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Howell, 1998", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["\n", FontColor->RGBColor[1, 0, 0]], StyleBox["Complimentary software to accompany our textbook:", FontColor->RGBColor[0, 1, 0]] }], "Text", Editable->False, Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ StyleBox[ "COMPLEX ANALYSIS: for Mathematics and Engineering, \n3rd Edition, 1997, \ ISBN: 0-7637-0270-6", FontSize->18, FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontSize->18, FontColor->RGBColor[0, 1, 1]], StyleBox[ "Jones & Bartlett Publishers, Inc.\n40 Tall Pine Drive, Sudbury, MA 01776\n\ Tele. (800) 832-0034, FAX: (508) 443-8000\nE-mail: mkt@jbpub.com, \ http://www.jbpub.com/", FontSize->18, FontColor->RGBColor[0, 1, 0]], StyleBox["\n", FontSize->14], StyleBox[ "This free software is compliments of the authors.\nmathews@fullerton.edu, \ howell@westmont.edu", FontSize->14, FontColor->RGBColor[1, 0, 1]] }], "Text"] }, Closed]], Cell[TextData[{ StyleBox["CHAPTER 7 ", FontSize->18], StyleBox["TAYLOR AND LAURENT SERIES\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 7.1 ", FontSize->18], ButtonBox["Uniform Convergence", ButtonData:>"Section 7.1", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 7.2 ", FontSize->18], ButtonBox["Taylor series Representations", ButtonData:>"Section 7.2", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 7.3 ", FontSize->18], ButtonBox["Laurent series Representations", ButtonData:>"Section 7.3", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 7.4 ", FontSize->18], ButtonBox["Singularities, Zeros and Poles", ButtonData:>"Section 7.4", ButtonStyle->"Hyperlink"], StyleBox["\n", FontSize->18, FontColor->RGBColor[1, 0, 1]], StyleBox["Section 7.5 ", FontSize->18], ButtonBox["Applications of Taylor and Laurent Series\n", ButtonData:>"Section 7.5", ButtonStyle->"Hyperlink"], StyleBox["GoTo Chapter", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], StyleBox[" ", FontSize->18], ButtonBox["1", ButtonData:>{"C1.nb", "CHAPTER 1"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["2", ButtonData:>{"C2.nb", "CHAPTER 2"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["3", ButtonData:>{"C3.nb", "CHAPTER 3"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["4", ButtonData:>{"C4.nb", "CHAPTER 4"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["5", ButtonData:>{"C5.nb", "CHAPTER 5"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["6", ButtonData:>{"C6.nb", "CHAPTER 6"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["7", ButtonData:>{"C7.nb", "CHAPTER 7"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["8", ButtonData:>{"C8.nb", "CHAPTER 8"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["9", ButtonData:>{"C9.nb", "CHAPTER 9"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["10", ButtonData:>{"C10.nb", "CHAPTER 10"}, ButtonStyle->"Hyperlink"], StyleBox[", ", FontSize->18], ButtonBox["11", ButtonData:>{"C11.nb", "CHAPTER 11"}, ButtonStyle->"Hyperlink"], StyleBox[".\n", FontSize->18], StyleBox["GoTo ", FontSize->16, FontWeight->"Bold", FontColor->RGBColor[0, 1, 0]], ButtonBox["Contents", ButtonData:>{"Contents.nb", "CONTENTS"}, ButtonStyle->"Hyperlink"] }], "Text", CellTags->"CHAPTER"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["CHAPTER 7 ", FontSize->18], StyleBox["TAYLOR AND LAURENT SERIES", FontSize->18, FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, CellTags->"CHAPTER 7"], Cell[BoxData[ \(\(Clear[A, f, f1, f2, g, h, i, lim1, lim2, L, L1, L2, L3, L4, m, n, p, solset, S, SI, Sinf, S0, S7, S9, S11, S12, S13, S17, S21, S22, S25, x, y, z, Z1, Z2, Z3, Z4]; \)\)], "Input", InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Section 7.1 ", FontSize->18], StyleBox["Uniform Convergence", FontSize->18, FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, CellTags->"Section 7.1"], Cell[TextData[{ StyleBox["Definition 7.1, Page 209.", FontWeight->"Bold"], " The sequence ", Cell[BoxData[ \({\(S\_n\) \((z)\)}\)]], " converges uniformly to f(z) on the set T if for every ", Cell[BoxData[ \(\[Epsilon] > 0\)]], ", there exists a positive integer ", Cell[BoxData[ \(N\_\[Epsilon]\)]], " (which depends only on \[Epsilon]) such that\nif ", Cell[BoxData[ \(n \[GreaterEqual] N\)]], ", then ", Cell[BoxData[ \(\[VerticalSeparator]\( S\_n\) \((z)\) - f \((z)\) \[VerticalSeparator] \( < \ \[Epsilon]\)\)]], " for all ", Cell[BoxData[ \(z\ \[Epsilon]\ T\)]], ".\n\n", StyleBox["Theorem 7.1, Page 210.", FontWeight->"Bold"], " ", StyleBox["(Weierstrass M-test)", FontColor->RGBColor[1, 0, 1]], " Suppose the infinite series ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(u\_k\) \((z)\)\)}]]], " has the property that for each k, ", Cell[BoxData[ \(\[VerticalSeparator]\( u\_k\) \((z)\) \[VerticalSeparator] \( < \ M\)\_k\)]], " for all ", Cell[BoxData[ \(z\ \[Epsilon]\ T\)]], ". If ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(M\_k\)}]]], " converges, then ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(u\_k\) \((z)\)\)}]]], " converges uniformly on T.\n\n", StyleBox["Theorem 7.2, Page 211.", FontWeight->"Bold"], " Suppose the power series ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_k\) \((z - \[Alpha])\)\^k\)}]]], " has radius of convergence ", Cell[BoxData[ \(\[Rho]\ > 0\)]], ", Then for each r, ", Cell[BoxData[ \(\(0 < r < \ \[Rho]\ \)\)]], ", the series converges uniformly on the closed disk ", Cell[BoxData[ \(\(\(D\_r\)\&_\) \((\[Alpha])\) = { \(z : \)\ \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \( \[LessEqual] \ r\)\)}\)]], ".\n\n", StyleBox["Corollary 7.1, Page 211.", FontWeight->"Bold"], " For each r, ", Cell[BoxData[ \(\(0 < r < \ 1\ \)\)]], " , the geometric series converges uniformly on the closed disk ", Cell[BoxData[ \(\(\(D\_r\)\&_\) \((0)\) = { \(z : \)\ \[VerticalSeparator] \(z \[VerticalSeparator] \( \[LessEqual] \ r\)\)}\)]], ".\n\n", StyleBox["Theorem 7.3, Page 211.", FontWeight->"Bold"], " Suppose ", Cell[BoxData[ \({\(S\_k\) \((z)\)}\)]], " is a sequence of continuous functions defined on a set T containing \ the contour C. If ", Cell[BoxData[ \({\(S\_k\) \((z)\)}\)]], " converges uniformly to f(z) on the set T, then\n", StyleBox["(i)", FontWeight->"Bold"], "\t", Cell[BoxData[ RowBox[{"f", StyleBox[\((z)\), FontSize->12]}]]], " is continuous on T, and\n", StyleBox["(ii)", FontWeight->"Bold"], "\t", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{\(lim\_\(k \[Rule] \[Infinity]\)\), RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{\(S\_k\), StyleBox[\((z)\), FontSize->12], RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}], " ", "=", " ", RowBox[{ RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{\(lim\_\(k \[Rule] \[Infinity]\)\), RowBox[{\(S\_k\), StyleBox[\((z)\), FontSize->12], RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}], " ", "=", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{"f", StyleBox[\((z)\), FontSize->12], RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}]}], " "}]]], ".