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"\tSuppose that we have a series ", Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity] \[Xi]\_n\)]], " where ", Cell[BoxData[ \(\[Xi]\_n = \(c\_n\) \((z - \[Alpha])\)\^n\)]], ". If ", Cell[BoxData[ \(\[Alpha]\)]], " and the collection of ", Cell[BoxData[ \(c\_n\)]], " are fixed complex numbers, we will get different series by selecting \ different values for z. For example, if ", Cell[BoxData[ \(\[Alpha] = 2\)]], " and ", Cell[BoxData[ \(c\_n = 1\/\(n!\)\)]], " for all n, we get the series ", Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( 1\/\(n!\)\) \((\[ImaginaryI]\/2 - 2)\)\^n\)]], " if ", Cell[BoxData[ \(z = \[ImaginaryI]\/2\)]], " and ", Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( 1\/\(n!\)\) \((2 + \[ImaginaryI])\)\^n\)]], " if ", Cell[BoxData[ \(z = 4 + \[ImaginaryI]\)]], ". Note that when ", Cell[BoxData[ \(\[Alpha] = 0\)]], " and ", Cell[BoxData[ \(c\_n = 1\)]], " for all n, we get a geometric series. The collection of points for which \ the series ", Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " converges will be the domain of a function ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], ", which we call a ", StyleBox["power series function", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], StyleBox[".", FontColor->RGBColor[0, 0, 1]], StyleBox[" ", FontColor->RGBColor[0, 0, 1]], "Technically, this series is undefined if ", Cell[BoxData[ \(z = \[Alpha]\)]], " and n=0, since ", Cell[BoxData[ \(0\^0\)]], " is undefined. We get around this difficulty by stipulating that the \ series ", Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " is really compact notation for ", Cell[BoxData[ \(c\_0 + \[Sum]\+\(n = 1\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], ". In this section we list some results that will be useful in helping us \ establish properties of functions defined by power series." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nDefinition (", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ButtonBox["Power Series", ButtonData:>{ URL[ "http://mathworld.wolfram.com/PowerSeries.html"], None}, ButtonStyle->"Hyperlink"], StyleBox["), Page 151.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " The function ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " is called a ", StyleBox["power series", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], ", with center ", Cell[BoxData[ \(z = \[Alpha]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nTheorem 4.14, Page 151.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " . Then the set of points z for which the series converges is one of the \ following:\n\n", StyleBox["(i)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " The single point ", Cell[BoxData[ \(z = \[Alpha]\)]], ". \n\n", StyleBox["(ii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " The disk ", Cell[BoxData[ \(\(D\_\[Rho]\) \((\[Alpha])\) = {z : \ \(\(|\)\(z - \ \[Alpha]\)\(|\)\(\(<\)\(\ \)\(\[Rho]\)\)\)}\)]], ", along with part (either none, or some or all) of the circle ", Cell[BoxData[ \(\(C\_\[Rho]\) \((\[Alpha])\) = {z : \ \(\(|\)\(z - \[Alpha]\)\(|\)\) \ = \ \ \[Rho]}\)]], ". \n\n", StyleBox["(iii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " The entire complex plane." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 4.11, see text Page 151.