(************** 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[ 42385, 1449]*) (*NotebookOutlinePosition[ 43058, 1472]*) (* CellTagsIndexPosition[ 43014, 1468]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Calculus (one variable)", "Title", Evaluatable->False, TextAlignment->Center, TextJustification->0, AspectRatioFixed->True, Background->RGBColor[0, 1, 1]], Cell["\<\ This notebook is by Steven Amgott. Please send any questions or\ \[NonBreakingSpace]comments to samgott1@swarthmore.edu. Feel free to use and \ distribute this notebook, but keep this author information in any copy you \ use or distribute.\ \>", "SmallText"], Cell[TextData[{ "Buttons to create many of the symbols in this notebook are available in \ the ", StyleBox["BasicInput Palette", FontColor->RGBColor[0, 0, 1]], ". This palette is usually opened at the time ", StyleBox["Mathematica", FontSlant->"Italic"], " is started. If it is not up on the screen, it can be displayed using the \ ", StyleBox["Palette", FontColor->RGBColor[0, 0, 1]], " command in the ", StyleBox["File", FontColor->RGBColor[0, 0, 1]], " menu. ", StyleBox["Take a brief look at this palette", FontColor->RGBColor[1, 0, 1]], " to see what symbols it contains before continuing with this notebook.\n\n\ To open the closed group of cells for any of the sections or commands, you \ can ", StyleBox["double-click the bracket with a triangle at its bottom at the \ right edge of the cell", FontColor->RGBColor[1, 0, 1]], ".\n\nIf this is your first experience with ", StyleBox["Mathematica", FontSlant->"Italic"], ", or if you need a brief refresher before continuing, you might first go \ through the ", Cell[BoxData[ FormBox[ ButtonBox[\(Introduction\ to\ Mathematica\), ButtonData:>{"Introduction_to_Mathematica.nb", None}, ButtonStyle->"Hyperlink"], TraditionalForm]]], " notebook contained in the same folder as this notebook. In particular, \ remember that to evaluate a cell you need to press the ", StyleBox["enter", FontColor->RGBColor[0, 0, 1]], " key on the number pad at the lower right side of the keyboard and NOT the \ ", StyleBox["return", FontColor->RGBColor[0, 0, 1]], " key on a Macintosh or the ", StyleBox["Enter", FontColor->RGBColor[0, 0, 1]], " key next to the single and double quote marks on a Windows machine! (The \ latter only produces a line break. You can, however, use the ", StyleBox["Shift", FontColor->RGBColor[0, 0, 1]], " and ", StyleBox["Enter", FontColor->RGBColor[0, 0, 1]], " keys on a Windows machine, or the ", StyleBox["shift", FontColor->RGBColor[0, 0, 1]], " and ", StyleBox["return", FontColor->RGBColor[0, 0, 1]], " keys on a Mac, simultaneously to evaluate a cell. In particular, that is \ how you evaluate a cell on a notebook computer without a number pad.)" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[{ "What ", StyleBox["Mathematica", FontSlant->"Italic"], " can do" }], "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can perform many of the basic operations of Calculus. It can take Limits \ (including one-sided limits), compute Derivatives, find antiderivatives and \ Integrals (including both symbolic and numerical integration), calculate \ finite and some infinite Sums, and find Taylor polynomials. Examples of each \ of these commands are in the cells below." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Limits", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Suppose you want to find the limit as x\[Rule]0 of the function ", Cell[BoxData[ FormBox[ FractionBox[\(Sin[x]\), StyleBox["x", FontSlant->"Plain"]], TraditionalForm]]], ". You can enter it into ", StyleBox["Mathematica", FontSlant->"Italic"], " using the following cell. Remember, you ", StyleBox["evaluate a cell by clicking anywhere in the cell, and pressing \ the ", FontColor->RGBColor[1, 0, 1]], StyleBox["enter", FontColor->RGBColor[0, 0, 1]], StyleBox[" key", FontColor->RGBColor[1, 0, 1]], "." }], "Text"], Cell[BoxData[{ \(Clear[f, x]\), "\[IndentingNewLine]", \(f[x_] := Sin[x]\/x\)}], "Input"], Cell["Notice that f[0] does not exist.", "Text"], Cell[BoxData[ \(f[0]\)], "Input"], Cell["\<\ Let's start by looking at a graph of the function.\ \>", "Text"], Cell[BoxData[ \(\(Plot[f[x], {x, \(-5\), 5}];\)\)], "Input", AspectRatioFixed->True], Cell["\<\ You can \"zoom in\" a bit to get a better idea of the limit. It looks like \ it might be 1.