\n\n", StyleBox["Corollary 7.2, Page 212.", FontWeight->"Bold"], " If the series ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_k\) \((z - \[Alpha])\)\^k\)}]]], " converges uniformly to ", Cell[BoxData[ RowBox[{"f", StyleBox[\((z)\), FontSize->12]}]]], " on the set T, and C is a contour contained in T, then\n ", Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{\(c\_k\), \(\((z - \[Alpha])\)\^k\), RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}], " ", "=", " ", RowBox[{ RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], RowBox[{\(c\_k\), \(\((z - \[Alpha])\)\^k\), RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}], " ", "=", RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], StyleBox["C", FontSize->14], " "], StyleBox[" ", FontSize->14], RowBox[{"f", StyleBox[\((z)\), FontSize->12], RowBox[{ StyleBox["\[DifferentialD]", FontSize->14], "z"}]}]}]}]}], " "}]]], ". " }], "Text"], Cell[TextData[{ "Load the SymbolicSum package. Make sure this is done only once in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example 7.1, Page 209.", FontWeight->"Bold"], " The sequence ", Cell[BoxData[ \({\(S\_n\) \((z)\)} = \ {\[ExponentialE]\^z\ + 1\/n}\)]], " converges uniformly to ", Cell[BoxData[ \(\(\ \[ExponentialE]\^z\)\)]], " on the entire complex plane.\nSolution. The solution is given in the \ textbook." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["Example 7.2, Page 212.", FontWeight->"Bold"], " Show that Log(1 - z) = ", StyleBox["-", FontSize->14], " ", Cell[BoxData[ FormBox[ RowBox[{ UnderoverscriptBox["\[Sum]", StyleBox[\(n = 1\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], FractionBox[ StyleBox[\(z\^n\), FontSize->14], StyleBox["n", FontSize->14]]}], TraditionalForm]]], " for all z ", StyleBox["\[Element]", FontFamily->"Symbol"], " D." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, n, S17, S, Sinf, z]; \nf[z_]\ = \ Log[1 - z]; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 17}]]; \n Print["\< f[z] = \>", f[z]]; \n Print[\*"\"\<\!\(S\_17\)[z] = \>\"", S[z]]; \)], "Input", AspectRatioFixed->True], Cell["Sum up the terms to verify that we have things right.", "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S17[z_]\ \ \ = \ \[Sum]\+\(n = 1\)\%17\ \(-\ z\^n\)\/n; \n Sinf[z_]\ = \ \[Sum]\+\(n = 1\)\%\[Infinity]\ \(-\ z\^n\)\/n; \n Print[\*"\"\<\!\(S\_17\)[z] = \>\"", S17[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 1\)\%\[Infinity]\ \)\!\(\(-\\\ z\^n\)\/n\) = \>\"", Sinf[z]]; \nPrint["\< f[z] = \>", f[z]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(\(Plot[{f[x], S17[x]}, {x, \(-0.99999\), 0.99999}, PlotRange \[Rule] {{\(-1\), 1}, {\(-6\), 1}}, Ticks \[Rule] {Range[\(-1\), 1, 0.5], Range[\(-6\), 1, 1]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)\)], "Input"], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 7", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 7.2 ", StyleBox["Taylor series Representations", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, FontSize->18, CellTags->"Section 7.2"], Cell[TextData[{ " In Section 4.2 we saw that functions defined by power series have \ derivatives of all orders. In Section 6.5 we saw that analytic functions \ also have derivatives of all orders. It seems natural, therefore, that there \ would be some connection between analytic functions and power series. As you \ might guess, the connection exists via the Taylor and Maclaurin series of \ analytic functions.\n\n", StyleBox["Definition 7.2, Page 214.", FontWeight->"Bold"], " If ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic at ", Cell[BoxData[ \(z = \[Alpha]\)]], ", the series ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], RowBox[{ FractionBox[ StyleBox[ RowBox[{ SuperscriptBox["f", TagBox[\((k)\), Derivative], MultilineFunction->None], \((\[Alpha])\)}], FontSize->12], StyleBox[\(k!\), FontSize->12]], \(\((z - \[Alpha])\)\^k\)}]}]]], " is called the ", StyleBox["Taylor series", FontColor->RGBColor[1, 0, 1]], " for ", Cell[BoxData[ \(f \((z)\)\)]], " centered around ", Cell[BoxData[ \(\[Alpha]\)]], ". When the center is ", Cell[BoxData[ \(\[Alpha] = 0\)]], ", the series is called the ", StyleBox["Maclaurin series", FontColor->RGBColor[1, 0, 1]], " for ", Cell[BoxData[ \(f \((z)\)\)]], ".\n \n", StyleBox["Theorem 7.4, Page 215.", FontWeight->"Bold"], " ", StyleBox["(Taylor's Theorem)", FontColor->RGBColor[1, 0, 1]], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in a domain G, and that ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\) = { \(z : \)\ \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \ \( < \ R\)\)}\)]], " is contained in G. Then the Taylor series converges to ", Cell[BoxData[ \(f \((z)\)\)]], " for all z in ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\)\)]], "; that is ", Cell[BoxData[ RowBox[{\(f \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], RowBox[{ FractionBox[ StyleBox[ RowBox[{ SuperscriptBox["f", TagBox[\((k)\), Derivative], MultilineFunction->None], \((\[Alpha])\)}], FontSize->12], StyleBox[\(k!\), FontSize->12]], \(\((z - \[Alpha])\)\^k\)}]}]}]]], " for all ", Cell[BoxData[ \(z\ \[Epsilon]\ \(D\_R\) \((\[Alpha])\)\)]], ".\n Furthermore, this representation is valid in the largest disk with \ center ", Cell[BoxData[ \(z = \[Alpha]\)]], " that is contained in G, and the convergence is uniform on any closed \ subdisk ", Cell[BoxData[ \(\(\(D\_r\)\&_\) \((\[Alpha])\) = { \(z : \)\ \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \( \[LessEqual] \ r\)\)}\)]], " for ", Cell[BoxData[ \(\(0 < r < \ R\ \)\)]], " . \n\n", StyleBox["Corollary 7.5, Page 217.", FontWeight->"Bold"], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in the domain G that contains the point ", Cell[BoxData[ \(z = \[Alpha]\)]], ". Let ", Cell[BoxData[ \(z\_0\)]], " be a singular point of minimum distance to ", Cell[BoxData[ \(\[Alpha]\)]], ". If ", Cell[BoxData[ \(\(\[VerticalSeparator]\( z\_0 - \[Alpha] \[VerticalSeparator] \)\) = \ R\)]], ", then\n", StyleBox["(i)\t", FontWeight->"Bold"], "the Taylor series converges to ", Cell[BoxData[ \(f \((z)\)\)]], " on all of ", Cell[BoxData[ \(\(D\_R\) \((\[Alpha])\)\)]], ", and\n", StyleBox["(ii)\t", FontWeight->"Bold"], "if ", Cell[BoxData[ \(S > R\)]], ", the Taylor series does not converge to ", Cell[BoxData[ \(f \((z)\)\)]], " on all of ", Cell[BoxData[ \(\(D\_S\) \((\[Alpha])\)\)]], ".\n\n", StyleBox["Theorem 7.5, Page 218.", FontWeight->"Bold"], " ", StyleBox["(Uniqueness of Power Series)", FontColor->RGBColor[1, 0, 1]], " Suppose that in some disk ", Cell[BoxData[ \(\(D\_r\) \((\[Alpha])\)\)]], " we have ", Cell[BoxData[ RowBox[{ RowBox[{\(f \((z)\)\), " ", "=", " ", RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(a\_n\) \((z - \[Alpha])\)\^n\)}], " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(b\_n\) \((z - \[Alpha])\)\^n\)}]}]}], " "}]]], ". \nThen ", Cell[BoxData[ \(a\_n = \(b\_n\ \ for\ \ \ n = 0\), 1, 2, \[CenterEllipsis]\)]], ".\n\n", StyleBox["Theorem 7.6, Page 219.", FontWeight->"Bold"], " Let ", Cell[BoxData[ \(f \((z)\)\)]], " and ", Cell[BoxData[ \(g \((z)\)\)]], " have the power series representations\n", Cell[BoxData[{ RowBox[{\(f \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(a\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_\(r\_1\)\) \((\[Alpha])\)\ \ and\)}]}], RowBox[{\(g \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(b\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_\(r\_2\)\) \(\((\[Alpha])\).\)\)}]}]}]], "\nIf ", Cell[BoxData[ \(r\ = \ min {r\_1, r\_2}\)]], ", and ", Cell[BoxData[ \(\[Beta]\)]], " is any complex constant, then\n", StyleBox["(i)", FontWeight->"Bold"], "\t", Cell[BoxData[ RowBox[{\(\[Beta]f \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(\[Beta]a\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_\(r\_1\)\) \((\[Alpha])\)\)}]}]]], ",\n\n", StyleBox["(ii)", FontWeight->"Bold"], "\t", Cell[BoxData[ RowBox[{\(f \((z)\) + g \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\((a\_n + b\_n)\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_r\) \((\[Alpha])\)\)}]}]]], ", and\n\n", StyleBox["(iii)", FontWeight->"Bold"], "\t", Cell[BoxData[ RowBox[{\(f \((z)\)*g \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_r\) \((\[Alpha])\)\)}]}]]], ", where\n\t", Cell[BoxData[ \(c\_n\)]], " = ", Cell[BoxData[ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(k = 0\), FontSize->10], StyleBox["n", FontSize->10]], \(\(a\_k\) b\_\(n - k\)\)}]]], ".