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\tAnother way to phrase case (ii) of Theorem 4.14 is to say that the power \ series ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " converges if ", Cell[BoxData[ \(\(\(|\)\(z - \[Alpha]\)\(|\)\(\(<\)\(\ \)\(\[Rho]\)\)\)\)]], " and diverges if ", Cell[BoxData[ \(\(\(|\)\(z - \[Alpha]\)\(|\)\(\(>\)\(\ \)\(\[Rho]\)\)\)\)]], ". We call the number ", Cell[BoxData[ \(\[Rho]\)]], " the ", StyleBox["radius of convergence", FontSlant->"Italic"], " of the power series (see Figure 4.2). If we are in case (i) of Theorem \ 4.11, we say that the radius of convergence is zero, and that the radius of \ convergence is infinity if we are in case (iii)." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nTheorem 4.15, Page 153.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " For the power series function ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], ", we can find ", Cell[BoxData[ \(\[Rho]\)]], " , its radius of convergence, by any of the following methods:\n\n\t", StyleBox["(i) Cauchy's Root Test:", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(\[Rho] = 1\/\(lim\+\(n \[Rule] \[Infinity]\)\ \ \@\(\(|\)\(c\_n\)\(|\)\)\%n\)\)]], " (Provided that the limit exists.) \n\n\t", StyleBox["(ii) Cauchy-Hadamard Formula:", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(\[Rho] = 1\/\(\(lim\ sup\)\+\(n \[Rule] \[Infinity]\)\ \ \@\(\(\[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%n\)\)]], " (The limit superior always exists.) \n\n\t", StyleBox["(iii) d'Alembert's Ratio Test:", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(\[Rho] = lim\+\(n \[Rule] \[Infinity]\)\ \(\(|\)\(c\_n\/c\_\(n + \ 1\)\)\(|\)\)\)]], " (Provided that the limit exists.)\n\tIn cases (i) and (ii) we set ", Cell[BoxData[ \(\[Rho] = \[Infinity]\)]], " if the limit equals 0, and set ", Cell[BoxData[ \(\[Rho] = 0\)]], " if the limit equals ", Cell[BoxData[ \(\[Infinity]\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 4.12, see text Page 153.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["\nExample 4.21, Page 153.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the radius of convergence of ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\(\((\(n\ + \ 2\)\/\(3 n\ \ + \ 1\))\)\^n\) \((z - 4)\)\^n\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 4.21.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the formula for the coefficients. ", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[c];\)\ \), "\[IndentingNewLine]", \(\(Clear[L, n, Pn, R];\)\ \), "\[IndentingNewLine]", \(\(c\_n_\ = \ \((\(n + 2\)\/\(3 n + 1\))\)\^n;\)\ \), "\ \[IndentingNewLine]", \(\(p\_n_ = \((c\_n)\)\^\(1/n\);\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(c\_n\) = \>\"", c\_n];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\@\(\(\[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%\ n\) = \>\"", p\_n];\)\ \), "\[IndentingNewLine]", \(\(p\_n_ = PowerExpand[p\_n];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(\@\(\(\[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%\ n\) = \>\"", p\_n];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Use the Cauchy root test and find the limit and then the radius of \ convergence R.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(L\ = \ Limit[\ p\_n, \ n \[Rule] \[Infinity]];\)\ \), "\n", \(\(R\ = \(\(\ \)\(1\)\)\/L;\)\ \), "\n", \(\(Print[\*"\"\< L = \!\(lim\+\(n \[Rule] \[Infinity]\)\) \!\(\@\(\(\ \[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%n\) = \>\"", L];\)\ \), "\n", \(\(Print["\", L];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", R];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThe series ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\(\((\(n\ + \ 2\)\/\(3 n\ \ + \ 1\))\)\^n\) \((z - 4)\)\^n\)]], " will converge in the disk ", Cell[BoxData[ \(\(\(|\)\(z - 4\)\(|\)\(\(<\)\(3\)\)\)\)]], ". ", "\n", "\nWe can plot some of the partial sums and see that they converge. \ Convergence will be faster if we choose a smaller disk ", Cell[BoxData[ \(\(\(|\)\(z - 4\)\(|\)\(\(<\)\(r\)\)\)\)]], " the following graphs use the smaller disk with ", Cell[BoxData[ \(r = 1\)]], ". " }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[g, k, m, n, s];\)\ \), "\[IndentingNewLine]", \(\(s[z_, n_] = \[Sum]\+\(k = 0\)\%n\(\((\(k + 2\)\/\(3 k + 1\))\)\^k\) \((z \ - 4)\)\^k;\)\ \), "\[IndentingNewLine]", \(\(For\ [m = 3, m \[LessEqual] 5, \(m++\), \[IndentingNewLine]g[z_] = s[z + 4, m]; \ \[IndentingNewLine]PolarMap[ g\ , {0, 1.0, 0.2}, {0, 2 \[Pi], \[Pi]\/12}, \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {\(-1.5\), 1.5}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-\(3\/2\)\), 3\/2, 1\/2]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Cyan, Blue}]; \ \[IndentingNewLine]Print["\", m, "\<] = \>", s[z, m]];\ \ ];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe see that the series ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\(\((\(n\ + \ 2\)\/\(3 n\ \ + \ 1\))\)\^n\) \((z - 4)\)\^n\)]], " is converging in the smaller disk ", Cell[BoxData[ RowBox[{ RowBox[{"|", \(z - 4\), "|", RowBox[{"<", " ", RowBox[{"r", " ", StyleBox["where", FontFamily->"Times New Roman"], " ", "r"}]}]}], "=", \(1 < 3\)}]]], ". " }], "Text"] }, Closed]], Cell[TextData[{ StyleBox["\nExample 4.22, Page 153.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the radius of convergence of ", Cell[BoxData[ \(f \((z)\) = 1 + 4 z + 5\^2\ z\^2 + \(4\^3\) z\^3 + 5\^4\ z\^4 + \(4\^5\) z\^5 + 5\^6\ z\^6 + ... \)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 4.22.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["\<\ Enter the formula for the coefficients. There is a formula for the \ even subscripts and a different formula for the odd subscripts.\ \>", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[c];\)\ \), "\[IndentingNewLine]", \(\(Clear[k, limsup, n, \[Rho], R];\)\ \), "\[IndentingNewLine]", \(\(c\_\(even, k_\) = 5\^k;\)\ \), "\[IndentingNewLine]", \(\(c\_\(odd, k_\) = 4\^k;\)\ \), "\[IndentingNewLine]", \(\(table = Flatten[Table[{c\_\(even, 2 k\), c\_\(odd, 2 k + 1\)}, {k, 0, 4}]];\)\ \), "\[IndentingNewLine]", \(\(table = Append[table, "\<...\>"];\)\ \), "\n", \(\(Print[\*"\"\<{\!\(c\_n\)} = \>\"", table];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\< \!\(c\_\(2 k\)\) = \>\"", c\_\(even, 2 k\)\ ];\)\ \), "\n", \(\(Print[\*"\"\< \!\(c\_\(2 k + 1\)\) = \>\"", c\_\(odd, 2 k + 1\)];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Use the Cauchy-Hadamard formula and find the radius of convergence R.\ \>", \ "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(lim1 = Limit[\((c\_\(even, 2 k\))\)\^\(1\/\(2 k\)\), k \[Rule] \[Infinity]];\)\ \), "\[IndentingNewLine]", \(\(lim2 = Limit[\((c\_\(odd, 2 k + 1\))\)\^\(1\/\(2 k + 1\)\), k \[Rule] \[Infinity]];\)\ \), "\n", \(\(\[Rho] = Max[lim1, lim2];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\< \!\(lim\+\(k \[Rule] \[Infinity]\)\) \!\(\@\(\(\ \[VerticalBar]\)\(c\_\(2 k\)\)\(\[VerticalBar]\)\)\%\(2 k\)\) = \>\"", \ \*"\"\<\!\(lim\+\(k \[Rule] \[Infinity]\)\) \>\"", \((c\_\(even, 2 \ k\))\)\^\(1\/\(2 k\)\), "\< = \>", lim1];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(lim\+\(k \[Rule] \[Infinity]\)\) \!\(\@\(\(\ \[VerticalBar]\)\(c\_\(2 k + 1\)\)\(\[VerticalBar]\)\)\%\(2 k + 1\)\) = \ \>\"", \*"\"\<\!\(lim\+\(k \[Rule] \[Infinity]\)\) \>\"", \((c\_\(odd, 2 k + \ 1\))\)\^\(1\/\(2 k + 1\)\), "\< = \>", lim2];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\[Rho] = \!\(limsup\+\(n \[Rule] \[Infinity]\)\) \ \!\(\@\(\(\[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%n\) = Max[\>\"", lim1, "\<,\>", lim2, "\<]\>"];\)\ \), "\n", \(\(Print[\*"\"\<\[Rho] = \!\(limsup\+\(n \[Rule] \[Infinity]\)\) \ \!