\ \>", "Text"], Cell[BoxData[ \(\(Plot[f[x], {x, \(- .1\), .1}];\)\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " is able to find the limit using the command ", StyleBox["Limit[f[x],x\[Rule]x0]", FontColor->RGBColor[1, 0, 0]], ". (x0 is the point where you want the limit evaluated, and f[x] is the \ function.) You can use variables other than x, as in the examples below. By \ the way, you can create the arrow either by using a palette (such as ", StyleBox["Basic Input", FontColor->RGBColor[0, 0, 1]], ") or by typing the two symbols - and >, one right after the other. Newer \ versions of ", StyleBox["Mathematica", FontSlant->"Italic"], " will even change the two symbols into the single arrow symbol.." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Limit[f[x], x \[Rule] 0]\)], "Input", AspectRatioFixed->True], Cell["Two more limits:", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(Clear[t]\), "\[IndentingNewLine]", \(Limit[\[ExponentialE]\^\(-t\^2\), t \[Rule] \[Infinity]]\)}], "Input", AspectRatioFixed->True], Cell[BoxData[{ \(Clear[g, x]\), "\[IndentingNewLine]", \(\(g[x_] = \(x\^2 - 3\ x - 4\)\/\(x\^3 + x - 2\);\)\), "\n", \(Limit[g[x], x \[Rule] 1]\)}], "Input", AspectRatioFixed->True], Cell["\<\ This last limit is actually wrong, as you can see in the following graph.\ \>", "Text"], Cell[BoxData[ \(\(Plot[g[x], {x, 0, 2}];\)\)], "Input"], Cell[TextData[{ StyleBox["The reason for this is ", FontVariations->{"CompatibilityType"->0}], StyleBox["Mathematica", FontSlant->"Italic"], " always computes limits as if they are one-sided limits, from the right \ side, unless you specify otherwise. To specify a one-sided limit, include \ the ", StyleBox["Direction", FontColor->RGBColor[1, 0, 0]], " option in the command. Use ", StyleBox["1", FontColor->RGBColor[1, 0, 0]], " for a limit from the left side, and ", StyleBox["-1", FontColor->RGBColor[1, 0, 0]], " for a limit from the right side. Only if both one-sided limits exist and \ are equal is there a (two-sided) limit." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Limit[g[x], x \[Rule] 1, Direction \[Rule] \(-1\)]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Limit[g[x], x \[Rule] 1, Direction \[Rule] 1]\)], "Input", AspectRatioFixed->True], Cell["\<\ In this case the one-sided limits are different, so no (two-sided) limit \ exists.\ \>", "Text"], Cell[TextData[{ "Exercise: Do you think ", StyleBox["Mathematica", FontSlant->"Italic"], " is able to find the limit as x\[Rule]\[Infinity] of ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ StyleBox["n", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], " "}]], " ", SuperscriptBox["\[ExponentialE]", RowBox[{"-", StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}]]}], TraditionalForm]]], " without putting in a specific value of n? Try it and see. What about \ the limit as x\[Rule]\[Infinity] of ", Cell[BoxData[ FormBox[ FractionBox[\(ln(x)\), SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], StyleBox["n", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]]], TraditionalForm]]], "? If there are any differences, can you explain them?" }], "Text", CellFrame->True, Background->GrayLevel[0.833326]] }, Closed]], Cell[CellGroupData[{ Cell["Derivatives", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The basic syntax for the derivative command is ", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ SubscriptBox["\[PartialD]", RowBox[{"x", StyleBox[" ", FontColor->RGBColor[0, 0, 1]]}]], \(g[x]\)}], FontColor->RGBColor[1, 0, 0]], TraditionalForm]]], ". A button to create this symbol is available in the ", StyleBox["BasicInput Palette", FontColor->RGBColor[0, 0, 1]], ". You can use variables other than x, as in the examples below. ", StyleBox["Evaluate a cell by clicking anywhere in the cell, and pressing \ the ", FontColor->RGBColor[1, 0, 1]], StyleBox["enter", FontColor->RGBColor[0, 0, 1]], StyleBox[" key", FontColor->RGBColor[1, 0, 1]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[PartialD]\_x Sin[x]\)], "Input", AspectRatioFixed->True], Cell[BoxData[{ \(Clear[h, t]\), "\[IndentingNewLine]", \(\(h[t_] = Cos[t\^2];\)\), "\n", \(\[PartialD]\_t h[t]\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can still use the older version 2.2 syntax ", StyleBox["D[f[x],x]", FontColor->RGBColor[1, 0, 0]], " to find the derivative of f[x] with respect to x." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(D[3 x\^2 + 2 x - 1, x]\)], "Input"], Cell[BoxData[ \(D[x^2 + 3\ x, x]\)], "Input", FontSize->14], Cell[TextData[{ StyleBox["You could also do this last one by using the palette symbol ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Placeholder]\ \[Placeholder]\)], FontSize->12, FontColor->RGBColor[1, 0, 0]], StyleBox[". ", FontSize->12], StyleBox["Evaluate the next cell", FontSize->12, FontColor->RGBColor[1, 0, 1]], StyleBox[" to try and see the derivative (again).", FontSize->12] }], "Text", FontSize->14], Cell[BoxData[ \(\[PartialD]\_x\ x\^2 + 3\ x\)], "Input", FontSize->14], Cell[TextData[{ StyleBox["Exercise: What went wrong with this calculation? Make a new \ (input) cell after this one, and click on the palette symbol ", FontSize->12], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Placeholder]\ \[Placeholder]\)], FontSize->12, FontColor->RGBColor[1, 0, 0]], StyleBox[" to paste it into that cell. Then put in the necessary ", FontSize->12], StyleBox["Mathematica", FontSize->12, FontSlant->"Italic"], StyleBox[" syntax to find the correct derivative of ", FontSize->12], Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["x", FontSlant->"Plain"], "2"], TraditionalForm]], FontSize->12], StyleBox[" + 3 x. Check it against the correct answer we got the first \ time.", FontSize->12] }], "Text", CellFrame->True, FontSize->14, Background->GrayLevel[0.833326]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also use the \"prime\" notation for a derivative. Note that ", StyleBox["Log", FontColor->RGBColor[1, 0, 0]], " is the way ", StyleBox["Mathematica", FontSlant->"Italic"], " refers to ln, the natural logarithm." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{\(Clear[g, x]\), "\[IndentingNewLine]", \(g[x_] = Log[x];\), "\n", RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], "[", "x", "]"}]}], "Input", AspectRatioFixed->True], Cell[TextData[{ "You might run into one minor problem if you want the value of a derivative \ at a point. If you define a function f[x] = ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["x", FontSlant->"Plain"], "3"], TraditionalForm]]] }], "Text"], Cell[BoxData[{ \(Clear[f, x]\), "\[IndentingNewLine]", \(f[x_] := x\^4\)}], "Input"], Cell["\<\ and then try to evaluate the derivative of this function at the point x = 3 \ using\ \>", "Text"], Cell[BoxData[ \(\[PartialD]\_x\ f[3]\)], "Input"], Cell[TextData[{ "you get the wrong answer. This is because ", StyleBox["Mathematica", FontSlant->"Italic"], " has substituted 3 in for x before taking the derivative, and then is \ taking the derivative of a constant (in this case, 81) and getting 0 as the \ answer." }], "Text"], Cell["\<\ There are several ways to get around this. You could define a new function \ df[x] to be the derivative, and then substitute 3 into that function.\ \>", "Text"], Cell[BoxData[{ \(Clear[df, x]\), "\[IndentingNewLine]", \(df[x_] = \[PartialD]\_x\ f[x]\), "\[IndentingNewLine]", \(df[3]\)}], "Input"], Cell[TextData[{ "Note that using delayed execution (", StyleBox[":=", FontColor->RGBColor[1, 0, 0]], ") does not work in this case, as then 3 is being substituted in as the \ differentiation variable, as well as the point where the derivative is to be \ evaluated." }], "Text"], Cell[BoxData[{ \(Clear[df, x]\), "\[IndentingNewLine]", \(df[x_] := \[PartialD]\_x\ f[x]\), "\[IndentingNewLine]", \(df[3]\)}], "Input"], Cell[TextData[{ "Another (perhaps better) way to do this is to use ", StyleBox["Mathematica", FontSlant->"Italic"], "'s ", StyleBox["ReplaceAll", FontColor->RGBColor[1, 0, 0]], " command, which can also be given in \"slash-dot\" notation ", StyleBox["/. ", FontColor->RGBColor[1, 0, 0]], "." }], "Text"], Cell[BoxData[ \(\[PartialD]\_x\ x\^4 /. x \[Rule] 3\)], "Input"], Cell[TextData[{ "(", StyleBox["expr", FontSlant->"Italic"], "/.x\[Rule]3 can be read as \"evaluate ", StyleBox["expr ", FontSlant->"Italic"], "and then replace all instances of x in the evaluated expression by 3.\")" }], "Text"], Cell["\<\ This will also work for previously defined functions, such as f[x] above.\ \>", "Text"], Cell[BoxData[ \(\[PartialD]\_x\ f[x] /. x \[Rule] 3\)], "Input"], Cell["\<\ Finally, if you are working with a previously defined function such as f[x], \ you can use the \"prime\" notation.