\nIdentity (iii) is known as the ", StyleBox["Cauchy product", FontColor->RGBColor[1, 0, 1]], " of the series for ", Cell[BoxData[ \(f \((z)\)\)]], " and ", Cell[BoxData[ \(g \((z)\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["Example 7.3, Page 217.", FontWeight->"Bold"], " Show that ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[\(\((1\ - \ z)\)\^2\), FontSize->14]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ UnderoverscriptBox["\[Sum]", StyleBox[\(n = 0\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], StyleBox[\(\((n + 1)\) z\^\(\ n\)\), FontSize->14]}], TraditionalForm]]], " is valid for z ", StyleBox["\[Element]", FontFamily->"Symbol"], " D." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, n, S, S17, Sinf, z]; \nf[z_]\ = \ 1\/\((1 - z)\)\^2; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 17}]]; \n Print["\< f[z] = \>", f[z]]; \n Print[\*"\"\<\!\(S\_17\)[z] = \>\"", S[z]]; \)], "Input"], Cell[TextData["Sum up the terms to verify that we have things right."], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S17[z_]\ \ \ = \ \[Sum]\+\(n = 0\)\%17\((n + 1)\)\ z\^n; \n Sinf[z_]\ = \ \[Sum]\+\(n = 0\)\%\[Infinity]\((n + 1)\)\ z\^n; \n Print[\*"\"\<\!\(S\_17\)[z] = \>\"", S17[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 0\)\%\[Infinity]\ \)(n+1) \!\(z\^n\) = \>\"", Sinf[z]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 7.4, Page 218.", FontWeight->"Bold"], " ", StyleBox["(a)", FontWeight->"Bold"], " Show that ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[ RowBox[{"1", " ", "-", " ", FormBox[\(z\^2\), "TraditionalForm"]}], FontSize->14]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ UnderoverscriptBox["\[Sum]", StyleBox[\(n = 0\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], " ", StyleBox[\(z\^\(\ 2 n\)\), FontSize->14]}], TraditionalForm]]], " is valid for z ", StyleBox["\[Element]", FontFamily->"Symbol"], " D." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, n, S25, S, Sinf, z]; \nf[z_]\ = \ 1\/\(1 - z\^2\); \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 50}]]; \n Print["\< f[z] = \>", f[z]]; \n Print[\*"\"\<\!\(S\_50\)[z] = \>\"", S[z]]; \)], "Input"], Cell[TextData["Sum up the terms to verify that we have things right."], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S25[z_]\ \ \ = \ \[Sum]\+\(n = 0\)\%25 z\^\(2 n\); \n Sinf[z_]\ = \ \ \[Sum]\+\(n = 0\)\%\[Infinity] z\^\(2 n\); \n Print[\*"\"\<\!\(S\_25\)[z] = \>\"", S25[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 0\)\%\[Infinity]\ \)\!\(z\^\(2 n\)\) = \>\"", Sinf[z]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["(b)", FontWeight->"Bold"], " Show that ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[ RowBox[{"1", " ", "+", " ", FormBox[\(z\^2\), "TraditionalForm"]}], FontSize->14]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ RowBox[{ UnderoverscriptBox["\[Sum]", StyleBox[\(n = 0\), FontSize->12], StyleBox["\[Infinity]", FontSize->12]], StyleBox[" ", FontSize->14], StyleBox[\(\(\((\(-1\))\)\^n\) z\^\(\ 2 n\)\), FontSize->14]}], TraditionalForm]]], " is valid for z ", StyleBox["\[Element]", FontFamily->"Symbol"], " D." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, n, S, S25, Sinf, z]; \nf[z_]\ = \ 1\/\(1 + z\^2\); \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 50}]]; \n Print["\< f[z] = \>", f[z]]; \n Print[\*"\"\<\!\(S\_50\)[z] = \>\"", S[z]]; \)], "Input"], Cell[TextData["Sum up the terms to verify that we have things right."], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S25[z_]\ \ \ = \ \[Sum]\+\(n = 0\)\%25\(\((\(-1\))\)\^n\) z\^\(2 n\); \nSinf[z_]\ = \ \ \[Sum]\+\(n = 0\)\%\[Infinity]\(\((\(-1\))\)\^n\) z\^\(2 n\); \n Print[\*"\"\<\!\(S\_25\)[z] = \>\"", S25[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 0\)\%\[Infinity]\ \)\!\(\((\(-1\))\)\^n\)\!\(z\^\(2 n\)\) = \>\"", Sinf[z]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 7.5, Page 219.", FontWeight->"Bold"], " Find the Maclaurin series of f(z) = ", Cell[BoxData[ \(TraditionalForm\`sin\^3\)], FontSize->14], "(z)." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, f1, f2, n, S, z]; \nf[z_]\ = \ Sin[z]\^3; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 25}]]; \n Print["\< f[z] = \>", f[z]]; \n Print[\*"\"\<\!\(S\_25\)[z] = \>\"", S[z]]; \)], "Input"], Cell["Sum up the terms to verify that we have things right.", "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(f1[z_]\ = PowerExpand[ \[Sum]\+\(n = 0 \)\%\[Infinity]\(\((\(-1\))\)\^n\ 3\)\/\(4 \(\((2 n + 1)\)!\)\)\ z\^\(2 n + 1\)]; \n f2[z_]\ = \ PowerExpand[ \[Sum]\+\(n = 0 \)\%\[Infinity]\(\(-\ \((\(-1\))\)\^n\)\ 3\ 9\^n\)\/\(4 \(\((2 n + 1)\)!\)\)\ z\^\(2 n + 1\)]; \n f[z_]\ = \ Factor[f1[z]\ + \ f2[z], \ Trig \[Rule] True]; \n Print[\*"\"\<\!\(f\_1\)[z] = \!\(\[Sum]\+\(n = \ 0\)\%\[Infinity]\)\!\(\(\((\(-1\))\)\^n\\\ 3\)\/\(4 \(\((2 n + 1)\)!\)\)\) \ \!\(z\^\(2 n + 1\)\) = \>\"", f1[z]]; \n Print[\*"\"\<\!\(f\_2\)[z] = \!\(\[Sum]\+\(n = \ 0\)\%\[Infinity]\)\!\(\(\(-\\\ \((\(-1\))\)\^n\)\\\ 3\\\ 9\^n\)\/\(4 \(\((2 \ n + 1)\)!\)\)\) \!\(z\^\(2 n + 1\)\) = \>\"", f2[z]]; \n Print[\*"\"\< f[z] = \!\(f\_1\)[z] + \!\(f\_2\)[z] = \>\"", \ f1[z]\ + \ f2[z]]; \nPrint["\< f[z] = \>", f[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.6, Page 220.", FontWeight->"Bold"], " Use the Cauchy product of series to show that \n", Cell[BoxData[ RowBox[{ FractionBox[ StyleBox["1", FontSize->12], StyleBox[\(\((1\ - \ z)\)\^2\), FontSize->12]], "=", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->10]], \(\((n + 1)\) z\^n\ \ \ for\ \ z\ \[Epsilon]\ \(D\_1\) \((0)\)\)}]}]]], ".\nSolution. The solution is given in the textbook." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 7", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 7.3 ", StyleBox["Laurent series Representations", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, FontSize->18, CellTags->"Section 7.3"], Cell[TextData[{ " The main result of this section specifies how functions analytic in \ an annulus can be expanded in a Laurent series. Often times this just means \ including terms with negative powers of z .\n\n", StyleBox["Theorem 7.7, Page 225.", FontWeight->"Bold"], " Suppose the Laurent series ", Cell[BoxData[ RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], RowBox[{ StyleBox["n", FontSize->10], StyleBox["=", FontSize->10], StyleBox[\(-\[Infinity]\), FontSize->12]}], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z - \[Alpha])\)\^n\)}], " "}]]], " converges on an annulus ", Cell[BoxData[ \(A \((r, R, \[Alpha])\) = { z : \ \(r\ < \) \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \( < \ R\)\)}\)]], ". Then the series converges uniformly on any subannulus ", Cell[BoxData[ \(\(A\&_\) \((s, t, \[Alpha])\) = { z : \ \(s\ \[LessEqual] \) \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \( \[LessEqual] \ t\)\)}\)]], " where ", Cell[BoxData[ \(r\ < s < t < \ R\)]], ".