if you prefer to get decimal approximations instead of complicated \
exact expressions."
}], "Text"],

Cell["\<\
If you want to restrict the solution set, then you have to use \
Reduce.  For example, here are all of the solutions to a cubic \
equation.\
\>", "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Solve[2  x\^3 - x\^2 - x \[Equal] 0, x]\)}], "Input"],

Cell["and here are the integer solutions.", "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Reduce[2  x\^3 - x\^2 - x \[Equal] 0\ , x, Integers]\)}], "Input"],

Cell["and here are the solutions between -1 and 0.", "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Reduce[2  x^2 - x - 1 \[Equal] 0\  && \(-1\) \[LessEqual] x && x < 0, 
      x]\)}], "Input"],

Cell["\<\
You can also restrict solutions to Rationals, Algebraics, Reals, \
Complexes, Primes, and Booleans.\
\>", "Text"],

Cell["\<\
You can use Reduce along with Exists and ForAll (these are called \
logical quantifiers) to answer questions about the existence of solutions and \
a variety of other questions.  For example, we can check to see the \
conditions on the coefficients, a, b, and c,  for which a quadratic equation \
has a root equal to 3.\
\>", "Text"],

Cell[BoxData[{
    \(Clear[a, b, c, x]\), "\[IndentingNewLine]", 
    \(Reduce[
      Exists[x, a\ x\^2 + b\ x + c \[Equal] 0\  && \ x\  \[Equal] \ 3], {a, 
        b, c}]\)}], "Input"],

Cell[TextData[{
  StyleBox["Exercise 2.2:",
    FontWeight->"Bold",
    FontColor->RGBColor[0, 0.500008, 0.250004]],
  " For what values of the line's slope will a line through the origin \
intersect the circle of radius 2 centered at (3, 5)?"
}], "Text",
  Background->GrayLevel[0.750011]],

Cell[TextData[{
  "If ",
  StyleBox["NSolve",
    FontFamily->"Courier"],
  " fails to solve your equations, then you can resort to ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  ".  Unlike ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  " and ",
  StyleBox["Reduce",
    FontFamily->"Courier"],
  ", though, ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  " can only find numerical solutions.  Your equations cannot include any \
arbitrary parameters.  ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  " requires that you provide one or two intial guesses and it will use them \
to find one solution.  If you provide one initial guess, ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  " uses Newton's method.  If you provide two initial guesses, ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  " approximates the derivative in Newton's method.  If your function \
involves only real coefficients and you want to find a complex root, you must \
start with a complex initial value.  The first arguemnt in ",
  StyleBox["FindRoot",
    FontFamily->"Courier"],
  " is the equation and the second arguement is a list containing the \
variable and the initial guess(es).  The example below finds a root of the \
equation sin(",
  Cell[BoxData[
      \(TraditionalForm\`x\^2 + 1\)]],
  ") = ",
  Cell[BoxData[
      \(TraditionalForm\`1\/3\)]],
  "with initial guess ",
  StyleBox["x",
    FontSlant->"Italic"],
  " = 1."
}], "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(FindRoot[Sin[x\^2 + 1] \[Equal] 1\/3, {x, 1}]\)}], "Input"],

Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " has a collection of commands that you can use to perform algebraic \
transformations of expressions.  Sometimes you have to resort to using these \
to get an equation into a form that ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  " can handle.  Fortunately, this rarely happens.  But you should take a \
quick look at these commands in ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  "'s Help Browser, just so you know what is available.  Look at ",
  StyleBox["Simplify",
    FontFamily->"Courier"],
  ", ",
  StyleBox["Expand",
    FontFamily->"Courier"],
  ",  ",
  StyleBox["ExpandNumerator",
    FontFamily->"Courier"],
  ", ",
  StyleBox["ExpandDenominator",
    FontFamily->"Courier"],
  ", ",
  StyleBox["RootReduce",
    FontFamily->"Courier"],
  ", ",
  StyleBox["ToRadicals",
    FontFamily->"Courier"],
  ", ",
  StyleBox["TrigExpand",
    FontFamily->"Courier"],
  ", ",
  StyleBox["TrigFactor",
    FontFamily->"Courier"],
  ", ",
  StyleBox["TrigReduce",
    FontFamily->"Courier"],
  ", ",
  StyleBox["TrigToExp",
    FontFamily->"Courier"],
  ", and ",
  StyleBox["ExpToTrig",
    FontFamily->"Courier"],
  ". "
}], "Text"],

Cell[TextData[{
  StyleBox["Exercise 2.3:",
    FontWeight->"Bold",
    FontColor->RGBColor[0, 0.500008, 0.250004]],
  " Suppose that a sphere of radius 10 cm and density 0.2 g/",
  Cell[BoxData[
      \(TraditionalForm\`cm\^3\)]],
  "is dropped into water with density 1.0 g/",
  Cell[BoxData[
      \(TraditionalForm\`cm\^3\)]],
  ".  The sphere will sink to the depth at which the mass of the displaced \
water equals the mass of the entire sphere.  To what depth will the sphere \
sink?"
}], "Text",
  Background->GrayLevel[0.750011]]
}, Closed]],

Cell[CellGroupData[{

Cell["Solving systems of equations", "Section"],

Cell[TextData[{
  "You can solve systems of equations using ",
  StyleBox["Solve",
    FontFamily->"Courier"],
  " or ",
  StyleBox["NSolve",
    FontFamily->"Courier"],
  ".  Both of these commands have two arguments.  The first argument is the \
equation or list of equations and the second argument is the variable or \
variables to be solved for.  Lists are enclosed in braces, { }.  Here is an \
example in which we solve a list of two equaitons for a list of two \
variables.  Geometrically, we are checking to see if a line intersects a \
quartic function."
}], "Text"],

Cell[BoxData[{
    \(Clear[x, y]\), "\[IndentingNewLine]", 
    \(Solve[{2  x + y \[Equal] 1, y == x\^4 - 3  x\^3 + x - 1}, {x, 
        y}]\)}], "Input"],

Cell["At how many points do the curves intersect?", "Text"],

Cell[TextData[{
  StyleBox["Exercise 2.4:",
    FontWeight->"Bold",
    FontColor->RGBColor[0, 0.500008, 0.250004]],
  " Where does ",
  StyleBox["y",
    FontSlant->"Italic"],
  " = ",
  Cell[BoxData[
      \(TraditionalForm\`x\^4\)]],
  " intersect the circle of radius 3 centered at (0, 5)?"
}], "Text",
  Background->GrayLevel[0.750011]]
}, Closed]],

Cell[CellGroupData[{

Cell["Solving inequalities", "Section"],

Cell["You can use Reduce to solve inequalities, too.", "Text"],

Cell[BoxData[{
    \(Clear[x]\), "\[IndentingNewLine]", 
    \(Reduce[1 < Tan[x] < \@3\  && \ x \[Element] Reals, x]\)}], "Input"],

Cell[TextData[{
  StyleBox["Exercise 2.5:",
    FontWeight->"Bold",
    FontColor->RGBColor[0, 0.500008, 0.250004]],
  " Find the ordered pairs of integers above the parabola ",
  StyleBox["y",
    FontSlant->"Italic"],
  " = ",
  Cell[BoxData[
      \(TraditionalForm\`x\^2\)]],
  "and below the line ",
  StyleBox["y",
    FontSlant->"Italic"],
  " = 4."
}], "Text",
  Background->GrayLevel[0.750011]]
}, Closed]]
}, Open  ]]
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