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Cell[TextData[{
  Cell[BoxData[{
      StyleBox[\(MATH\ 257\ Calculus\ III\t\t\t\tWeek\ of\ April\ 14, \ 
        2003\),
        FontSize->14], "\n", 
      StyleBox[\(Lab\ 3 : \ Power\ Series\),
        "Title"], "\n", 
      StyleBox[\(Name\ 1\),
        "Section"], "\n", 
      StyleBox[\(Name\ 2\),
        "Section"], "\n", 
      StyleBox[\(\(Section\)\(:\)\),
        "Section"]}], "Input"],
  "\n"
}], "Text"],

Cell[TextData[{
  "I.  Laboratory Objectives\n\tUse ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " to explore\n\t\tPower Series\n\t\tUsing the Ratio Test in Mathematica\n\t\
\tGraphing Power Series partial sums"
}], "Text"],

Cell[TextData[StyleBox["II.  Formatting and Syntax Information", "Text"]], \
"Text"],

Cell[TextData[{
  Cell[BoxData[
      \(InequalitySolve[f[x] < 1, \ x]\)]],
  " This solves inequalities.  It's nifty.\n\t\t\t\t\t\n",
  StyleBox["PlotStyle\[Rule]RGBColor[1,0,0]\t",
    FontFamily->"Courier"],
  "Use this in any Plot[ ] command.\n\t\t\t\t\tMakes graphs in different \
colors."
}], "Text",
  FontFamily->"Helvetica",
  FontSize->14],

Cell[TextData[StyleBox["Power Series",
  FontWeight->"Bold",
  FontColor->RGBColor[0, 0, 1]]], "Subtitle"],

Cell[TextData[{
  StyleBox["The general form for a Power Series about the point x = a:", 
    "Text",
    FontFamily->"Times",
    FontSize->14],
  "\n\n ",
  Cell[BoxData[
      \(TraditionalForm\`\[Sum]\+\(n = 0\)\%\[Infinity]\(\( c\_n\)(x - \
a)\)\^n\)]],
  "\n \n",
  StyleBox[" This is an infinite polynomial written in terms of powers of (x \
- a).", "Text",
    FontFamily->"Times",
    FontSize->14]
}], "Section"],

Cell[TextData[{
  StyleBox["Given a function, expanding it into a power series can be hard to \
do.  This lab will explore several tricks for doing this.\n\n",
    FontSize->14],
  StyleBox["Trick 1: Use geometric series.", "Section",
    FontColor->RGBColor[0, 0, 1]],
  StyleBox["\nIf the power series is in the form of a geometric series a ",
    FontSize->14],
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(\(+\ ar\)\  + \ ar\^2 + \ ar\^3 +  ... \)\(\ \)\),
          FontSize->14], TraditionalForm]]],
  StyleBox["then we can just use the formula for a convergent geometric \
series:  ",
    FontSize->14],
  Cell[BoxData[
      FormBox[
        StyleBox[\(\(a\/\(1 - r\)\)\(.\)\),
          "Text",
          FontSize->14,
          FontSlant->"Italic"], TraditionalForm]]],
  "\n\n",
  StyleBox["In other words, we know that\n",
    FontSize->16],
  Cell[BoxData[
      \(TraditionalForm\`1\  + \ x\  + \ x\^2\  + \ x\^3\  + \ 
            x\^4 + \  ... \  = \ 1\/\(1 - x\)\)],
    FontSize->14],
  StyleBox[" for  ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(x\)\(|\)\(\(\[LessEqual]\)\(\ \
\)\(1\)\)\)\)],
    FontSize->14],
  StyleBox[".  \n",
    FontSize->14],
  StyleBox["Example: ",
    FontSize->16,
    FontWeight->"Bold"],
  StyleBox["  what is the power series expansion of ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`2\/\(1 + x\/4\)\)],
    FontSize->14],
  StyleBox["?  Well, write it as ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`2\/\(1 - \((\(-\(x\/4\)\))\)\)\)],
    FontSize->14],
  StyleBox["and then we can see that ",
    FontSize->14],
  StyleBox["a = 2",
    FontSize->14,
    FontSlant->"Italic"],
  StyleBox[" and ",
    FontSize->14],
  StyleBox["r = -",
    FontSize->14,
    FontSlant->"Italic"],
  Cell[BoxData[
      \(TraditionalForm\`x\/4\)]],
  StyleBox[".  So...",
    FontSize->14]
}], "Text",
  TextAlignment->Left,
  TextJustification->0],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(Simplify[
        2\/\(1 + x\/4\)] == \[Sum]\+\(n = 0\)\%\[Infinity] 
          2 \((\(-\(x\/4\)\))\)\^n\)\)\)], "Input"],