\ \>", "Text"], Cell[BoxData[ \(\(f'\)[3]\)], "Input"], Cell[TextData[{ "You can find higher order derivatives by including the order in the \ command. For example, to find the 2nd derivative of Cos[", Cell[BoxData[ \(TraditionalForm\`t\^2\)]], "], use: " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[PartialD]\_{t, 2}Cos[t\^2]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In the older version 2.2 syntax, this is ", StyleBox["D[g[t],{t,2}]", FontColor->RGBColor[1, 0, 0]], ". Be sure to use braces ", StyleBox["{ }", FontColor->RGBColor[1, 0, 0]], " around the variable and order of the derivative." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(D[Cos[t^2], {t, 2}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can differentiate expressions which have other letters in addition to \ the variable of differentiation. (In fact, ", StyleBox["Mathematica", FontSlant->"Italic"], " is computing a ", StyleBox["partial derivative", FontVariations->{"Underline"->True}], ", which is defined in a multi-variable calculus class.) Make sure to \ leave a space or put a ", StyleBox["*", FontColor->RGBColor[1, 0, 0]], " between letters that are being multiplied (such as the use of ", StyleBox["a x", FontColor->RGBColor[1, 0, 0]], ", instead of ", StyleBox["ax", FontColor->RGBColor[1, 0, 0]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], " in the following example). " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[PartialD]\_x \[ExponentialE]\^\(a\ x\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also calculate implicit derivatives, in which one (or more) variables \ is assumed to be a function of another one. For instance, if y is a function \ of x, the following cell computes an implicit derivative of the expression \ with respect to x (assuming z is just a constant)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[PartialD]\_x\((x\^2 + y[x]\^2 + z\^2)\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Note that we specified that y is a function of x by using ", StyleBox["y[x]", FontColor->RGBColor[1, 0, 0]], ". Compare the above output to that of" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[PartialD]\_x\((x\^2 + y\^2 + z\^2)\)\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "If every variable that appears is assumed to be a function of one \ variable, we can use ", StyleBox["Dt", FontColor->RGBColor[1, 0, 0]], " to find the implicit derivative. In the next cell we assume both y and z \ are functions of x. This is indicated by the x just before the final bracket \ in the syntax." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Dt[x\^2 + y\^2 + z\^2, x]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["An example using derivatives (optimization)", "Subsection"], Cell[TextData[{ "Suppose you want to find the maximum and the minimum of the function f[x] \ = ", Cell[BoxData[ \(x\^4 - 3\ x\^3 + x\^2 + \ 3\ x\ + 1\)]], " on the interval [-2,3]. You need to find the critical points (where \ f'[x] = 0 or undefined), mix in the endpoints, and check the function values \ at each of these. Here goes. First enter and graph the function on the \ interval." }], "Text"], Cell[BoxData[{ \(Clear[f, x, xmin, xmax]\), "\[IndentingNewLine]", \(f[x_] := x\^4 - 3\ x\^3 + x\^2 + \ 3\ x\ + 1\), "\[IndentingNewLine]", \(\(xmin = \(-2\);\)\), "\[IndentingNewLine]", \(\(xmax = 3;\)\), "\[IndentingNewLine]", \(\(Plot[f[x], {x, xmin, xmax}];\)\)}], "Input"], Cell[TextData[{ "It looks like a maximum is at an endpoint (probably the left one) and the \ minimum is at the leftmost critical point. If you click within the graph to \ get a box around it and hold down the ", StyleBox["Ctrl", FontColor->RGBColor[0, 0, 1]], " key (Windows) or the ", StyleBox["Apple", FontColor->RGBColor[0, 0, 1]], " key (Mac), you get a cursor in the graph window. The coordinates of the \ point where the cursor is located show up in the lower left corner of the \ notebook window. In this case, the minimum seems to be approximately at the \ point (-0.4,0.2)." }], "Text"], Cell["\<\ In order to find the maximum and minimum using calculus, you should first \ take the derivative.\ \>", "Text"], Cell[BoxData[ \(\(f'\)[x]\)], "Input"], Cell["\<\ Since this is a polynomial, the only critical points are where it is zero.