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Notebook[{
Cell["\<\
Examples for Section 8.11
Applications of Power Series\
\>", "Subsection"],

Cell["p. 697", "Subsection"],

Cell[CellGroupData[{

Cell["Binomial Series", "Subsection"],

Cell[CellGroupData[{

Cell["7.", "Subsubsection"],

Cell[TextData[{
  "Find the first four terms of the binomial series for ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\  = \ \((1 + x\^3)\)\^\(\(-1\)/2\)\)]],
  ".\n\n",
  StyleBox["Solution.",
    FontSlant->"Italic"],
  " The binomial formula is: \n",
  Cell[BoxData[
      FormBox[
        RowBox[{\(\((1 + u)\)\^m\), "=", 
          RowBox[{
            RowBox[{"1", "+", \(m\ u\), " ", "+", " ", 
              RowBox[{
                RowBox[{"(", GridBox[{
                      {"m"},
                      {"2"}
                      }], ")"}], \(u\^2\)}], "+", " ", 
              RowBox[{
                RowBox[{"(", GridBox[{
                      {"m"},
                      {"3"}
                      }], ")"}], \(u\^3\)}], "+"}], "..."}]}], 
        TraditionalForm]]],
  "\nWhen ",
  Cell[BoxData[
      \(TraditionalForm\`\(m = \)\)]],
  "-1/2 and ",
  Cell[BoxData[
      \(TraditionalForm\`u = x\^3\)]],
  " we get: \n",
  Cell[BoxData[
      FormBox[
        RowBox[{\(\((1 + x\^3)\)\^\(\(-1\)/2\)\), "=", 
          RowBox[{
            RowBox[{
              RowBox[{"1", "+", \(\((\(-\(1\/2\)\))\) x\^3\), "+", 
                RowBox[{
                  RowBox[{"(", GridBox[{
                        {\(\(-1\)/2\)},
                        {"2"}
                        }], ")"}], \(\((x\^3)\)\^2\)}], "+", 
                RowBox[{
                  RowBox[{"(", GridBox[{
                        {\(\(-1\)/2\)},
                        {"3"}
                        }], ")"}], \(\((x\^3)\)\^3\)}], "+"}], "..."}], " ", 
            "=", \(1 - \(1\/2\) x\^3 + 
                  \(\(\ \((\(-1\)/2)\) \((\(-3\)/2)\)\)\/2\) x\^6 + 
                  \(\(\ \((\(-1\)/2)\) \((\(-3\)/2)\) \((\(-5\)/2)\)\)\/\(3*2
                          \)\) x\^9 +  ...  = 
              1 - \(1\/2\) x\^3 + \(\(\ 3\)\/8\) x\^6 - \(5\/16\) x\^9 +  ... 
                \)}]}], TraditionalForm]]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[\((1 + x\^3)\)\^\(\(-1\)/2\), {x, 0, 9}]\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      "1", "-", \(x\^3\/2\), "+", \(\(3\ x\^6\)\/8\), "-", 
        \(\(5\ x\^9\)\/16\), "+", 
        InterpretationBox[\(O[x]\^10\),
          SeriesData[ x, 0, {}, 0, 10, 1]]}],
      SeriesData[ x, 0, {1, 0, 0, 
        Rational[ -1, 2], 0, 0, 
        Rational[ 3, 8], 0, 0, 
        Rational[ -5, 16]}, 0, 10, 1]]], "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["8.", "Subsubsection"],

Cell[TextData[{
  "Find the first four terms of the binomial series for ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\  = \ \((1 + x\^2)\)\^\(\(-1\)/3\)\)]],
  ".\n\n",
  StyleBox["Solution",
    FontSlant->"Italic"],
  ".  Let's let ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " do more of the work."
}], "Text"],

Cell[BoxData[
    \(binom[m_, k_]\  := \ 
      \(\[Product]\+\(i = 0\)\%\(k - 1\)\((m - i)\)\)\/\(k!\)\)], "Input"],