\n\n", StyleBox["Theorem 7.8, Page 225.", FontWeight->"Bold"], " ", StyleBox["(Laurent's Theorem)", FontColor->RGBColor[1, 0, 1]], " Suppose ", Cell[BoxData[ \(0 \[LessEqual] r < R\)]], ", and that ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in the annulus ", Cell[BoxData[ \(A = \(A \((r, R, \[Alpha])\) = { z : \ \(r\ < \) \[VerticalSeparator] \(z - \[Alpha] \[VerticalSeparator] \( < \ R\)\)}\)\)]], ". If ", Cell[BoxData[ \(\[Rho]\)]], " is any number such that ", Cell[BoxData[ \(r\ < \[Rho] < \ R\)]], ", then for all ", Cell[BoxData[ \(z\_0\ \[Epsilon]\ A \((r, R, \[Alpha])\), \ \ f \((z)\)\)]], " has the Laurent series representation \n", Cell[BoxData[ RowBox[{\(f \((z\_0)\)\), "=", RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], RowBox[{ StyleBox["n", FontSize->10], StyleBox["=", FontSize->10], StyleBox[\(-\[Infinity]\), FontSize->12]}], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z\_0 - \[Alpha])\)\^n\)}], "=", RowBox[{ RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 1\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_\(-n\)\) \((z\_0 - \[Alpha])\)\^\(-n\)\)}], "+", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z\_0 - \[Alpha])\)\^n\)}]}]}]}]]], "\nwhere for ", Cell[BoxData[ \(n = 0, 1, 2, \[CenterEllipsis]\)]], " the coefficients ", Cell[BoxData[ \(c\_n\ \ and\ \ c\_\(-n\)\)]], " are given by\n", Cell[BoxData[ FormBox[ RowBox[{\(c\_\(-n\)\), "=", RowBox[{ StyleBox[\(1\/\(2 \[Pi]\ i\)\), FontSize->16], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], RowBox[{ SubscriptBox[ SuperscriptBox[ StyleBox["C", FontSize->14], "+"], "\[Rho]"], StyleBox["(", FontSize->12], StyleBox["\[Alpha]", FontSize->12], StyleBox[")", FontSize->12]}], " "], StyleBox[" ", FontSize->14], StyleBox[ RowBox[{ FractionBox[ StyleBox[\(f(z)\), FontSize->14], SuperscriptBox[ RowBox[{"(", RowBox[{ StyleBox["z", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["-", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["\[Alpha]", FontSize->12]}], ")"}], \(\(-n\) + 1\)]], \(\[DifferentialD]z\)}], FontSize->14]}]}]}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{\(c\_n\), "=", RowBox[{ StyleBox[\(1\/\(2 \[Pi]\ i\)\), FontSize->16], RowBox[{ SubsuperscriptBox[ StyleBox["\[Integral]", FontSize->18], RowBox[{ SubscriptBox[ SuperscriptBox[ StyleBox["C", FontSize->14], "+"], "\[Rho]"], StyleBox["(", FontSize->12], StyleBox["\[Alpha]", FontSize->12], StyleBox[")", FontSize->12]}], " "], StyleBox[" ", FontSize->14], StyleBox[ RowBox[{ FractionBox[ StyleBox[\(f(z)\), FontSize->14], SuperscriptBox[ RowBox[{"(", RowBox[{ StyleBox["z", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["-", FontSize->14], StyleBox[" ", FontSize->14], StyleBox["\[Alpha]", FontSize->12]}], ")"}], \(n + 1\)]], \(\[DifferentialD]z\)}], FontSize->14]}]}]}], TraditionalForm]]], ".\n\n", StyleBox["Theorem 7.9, Page 228.", FontWeight->"Bold"], " ", StyleBox["(Uniqueness and differentiation of Laurent Series)", FontColor->RGBColor[1, 0, 1]], " Suppose that ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in the annulus ", Cell[BoxData[ \(A \((r, R, \[Alpha])\)\)]], ", and has the Laurent series representation ", Cell[BoxData[ RowBox[{\(f \((z)\)\), "=", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], RowBox[{ StyleBox["n", FontSize->10], StyleBox["=", FontSize->10], StyleBox[\(-\[Infinity]\), FontSize->12]}], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ all\ \ \ z\ \[Epsilon]\ A \((r, R, \[Alpha])\)\)}]}]]], ".\n", StyleBox["(i)", FontWeight->"Bold"], "\tIf ", Cell[BoxData[ RowBox[{\(f \((z)\)\), "=", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], RowBox[{ StyleBox["n", FontSize->10], StyleBox["=", FontSize->10], StyleBox[\(-\[Infinity]\), FontSize->12]}], StyleBox["\[Infinity]", FontSize->12]], \(\(b\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ all\ \ \ z\ \[Epsilon]\ A \((r, R, \[Alpha])\)\)}]}]]], ", then ", Cell[BoxData[ \(b\_n = c\_n\)]], " for all n. (In other words, the Laurent series for ", Cell[BoxData[ \(f \((z)\)\)]], " in a given annulus is unique.)\n", StyleBox["(ii)", FontWeight->"Bold"], "\tFor all ", Cell[BoxData[ \(z\ \[Epsilon]\ A \((r, R, \[Alpha])\)\)]], ", the derivatives for ", Cell[BoxData[ \(f \((z)\)\)]], " may be obtained by termwise differentiation of its Laurent series. " }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Load the SymbolicSum package. Make sure this is done only once in a ", StyleBox["Mathematica", FontSlant->"Italic"], " session." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Example on, Page 224.", FontWeight->"Bold"], " Find the Laurent series for f(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[ FormBox[\(e\^z\), "TraditionalForm"], FontSize->14], StyleBox[\(z\^3\), FontSize->14]], TraditionalForm]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, L, S, S9, S11, S12, S21, S22, Sinf, z]; \n f[z_]\ = \ \[ExponentialE]\^z\/z\^3; \n S[z_]\ = \ Normal[Series[\[ExponentialE]\^z, {z, 0, 9}]]\/z\^3; \n L[z_]\ = \ Normal[Series[f[z], {z, 0, 6}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\", L[z]]; \)], "Input", AspectRatioFixed->True], Cell["The latter is a Laurent series is valid for 0 < |z|.", "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData["Sum up the terms to verify that we have things right."], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S9[z_]\ = \ \[Sum]\+\(n = 0\)\%9 z\^\(n - 3\)\/\(n!\); \n Sinf[z_]\ = \ \ \[Sum]\+\(n = 0\)\%\[Infinity] z\^\(n - 3\)\/\(n!\); \n Print[\*"\"\<\!\(S\_9\)[z] = \>\"", S9[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 0\)\%\[Infinity]\ \)\!\(z\^\(n - 3\)\/\(n!\)\) = \>\"", Sinf[z]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[x], \*"\"\< and s = \!\(S\_9\)[x]\>\""]; \n Plot[{f[x], S9[x]}, {x, 0.1, 8}, PlotRange \[Rule] {{0, 8}, {0, 10}}, Ticks \[Rule] {Range[0, 8, 1], Range[0, 10, 1]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.7, Page 229.", FontWeight->"Bold"], " Find three different Laurent series representations for\nf(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["3", FontSize->14], StyleBox[ RowBox[{"2", " ", "+", " ", "z", " ", "-", " ", FormBox[\(z\^2\), "TraditionalForm"]}], FontSize->14]], TraditionalForm]]], " involving powers of z." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[A, f, f1, f2, L1, L2, L3, L4, S0, S11, S12, S21, S22, SI, z]; \n f[z_]\ = \ 3\/\(2\ + \ z\ - \ z\^2\); \nPrint["\", f[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica ", FontSlant->"Italic"], " easily finds the Maclaurin series involving positive powers of z:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(S0[z_]\ = \ Normal[Series[f[z], {z, 0, 7}]]; \n Print[\*"\"\<\!\(S\_0\)[z] = \>\"", S0[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The above series converges for |z| < 1, this requires looking at the sum \ of the two \n\"geometric series that form the parts of S", StyleBox["0", FontVariations->{"CompatibilityType"->"Subscript"}], ".\"" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Mathematica ", FontSlant->"Italic"], " can also find \"the other series\" that involves negative powers of z, \ this involves the mental thinking that you substitute Z = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " in the original series to get a new function of Z, then expand it about \ Z = 0 and then substitute Z = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(SI[z_]\ = \ Normal[Series[f[z], {z, \[Infinity], 10}]]; \n Print[\*"\"\<\!\(S\_I\)[z] = \>\"", SI[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The above series converges for 2 < |z|, this requires looking at the sum \ of the two \"geometric series that form the parts of ", Cell[BoxData[ \(S\_I\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[ "This leaves an unresolved question: \"What happens in the annulus 1 < |z| < \ 2 ? \nThis requires a little work. First split up the functions up into \ their partial fraction form, \nand then do expansions about z=0 and z=\ \[Infinity] for each part."