\(\@\(\(\[VerticalBar]\)\(c\_n\)\(\[VerticalBar]\)\)\%n\) = \>\"", \ \[Rho]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\\"", 1\/\[Rho]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\n", "The series ", Cell[BoxData[ \(f \((z)\) = \(\[Sum]\+\(n = 1\)\%\[Infinity]\( c\_n\) z\^n = 1 + 4 z + 5\^2\ z\^2 + \(4\^3\) z\^3 + 5\^4\ z\^4 + \(4\^5\) z\^5 + 5\^6\ z\^6 + ... \)\)]], " has radius of convergence ", Cell[BoxData[ \(1\/5\)]], " by the Cauchy-Hadamard formula because ", Cell[BoxData[ \(\(lim\ sup\)\+\(n \[Rule] \[Infinity]\)\ \@\(\(\[VerticalBar]\)\(c\_n\ \)\(\[VerticalBar]\)\)\%n\ = 5\)]], ". ", "\n", "\nWe can let ", StyleBox["Mathematica", FontSlant->"Italic"], " find the sum of this series." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(f0[ z_] = \[Sum]\+\(k = 0\)\%\[Infinity]\( 5\^\(2\ k\)\) z\^\(2\ k\);\)\ \), "\n", \(\(f1[ z_] = \[Sum]\+\(k = 0\)\%\[Infinity]\( 4\^\(2\ k + 1\)\) z\^\(2\ k + 1\);\)\ \), "\n", \(\(f[z_] = f0[z] + f1[z];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(f\_1\)[z] = \!\(\[Sum]\+\(k = \ 0\)\%\[Infinity]\)\!\(5\^\(2\\\ k\)\)\!\(z\^\(2\\\ k\)\) = \>\"", f0[z]];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\<\!\(f\_2\)[z] = \!\(\[Sum]\+\(k = \ 0\)\%\[Infinity]\)\!\(4\^\(2\\\ k + 1\)\)\!\(z\^\(2\\\ k + 1\)\) = \>\"", f1[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\[IndentingNewLine]", \(\(Print[\*"\"\\""];\)\ \), "\ \[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nThus we see that f[z] is composed of two \"infinite geometric series\" \ ", Cell[BoxData[ \(f\_1[z]\ \ and\ \ \ f\_2[z]\)]], ", whose radius of convergence are ", Cell[BoxData[ RowBox[{\(R\_1\), "=", RowBox[{ RowBox[{\(1\/5\), " ", StyleBox["and", FontFamily->"Times New Roman"], " ", \(R\_2\)}], "=", \(1\/4\)}]}]]], ", respectively. Hence the radius of convergence of f[z] must be ", Cell[BoxData[ \(R = \(min {1\/5, 1\/4} = 1\/5\)\)]], ". \n\nWe can use ", StyleBox["Mathematica", FontSlant->"Italic"], " to compute the first few terms in the Maclaurin series for f[z] and \ compare them with the formula that was given." }], "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(s[z_] = Series[f[z], {z, 0, 10}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\(Print["\", s[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "\nWe can plot some of the partial sums and see that they converge. \ Convergence will be faster if we choose a smaller disk ", Cell[BoxData[ \(\(\(|\)\(z\)\(|\)\(\(<\)\(r\)\)\)\)]], " the following graphs use the smaller disk with ", Cell[BoxData[ \(r = 0.16 < 1\/5\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[g, k, m, n];\)\ \), "\[IndentingNewLine]", \(For\ [m = 3, m \[LessEqual] 7, \(m++\), \[IndentingNewLine]g[z_] = Normal[Series[f[z], {z, 0, m}]]; \ \[IndentingNewLine]PolarMap[ g\ , {0, 0.16, 0.02}, {0, 2 \[Pi], \[Pi]\/12}, \[IndentingNewLine]PlotRange \[Rule] {{0, 2.8}, {\(-1.4\), 1.4}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}]; \ \[IndentingNewLine]Print["\", m, "\<] = \>", g[z]];\ \ ]\), "\[IndentingNewLine]", \(\(PolarMap[ f, {0, 0.16, 0.02}, {0, 2 \[Pi], \[Pi]\/12}, \[IndentingNewLine]PlotRange \[Rule] {{0, 2.8}, {\(-1.4\), 1.4}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-1\), 1, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {DarkGreen, Green}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 4.23, Page 153.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Find the radius of convergence of ", Cell[BoxData[ \(f \((z)\)\ = \ \[Sum]\+\(n = 0\)\%\[Infinity] z\^n\/\(n!\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 4.23.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell["Enter the formula for the coefficients.", "Text"], Cell[BoxData[{ \(\[IndentingNewLine]\(Remove[c, q];\)\ \), "\[IndentingNewLine]", \(\(Clear[L, n, R, z];\)\ \), "\n", \(\(c\_n_\ = \ 1\/\(n!