Cell[BoxData[
    \(bseries[u_, m_]\  := 
      \[Sum]\+\(k = 0\)\%\[Infinity]\((binom[m, k] u\^k)\)\)], "Input"],

Cell[BoxData[
    \(bpoly[u_, m_, N_]\  := \[Sum]\+\(k = 0\)\%N\((binom[m, k] u\^k)\)\)], 
  "Input"],

Cell[CellGroupData[{

Cell[BoxData[
    \(bseries[x\^2, \(-1\)/3]\)], "Input"],

Cell[BoxData[
    \(1\/\((1 + x\^2)\)\^\(1/3\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(bpoly[x\^2, \(-1\)/3, 5]\)], "Input"],

Cell[BoxData[
    \(1 - x\^2\/3 + \(2\ x\^4\)\/9 - \(14\ x\^6\)\/81 + \(35\ x\^8\)\/243 - 
      \(91\ x\^10\)\/729\)], "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["10.", "Subsubsection"],

Cell[TextData[{
  "Find the first four terms of the binomial series for ",
  Cell[BoxData[
      \(TraditionalForm\`f(x)\  = \ \((1 - 2\/x)\)\^\(1/3\)\)]],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(bseries[\(-\(2\/x\)\), 1/3]\)], "Input"],

Cell[BoxData[
    \(\((\(\(-2\) + x\)\/x)\)\^\(1/3\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(bpoly[\(-\(2\/x\)\), 1/3, 5]\)], "Input"],

Cell[BoxData[
    \(1 - 704\/\(729\ x\^5\) - 160\/\(243\ x\^4\) - 40\/\(81\ x\^3\) - 
      4\/\(9\ x\^2\) - 2\/\(3\ x\)\)], "Output"]
}, Open  ]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Initial Value Problems", "Subsection"],

Cell[CellGroupData[{

Cell["24.", "Subsubsection"],

Cell[TextData[{
  "Find a power series solution for the initial value problem ",
  Cell[BoxData[
      \(TraditionalForm\`\((1 + x\^2)\) y'\  + \ 2  x\ y\  = \ 0\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`y(0)\  = \ 3\)]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(DSolve[{\((1 + x\^2)\) \(y'\)[x]\  + \ 2\ x\ y[x] == 0, \ y[0] == 3}, 
      y[x], x]\)], "Input"],

Cell[BoxData[
    \({{y[x] \[Rule] 3\/\(1 + x\^2\)}}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[3\/\(1 + x\^2\), {x, 0, 8}]\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      "3", "-", \(3\ x\^2\), "+", \(3\ x\^4\), "-", \(3\ x\^6\), "+", 
        \(3\ x\^8\), "+", 
        InterpretationBox[\(O[x]\^9\),
          SeriesData[ x, 0, {}, 0, 9, 1]]}],
      SeriesData[ x, 0, {3, 0, -3, 0, 3, 0, -3, 0, 3}, 0, 9, 1]]], "Output"]
}, Open  ]],

