(************** 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). 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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[ 22707, 924]*) (*NotebookOutlinePosition[ 23774, 959]*) (* CellTagsIndexPosition[ 23730, 955]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Sums, Products and Limits", "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[{ CounterBox["Section"], ". Sums" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The ", StyleBox["Sum[]", "Input"], " function uses the standard iterator notation, (the same that ", StyleBox["Table[]", "Input"], " uses)." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sum[x^n, {n, 1, 5}]\)], "Input", AspectRatioFixed->True], Cell[TextData[ "To get the typeset version of the above sum, click on the sum symbol in the \ BasicInput palette and use \[TabKey] to move between the fields or type\n\t\ \[EscapeKey]sum\[EscapeKey] \[ControlKey]\[LeftModified]=\[RightModified] n=1 \ \[ControlKey]\[LeftModified]5\[RightModified] 5 \[ControlKey]\[LeftModified]\ \[SpaceKey]\[RightModified] x \[ControlKey]\[LeftModified]6\[RightModified] \ n"], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%5 x\^n\)], "Input", AspectRatioFixed->True], Cell["Try typing the above input yourself in the two ways.", "Text", TextAlignment->Left, TextJustification->1], Cell["\<\ If the summand has more that one term, they must be enclosed in \ parentheses.\ \>", "Text"], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%5\((x\^n + n\ y\^n)\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Here are the first six nonzero terms in the Taylor series expansion of ", Cell[BoxData[ \(TraditionalForm\`cos\ x\)]], " about ", Cell[BoxData[ \(TraditionalForm\`x\ = \ 0\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sum[\(\((\(-1\))\)\^\(n/2\)\ x\^n\)\/\(n!\), {n, 0, 10, 2}]\)], "Input",\ AspectRatioFixed->True], Cell[TextData[ "To get the typeset version, type\n\t\[EscapeKey]sum\[EscapeKey] \ \[ControlKey]\[LeftModified]=\[RightModified] n=0 \ \[ControlKey]\[LeftModified]\[ReturnKey]\[RightModified] \[EscapeKey]D\ \[EscapeKey] \[SpaceKey] n=2 \[ControlKey]\[LeftModified]5\[RightModified] 10 \ \[ControlKey]\[LeftModified]\[SpaceKey]\[RightModified] \nand then type the \ summand in the usual fashion."], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ RowBox[{ UnderoverscriptBox["\[Sum]", GridBox[{ {\(n = 0\)}, {\(\[CapitalDelta]\[MediumSpace]n = 2\)} }], "10"], \(\(\((\(-1\))\)\^\(n/2\)\ x\^n\)\/\(n!\)\)}]], "Input",\ AspectRatioFixed->True], Cell["\<\ I've put the above typeset command in for completeness, and not \ usefulness. Try typing this yourself if you feel like it, or just evaluate \ the input. \ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[TextData[{ "By evaluating the sum at the point ", Cell[BoxData[ \(TraditionalForm\`x\ = \ 0.1\)]], " we can see how well it approximates the value of ", Cell[BoxData[ \(TraditionalForm\`cos\ x\)]], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(N[% /. x \[Rule] 0.1]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Cos[0.1]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can evaluate many infinite and finite sums in closed form. Below are two \ geometric sums and the Riemann zeta function of 2." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 2\^\(-n\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(i = 1\)\%n\((2\/3)\)\^\(-i\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 1\/n\^2\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The Riemman zeta function is a built-in function in ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Zeta[2]\)], "Input"], Cell["Mathematica also knows when a sum diverges.