\ \>", "Text"], Cell[BoxData[{ \(Clear[solutions]\), "\[IndentingNewLine]", \(solutions = Solve[\(f'\)[x] \[Equal] 0, x]\)}], "Input"], Cell["\<\ If you wish, you can extract just the x values from this list, using the \ following syntax.\ \>", "Text"], Cell[BoxData[{ \(Clear[critPts]\), "\[IndentingNewLine]", \(critPts = x /. solutions\)}], "Input"], Cell["You might want numerical approximations to these points.", "Text"], Cell[BoxData[ \(N[critPts]\)], "Input"], Cell[TextData[{ "You can have ", StyleBox["Mathematica", FontSlant->"Italic"], " find the function values at each of these points in either of two ways." }], "Text"], Cell[BoxData[ \(f[critPts]\)], "Input"], Cell[BoxData[ \(f[x] /. solutions\)], "Input"], Cell["Once again, numerical approximations are useful.", "Text"], Cell[BoxData[ \(N[f[critPts]]\)], "Input"], Cell["You also need to find the function values at the endpoints.", "Text"], Cell[BoxData[ \(f[{xmin, xmax}]\)], "Input"], Cell["\<\ The maximum (39) is at the left endpoint (-2), and the minimum (approximately \ 0.166578) is at the leftmost critical point (approximately -0.443), just as \ you might have expected from the graph.\ \>", "Text"], Cell[CellGroupData[{ Cell["Automating this procedure (advanced)", "Subsubsection"], Cell[TextData[{ "You can make a list of all the possibilities for the locations of the \ maximum and minimum by taking the ", StyleBox["Union", FontColor->RGBColor[1, 0, 0]], " of the list of critical points and the list of endpoints. Using the \"x\ \[Rule]\" form (as in ", StyleBox["solutions", FontColor->RGBColor[1, 0, 0]], ") will be a bit easier in one of the steps below (where ", StyleBox["points", FontColor->RGBColor[1, 0, 0]], " is defined), but the procedure could be done with the other form (as in \ ", StyleBox["critPts", FontColor->RGBColor[1, 0, 0]], ") instead. ", StyleBox["Make sure you have evaluated the cells in the example section \ above during your current ", FontColor->RGBColor[1, 0, 1]], StyleBox["Mathematica", FontSlant->"Italic", FontColor->RGBColor[1, 0, 1]], StyleBox[" session before evaluating the cells in this subsection.", FontColor->RGBColor[1, 0, 1]] }], "Text"], Cell[BoxData[{ \(Clear[locations]\), "\[IndentingNewLine]", \(locations = Union[solutions, {{x \[Rule] xmin}, {x \[Rule] xmax}}]\)}], "Input"], Cell["\<\ You can use the next cell to see the function values at these points.\ \>", "Text"], Cell[BoxData[{ \(Clear[fVals]\), "\[IndentingNewLine]", \(fVals = f[x] /. locations\)}], "Input"], Cell["You need to find the largest and smallest values among these", "Text"], Cell[BoxData[{ \(Max[fVals]\), "\[IndentingNewLine]", \(Min[fVals]\)}], "Input"], Cell["and their numerical approximations", "Text"], Cell[BoxData[{ \(N[Max[fVals]]\), "\[IndentingNewLine]", \(N[Min[fVals]]\)}], "Input"], Cell[TextData[{ "You can also get ", StyleBox["Mathematica", FontSlant->"Italic"], " to produce the coordinates of the points where the maximum and minimum \ are. First here are all the possible points on the graph." }], "Text"], Cell[BoxData[{ \(Clear[points]\), "\[IndentingNewLine]", \(points = {x, f[x]} /. locations\)}], "Input"], Cell[TextData[{ "You can find the positions within the list of function values of the \ maximum and minimum function values using ", StyleBox["Position", FontColor->RGBColor[1, 0, 0]], ". They should be 1 and 4 respectively." }], "Text"], Cell[BoxData[{ \(Clear[maxPosition, minPosition]\), "\[IndentingNewLine]", \(maxPosition = Position[fVals, Max[fVals]]\), "\[IndentingNewLine]", \(minPosition = Position[fVals, Min[fVals]]\)}], "Input"], Cell[TextData[{ "You can now ", StyleBox["Extract", FontColor->RGBColor[1, 0, 0]], " the points in these positions. Here it is for the maximum." }], "Text"], Cell[BoxData[ \(Extract[points, maxPosition]\)], "Input"], Cell["\<\ The maximum is 39, and is located at x = -2. Now for the minimum.