Cell[TextData[{
  "To get a power series solution we write\n",
  Cell[BoxData[
      \(TraditionalForm\`y = c\_0 + \(c\_1\) x + \(c\_2\) x\^2 + \  ... \)]],
  "\nNote that the initial condition gives: ",
  Cell[BoxData[
      \(TraditionalForm\`3\  = \ \(y(0)\  = \ c\_0\)\)]],
  ". The derivative is \n",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            FormBox[\(y' = c\_1 + 2 \( c\_2\) x\),
              "TraditionalForm"], "+", \(3 \( c\_3\) x\^2\), "+"}], "..."}], 
        TraditionalForm]]],
  "\nso\n",
  Cell[BoxData[
      FormBox[
        RowBox[{\(\((1 + x\^2)\) y' + 2\ x\ y\), " ", "=", " ", 
          RowBox[{
            RowBox[{
              RowBox[{"(", 
                RowBox[{
                  RowBox[{
                    FormBox[\(c\_1 + 2 \( c\_2\) x\),
                      "TraditionalForm"], "+", \(3 \( c\_3\) x\^2\), "+"}], 
                  "..."}], ")"}], " ", "+", " ", 
              RowBox[{\(x\^2\), " ", 
                RowBox[{"(", 
                  RowBox[{
                    RowBox[{
                      FormBox[\(c\_1 + 2 \( c\_2\) x\),
                        "TraditionalForm"], "+", \(3 \( c\_3\) x\^2\), "+"}], 
                    "..."}], ")"}]}], " ", "+", 
              \(2\ x\ \ 
                \((c\_0 + \(c\_1\) x + \(c\_2\) x\^2 + \  ... )\)\)}], "=", 
            "  ", 
            RowBox[{
              RowBox[{
                RowBox[{
                  RowBox[{
                    FormBox[\(c\_1 + \((2  c\_2 + 2  c\_0)\) x\),
                      "TraditionalForm"], "+", 
                    \(\((3  c\_3 + c\_1 + 2  c\_1)\) x\^2\), "+", 
                    \(\((4  c\_4 + 2  c\_2 + 2  c\_2)\) x\^3\), "+"}], 
                  "..."}], "+", 
                \(\((\((n + 1)\) c\_\(n + 1\) + \((n - 1)\) c\_\(n - 1\) + 
                      2  c\_\(n - 1\))\) x\^n\), "+"}], " ", "..."}]}]}], 
        TraditionalForm]]],
  "\nThis series must equal 0, so we get:\n  ",
  Cell[BoxData[
      \(TraditionalForm\`c\_1 = 0\)]],
  "\n  ",
  Cell[BoxData[
      \(TraditionalForm\`2  c\_2 + 2  c\_0 = 0\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`c\_2 = \(\(-c\_0\) = \(-3\)\)\)]],
  "\n  ",
  Cell[BoxData[
      \(TraditionalForm\`3  c\_3 + 3  c\_1 = \ 0\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`c\_3 = \(\(-c\_1\) = \ 0\)\)]],
  "\n  ",
  Cell[BoxData[
      \(TraditionalForm\`4  c\_4 + \ 4  c\_2 = \ 0\)]],
  " or ",
  Cell[BoxData[
      \(TraditionalForm\`c\_4 = \ \(\(-c\_2\) = \ 3\)\)]],
  "\nThe general term is:\n",
  Cell[BoxData[
      FormBox[
        RowBox[{" ", 
          FormBox[\(n\ c\_n + n\ c\_\(n - 2\)\),
            "TraditionalForm"]}], TraditionalForm]]],
  "=0 or ",
  Cell[BoxData[
      \(TraditionalForm
      \`c\_n = \ 
        \(\(-c\_\(n - 2\)\) = 
          \(\(c\_\(n - 4\) =. \) ..  = 
            \(\(\((\(-1\))\)\^\(n/2\)\) c\_0 = \ 
              \(\((\(-1\))\)\^\(n/2\)\) 3\)\)\)\)]],
  " if ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " is even and ",
  Cell[BoxData[
      \(TraditionalForm\`c\_n = 0\)]],
  " if ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " is odd. The solution is thus ",
  Cell[BoxData[
      \(TraditionalForm
      \`y\  = \ \(3\  - \ 3  x\^2 + \ 3  x\^4 - \  ...  = \ 
          \(3 \((1 - x\^2 + x\^4 -  ... )\)\  = \ 3\/\(1 + x\^2\)\)\)\)]]
}], "Text"]
}, Open  ]],

