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Cell[CellGroupData[{
Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " Tutorial"
}], "Title"],

Cell[CellGroupData[{

Cell["Algebraic Manipulations", "Section"],

Cell[BoxData[
    \(\(Expression1 = 
        EnzS[t]\ \((1 + \(k1b + k2f\)\/\(\(k1f\)\(\ \)\))\) \[Equal] 
          S[t];\)\)], "Input",
  CellLabel->"In[11]:="],

Cell[TextData[{
  "In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", every object (from equations, to numerical solutions and graphics) is \
represented as an expression. All expressions have a 'Head', that tells you \
what type of expression the object is. In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", one can find out information of commands by typing ",
  StyleBox["?Command",
    FontWeight->"Bold"],
  ". So, to find out what head does, type and evaluate:"
}], "Text"],

Cell[BoxData[
    \(\(?Head\)\)], "Input",
  CellLabel->"In[17]:="],

Cell[TextData[{
  "Now find out what head ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  " has."
}], "Text"],

Cell[BoxData[""], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", functions can be written in the ",
  StyleBox["Prefix",
    FontWeight->"Bold"],
  " form as well as the ",
  StyleBox["Postfix",
    FontWeight->"Bold"],
  " form. As an example for showing the two (equivalent) ways to input \
expressions, evaluate and compare the following:"
}], "Text"],

Cell[BoxData[{
    \(Head[Expression1]\[IndentingNewLine]\), "\[IndentingNewLine]", 
    \(Expression1 // Head\)}], "Input"],

Cell[TextData[{
  "What does it mean that objects are expressions? To look at how ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  " is represented, use the command ",
  StyleBox["FullForm",
    FontWeight->"Bold"],
  ":"
}], "Text"],

Cell[BoxData[
    \(Expression1 // FullForm\)], "Input"],

Cell[TextData[{
  "As another (visual) representation of the same expression, use ",
  StyleBox["TreeForm",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(Expression1 // TreeForm\)], "Input"],

Cell[TextData[{
  "How do we extract elements of expressions? Use commands ",
  StyleBox["Last",
    FontWeight->"Bold"],
  " and ",
  StyleBox["First",
    FontWeight->"Bold"],
  ". Find out what they do below, and find out what are the ",
  StyleBox["Last",
    FontWeight->"Bold"],
  " and ",
  StyleBox["Head",
    FontWeight->"Bold"],
  " of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  "."
}], "Text"],

Cell[BoxData[""], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "The command ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " is useful in cancelling factors in a common denominator. Find out what it \
does, and factor the left-hand-side of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ", using combination of ",
  StyleBox["First",
    FontWeight->"Bold"],
  " and ",
  StyleBox["Together",
    FontWeight->"Bold"],
  ". "
}], "Text"],

Cell[BoxData[""], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "For comparison, we see that applying ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " on ",
  StyleBox["Expression 1",
    FontWeight->"Bold"],
  " fails to get the desired result."
}], "Text"],

Cell[BoxData[
    \(Expression1 // Together\)], "Input"],

Cell[TextData[{
  "The problem arises from the fact that the term ",
  Cell[BoxData[
      \(\((1 + \(k1b + k2f\)\/k1f)\)\)]],
  "is a subexpression, i.e., not at the first level of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ". To obtain the desired result, we should map ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " to all subexpressions. Find out what ",
  StyleBox["MapAll",
    FontWeight->"Bold"],
  " does, and apply it to ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ". Complete the following:"
}], "Text"],

Cell[BoxData[
    \(\(?MapAll\)\)], "Input"],

Cell[BoxData[
    \(\(\(Together\)\(//@\)\(\ \ \ \ \ \)\)\)], "Input",
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Cell["\<\
The command Thread is useful in writing down a system of equations, \
equating a list of left-hand-side to a list of right-hand-side. Simply \
equating the lists do not work:\
\>", "Text"],

Cell[BoxData[{
    \(\(LHS = {a, b, c};\)\), "\[IndentingNewLine]", 
    \(\(\(RHS = {1, 2, 3};\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(LHS \[Equal] RHS\)}], "Input"],

Cell[TextData[{
  "To equate the individual terms, one needs to apply ",
  StyleBox["Thread",
    FontWeight->"Bold"],
  "; see what happens:"
}], "Text"],

Cell[BoxData[
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Cell[CellGroupData[{

Cell["Perturbation Analysis", "Subsection"],

Cell["\<\
Suppose one has algebraic equation obtained from perturbative \
analysis, e.g.,\
\>", "Text"],

Cell[BoxData[
    \(\(PertEqn = \((La\  + \ Lb\ \[Epsilon])\)*\((Lc + Ld\ \[Epsilon])\) + 
            Lf\ \[Epsilon]\^2 \[Equal] \ 
          Ra\  + \ \((Rb + Ra)\)*\[Epsilon]\  + \ 
            Rc\ \[Epsilon]\^2;\)\)], "Input"],

Cell["\<\
One would like to equate equal powers of left and right-hand sides.\
\
\>", "Text"],

Cell["Firstly, obtain the LHS of PertEqn below:", "Text"],

Cell[BoxData[{
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Cell[TextData[{
  "The terms of each order in ",
  "\[Epsilon] ",
  "can be obtained using the command ",
  StyleBox["CoefficientList",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(\(?CoefficientList\)\)], "Input"],