Cell[TextData[{
  StyleBox["Evaluate the above expression.  The double equal sign \"==\" is a \
command that asks ",
    FontSize->14],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic"],
  StyleBox[", \"Is this true?\"  ",
    FontSize->14],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic"],
  StyleBox[" will respond \"True\" or \"False\".  So this can be a great way \
to check if your power series expansion is correct. (It's a good idea to put \
the Simplify[] command in there to make sure the Left-Hand-Side is as simple \
as possible.)",
    FontSize->14]
}], "Text"],

Cell[TextData[{
  "\n",
  StyleBox["Trick 2: Finding the radius and interval of convergence:", 
    "Section",
    FontSize->16,
    FontColor->RGBColor[0, 0, 1]],
  StyleBox["\nYou can find the radius of convergence of power series by hand, \
but you can also make ",
    FontSize->16],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic"],
  StyleBox[" do it for you using the Ratio Test.  Here's how:  we'll find the \
interval of convergence for the series ",
    FontSize->14],
  Cell[BoxData[
      \(\[Sum]\+\(n = 0\)\%\[Infinity] 2 \((\(-x\)/4)\)\^n\)],
    FontSize->14],
  StyleBox["that we saw above. \n\nIn preparation for using the Ratio Test, \
let's define the nth term of the series ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`a\_n\)],
    FontSize->14]
}], "Text"],

Cell[BoxData[
    \(\(\(a[n_, x_] = 2 \((\(-\(x\/4\)\))\)^n\)\(\[IndentingNewLine]\)
    \)\)], "Input"],

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  StyleBox["Now we can form the ratio of ",
    FontSize->14],
  Cell[BoxData[
      \(TraditionalForm\`a\_\(n + 1\)\/a\_\(\(n\)\(\ \)\)\)],
    FontSize->14],
  StyleBox["giving it the name \"ratio.\"",
    FontSize->14],
  "\n"
}], "Text"],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(ratio = a[n + 1, x]\/a[n, x]\)\)\)], "Input"],

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  "\n",
  StyleBox["Next take the limit of this ratio as n\[Rule] \[Infinity].",
    FontSize->14]
}], "Text"],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(Limit[ratio, 
      n \[Rule] \[Infinity]]\)\)\)], "Input"],

Cell[TextData[{
  StyleBox["By the Ratio Test, the series will converge when the ",
    FontSize->14],
  StyleBox["Absolute Value",
    FontSize->14,
    FontWeight->"Bold"],
  StyleBox[" of this answer is less than 1.  To solve the inequality ",
    FontSize->14],
  Cell[BoxData[
      \(Abs[\(-x\)\/4]\)]],
  StyleBox[" < 1, we must load the package \"Inequality Solve\" in the \
Algebra package.  This is done below.",
    FontSize->14]
}], "Text"],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(<< Algebra`InequalitySolve`\)\)\)], "Input"],

Cell[TextData[{
  "\n",
  StyleBox["To solve the inequality we use the syntax below:",
    FontSize->14]
}], "Text"],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(InequalitySolve[Abs[\(-\(x\/4\)\)] < 1, 
      x]\)\)\)], "Input"],

Cell[TextData[{
  "\n",
  StyleBox["So now we know that ",
    FontSize->16],
  Cell[BoxData[
      \(2\/\(1 + x\/4\) = \ \[Sum]\+\(n = 0\)\%\[Infinity] 
            2 \((\(-\(x\/4\)\))\)\^n\)],
    FontSize->14],
  StyleBox[" is a convergent power series expansion on the interval -4 < x < \
4.  We DON\"T know what is happening at the end points, at x = -4 and at x = \
4.  You would have to check these separately by hand.\n\nWe can also check \
our work by making a plot of our original function with a partial sum of our \
power series.  I like to use the command PlotStyle->{RGBColor[0,0,0], \
RGBColor[1,0,0] ...} to make the graphs on the same axis in different colors. \
 [0,0,0] is black, [1,0,0] is red, and you can experiment with others.  So, \
to check out power series and interval of convergence for ",
    FontSize->14],
  Cell[BoxData[
      \(2\/\(1 + x\/4\)\)],
    FontSize->14],
  StyleBox[" we can try graphing it with, say, the 6th partial sum:",
    FontSize->14]
}], "Text"],

Cell[BoxData[
    \(\(\(\[IndentingNewLine]\)\(Plot[{2\/\(1 + x\/4\), \ \[Sum]\+\(n = \
0\)\%6 2 \((\(-\(x\/4\)\))\)\^n}, \ {x, \(-5\), 
        5}, \[IndentingNewLine]PlotStyle \[Rule] {RGBColor[0, 0, 0], \ 
          RGBColor[1, 0, 0]}]\)\)\)], "Input"],