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(A\ = \ MapAll[Cancel, Apart[f[z]]]; \n f1[z_]\ = \ A\[LeftDoubleBracket]2\[RightDoubleBracket]; \n f2[z_]\ = \ A\[LeftDoubleBracket]1\[RightDoubleBracket]; \n Print["\< f[z] = \>", f[z], "\< = \>", A]; \n Print[\*"\"\<\!\(f\_1\)[z] = \>\"", f1[z]]; \n Print[\*"\"\<\!\(f\_2\)[z] = \>\"", f2[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Now form Laurent expansions for the two parts ", Cell[BoxData[ FormBox[ StyleBox[\(f\_1\), FontSize->14], TraditionalForm]]], "(z) and ", Cell[BoxData[ FormBox[ StyleBox[\(f\_2\), FontSize->14], TraditionalForm]]], "(z) of f(z)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(S11[z_]\ = \ Normal[Series[f1[z], {z, 0, 12}]]; \n S12[z_]\ = \ Normal[Series[f1[z], {z, \[Infinity], 11}]]; \n S21[z_]\ = \ Normal[Series[f2[z], {z, 0, 12}]]; \n S22[z_]\ = \ Normal[Series[f2[z], {z, \[Infinity], 12}]]; \n Print[\*"\"\<\!\(S\_11\)[z] = \>\"", S11[z]]; \n Print[\*"\"\<\!\(S\_12\)[z] = \>\"", S12[z]]; \n Print[\*"\"\<\!\(S\_21\)[z] = \>\"", S21[z]]; \n Print[\*"\"\<\!\(S\_22\)[z] = \>\"", S22[z]]; \)], "Input", AspectRatioFixed->False], Cell[TextData[{ "The Laurent series L", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) = S", StyleBox["11", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) + S", StyleBox["21", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is valid for |z| < 1." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(L1[z_]\ = \ S11[z]\ + \ S21[z]; \n Print[\*"\"\<\!\(L\_1\)[z] = \>\"", L1[z]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[x], \*"\"\< and s = \!\(L\_1\)[z]\>\""]; \n Plot[{f[x], L1[x]}, {x, 0.0, 0.999999}, PlotRange \[Rule] {{0, 1}, {1, 2}}, Ticks \[Rule] {Range[0, 1, 0.2], Range[1, 2, 0.2]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The following Laurent series L", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) = S", StyleBox["12", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) + S", StyleBox["21", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is valid for 1 < |z| < 2." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(L2[z_]\ = \ S12[z]\ + \ S21[z]; \n Print[\*"\"\<\!\(L\_2\)[z] = \>\"", L2[z]]; \)], "Input", AspectRatioFixed->False], Cell[BoxData[ \(Print["\", f[x], \*"\"\< and s = \!\(L\_2\)[x]\>\""]; \n Plot[{f[x], L2[x]}, {x, 1.0001, 1.9999}, \n\ \ PlotRange \[Rule] {{1, 2}, {0, 7}}, Ticks \[Rule] {Range[1, 2, 0.2], Range[0, 7, 1]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "So L", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is an approximation to f[z] valid for 1 < |z| < 2.\nNotice that L", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " a good approximation for 1 < |z| < 2,\nfor example it is not accurate at \ z = 1.5" }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Print["\< f[1.5] = \>", f[1.5]]; \n Print[\*"\"\<\!\(L\_1\)[1.5] = \>\"", L1[1.5]]; \n Print["\< f[1.5] = \>", f[1.5]]; \n Print[\*"\"\<\!\(L\_2\)[1.5] = \>\"", L2[1.5]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "The Laurent series L", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) = S", StyleBox["12", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) + S", StyleBox["22", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is valid for 2 < |z| ." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(L3[z_]\ = \ S12[z]\ + \ S22[z]; \n Print[\*"\"\<\!\(L\_3\)[z] = \>\"", L3[z]]; \)], "Input", AspectRatioFixed->False], Cell[BoxData[ \(Print["\", f[x], \*"\"\< and s = \!\(L\_3\)[x]\>\""]; \n Plot[{f[x], L3[x]}, {x, 2.0001, 4.0}, \n\ \ PlotRange \[Rule] {{2, 4}, {\(-6\), 0}}, Ticks \[Rule] {Range[2, 4, 0.2], Range[\(-6\), 0, 1]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "So L", StyleBox["3", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) is an approximation to f[z] valid for 2< |z|.\nNotice that L", StyleBox["1", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) and L", StyleBox["2", FontVariations->{"CompatibilityType"->"Subscript"}], "(z) are ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " good approximations for 2< |z|,\nfor example they are not accurate at z \ = 4.0" }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Print["\< f[4.0] = \>", f[4.0]]; \n Print[\*"\"\<\!\(L\_1\)[4.0] = \>\"", L1[4.0]]; \n Print["\< f[4.0] = \>", f[4.0]]; \n Print[\*"\"\<\!\(L\_2\)[4.0] = \>\"", L2[4.0]]; \n Print["\< f[4.0] = \>", f[4.0]]; \n Print[\*"\"\<\!\(L\_3\)[4.0] = \>\"", L3[4.0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[ "The fourth possibility does not converge for any values of z and should not \ be considered, however you could write down a few terms and think about why \ this series diverges."], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(L4[z_]\ = \ S11[z]\ + \ S22[z]; \n Print[\*"\"\<\!\(L\_4\)[z] = \>\"", L4[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "In the above series, when |z| < ", Cell[BoxData[ FractionBox[ StyleBox["1", FontSize->12], StyleBox["2", FontSize->12]]]], " the portion of the series with positive exponents will\nconverge but the \ portion with negative exponents will diverge. Similarly, when |z| > 1 the \n\ portion of the series with negative exponents will converge, but the portion \ of the series\nwith positive exponents will diverge. Hence, the series \ diverges for all z." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\< f[0.5] = \>", f[0.5]]; \n Print[\*"\"\<\!\(L\_4\)[0.5] = \>\"", L4[0.5]]; \n Print["\< f[1.5] = \>", f[1.5]]; \n Print[\*"\"\<\!\(L\_4\)[1.5] = \>\"", L4[1.5]]; \n Print["\< f[4.0] = \>", f[4.0]]; \n Print[\*"\"\<\!\(L\_4\)[4.0] = \>\"", L4[4.0]]; \)], "Input"], Cell[TextData[{ "\n", StyleBox["Example 7.8, Page 229.", FontWeight->"Bold"], " Find the Laurent series for f(z) = ", Cell[BoxData[ FractionBox[ StyleBox[\(cos \((z)\)\ - \ 1\), FontSize->14], StyleBox[\(z\^4\), FontSize->14]]]], "." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, L, S, S7, Sinf, z]; \nf[z_]\ = \ \(Cos[z]\ - \ 1\)\/z\^4; \n S[z_]\ = \ Normal[Series[Cos[z] - 1, \ {z, 0, 14}]]\/z\^4; \n L[z_]\ = \ Normal[Series[f[z], {z, 0, 10}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\", L[z]]; \)], "Input", AspectRatioFixed->False], Cell["The latter Laurent series is valid for 0 < |z| .", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData["Sum up the terms to verify that we have things right."], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S7[z_]\ = \ \[Sum]\+\(n = 1\)\%7 \(\(\((\(-1\))\)\^n\) z\^\(2 n - 4\)\)\/\(\((2 n)\)!\); \n Sinf[z_]\ = \ \ \[Sum]\+\(n = 1\)\%\[Infinity]\(\(\((\(-1\))\)\^n\) z\^\(2 n - 4\)\)\/\(\((2 n)\)!\); \n Print[\*"\"\<\!\(S\_7\)[z] = \>\"", S7[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 1\)\%\[Infinity]\ \)\!\(\(\(\((\(-1\))\)\^n\) z\^\(2 n - 4\)\)\/\(\((2 n)\)!\)\) = \>\"", Sinf[z]]; \n Print[\*"\"\<\!\(S\_\[Infinity]\)[z] = \!\(\[Sum]\+\(n = 1\)\%\[Infinity]\ \)\!\(\(\(\((\(-1\))\)\^n\) z\^\(2 n - 4\)\)\/\(\((2 n)\)!\)\) = \>\"", PowerExpand[Sinf[z]]]; \)], "Input"], Cell[BoxData[ \(Print["\", f[0.2]]; \nPrint["\", L[0.2]]; \n Print["\", f[2.0]]; \nPrint["\", L[2.0]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[x], "\< and s = L[x]\>"]; \n Plot[{f[x], L[x]}, {x, 0.0001, 5.0}, \n\ \ PlotRange \[Rule] {{0, 5}, {\(-5\), 0.1}}, Ticks \[Rule] {Range[0, 5, 1], Range[\(-5\), 0, 1]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.9, Page 230.", FontWeight->"Bold"], " Find the Laurent series for f(z) = ", Cell[BoxData[ StyleBox[\(\[ExponentialE]\^\(\(-1\)\/z\^2\)\), FontSize->14]]], " \ncentered at ", Cell[BoxData[ StyleBox[\(z\_0\), FontSize->14]]], " = 0." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(Clear[f, L, L1, S1, z]; \n f[z_]\ = \ \[ExponentialE]\^\(\(-1\)\/z\^2\); \n L[z_]\ = \ Normal[Series[f[z], {z, \[Infinity], 20}]]; \n Print["\", f[z]]; \nPrint["\", L[z]]; \)], "Input", AspectRatioFixed->False], Cell[TextData[ "Note that we had to expand the function about \[Infinity] in order to get \ the negative powers of z.\nAlso, can get the Laurent expansion by series \ substitution:"], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[BoxData[ \(S1[z_]\ = \ Normal[Series[\[ExponentialE]\^z, {z, 0, 10}]]; \n L1[z_]\ = \ S1[\(-1\)\/z\^2]; \n Print[\*"\"\< \!\(\[ExponentialE]\^z\) = \>\"", S1[z]]; \n Print[\*"\"\<\!\(L\_1\)[z] = \>\"", L1[z]]; \)], "Input", AspectRatioFixed->False], Cell["The above Laurent series are valid for 0 < |z| .", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[x], \*"\"\< and s = \!\(L\_1\)[x]\>\""]; \n Plot[{f[x], L1[x]}, {x, 0.3, 2.0}, \n\ \ PlotRange \[Rule] {{0, 2}, {\(-0.1\), 1}}, Ticks \[Rule] {Range[0, 2, 0.5], Range[0, 1, 0.2]}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, AxesLabel \[Rule] {"\", "\"}]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 7", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 7.4 ", StyleBox["Singularities, Zeros and Poles", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, FontSize->18, CellTags->"Section 7.4"], Cell[TextData[{ StyleBox["What is a singular point ?", FontWeight->"Bold"], "\n The point ", StyleBox["\[Alpha]", FontFamily->"Symbol"], " is called a ", StyleBox["singular point", FontColor->RGBColor[1, 0, 1]], ", or ", StyleBox["singularity", FontColor->RGBColor[1, 0, 1]], ", of the complex function f(z)\nif f(z) is not analytic at z = ", StyleBox["\[Alpha]", FontFamily->"Symbol"], ", but every neighborhood D", StyleBox["R", FontVariations->{"CompatibilityType"->"Subscript"}], "(", StyleBox["\[Alpha]", FontFamily->"Symbol"], ") of ", StyleBox["\[Alpha]", FontFamily->"Symbol"], " contains\nat least one point at which f(z) is analytic.\n\n", StyleBox["Theorem 7.10, Page 234.", FontWeight->"Bold"], " A function ", Cell[BoxData[ \(f \((z)\)\)]], " analytic in ", Cell[BoxData[ \(\(\ \(D\_R\) \((\[Alpha])\)\)\)]], " has a zero of order k at the point ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " if and only if its Taylor series given by ", Cell[BoxData[ RowBox[{\(f \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z - \[Alpha])\)\^n\)}]}]]], " has ", Cell[BoxData[ \(c\_0 = \(c\_1 = \(\[CenterEllipsis] = \(c\_\(k - 1\) = 0\ \ \ and\ \ \ c\_k \[NotEqual] 0\)\)\)\)]], ".\n\n", StyleBox["Theorem 7.11, Page 235.", FontWeight->"Bold"], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in ", Cell[BoxData[ \(\(\ \(D\_R\) \((\[Alpha])\)\)\)]], ". Then ", Cell[BoxData[ \(f \((z)\)\)]], " has a zero of order k at the point ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " if and only if it can be expressed in the form ", Cell[BoxData[ \(f \((z)\)\ = \(\((z - \[Alpha])\)\^k\) g \((z)\)\)]], ", where ", Cell[BoxData[ \(g \((z)\)\)]], " is analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " and ", Cell[BoxData[ \(g \((\[Alpha])\) \[NotEqual] 0\)]], ".\n\n", StyleBox["Theorem 7.12, Page 236.", FontWeight->"Bold"], " A function ", Cell[BoxData[ \(f \((z)\)\)]], " analytic in the punctured disk ", Cell[BoxData[ \(\(\ \(D\_R\^*\) \((\[Alpha])\)\)\)]], " has a pole of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " if and only if it can be expressed in the form ", Cell[BoxData[ RowBox[{\(f \((z)\)\), " ", "=", FractionBox[ StyleBox[\(h \((z)\)\), FontSize->12], StyleBox[ SuperscriptBox[ RowBox[{"(", RowBox[{"z", StyleBox[" ", FontSize->12], "-", " ", "\[Alpha]"}], ")"}], "k"], FontSize->12]]}]]], ", where ", Cell[BoxData[ \(h \((z)\)\)]], " is analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " and ", Cell[BoxData[ \(h \((\[Alpha])\) \[NotEqual] 0\)]], ".\n\n", StyleBox["Theorem 7.13, Page 236.", FontWeight->"Bold"], " \n", StyleBox["(i)", FontWeight->"Bold"], "\tIf ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic and has a zero of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ", \n\tthen ", Cell[BoxData[ RowBox[{\(g \((z)\)\), " ", "=", FractionBox[ StyleBox["1", FontSize->12], StyleBox[\(f \((z)\)\), FontSize->12]]}]]], " has a pole of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ".\n", StyleBox["(ii)", FontWeight->"Bold"], "\tIf ", Cell[BoxData[ \(f \((z)\)\)]], " has a pole of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ", \n\tthen ", Cell[BoxData[ RowBox[{\(h \((z)\)\), " ", "=", FractionBox[ StyleBox["1", FontSize->12], StyleBox[\(f \((z)\)\), FontSize->12]]}]]], " has a removable singularity at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ". \n\tIf we define ", Cell[BoxData[ \(h \((\[Alpha])\) = 0\)]], ", then ", Cell[BoxData[ \(h \((z)\)\)]], " has a zero of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], "." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ StyleBox["What is a pole ?", FontWeight->"Bold"], "\n The following example will help this concept. \nConsider the \ function f(z) = cot(z). The leading term in the series expansion \nS(z) \ is ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " and S(z) goes to \[Infinity] as z\[Rule]0 in the same manner as \ cot(z)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ Cot[z]; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 13}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", f[x]]; \n Print["\", S[x]]; \n graph1\ = \ Plot[{f[z], S[z]}, {z, 0.01, 3.1}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, DisplayFunction \[Rule] Identity]; \ \n graph2\ = \ Plot[{f[z], S[z]}, {z, \(-3.1\), \(-0.01\)}, PlotStyle \[Rule] {RGBColor[1, 0, 0], RGBColor[0, 1, 0]}, DisplayFunction \[Rule] Identity]; \ \n Show[graph1, graph2, PlotRange \[Rule] {{\(-3.1\), 3.1}, {\(-10\), 10}}, Ticks \[Rule] {Range[\(-3\), 3, 1], Range[\(-10\), 10, 5]}, DisplayFunction \[Rule] $DisplayFunction]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example of a removable singularity, Page 233.", FontWeight->"Bold"], "\nThe function f(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(sin(z)\), FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " has a removable singularity at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ Sin[z]\/z; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 13}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\< f[0] = \>", f[0]]; \n Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \>\"", f[x], "\< = \>", Limit[f[x], x \[Rule] 0]]; \nPrint["\< S[0] = \>", S[0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example of a removable singularity, Page 233.", FontWeight->"Bold"], "\nThe function f(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(cos(z)\ - \ 1\), FontSize->14], StyleBox[\(z\^2\), FontSize->14]], TraditionalForm]]], " has a removable singularity at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ \(Cos[z]\ - 1\)\/z\^2; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 11}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\< f[0] = \>", f[0]]; \n Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) \>\"", f[x], "\< = \>", Limit[f[x], x \[Rule] 0]]; \nPrint["\< S[0] = \>", S[0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example of a pole of order 2, Page 234.", FontWeight->"Bold"], "\nThe function f(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(sin(z)\), FontSize->14], StyleBox[\(z\^3\), FontSize->14]], TraditionalForm]]], " has a pole of order 2 at the origin." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ Sin[z]\/z\^3; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 11}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example of a simple pole, Page 234.", FontWeight->"Bold"], "\nThe function f(z) = ", Cell[BoxData[ FractionBox[ StyleBox[\(\[ExponentialE]\^z\), FontSize->14], StyleBox["z", FontSize->14]]]], " has a simple pole at the origin." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ \[ExponentialE]\^z\/z; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 8}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Examples of an essential singularity, Page 234.", FontWeight->"Bold"], "\nThe function f(z) = ", Cell[BoxData[ \(TraditionalForm\`z\^2\)], FontSize->14], "sin(", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], ") has an essential singularity at the origin." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[g, h, p, S, z]; \ng[z_]\ = \ Sin[z]; \n S[z_]\ = \ Normal[Series[Sin[z], {z, 0, 17}]]; \nh[z_]\ = \ S[1\/z]; \n f[z_]\ = \ z\^2\ g[1\/z]; \np[z_]\ = \ Simplify[z\^2\ h[z]]; \n Print["\", f[z]]; \nPrint["\", p[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.10, Page 234.", FontWeight->"Bold"], " Show that f(z) = z sin(", Cell[BoxData[ FormBox[ StyleBox[\(z\^2\), FontSize->14], TraditionalForm]]], ") has a zero of order 3 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, i, L, m, p, S, z]; \nf[z_]\ = \ z\ Sin[z\^2]; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 23}]]; \n L\ = \ CoefficientList[S[z], z]; \ni = 1; \ m\ = \ Length[L]; \n While[L[\([i]\)] == 0\ && \ i < m, \n\ \ term\ = \ z\^i; \ i = i + 1]; \ \nfact\ = \ Simplify[S[z]\/term]; \np[z_]\ = \ term\ fact; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\", p[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus, f(z) = z sin(", Cell[BoxData[ FormBox[ StyleBox[\(z\^2\), FontSize->14], TraditionalForm]]], ") has a zero of order 3 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.11, Page 236.", FontWeight->"Bold"], " Show that f(z) = ", Cell[BoxData[ \(TraditionalForm\`z\^3\)]], "sin(z) has a zero of order 4 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, i, L, m, p, S, z]; \nf[z_]\ = \ z\^3\ Sin[z]; \n S[z_]\ = \ Normal[Series[f[z], {z, 0, 16}]]; \n L\ = \ CoefficientList[S[z], z]; \ni = 1; \ m\ = \ Length[L]; \n While[L\[LeftDoubleBracket]i\[RightDoubleBracket] == 0\ && \ i < m, \n \ \ \ term\ = \ z\^i; \ i = i + 1]; \ \n fact\ = \ Simplify[S[z]\/term\ ]; \np[z_]\ = \ term\ fact; \n Print["\", f[z]]; \nPrint["\", S[z]]; \n Print["\", p[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus, f(z) = ", Cell[BoxData[ \(TraditionalForm\`z\^3\)]], "sin(z) has a zero of order 4 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.12, Page 237.", FontWeight->"Bold"], " Locate the zeros and poles of h(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(tan(z)\), FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], ",\nand determine their order." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[h, i, S, z]; \nh[z_]\ = \ Tan[z]\/z; \n S[z_]\ = \ Normal[Series[h[z], {z, 0, 13}]]; \n Print["\", h[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus, h(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(tan \((z)\)\), FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " has a removable singularity at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\< h[z] = \>", h[z]]; \n For[i = 1, i \[LessEqual] 5, \(i++\), \n\ \ \ Print["\", i, "\<\[Pi]] = \>", h[i\ \[Pi]]]\ ]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "And h(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(tan \((z)\)\), FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " has simple zeros at z = n", StyleBox["\[Pi]", FontFamily->"Symbol"], " where n = \[PlusMinus]1, \[PlusMinus]2, ..." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\< h[z] = \>", h[z]]; \n For[i = 1, i <= 5, \(i++\), Print["\", 2 i - 1, "\", Limit[h[z], z \[Rule] \(\((2 i - 1)\) \[Pi]\)\/2]\ ]\ ]; \)], "Input"], Cell[TextData[{ "And h(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(tan \((z)\)\), FontSize->14], StyleBox["z", FontSize->14]], TraditionalForm]]], " has simple poles at z = (n+", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->12], StyleBox["2", FontSize->12]], TraditionalForm]]], ")", StyleBox["\[Pi]", FontFamily->"Symbol"], " where n is an integer." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.13, Page 237.", FontWeight->"Bold"], " Locate the poles of g(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[ RowBox[{ RowBox[{"5", FormBox[\(z\^4\), "TraditionalForm"]}], "+", " ", RowBox[{"26", FormBox[\(z\^2\), "TraditionalForm"]}], " ", "+", " ", "5"}], FontSize->14]], TraditionalForm]]], Cell[BoxData[ StyleBox[ RowBox[{ FormBox["", "TraditionalForm"], FormBox["", "TraditionalForm"]}], FontSize->14]]], ",\nand specify their order." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, zZ1, Z2, Z3, Z4]; \n g[z_]\ = \ 1\/\(5 z\^4\ + \ 26 z\^2\ + \ 5\); \n solset\ = \ Solve[Denominator[g[z]] == 0, z]; \n Print["\", g[z]]; \nPrint["\"]; \n Print[solset]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Z1\ = \ solset\[LeftDoubleBracket]1, 1, 2\[RightDoubleBracket]; \n Z2\ = \ solset\[LeftDoubleBracket]2, 1, 2\[RightDoubleBracket]; \n Z3\ = \ solset\[LeftDoubleBracket]3, 1, 2\[RightDoubleBracket]; \n Z4\ = \ solset\[LeftDoubleBracket]4, 1, 2\[RightDoubleBracket]; \n Print["\< g[z] = \>", g[z]]; \n Print["\", Z1, "\<] = \>", Limit[g[z], z \[Rule] Z1]]; \n Print["\< g[\>", Z2, "\<] = \>", Limit[g[z], z \[Rule] Z2]]; \n Print["\", Z3, "\<] = \>", Limit[g[z], z \[Rule] Z3]]; \n Print["\< g[\>", Z4, "\<] = \>", Limit[g[z], z \[Rule] Z4]]; \)], "Input"], Cell[TextData[{ "Thus g(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox["1", FontSize->14], StyleBox[ RowBox[{ RowBox[{"5", FormBox[\(z\^4\), "TraditionalForm"]}], "+", " ", RowBox[{"26", FormBox[\(z\^2\), "TraditionalForm"]}], " ", "+", " ", "5"}], FontSize->14]], TraditionalForm]]], ", has four simple poles." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.14, Page 237.", FontWeight->"Bold"], " Locate the zeros and poles of g(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(\[Pi]\ \(cot(\[Pi]z)\)\), FontSize->14], StyleBox[\(z\^2\), FontSize->14]], TraditionalForm]]], ",\nand determine their order." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[g, S, z]; \ng[z_]\ = \ \(\[Pi]\ Cot[\[Pi]\ z]\)\/z\^2; \n S[z_]\ = \ Normal[Series[g[z], {z, 0, 9}]]; \n Print["\", g[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus g(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(\[Pi]\ \(cot(\[Pi]z)\)\), FontSize->14], StyleBox[\(z\^2\), FontSize->14]], TraditionalForm]]], ", has a pole of order n = 3 at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Print["\", g[z]]; \n Print["\", Limit[g[z], z \[Rule] 1]]; \n Print["\", Limit[g[z], z \[Rule] 2]]; \n Print["\", Limit[g[z], z \[Rule] 3]]; \n Print["\", Limit[g[z], z \[Rule] 4]]; \)], "Input"], Cell[TextData[{ "Thus g(z) = ", Cell[BoxData[ FormBox[ FractionBox[ StyleBox[\(\[Pi]\ \(cot(\[Pi]z)\)\), FontSize->14], StyleBox[\(z\^2\), FontSize->14]], TraditionalForm]]], " has simple poles at z = \[PlusMinus]1, \[PlusMinus]2, ..." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 7", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Section 7.5 ", StyleBox["Applications of Taylor and Laurent Series", FontColor->RGBColor[1, 0, 1]] }], "Text", Evaluatable->False, AspectRatioFixed->False, FontSize->18, CellTags->"Section 7.5"], Cell[TextData[{ " In this section we show how Taylor and Laurent series can be used to \ derive important properties of analytic functions.