Cell[CellGroupData[{

Cell["31.", "Subsubsection"],

Cell[TextData[{
  "Find a power series solution for the initial value problem ",
  Cell[BoxData[
      \(TraditionalForm\`\(y'\)'\  + \ \(x\^2\) y\  = \ x\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`y' \((0)\) = b\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = a\)]],
  ".\n\n",
  StyleBox["Solution.",
    FontSlant->"Italic"],
  " We look for a solution of the form ",
  Cell[BoxData[
      \(TraditionalForm
      \`y = \(a\_0 + \ \(a\_1\) x + \(a\_2\) x\^2 +  ... \) + 
            \(a\_n\) x\^n +  ... \)]],
  ".\nTaking derivatives:\n",
  Cell[BoxData[
      \(TraditionalForm
      \`y'\  = \ 
        \(a\_1 + 2 \( a\_2\) x\  + 3 \( a\_3\) x\^2 +  ... \) + 
            n\ \(a\_n\) x\^\(n - 1\) +  ... \)]],
  "\ny'' = ",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            RowBox[{
              RowBox[{
                FormBox[\(2  a\_2\),
                  "TraditionalForm"], "+", \(6 \( a\_3\) x\), " ", "+", " ", 
                \(12 \( a\_4\) x\^2\), "+"}], "..."}], "+", 
            \(\(n(n - 1)\) \(a\_n\) x\^\(n - 2\)\), "+"}], "..."}], 
        TraditionalForm]]],
  ".\nPlug these into the equation ",
  Cell[BoxData[
      \(TraditionalForm\`\(y'\)'\  + \ \(x\^2\) y\  = \ x\)]],
  " gives: \n",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            RowBox[{"(", 
              RowBox[{
                RowBox[{
                  RowBox[{
                    RowBox[{
                      FormBox[\(2  a\_2\),
                        "TraditionalForm"], "+", \(6 \( a\_3\) x\), " ", "+", 
                      "  ", \(12 \( a\_4\) x\^2\), "+"}], "..."}], "+", 
                  \(\((n + 2)\) \((n + 1)\) \(a\_\(n + 2\)\) x\^n\), "+"}], 
                "..."}], ")"}], "+", 
            \(\(x\^2\)(
              \(a\_0 + \ \(a\_1\) x + \(a\_2\) x\^2 +  ... \) + 
                  \(a\_\(n - 2\)\) x\^\(n - 2\) +  ... )\)}], "=", "x"}], 
        TraditionalForm]]],
  " \nCollecting terms: \n",
  Cell[BoxData[
      FormBox[
        RowBox[{
          RowBox[{
            RowBox[{
              RowBox[{
                FormBox[
                  RowBox[{
                    RowBox[{
                      RowBox[{
                        RowBox[{
                          RowBox[{
                            RowBox[{
                              
                              FormBox[
                                \(2  a\_2 + 3*2 \( a\_3\) x + 
                                  \((\(a\_0 + \)\ \)\),
                                "TraditionalForm"], "4", "*", "3", 
                              \(a\_4\)}], ")"}], \(x\^2\)}], "+"}], "..."}], 
                    "+", \((
                      \((n + 2)\) \((n + 1)\) a\_\(n + 2\) + a\_\(n - 2\))
                      \)}],
                  "TraditionalForm"], \(x\^\(n + 2\)\)}], "+"}], "..."}], 
          "=", "x"}], TraditionalForm]]],
  " \nAlong with ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = \(a\_0 = a\)\)]],
  " and y'(0)= ",
  Cell[BoxData[
      \(TraditionalForm\`a\_1 = b\)]],
  ", this implies that ",
  Cell[BoxData[
      \(TraditionalForm\`a\_2 = 0\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`3*2  a\_3 = 1\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`a\_0 + 4*3  a\_4 = 0\)]],
  ", ... , ",
  Cell[BoxData[
      \(TraditionalForm
      \`\((n + 2)\) \((n + 1)\) a\_\(n + 2\) + a\_\(n - 2\)\  = \ 0\)]],
  ", ... . Therefore, ",
  Cell[BoxData[
      \(TraditionalForm
      \`a\_4 = \ \(\(-\(1\/\(4*3\)\)\) a\_0 = \(-\(1\/\(4*3\)\)\) a\)\)]],
  ",  ",
  Cell[BoxData[
      \(TraditionalForm
      \`a\_5 = \ 
        \(\(-\ \(1\/\(5*4\)\)\) a\_1 = \ \(-\ \(1\/\(5*4\)\)\) b\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm\`a\_6 = \(\(-\(1\/\(6*5\)\)\) a\_2 = 0\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`a\_7 = \(\(-\(1\/\(7*6\)\)\) a\_3 = 
          \(-\(1\/\(7*6\)\)\) \((1\/6)\)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`a\_8 = \(\(-\ \(1\/\(8*7\)\)\) a\_4 = 
          \(\(-\(1\/\(8*7\)\)\) \((\(-\(1\/\(4*3\)\)\) a)\) = 
            \(1\/\(8*7*4*3\ \)\) a\)\)\)]],
  ", ",
  Cell[BoxData[
      \(TraditionalForm
      \`a\_9 = \(\(-\ \(1\/\(9*8\)\)\) a\_5 = 
          \(\(-\(1\/\(9*8\)\)\) \((\(-\(1\/\(5*4\)\)\) a\_1)\) = 
            \(1\/\(9*8*5*4\ \)\) b\)\)\)]],
  ", etc. The solution then looks like \n",
  Cell[BoxData[
      \(TraditionalForm
      \`y\  = \ a\  + \ b\ x\  + \ \(1\/6\) x\^3 - \ \((a\/\(4*3\))\) x\^4 - 
            \ \((b\/\(5*4\))\) x\^5 - \((1\/\(7*6*6\))\) x\^7 + \ 
            \(a\/\(8*7*4*3\)\) x\^8 + \(b\/\(9*9*5*4\ \)\) x\^9 +  ... \)]],
  "\n"
}], "Text"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Approximating Integrals", "Subsection"],