\ \>", "Text"], Cell[BoxData[ \(Extract[points, minPosition]\)], "Input"], Cell["A numerical approximation might be nice.", "Text"], Cell[BoxData[ \(N[Extract[points, minPosition]]\)], "Input"], Cell[TextData[{ "That's it! You're done. (Note: The above steps could be combined into a \ single ", StyleBox["Mathematica", FontSlant->"Italic"], " command, but it would then be more difficult to see exactly how ", StyleBox["Mathematica", FontSlant->"Italic"], " does all of this.)" }], "Text"] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Integrals", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The syntax for integrals is ", Cell[BoxData[ \(TraditionalForm\`\[Integral]f[x] \[DifferentialD]x\)], FontColor->RGBColor[1, 0, 0]], StyleBox[" ", FontColor->RGBColor[0.499992, 0.0658274, 0.17203]], "for an indefinite integral (antiderivative), and ", Cell[BoxData[ FormBox[ StyleBox[\(\[Integral]\_a\%b f[x] \[DifferentialD]x\), FontColor->RGBColor[1, 0, 0]], TraditionalForm]]], " for a definite integral. The version 2 syntax for the integrate command \ is ", StyleBox["Integrate[f[x],x]", FontColor->RGBColor[1, 0, 0]], " for an indefinite integral, and ", StyleBox["Integrate[f[x],{x,a,b}]", FontColor->RGBColor[1, 0, 0]], " for a definite integral. ", StyleBox["NIntegrate[f[x],{x,a,b}]", FontColor->RGBColor[1, 0, 0]], " finds a numerical approximation to the integral. You can use variables \ other than x, as in the examples below. ", StyleBox["Evaluate a cell by clicking anywhere in the cell, and pressing \ the ", FontColor->RGBColor[1, 0, 1]], StyleBox["enter", FontColor->RGBColor[0, 0, 1]], StyleBox[" key", FontColor->RGBColor[1, 0, 1]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["Here is an indefinite integral, or antiderivative.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]Sin[t] \[DifferentialD]t\)], "Input", AspectRatioFixed->True], Cell["In the version 2 notation, this would be", "Text"], Cell[BoxData[ \(Integrate[Sin[t], t]\)], "Input"], Cell[TextData[{ "Notice that ", StyleBox["Mathematica", FontSlant->"Italic"], " does not put in a \"+c,\" as you learned to do in class. You will have \ to (at least mentally) do that yourself." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ As with derivatives, if you wish to evaluate an antiderivative (with constant \ \"c\" equal to 0), it is probably best to use \"slash-dot.\"\ \>", "Text"], Cell[BoxData[ \(\[Integral]Sin[t] \[DifferentialD]t /. t \[Rule] \[Pi]\)], "Input", AspectRatioFixed->True], Cell["\<\ To find a definite integral, you need to include the bounds of integration.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%\[Pi] Sin[t] \[DifferentialD]t\)], "Input", AspectRatioFixed->True], Cell["In the version 2 notation, this would be", "Text"], Cell[BoxData[ \(Integrate[Sin[t], {t, 0, Pi}]\)], "Input"], Cell["Here is another integral.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%2\( u\ E\^\(u\^2\)\) \[DifferentialD]u\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can get a numerical approximation to the exact answer by using ", StyleBox["N[ ]", FontColor->RGBColor[1, 0, 0]], " around the integral." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(N[\[Integral]\_0\%2 u\ \(\[ExponentialE]\^\(u\^2\)\) \[DifferentialD]u]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can also find definite integrals numerically, using ", StyleBox["NIntegrate", FontColor->RGBColor[1, 0, 0]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(NIntegrate[u\ \[ExponentialE]\^\(u\^2\), {u, 0, 2}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Although the answers might look the same, they were found by very \ different methods. The first one was found by getting an exact answer, using \ symbolic integration techniques, and then approximating that answer. The \ second was found by using a numerical method of integration (such as \ Simpson's rule), bypassing symbolic integration techniques altogether. If \ you use ", StyleBox["N[ ]", FontColor->RGBColor[1, 0, 0]], ", and ", StyleBox["Mathematica", FontSlant->"Italic"], " is unable to find a symbolic integral, it will then turn to numerical \ methods. This will take longer than simply using ", StyleBox["NIntegrate", FontColor->RGBColor[1, 0, 0]], " to get the numerical result. This is illustrated by the following two \ cells. The ", StyleBox["Timing", FontColor->RGBColor[1, 0, 0]], " command gives us an idea of how long the expression takes to evaluate, \ along with the answer." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Timing[N[\[Integral]\_0\%2 Sin[x\^4] \[DifferentialD]x]]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Timing[NIntegrate[Sin[x\^4], {x, 0, 2}]]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " is also able to compute improper integrals, such as in the following four \ cells. If the integral does not converge, ", StyleBox["Mathematica", FontSlant->"Italic"], " will say so." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%\[Infinity] t\ \(\[ExponentialE]\^\(-t\)\) \[DifferentialD]t\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%\[Infinity]\( 1\/s\) \[DifferentialD]s\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%2\( 1\/\@z\) \[DifferentialD]z\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Integral]\_0\%2\( 1\/z\) \[DifferentialD]z\)], "Input", AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Sums", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The syntax for the sum command is ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\[Sum]\+\(n = a\)\%b\), FontColor->RGBColor[1, 0, 0]], StyleBox[\(f[n]\), FontColor->RGBColor[1, 0, 0]], " "}], TraditionalForm]]], ". The older syntax for this is ", StyleBox["Sum[f[n],{n,a,b}]", FontColor->RGBColor[1, 0, 0]], ". ", StyleBox["NSum[f[n],{n,a,b}]", FontColor->RGBColor[1, 0, 0]], " finds a numerical approximation to the sum. You can use variables other \ than n, as in the examples below. ", StyleBox["Evaluate a cell by clicking anywhere in the cell, and pressing \ the ", FontColor->RGBColor[1, 0, 1]], StyleBox["enter", FontColor->RGBColor[0, 0, 1]], StyleBox[" key", FontColor->RGBColor[1, 0, 1]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["Here is a finite sum.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%10 1\/n\^3\)], "Input", AspectRatioFixed->True], Cell["This is the same sum, using the older syntax.", "Text"], Cell[BoxData[ \(Sum[1/n^3, {n, 1, 10}]\)], "Input"], Cell[TextData[{ "If you would rather see a decimal approximation, that can be arranged, \ using ", StyleBox["N[ ]", FontColor->RGBColor[1, 0, 0]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(N[\[Sum]\+\(n = 1\)\%10 1\/n\^3]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "This can also be done, perhaps more quickly for some sums, using ", StyleBox["NSum", FontColor->RGBColor[1, 0, 0]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(NSum[1\/n\^3, {n, 1, 10}]\)], "Input", AspectRatioFixed->True], Cell["\<\ (Well, maybe not this one, but probably for sums with many more terms.)\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " can even find closed form answers for some partial sums." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%n k\^2\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(i = 0\)\%n 1\/2\^i\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "You can take a limit of this partial sum as n\[Rule]\[Infinity] to find \ the sum of the series ", Cell[BoxData[ \(\[Sum]\+\(i = 0\)\%\[Infinity] 1\/2\^i\)], AspectRatioFixed->True], ". To tell ", StyleBox["Mathematica", FontSlant->"Italic"], " to use the result of the last calculation it did in the next calculation, \ use ", StyleBox["%", FontColor->RGBColor[1, 0, 0]], " to refer to the previous result." }], "Text"], Cell[BoxData[ \(Limit[%, n \[Rule] \[Infinity]]\)], "Input"], Cell[TextData[{ "Can ", StyleBox["Mathematica", FontSlant->"Italic"], " find the sum of the series in one step? Try and see using the next \ cell." }], "Text"], Cell[BoxData[ \(\[Sum]\+\(i = 0\)\%\[Infinity] 1\/2\^i\)], "Input", AspectRatioFixed->True], Cell["Here are some other infinite sums", "Text"], Cell[BoxData[ \(\[Sum]\+\(k = 0\)\%\[Infinity] 1\/\(k!\)\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 1\/n\^5\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(r = 1\)\%\[Infinity] 1\/\(r\^3 + 1\)\)], "Input"], Cell["\<\ although you might not always recognize all the terms in the answer.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ When we try an infinite sum, we might not always get an answer.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 1\/\(\(n\^2\) Log[n\^3 + n\^2 + 1]\)\)], \ "Input"], Cell["We can get a numerical approximation, however.