Cell[TextData[{
  "\n",
  StyleBox["If the graphs look like they line up inside -4 < x < 4, then \
we're lookin' good! (Try a higher partial sum, like the forst 10 terms, to \
make it even more accurate.)",
    FontSize->16]
}], "Text"],

Cell[TextData[{
  "\n",
  StyleBox["Question 1:",
    FontSize->16,
    FontWeight->"Bold",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["  Consider the function ",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`5\/\(3 - x\)\)],
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[".\n(a) Find the values of a and r in the geometric series \
representation of this ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["(",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`a\/\(1 - r\)\)],
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["=",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      FormBox[
        RowBox[{\(\[Sum]\+\(n = 0\)\%\[Infinity]\), 
          SuperscriptBox[
            StyleBox[
              RowBox[{
                StyleBox["a",
                  FontSlant->"Italic"], "r"}]], "n"]}], TraditionalForm]],
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[")",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" and write the power series representation.  (Hint: factor out a \
1/3.)  Use ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" to check your work.",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]]
}], "Text"],

Cell[BoxData[
    \(\[IndentingNewLine]\)], "Input"],

Cell[TextData[{
  "\n",
  StyleBox["(b) Use the Ratio Test in ",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" to find the interval of convergence of this power series.",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]]
}], "Text"],

Cell[BoxData[
    \(\[IndentingNewLine]\)], "Input"],

Cell[TextData[{
  StyleBox["(c) Make a pretty plot (with color!) of the original function \
together with the 8th partial sum.",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  "\n"
}], "Text"],

Cell[BoxData[
    \(\[IndentingNewLine]\)], "Input"],

Cell[TextData[{
  "\n",
  StyleBox["Trick 3: Use derivatives!", "Section",
    FontColor->RGBColor[0, 0, 1]],
  "\n\n",
  StyleBox["Question 2:",
    FontSize->16,
    FontWeight->"Bold",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["  (a) Find a power series expansion for ",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`1\/\((1 - x)\)\^2\)],
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[". Write your answer in summation notation and use ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" to check if it's true.  ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["Hint: ",
    FontSize->14,
    FontWeight->"Bold",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" Use ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["Mathematica",
    FontSize->14,
    FontSlant->"Italic",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" to remind yourself what the derivative of ",
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    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`1\/\(1 - x\)\)],
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["is.  (Recall that ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["D[f[x],x]", "Input",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox["  will give you the derivative of f[x] with respect to x.)  And \
remember that ",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`1\/\(1 - x\) = \[Sum]\+\(n = 0\)\%\[Infinity] \
x\^n\)],
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  StyleBox["for ",
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  Cell[BoxData[
      \(TraditionalForm\`\(\(|\)\(x\)\(|\)\(\(<\)\(\ \)\(1\)\)\)\)],
    FontSize->16,
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[".",
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    FontColor->RGBColor[1, 0, 0]]
}], "Text"],

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    \(\[IndentingNewLine]\)], "Input"],

Cell[TextData[{
  StyleBox["(b) Find the radius of convergence of this power series.",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  "\n"
}], "Text"],

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    \(\[IndentingNewLine]\)], "Input"],

Cell[TextData[{
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  StyleBox["Trick 4:  Use antiderivatives!", "Section",
    FontColor->RGBColor[0, 0, 1]],
  "\n\n",
  StyleBox["Question 3: ",
    FontSize->14,
    FontWeight->"Bold",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" (a) Find a power series expansion of ln(1+x), writing your \
answer in summation notation.  Hint:  What's the derivative of Log[1+x]?  \
Look kinda familiar? You can write this in summation notation (only do the \
first 5 terms or so) and then find the antiderivative (using the \
Integrate[f[x],x] command) to get the power series of the original.",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]]
}], "Text"],

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some of the partial sums of its power series using the range {x, -1, 2}.  Be \
sure to indicate which colors correspond to which functions!",
    FontSize->14,
    FontColor->RGBColor[1, 0, 0]],
  "\n"
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  StyleBox["Question 4:  Name that Power Series:",
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  StyleBox["\nConsider the power series ",
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  Cell[BoxData[
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  StyleBox["Mathematica",
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    FontSlant->"Italic",
    FontColor->RGBColor[1, 0, 0]],
  StyleBox[" and the Ratio Test to compute the interval of convergence of \
this power series. WARNING:  you may need to use the ",
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  "\n"
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disk.\n\t(3)\tGo back to Blackboard and upload your final lab report to the \
",
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