Cell[CellGroupData[{

Cell["34.", "Subsubsection"],

Cell[TextData[{
  "The series for ",
  Cell[BoxData[
      \(TraditionalForm\`\(e\^\(-x\) - 1\)\/x\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(\((1 - x + \(1\/\(2!\)\) x\^2 - \(1\/\(3!\)\) x\^3 + \  ... )\)\  - 
            1\)\/x = 
        \(-1\) + \(1\/\(2!\)\) x - \(1\/\(3!\)\) x\^2 + \  ... \)]],
  " "
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[\(E\^\(-x\) - 1\)\/x, {x, 0, 5}]\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      \(-1\), "+", \(x\/2\), "-", \(x\^2\/6\), "+", \(x\^3\/24\), "-", 
        \(x\^4\/120\), "+", \(x\^5\/720\), "+", 
        InterpretationBox[\(O[x]\^6\),
          SeriesData[ x, 0, {}, 0, 6, 1]]}],
      SeriesData[ x, 0, {-1, 
        Rational[ 1, 2], 
        Rational[ -1, 6], 
        Rational[ 1, 24], 
        Rational[ -1, 120], 
        Rational[ 1, 720]}, 0, 6, 1]]], "Output"]
}, Open  ]],

Cell[TextData[{
  "Integrating from 0 to 0.2 gives:\n",
  Cell[BoxData[
      \(TraditionalForm
      \`\(-\((0.2)\)\)\  + \ \(1\/\(2*\(2!\)\)\) \((0.2)\)\^2 - \ 
          \(1\/\(3*\(3!\)\)\) \((0.2)\)\^3 + \ 
          \(1\/\(4*\(4!\)\)\) \((0.2)\)\^4 - \  ... \)]],
  " \nThis is an alternating series, so the error in using the first ",
  Cell[BoxData[
      \(TraditionalForm\`n\)]],
  " terms is less than the absolute value ofthe ",
  Cell[BoxData[
      \(TraditionalForm\`n + 1\)]],
  "term: \n",
  Cell[BoxData[
      \(TraditionalForm
      \`error\  < \ 
        \(1\/\(\((n + 1)\)*\(\((n + 1)\)!\)\)\) \((0.2)\)\^\(n + 1\)\)]],
  ". A table of these values is:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(Table[\ 
      \(1\/\(\((n + 1)\)*\(\((n + 1)\)!\)\)\) \((0.2)\)\^\(n + 1\), {n, 1, 5}]
      \n\)\)], "Input"],