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 1\/n\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can do a sum over two variables just as you would make a two \ dimensional table. The first iterator is the outer sum. Nesting ", StyleBox["Sum[]", "Input"], " functions is equivalent to including multiple iterators in a single ", StyleBox["Sum[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sum[Sum[x\^i\ y\^j, {j, 0, 2}], {i, 0, 2}]\)], "Input"], Cell[BoxData[ \(Sum[x\^i\ y\^j, {j, 0, 2}, {i, 0, 2}]\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(i = 0\)\%2\(\[Sum]\+\(j = 0\)\%2 x\^i\ y\^j\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "To get an numerical approximation of a sum, use ", StyleBox["NSum[]", "Input"], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(NSum[2\^\(-n\), {n, 1, \[Infinity]}]\)], "Input"], Cell[BoxData[ \(NSum[1\/n\^2, {n, 1, \[Infinity]}]\)], "Input"], Cell[TextData[{ "When a sum diverges, ", StyleBox["NSum[]", "Input"], " will give you a number, but not without warning you that the result may \ be worthless." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(NSum[1\/n, {n, 1, \[Infinity]}]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Find the closed form expressions of the following sums.", "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 0\)\%n\((1\/2)\)\^k, \ \ \ \ \[Sum]\+\(k = 1\)\%n\( k(1\/2)\)\^k, \ \ \ \ \[Sum]\+\(k = 1\)\%n k\^2, \ \ \ \ \[Sum]\+\(k = 1\)\%n\(\((\(-1\))\)\^k\) k\^4, \ \ \ \ \[Sum]\+\(k = 1\)\%n\( cos(\(\[Pi]\ k\)\/n) . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "In how many ways can you make change for a dollar using pennies, nickels, \ dimes and quarters? The answer is the coefficient of ", Cell[BoxData[ \(TraditionalForm\`x\^100\)]], " in the generating polynomial:" }], "Text", TextJustification->1], Cell[BoxData[ \(TraditionalForm \`\((1 + x + \[CenterEllipsis]x\^100)\) \((1 + x\^5 + \[CenterEllipsis] + x\^100)\) \((1 + x\^10 + \[CenterEllipsis] + x\^100)\) \(\((1 + x\^25 + x\^100)\) . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Products" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell["The syntax for products is identical to that for sums.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Product[1 + x\^n, {n, 1, 6}]\)], "Input"], Cell[TextData[ "For the typeset version:\n\t\[EscapeKey]prod\[EscapeKey] \[ControlKey]\ \[LeftModified]=\[RightModified] n=1 \[ControlKey]\[LeftModified]5\ \[RightModified] 6 \[ControlKey]\[LeftModified]\[SpaceKey]\[RightModified] \ (1+x \[ControlKey]\[LeftModified]6\[RightModified] n \[ControlKey]\ \[LeftModified]\[SpaceKey]\[RightModified] )\nIf the argument of the product \ has more than one term than it must be enclosed in parentheses."], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\[Product]\+\(n = 1\)\%6\((1 + x\^n)\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "With the ", StyleBox["SymbolicSum", "Input"], " package loaded, ", StyleBox["Mathematica", FontSlant->"Italic"], " is able to find closed-form expressions for many infinite and finite \ products." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Product]\+\(n = 1\)\%\[Infinity]\((1 + 1\/n\^2)\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Product]\+\(i = 2\)\%n\((1 - 1\/i\^2)\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If ", StyleBox["Mathematica", FontSlant->"Italic"], " cannot find a closed-form expression for an infinite product it will \ return the input." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(\[Product]\+\(n = 1\)\%\[Infinity]\((1 + 1\/n\^2 + 1\/n\^3)\)\)], "Input", AspectRatioFixed->True], Cell["\<\ However, you can still get a numerical approximate of the value of \ the infinite product.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(N[%]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If you only want a numerical approximation to start with, then use the ", StyleBox["NProduct[]", FontWeight->"Bold"], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(NProduct[\((1 + 1\/n\^2 + 1\/n\^3)\), {n, 1, \[Infinity]}]\)], "Input"], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Find the closed form expressions for the following products.", "Text"], Cell[BoxData[ \(TraditionalForm\`\[Product]\+\(k = 1\)\%n k, \ \ \ \ \[Product]\+\(k = 1\)\%n\((k + 1\/2)\), \ \ \ \ \[Product]\+\(k = 1\)\%n\((1 + 1\/k)\), \ \ \ \ \[Product]\+\(k = 1\)\%\(n - 1\)\(Sin[\[Pi]\ k/n] . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Limits" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ To evaluate an expression at a certain value of a formal variable, \ we use the substitution operator followed by a rule.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sin[x] /. x \[Rule] 0\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The syntax of rules in ", StyleBox["Mathematica", FontSlant->"Italic"], " is a little misleading. The following looks very close to the \ mathematical notation for the limit as ", Cell[BoxData[ \(TraditionalForm\`x\)]], " goes to zero. However, the following is only a substitution." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Sin[x]\/x /. x \[Rule] 0\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In order to evaluate an expression in the limit as a formal variable \ aproached a limit, we must use the ", StyleBox["Limit[]", "Input"], " function." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[Sin[x]\/x, x \[Rule] 0]\)], "Input", AspectRatioFixed->True], Cell["\<\ You can evaluate a limit as a formal variables goes to positive \ infinity.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[x\^2\ Exp[\(-x\)], x \[Rule] \[Infinity]]\)], "Input", AspectRatioFixed->True], Cell["Mathematica will also tell you if a limit diverges.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[\(x + Sin[x]\)\/x\^2, x \[Rule] 0]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If the limit does not exists, ", StyleBox["Mathematica", FontSlant->"Italic"], " will tell you if the fluctuation of the expression is bounded in an \ interval." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[Sin[x], x \[Rule] \[Infinity]]\)], "Input", AspectRatioFixed->True], Cell["\<\ It will also tell you if the expression fluctuates as the limit \ diverges.\ \>", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[x\ Sin[x], x \[Rule] \[Infinity]]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "By default ", StyleBox["Mathematica", FontSlant->"Italic"], " evaluates limits from above, (i.e. it approaches the specified value of \ the formal variable from the right), except for limits at \[Infinity] which \ it approaches from below. You can change this default with the ", StyleBox["Direction", "Input"], " option." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[TextData[{ "This is the default direction to evaluate the limit. ", StyleBox["Direction -> -1", "Input"], " means to evaluate the limit from above." }], "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[Exp[1\/x], x \[Rule] 0, Direction \[Rule] \(-1\)]\)], "Input", AspectRatioFixed->True], Cell["Here we evaluate the limit from below.", "Text", Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True], Cell[BoxData[ \(Limit[Exp[1\/x], x \[Rule] 0, Direction \[Rule] 1]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["Evaluate the following limits.", "Text"], Cell[BoxData[ \(TraditionalForm\`lim\+\(x \[Rule] \[Infinity]\)\((1 + 1\/x)\)\^x, \ \ \ \ lim \+\(x \[Rule] 0\)\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), \ \ \ \ lim \+\(x \[Rule] \[Infinity]\)\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + \(f . \)\)\)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "Check that ", Cell[BoxData[ \(TraditionalForm \`\(\[DifferentialD]\/\[DifferentialD]x\) x\^n = n\ x\^\(n - 1\)\)]], " using the definition of the derivative," }], "Text"], Cell[BoxData[ \(TraditionalForm \`\(\[DifferentialD]\/\[DifferentialD]x\) \(f(x)\) \[Congruent] lim\+\(\[CapitalDelta]\ x \[Rule] 0\)\ \(\(f(x + \[CapitalDelta]\ x) - f(x)\)\/\(\[CapitalDelta]\ x\) . \)\)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Exercise ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ "Use ", StyleBox["Mathematica", FontSlant->"Italic"], " to evaluate" }], "Text"], Cell[BoxData[ \(TraditionalForm \`lim\+\(n \[Rule] \[Infinity]\)\ \(1 - \((\(-1\))\)\^n\)\/n\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Solutions", "Subtitle", TextAlignment->Center, TextJustification->0], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Sums" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["We