\n\n", StyleBox["Theorem 7.14, Page 239.", FontWeight->"Bold"], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " and that ", Cell[BoxData[ \(f \((\[Alpha])\) = 0\)]], ". If ", Cell[BoxData[ \(f \((z)\)\)]], " is not identically zero, then there exists a punctured disk ", Cell[BoxData[ \(\(\ \(D\_r\^*\) \((\[Alpha])\)\)\)]], " in which ", Cell[BoxData[ \(f \((z)\)\)]], " has no zeros.\n\n", StyleBox["Corollary 7.9, Page 239.", FontWeight->"Bold"], " ", StyleBox["(L'Hopital's Rule)", FontColor->RGBColor[1, 0, 1]], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " and ", Cell[BoxData[ \(g \((z)\)\)]], " are analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ". If ", Cell[BoxData[ \(f \((\[Alpha])\) = 0\)]], " and ", Cell[BoxData[ \(g \((\[Alpha])\) = 0\)]], " but ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], \((\[Alpha])\)}], "\[NotEqual]", "0"}]]], ", then \n", Cell[BoxData[ RowBox[{ RowBox[{ SubscriptBox["lim", StyleBox[\(z \[Rule] \[Alpha]\), FontSize->10]], FractionBox[ StyleBox[\(f \((z)\)\), FontSize->12], RowBox[{" ", StyleBox[\(g \((z)\)\), FontSize->12]}]]}], " ", "=", " ", RowBox[{ RowBox[{ SubscriptBox["lim", StyleBox[\(z \[Rule] \[Alpha]\), FontSize->10]], FractionBox[ RowBox[{ SuperscriptBox[ StyleBox["f", FontSize->12], "\[Prime]", MultilineFunction->None], StyleBox[\((z)\), FontSize->12]}], RowBox[{" ", StyleBox[ RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], \((z)\)}], FontSize->12]}]]}], " ", "=", " ", FractionBox[ RowBox[{ SuperscriptBox[ StyleBox["f", FontSize->12], "\[Prime]", MultilineFunction->None], StyleBox[\((\[Alpha])\), FontSize->12]}], StyleBox[ RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], \((\[Alpha])\)}], FontSize->12]]}]}]]], ". \n\n", StyleBox["Theorem 7.15, Page 240.", FontWeight->"Bold"], " ", StyleBox["(Division of Power Series)", FontColor->RGBColor[1, 0, 1]], " Suppose ", Cell[BoxData[ \(f \((z)\)\)]], " and ", Cell[BoxData[ \(g \((z)\)\)]], " are analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " with power series representations\n", Cell[BoxData[{ RowBox[{\(f \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(a\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_R\) \((\[Alpha])\)\ \ and\)}]}], RowBox[{\(g \((z)\)\), " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(b\_n\) \((z - \[Alpha])\)\^n\ \ \ for\ \ \ z\ \[Epsilon]\ \(D\_R\) \(\((\[Alpha])\).\)\)}]}]}]], "\nIf ", Cell[BoxData[ \(g \((\[Alpha])\) \[NotEqual] 0\)]], ", then the quotient ", Cell[BoxData[ FractionBox[ StyleBox[\(f \((z)\)\), FontSize->12], RowBox[{" ", StyleBox[\(g \((z)\)\), FontSize->12]}]]]], " has the power series representation\n", Cell[BoxData[ RowBox[{ RowBox[{ FractionBox[ StyleBox[\(f \((z)\)\), FontSize->12], RowBox[{" ", StyleBox[\(g \((z)\)\), FontSize->12]}]], " ", "=", " ", RowBox[{ UnderoverscriptBox[ StyleBox["\[Sum]", FontSize->18], StyleBox[\(n = 0\), FontSize->10], StyleBox["\[Infinity]", FontSize->12]], \(\(c\_n\) \((z - \[Alpha])\)\^n\)}]}], " "}]]], ", \nwhere the coefficients satisfy the equations ", Cell[BoxData[ \(a\_n = \(b\_0\) c\_n + \(b\_1\) c\_\(n - 1\) + \[CenterEllipsis] + \(b\_\(n - 1\)\) c\_1 + \(b\_n\) c\_0\)]], ".\nIn other words, the series for the quotient ", Cell[BoxData[ FractionBox[ StyleBox[\(f \((z)\)\), FontSize->12], RowBox[{" ", StyleBox[\(g \((z)\)\), FontSize->12]}]]]], " can be obtained by the familiar process of dividing the series for ", Cell[BoxData[ StyleBox[\(f \((z)\)\), FontSize->12]]], " by the series for ", Cell[BoxData[ RowBox[{" ", StyleBox[\(g \((z)\)\), FontSize->12]}]]], " using the standard long division algorithm.\n\n", StyleBox["Theorem 7.16, Page 241.", FontWeight->"Bold"], " ", StyleBox["(Riemann's Theorem)", FontColor->RGBColor[1, 0, 1]], " Suppose that ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic in ", Cell[BoxData[ \(\(\ \(D\_r\^*\) \((\[Alpha])\)\)\)]], ". If ", Cell[BoxData[ \(f \((z)\)\)]], " is bounded in ", Cell[BoxData[ \(\(\ \(D\_r\^*\) \((\[Alpha])\)\)\)]], ", then either ", Cell[BoxData[ \(f \((z)\)\)]], " is analytic at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " or ", Cell[BoxData[ \(f \((z)\)\)]], " has a removable singularity at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], ".\n\n", StyleBox["Theorem 7.17, Page 241.", FontWeight->"Bold"], " The function ", Cell[BoxData[ \(f \((z)\)\)]], " has a pole of order k at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " if and only if ", Cell[BoxData[ RowBox[{ RowBox[{ SubscriptBox["lim", StyleBox[\(z \[Rule] \[Alpha]\), FontSize->10]], \(\[VerticalSeparator]\ \(f \((z)\) \[VerticalSeparator] \)\)}], "=", "\[Infinity]"}]]], ".\n\n", StyleBox["Theorem 7.18, Page 242.", FontWeight->"Bold"], " The function ", Cell[BoxData[ \(f \((z)\)\)]], " has an essential singularity at ", Cell[BoxData[ \(\(\ z = \[Alpha]\)\)]], " if and only if ", Cell[BoxData[ RowBox[{ SubscriptBox["lim", StyleBox[\(z \[Rule] \[Alpha]\), FontSize->10]], \(\[VerticalSeparator]\ \(f \((z)\) \[VerticalSeparator] \)\)}]]], " does not exist." }], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "\n", StyleBox["Example 7.15, Page 240.", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold", FontSlant->"Italic"], "Find the first few terms of the Maclaurin series \nfor f(z) = sec(z) \ and use it to compute ", Cell[BoxData[ FormBox[ StyleBox[\(f\^\((4)\)\), FontSize->14], TraditionalForm]]], "(0)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[f, S, z]; \nf[z_]\ = \ Sec[z]; \n S[z_]\ = \ Normal[Series[Sec[z], \ {z, 0, 12}]]; \n Print["\", f[z]]; \nPrint["\", S[z]]; \)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Print["\", Expand[\(\(\(\(f'\)'\)'\)'\)[z]]]; \n Print["\", \(\(\(\(S'\)'\)'\)'\)[z]]; \n Print["\", \(\(\(\(f'\)'\)'\)'\)[0]]; \n Print["\", \(\(\(\(S'\)'\)'\)'\)[0]]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 7.16, Page 242. ", FontWeight->"Bold"], "Show that the function g(z) = ", Cell[BoxData[ FormBox[ StyleBox[\(\[ExponentialE]\^\(\(-1\)\/z\^2\)\), FontSize->18], TraditionalForm]]], " is not continuous at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Clear[g, lim1, lim2, x, y, z]; \n g[z_]\ = \ \[ExponentialE]\^\(\(-1\)\/z\^2\); \n lim1\ = \ Limit[g[x\ + \ 0\ I], \ x \[Rule] 0]; \n lim2\ = \ Limit[g[0\ + \ I\ y], \ y \[Rule] 0]; \n val\ = \ MapAll[Together, ComplexExpand[g[x + I\ y], TargetFunctions \[Rule] {Re, Im}]]; \n Print["\< g[z] = \>", g[z]]; \n Print["\< g[x+Iy] = \>", g[x + I\ y]]; \n Print["\< g[x+Iy] = \>", val]; \n Print[\*"\"\<\!\(lim\+\(x \[Rule] 0\)\) g[x+0I] = \>\"", lim1]; \n Print[\*"\"\<\!\(lim\+\(y \[Rule] 0\)\) g[0+Iy] = \>\"", lim2]; \)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Since the limits are ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " the same, g(z) = ", Cell[BoxData[ FormBox[ StyleBox[\(\[ExponentialE]\^\(\(-1\)\/z\^2\)\), FontSize->18], TraditionalForm]]], " is ", StyleBox["not", FontColor->RGBColor[1, 0, 1]], " continuous at z = 0." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["GoTo ", FontSize->16, FontColor->RGBColor[0, 1, 0]], ButtonBox["Chapter 7", ButtonData:>"CHAPTER", ButtonStyle->"Hyperlink"] }], "Text"] }, Closed]] }, FrontEndVersion->"Microsoft Windows 3.0", ScreenRectangle->{{0, 640}, {0, 452}}, AutoGeneratedPackage->None, WindowToolbars->{}, CellGrouping->Manual, WindowSize->{472, 193}, WindowMargins->{{1, Automatic}, {Automatic, 5}}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of \ all cells in a given style. 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Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, LineSpacing->{1, 3}, CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles 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