Cell[BoxData[
    \({0.0100000000000000022`, 0.000444444444444444552`, 
      0.000016666666666666674`, 5.33333333333333658`*^-7, 
      1.48148148148148255`*^-8}\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "Thus, for ",
  Cell[BoxData[
      \(TraditionalForm\`n = 2\)]],
  ", the approimation ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(-\((0.2)\)\)\  + \ \(1\/\(2*\(2!\)\)\) \((0.2)\)\^2\)]],
  " has an error is less than ",
  Cell[BoxData[
      \(TraditionalForm\`10\^\(-3\)\)]],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    RowBox[{" ", 
      FormBox[\(\(-\((0.2)\)\)\  + \ \(1\/\(2*\(2!\)\)\) \((0.2)\)\^2\),
        "TraditionalForm"]}]], "Input"],

Cell[BoxData[
    \(\(-0.19`\)\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\[Integral]\_0\%0.2\(\( E\^\(-x\) - 1\)\/x\) \[DifferentialD]x\)], 
  "Input"],

Cell[BoxData[
    \(\(-0.190428296651325545`\)\)], "Output"]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["46.", "Subsubsection"],

Cell[TextData[{
  " Find a polynomial that approximates ",
  Cell[BoxData[
      \(TraditionalForm
      \`F(x)\  = \ 
        \[Integral]\_0\%x\(\( ln(1 + t)\)\/t\) \[DifferentialD]t\)]],
  " for ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " in ",
  Cell[BoxData[
      \(TraditionalForm\`\([0, 1]\)\)]],
  " with error less than ",
  Cell[BoxData[
      \(TraditionalForm\`10\^\(-3\)\)]],
  ".\n",
  StyleBox["Solution. ",
    FontSlant->"Italic"],
  " The power series for ",
  Cell[BoxData[
      \(TraditionalForm\`\(ln(1 + t)\)\/t\)]],
  " is ",
  Cell[BoxData[
      \(TraditionalForm
      \`\(t - \(1\/2\) t\^2 + \(1\/3\) t\^3 - \  ... \)\/t = 
        1 - \(1\/2\) t + \(1\/3\) t\^2 -  ... \)]]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Series[Log[1 + t]\/t, {t, 0, 5}]\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{
      "1", "-", \(t\/2\), "+", \(t\^2\/3\), "-", \(t\^3\/4\), "+", 
        \(t\^4\/5\), "-", \(t\^5\/6\), "+", 
        InterpretationBox[\(O[t]\^6\),
          SeriesData[ t, 0, {}, 0, 6, 1]]}],
      SeriesData[ t, 0, {1, 
        Rational[ -1, 2], 
        Rational[ 1, 3], 
        Rational[ -1, 4], 
        Rational[ 1, 5], 
        Rational[ -1, 6]}, 0, 6, 1]]], "Output"]
}, Open  ]],

Cell[TextData[{
  "Integrating from 0 to ",
  Cell[BoxData[
      \(TraditionalForm\`x\)]],
  " gives: ",
  Cell[BoxData[
      \(TraditionalForm
      \`x\  - \ \(1\/2\^2\) x\^2 + \ \(1\/3\^2\) x\^3 - \ \(1\/4\^2\) x\^4 + 
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  " \nSince ",
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  ", this series is alternating and the error of using the first ",
  Cell[BoxData[
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  " term: ",
  Cell[BoxData[
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        \( | \^\(n + 1\)\)\( \[LessEqual] 1\/\((n + 1)\)\^2\)\)]],
  ". the last inequality uses the given restriction that ",
  Cell[BoxData[
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  ". To get the error ",
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Cell[BoxData[
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Cell["The approximating polynomial is", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
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Cell[BoxData[
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      x\^18\/324 + x\^19\/361 - x\^20\/400 + x\^21\/441 - x\^22\/484 + 
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Cell[BoxData[
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