evaluate the following sums:", "Text"], Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 0\)\%n\((1\/2)\)\^k, \ \ \ \ \[Sum]\+\(k = 1\)\%n\( k(1\/2)\)\^k, \ \ \ \ \[Sum]\+\(k = 1\)\%n k\^2, \ \ \ \ \[Sum]\+\(k = 1\)\%n\(\((\(-1\))\)\^k\) k\^4, \ \ \ \ \[Sum]\+\(k = 1\)\%n\( Cos[\[Pi]\ k/n] . \)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(\[Sum]\+\(k = 0\)\%n\((1\/2)\)\^k\)], "Input"], Cell["We can simplify the answer a bit.", "Text"], Cell[BoxData[ \(Simplify[%]\)], "Input"], Cell["\<\ Note that you can copy the input from above by typing \[ControlKey]\ \[LeftModified]l\[RightModified]. (That's an el, not a one.)\ \>", "Text"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%n k \((1\/2)\)\^k\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%n k\^2\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%n\(\((\(-1\))\)\^k\) k\^4\)], "Input"], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%n Cos[\[Pi]\ k/n]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(Coefficient[ Expand[\n\t\t Sum[x\^n, {n, 0, 100}] Sum[x\^n, {n, 0, 100, 5}] Sum[x\^n, {n, 0, 100, 10}] Sum[x\^n, {n, 0, 100, 25}]], x, 100]\)], "Input"], Cell["\<\ There are 242 ways of making change for a dollar with pennies, \ nickels, dimes and quarters.\ \>", "Text", TextJustification->1] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Products" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell["We evaluate the following products,", "Text"], Cell[BoxData[ \(TraditionalForm\`\[Product]\+\(k = 1\)\%n k, \ \ \ \ \[Product]\+\(k = 1\)\%n\((k + 1\/2)\), \ \ \ \ \[Product]\+\(k = 1\)\%n\((1 + 1\/k)\), \ \ \ \ \[Product]\+\(k = 1\)\%\(n - 1\)\(sin(\(\[Pi]\ k\)\/n) . \)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(\[Product]\+\(k = 1\)\%n k\)], "Input"], Cell[BoxData[ \(\[Product]\+\(k = 1\)\%n\((k + 1\/2)\)\)], "Input"], Cell[BoxData[ \(\[Product]\+\(k = 1\)\%n\((1 + 1\/k)\)\)], "Input"], Cell[BoxData[ \(\[Product]\+\(k = 1\)\%\(n - 1\)Sin[\(\[Pi]\ k\)\/n]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ CounterBox["Section"], ". Limits" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(Limit[\((1 + 1\/x)\)\^x, x \[Rule] \[Infinity]]\)], "Input"], Cell[BoxData[ \(Limit[\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), x \[Rule] \[Infinity]]\)], "Input"], Cell[BoxData[ \(Limit[\(a\ x\^2 + b\ x + c\)\/\(d\ x\^2 + e\ x + f\), x \[Rule] 0]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[BoxData[ \(Limit[\(\((x + \[CapitalDelta])\)\^n - x\^n\)\/\[CapitalDelta], \ \[CapitalDelta] \[Rule] 0]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Solution ", CounterBox["Section"], ".", CounterBox["Subsubsection"] }], "Subsubsection"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " chokes on this simple limit." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Limit[\(1 - \((\(-1\))\)\^n\)\/n, n \[Rule] \[Infinity]]\)], "Input"], Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{\(\(1 - \((\(-1\))\)\^n\)\/n\), ",", RowBox[{"n", "\[Rule]", InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]}]}], "]"}]], "Output"] }, Open ]], Cell["\<\ It doesn't help to write expression in terms of the \ exponential.\ \>", "Text", TextJustification->1], Cell[CellGroupData[{ Cell[BoxData[ \(Limit[\(1 - Exp[I\ \[Pi]\ n]\)\/n, n \[Rule] \[Infinity]]\)], "Input"], Cell[BoxData[ RowBox[{"Limit", "[", RowBox[{\(\(1 - \[ExponentialE]\^\(\[ImaginaryI]\ n\ \[Pi]\)\)\/n\), ",", RowBox[{"n", "\[Rule]", InterpretationBox["\[Infinity]", DirectedInfinity[ 1]]}]}], "]"}]], "Output"] }, Open ]] }, Closed]] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1680}, {0, 963}}, WindowToolbars->"EditBar", CellGrouping->Automatic, WindowSize->{756, 777}, WindowMargins->{{Automatic, 1}, {Automatic, 0}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "Magnification"->1}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, CharacterEncoding->"XAutomaticEncoding", Magnification->1.5 ] (******************************************************************* Cached data follows. 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