Cell[TextData[{
  "Using ",
  StyleBox["CoefficientList",
    FontWeight->"Bold"],
  ", obtain a list of equations matching LHS and RHS, for each order in ",
  StyleBox["\[Epsilon]",
    FontWeight->"Bold"],
  ". Hint: use ",
  StyleBox["Thread",
    FontWeight->"Bold"],
  " to equate each term in the LHS and RHS lists. Complete the following:"
}], "Text"],

Cell[BoxData[{
    \(\(LHSCoefficientList = 
        CoefficientList[\ , \ ];\)\), "\[IndentingNewLine]", 
    \(\(\(\(\(RHSCoefficientList\)\(=\)\)\ \ \ \ ;\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(\(OrderEqn\  = \ \ Thread[\ \ \ \ ];\)\ \)}], "Input",
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Cell[TextData[{
  "\nNow, let's extract the RHS unknowns, based on the following hint using \
the command ",
  StyleBox["Cases",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(\(?Cases\)\)], "Input"],

Cell[TextData[{
  "As an example on using ",
  StyleBox["Cases",
    FontWeight->"Bold"],
  " to extract all symbols from a expression, evaluate below:"
}], "Text"],

Cell[BoxData[
    \(Cases[{a + b, \ c*d/f + f, \ 2.0*5 + g}, \ _Symbol, \ 
      Infinity]\)], "Input"],

Cell[TextData[{
  "Try a call to Cases, apply it to ",
  StyleBox["RHSCoefficientList",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(\(\(\(RHS\[UnderBracket]Unknowns\)\(=\)\)\ \ \ \ ;\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "Now, ",
  StyleBox["RHS\[UnderBracket]Unknowns",
    FontWeight->"Bold"],
  " contains repeated terms. Remove these redundant elements, by using the \
command Union:"
}], "Text"],

Cell[BoxData[
    \(\(?Union\)\)], "Input"],

Cell[TextData[{
  "Apply ",
  StyleBox["Union ",
    FontWeight->"Bold"],
  "to ",
  StyleBox["RHS\[UnderBracket]Unknowns",
    FontWeight->"Bold"],
  StyleBox[", and over-write the result:",
    FontVariations->{"CompatibilityType"->0}]
}], "Text"],

Cell[BoxData[
    \(\(\(\(\(RHS\[UnderBracket]Unknowns\)\(=\)\)\ \ \ \ \ ;\)\(\ \ \)\)\)], \
"Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "Let's now solve ",
  StyleBox["OrderEqn",
    FontWeight->"Bold"],
  ", with respect to ",
  StyleBox["RHS\[UnderBracket]Unknowns",
    FontWeight->"Bold"],
  ", using the command ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  ":"
}], "Text"],

Cell[BoxData[
    \(\(?Solve\)\)], "Input"],

Cell[BoxData[
    \(\(\(\(RHS\[UnderBracket]Soln = \ Solve[\ \ \ ];\)\[IndentingNewLine]
    % // ColumnForm\)\(\ \ \ \ \ \ \ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Numerical Integration", "Section"],

Cell[TextData[{
  "To numerically integrate ordinary differential equations, the function ",
  StyleBox["NDSolve",
    FontWeight->"Bold"],
  " is used."
}], "Text"],

Cell[BoxData[
    \(\(?NDSolve\)\)], "Input"],

Cell[TextData[{
  "Lets now solve the ODE system\n\n",
  Cell[BoxData[
      FormBox[
        StyleBox[\(dy\/dt = \(-y\) + 1\),
          FontColor->RGBColor[0, 0, 1]], TraditionalForm]]],
  "\n\nsubject to the initial condition ",
  StyleBox["y[0]=2",
    FontColor->RGBColor[0, 0, 1]],
  ". \n\nComplete the following"
}], "Text"],

Cell[BoxData[
    \(\(?D\)\)], "Input"],

Cell[BoxData[
    \(\(ODE = D[\ \ , ]\ \ \ ;\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell["\<\

Also, we need the initial condition. Complete below:\
\>", "Text"],

Cell[BoxData[
    \(\(\(\(IC\)\(\ \)\(=\)\)\ \ \ ;\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "\nNow, combine ",
  StyleBox["ODE",
    FontWeight->"Bold"],
  " and ",
  StyleBox["IC",
    FontWeight->"Bold"],
  " into a single List, and call it ",
  StyleBox["ODE\[UnderBracket]IC",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(\(\(\(ODE\[UnderBracket]IC\)\(=\)\)\ \ ;\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[TextData[{
  "Now, using ",
  StyleBox["NDSolve",
    FontWeight->"Bold"],
  ", compute the solution from t=0 to t=5."
}], "Text"],

Cell[BoxData[
    \(\(\(\(NumericSoln\)\(=\)\)\ \ ;\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell["Lets now plot the solution.", "Text"],

Cell[BoxData[
    \(\(?Plot\)\)], "Input"],

Cell["Complete the following:", "Text"],

Cell[BoxData[
    \(\(Plot[\ \(y[t] /. NumericSoln\)\(,\)\ \ \ ];\)\)], "Input",
  Background->RGBColor[0, 1, 1]]
}, Open  ]]
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(*******************************************************************
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*******************************************************************)