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(N[\[Sum]\+\(n = 1\)\%\[Infinity] 1\/\(n\^2\ Log[n\^3 + n\^2 + 1]\)]\)], \ "Input", AspectRatioFixed->True], Cell[TextData[{ "It might be faster to use ", StyleBox["NSum", FontColor->RGBColor[1, 0, 0]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(NSum[1\/\(n\^2\ Log[n\^3 + n\^2 + 1]\), {n, 1, \[Infinity]}]\)], "Input",\ AspectRatioFixed->True], Cell[TextData[{ "Note the \"0. \[ImaginaryI]\" in the answer. ", StyleBox["Mathematica", FontSlant->"Italic"], " is using stuff you almost certainly haven't learned yet in order to get \ the answer. You can clean it up using ", StyleBox["Chop", FontColor->RGBColor[1, 0, 0]], ", which replaces the approximate real number \"0.\" with the integer \ \"0\", getting rid of the last term. " }], "Text"], Cell[BoxData[ \(Chop[%]\)], "Input"], Cell[TextData[{ "Exercise: Find the sum of the series ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity]\), FractionBox["1", SuperscriptBox[ StyleBox["k", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "2"]]}], TraditionalForm]]], ". Does the answer surprise you at all?" }], "Text", CellFrame->True, Background->GrayLevel[0.833326]] }, Closed]], Cell[CellGroupData[{ Cell["Taylor Polynomials", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "The basic syntax for the Taylor command is ", StyleBox["Series[f[x],{x,a,n}]", FontColor->RGBColor[1, 0, 0]], ", where a is the point where the derivatives are evaluated, and n is the \ order of the desired polynomial. You can use variables other than x, as in \ the examples below. ", StyleBox["Evaluate a cell by clicking anywhere in the cell, and pressing \ the ", FontColor->RGBColor[1, 0, 1]], StyleBox["enter", FontColor->RGBColor[0, 0, 1]], StyleBox[" key", FontColor->RGBColor[1, 0, 1]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["Here are a few Taylor polynomials.", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Series[Sin[x], {x, 0, 7}]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Series[\[ExponentialE]\^\(t\^2\), {t, 0, 8}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Remember that ", StyleBox["Log[x]", FontColor->RGBColor[1, 0, 0]], " is the way ", StyleBox["Mathematica", FontSlant->"Italic"], " refers to ln(x)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Series[Log[z], {z, 1, 6}]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Actually, ", StyleBox["Mathematica", FontSlant->"Italic"], " returns something more than a polynomial. The term ", Cell[BoxData[ StyleBox[\(O[z - 1]\^7\), FontColor->RGBColor[1, 0, 0]]]], " tells us that in the Taylor series for ", StyleBox["Log[z]", FontColor->RGBColor[1, 0, 0]], " we are missing any terms of order 7 or more. We can get just the \ polynomial by using the command ", StyleBox["Normal", FontColor->RGBColor[1, 0, 0]], ", as in the cell below." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Normal[Series[Log[x], {x, 1, 6}]]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Note that ", StyleBox["Mathematica", FontSlant->"Italic"], " put the x at the end of the polynomial, rather than where we would \ probably put it." }], "Text"], Cell["\<\ To get a little idea of the difference between the \"series\" form and the \ \"polynomial\" form, evaluate the following two cells.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Series[Cos[x], {x, 0, 4}]\^2\)], "Input", AspectRatioFixed->True], Cell[BoxData[{ \(Normal[Series[Cos[x], {x, 0, 4}]]\^2\), "\n", \(Expand[%]\)}], "Input", AspectRatioFixed->True], Cell[TextData[{ "You should see that in the first case, when we square the series form, ", StyleBox["Mathematica", FontSlant->"Italic"], " returns terms only up to the order we originally specified. To actually \ square the Taylor polynomial, you need to use the ", StyleBox["Normal", FontColor->RGBColor[1, 0, 0]], " form. ", StyleBox["Mathematica", FontSlant->"Italic"], " is actually doing us a favor here, as the output of ", StyleBox["Series", FontColor->RGBColor[1, 0, 0]], " is an approximation only up to the order specified (in this example, 4) \ and so any additional calculations done with the output should only be \ thought of as being approximations up to that order. However, for other \ purposes (such as graphing the polynomial) we need to use ", StyleBox["Normal", FontColor->RGBColor[1, 0, 0]], " to turn the expression into an actual polynomial." }], "Text", Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1280}, {0, 971}}, WindowToolbars->"EditBar", WindowSize->{677, 637}, WindowMargins->{{Automatic, 43}, {Automatic, 106}} ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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