\);\)\ \), "\n", \(\(q\_n_\ = \ c\_\(n + 1\)\/c\_n;\)\ \), "\n", \(\(Print[\*"\"\< \!\(c\_n\) = \>\"", c\_n];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print[\*"\"\<|\!\(c\_\(n + 1\)\/c\_n\)| = \>\"", q\_n];\)\ \), "\n", \(\(q\_n_\ = \ FullSimplify[q\_n];\)\ \), "\n", \(\(Print[\*"\"\<|\!\(c\_\(n + 1\)\/c\_n\)| = \>\"", q\_n];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ Use d'Alembert's ratio test and find the limit and then the radius of \ convergence R.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(L\ = \ Limit[\ q\_n, \ n \[Rule] \[Infinity]];\)\ \), "\n", \(\(R\ = \ Limit[\ 1\/q\_n, \ n \[Rule] \[Infinity]];\)\ \), "\n", \(\(Print[\*"\"\< L = \!\(lim\+\(n \[Rule] \[Infinity]\)\) |\!\(c\_\(n + \ 1\)\/c\_n\)| = \!\(lim\+\(n \[Rule] \[Infinity]\)\) (\>\"", q\_n, "\<) = \>", L];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\", L];\)\ \), "\[IndentingNewLine]", \(\(Print["\<\>"];\)\ \), "\n", \(\(Print["\"\ , 1\/"\", "\< = \>", R];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell["\<\ The sum of the series is a well known function.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ RowBox[{"\[IndentingNewLine]", RowBox[{\(f[z_]\ = \[Sum]\+\(n = 0\)\%\[Infinity] z\^n\/\(n!\)\), StyleBox[";", FontSize->14]}], StyleBox[" ", FontSize->14]}], "\n", RowBox[{\(Print[\*"\"\\"", f[z]];\), " "}], "\[IndentingNewLine]", "Null"}], "Input"], Cell["\<\ We can plot some of the partial sums and see that they converge to \ f[z].\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[g, k, m, n, s];\)\ \), "\[IndentingNewLine]", \(\(terms[n_] = Table[z\^k\/\(k!\), {k, 0, n}];\)\ \), "\[IndentingNewLine]", \(For\ [m = 3, m \[LessEqual] 5, \(m++\), \[IndentingNewLine]g[z_] = ReplaceAll[terms[m], List \[Rule] Plus]; \ \[IndentingNewLine]PolarMap[ g\ , {0, 1.0, 0.2}, {0, 2 \[Pi], \[Pi]\/12}, \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {\(-1.5\), 1.5}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-\(3\/2\)\), 3\/2, 1\/2]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Cyan, Blue}]; \ \[IndentingNewLine]Print["\", m, "\<] = \>", g[z]];\ \ ]\), "\[IndentingNewLine]", \(\(PolarMap[ f\ , {0, 1.0, 0.2}, {0, 2 \[Pi], \[Pi]\/12}, \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {\(-1.5\), 1.5}}, AspectRatio \[Rule] 1, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-\(3\/2\)\), 3\/2, 1\/2]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Red, Magenta}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"], Cell[BoxData[{ \(\[IndentingNewLine]\(Clear[g, k, m, n, s];\)\ \), "\[IndentingNewLine]", \(\(terms[n_] = Table[z\^k\/\(k!\), {k, 0, n}];\)\ \), "\[IndentingNewLine]", \(For\ [m = 3, m \[LessEqual] 5, \(m++\), \[IndentingNewLine]g[z_] = ReplaceAll[terms[m], List \[Rule] Plus]; \ \[IndentingNewLine]CartesianMap[ g\ , {\(-1\), 1, 0.2}, {\(-1\), 1, 0.2}, \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {\(-2.5\), 2.5}}, AspectRatio \[Rule] 5\/3, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-2\), 2, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Green, DarkGreen}]; \ \[IndentingNewLine]Print["\", m, "\<] = \>", g[z]];\ \ ]\), "\[IndentingNewLine]", \(\(CartesianMap[ f\ , {\(-1\), 1, 0.2}, {\(-1\), 1, 0.2}, \[IndentingNewLine]PlotRange \[Rule] {{0, 3}, {\(-2.5\), 2.5}}, AspectRatio \[Rule] 5\/3, \[IndentingNewLine]Ticks \[Rule] {Range[0, 3, 1], Range[\(-2\), 2, 1]}, AxesLabel \[Rule] {"\", "\"}, PlotStyle \[Rule] {Red, Magenta}];\)\ \), "\[IndentingNewLine]", \(\(Print["\", f[z]];\)\ \), "\[IndentingNewLine]", \(\)}], "Input"] }, Closed]], Cell[TextData[{ StyleBox["\nTheorem 4.16, Page 154.", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " Suppose the function ", Cell[BoxData[ \(f \((z)\) = \[Sum]\+\(n = 0\)\%\[Infinity]\( c\_n\) \((z - \[Alpha])\)\^n\)]], " has radius of convergence ", Cell[BoxData[ \(\[Rho] > 0\)]], ". Then\n\n", StyleBox["(i)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " f(z) is infinitely differentiable for all ", Cell[BoxData[ \(z\ \[Epsilon]\ \(D\_\[Rho]\) \((\[Alpha])\)\)]], ", in fact\n\n", StyleBox["(ii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(\(f\^\((k)\)\) \((z)\) = \(\[Sum]\+\(n = 0\)\%\[Infinity] n \((n - 1)\) ... \) \((n - k + 1)\) \(c\_n\) \((z - \[Alpha])\)\^\(n - k\)\)]], " for all k, and\n\n", StyleBox["(iii)", FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], " ", Cell[BoxData[ \(c\_n = \(\(f\^\((n)\)\) \((\[Alpha])\)\)\/\(n!\)\)]], " is the formula for the coefficient ", Cell[BoxData[ \(c\_n\)]], " in the series. " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[StyleBox["Proof of Theorem 4.13, see text Page 154.", FontWeight->"Bold", FontColor->RGBColor[0, 1, 1]]], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "\n", StyleBox["Example 4.24, Page 156.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Show that ", Cell[BoxData[ \(f \((z)\) = \(\[Sum]\+\(n = 0\)\%\[Infinity]\((n + 1)\) z\^n = 1\/\((1\ - \ z)\)\^2\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 4.24.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Use the fact that ", Cell[BoxData[ \(S \((z)\) = \(\[Sum]\+\(n = 0\)\%\[Infinity] z\^n = 1\/\(1\ - \ z\)\)\)]], " and ", Cell[BoxData[ \(S' \((z)\) = f \((z)\)\)]], ". " }], "Text"], Cell[BoxData[{ RowBox[{"\[IndentingNewLine]", \(Remove[z];\), " "}], "\[IndentingNewLine]", RowBox[{\(Clear[f, n, s, S];\), " "}], "\n", RowBox[{ RowBox[{\(S[z_]\ \ = \ \ \[Sum]\+\(n = 0\)\%\[Infinity] z\^n\), StyleBox[";", FontSize->14]}], StyleBox[" ", FontSize->14]}], "\n", RowBox[{\(Print[\*"\"\< S[z] = \!\(\[Sum]\+\(n = \ 0\)\%\[Infinity]\)\!\(z\^n\) = \>\"", S[z]];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print["\<\>"];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print["\"];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print["\<\>"];\), " "}], "\n", RowBox[{\(Print["\", \(S'\)[z]];\), " "}]}], "Input"], Cell["\<\ Or sum the infinite series directly.\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[{ RowBox[{"\[IndentingNewLine]", RowBox[{\(f[z_]\ = \ \[Sum]\+\(n = 0\)\%\[Infinity]\((n + 1)\) z\^n\), StyleBox[";", FontSize->14]}], StyleBox[" ", FontSize->14]}], "\n", RowBox[{\(Print["\", \(S'\)[z]];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print[\*"\"\\"", f[z]\ ];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print["\"];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print[\*"\"\\"", \[Sum]\+\(n = 1\)\%\[Infinity]\ n\ z\^\(n - 1\)];\), " "}], "\[IndentingNewLine]", RowBox[{\(Print[\*"\"\\"", \[Sum]\+\(n = 0\)\%\[Infinity]\ n\ z\^\(n - 1\)];\), " "}]}], "Input"] }, Closed]], Cell[TextData[{ "\n", StyleBox["Example 4.25, Page 156.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], " Differentiate the Bessel function ", Cell[BoxData[ \(\(J\_0\) \((z)\)\)]], " termwise and get ", Cell[BoxData[ \(J\_0' \((z)\)\)]], " = ", Cell[BoxData[ \(\(-\ J\_1\) \((z)\)\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell[TextData[StyleBox["Solution 4.25.", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]]], "Text", Evaluatable->False, AspectRatioFixed->False], Cell[TextData[{ "Look at some of the coefficients of and observe the relationship between \ the coefficients of ", Cell[BoxData[ \(\(J\_0\) \((z)\), \ J\_0' \((z)\)\ \ and\ \ \ \(J\_1\) \((z)\)\)]], ". 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