(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 369036, 12282] NotebookOptionsPosition[ 310540, 10611] NotebookOutlinePosition[ 337965, 11264] CellTagsIndexPosition[ 336097, 11205] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell["Computing with Harmonic Functions", "Title"], Cell["Sheldon Axler", "Subtitle"], Cell["\<\ 1 January 2009\t\t\[Copyright] Sheldon Axler\ \>", "Text", CellChangeTimes->{{3.43844437540625*^9, 3.438444387484375*^9}, { 3.438803236671875*^9, 3.438803237125*^9}, {3.438869726734375*^9, 3.438869730921875*^9}, {3.43920845503125*^9, 3.4392084568125*^9}}, TextAlignment->Center], Cell[CellGroupData[{ Cell["Getting Started", "Section", CellTags->"Getting Started"], Cell[CellGroupData[{ Cell["Harmonic Function Theory", "Subsection", CellTags->"Loading the Package"], Cell[TextData[{ "This notebook is the manual for HFT7.m (version 7.00), a ", StyleBox["Mathematica", FontSlant->"Italic"], " package for computing with harmonic functions. The HFT7.m package allows \ the user to make calculations that would take a prohibitive amount of time if \ done without a computer. For example, the Poisson integral of any polynomial \ can be computed exactly." }], "Text", CellChangeTimes->{ 3.43880325990625*^9, 3.438803431265625*^9, {3.43880590521875*^9, 3.438805906625*^9}, {3.438869752046875*^9, 3.438869762984375*^9}}], Cell[TextData[{ "Earlier versions of this software focused on algorithms arising from the \ material in the book ", StyleBox["Harmonic Function Theory", FontSlant->"Italic"], " [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "] by Sheldon Axler, Paul Bourdon, and Wade Ramey. That book is still the \ source for many of the algorithms used in the HFT7.m package, but the goal of \ the package has expanded to include additional symbolic manipulations \ involving harmonic functions. 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To obtain the most recent versions, retrieve them from ", ButtonBox["http://www.axler.net", BaseStyle->"Hyperlink", ButtonData:>{ URL["http://math.sfsu.edu/axler"], None}], StyleBox[".", "html"] }], "Text", CellChangeTimes->{{3.438801338484375*^9, 3.438801339125*^9}, { 3.43880329484375*^9, 3.43880330553125*^9}, {3.438805974875*^9, 3.438805977765625*^9}, {3.43886979228125*^9, 3.438869804171875*^9}}, TextAlignment->AlignmentMarker], Cell[TextData[{ "Comments, suggestions, and bug reports about this notebook, the HFT7.m \ package, or the book ", StyleBox["Harmonic Function Theory", FontSlant->"Italic"], " are welcome; please send them via e-mail to axler@sfsu.edu." }], "Text", CellChangeTimes->{3.43880331184375*^9, 3.438869812921875*^9}, TextAlignment->AlignmentMarker] }, Closed]], Cell[CellGroupData[{ Cell["Loading the Package", "Subsection", CellTags->"Loading the Package"], Cell[TextData[{ "To begin a session with the HFT7.m package, first start ", StyleBox["Mathematica", FontSlant->"Italic"], ". 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However, users can be reminded about the basic use and syntax of \ each function created by HFT7.m by entering ? followed by the function name:\ \>", "Text", CellChangeTimes->{{3.43880336240625*^9, 3.438803370046875*^9}, { 3.43886986503125*^9, 3.43886987703125*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"?", "Volume"}]], "Input"], Cell[BoxData[ StyleBox["\<\"Volume[n] gives the volume of the unit ball in n-dimensional \ real Euclidean space.\"\>", "MSG"]], "Print", "PrintUsage", CellChangeTimes->{3.439235052234375*^9}, CellTags->"Info3439206252-2431977"] }, Closed]], Cell[TextData[{ "The ", StyleBox["Mathematica", FontSlant->"Italic"], " statement Names[\"HFT`*\"] will produce a complete list of the functions \ created by HFT7.m:" }], "Text", CellChangeTimes->{3.43880337515625*^9, 3.43886987978125*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Names", "[", "\"\\"", "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{"\<\"Annulus\"\>", ",", "\<\"AntiLaplacian\"\>", ",", "\<\"Ball\"\>", ",", "\<\"BasisH\"\>", ",", "\<\"BergmanKernel\"\>", ",", "\<\"BergmanKernelH\"\>", ",", "\<\"BergmanProjection\"\>", ",", "\<\"BiDirichlet\"\>", ",", "\<\"Dimension\"\>", ",", "\<\"DimensionH\"\>", ",", "\<\"Dirichlet\"\>", ",", "\<\"Divergence\"\>", ",", "\<\"ExpandNorm\"\>", ",", "\<\"ExteriorNeumann\"\>", ",", "\<\"ExteriorSphere\"\>", ",", "\<\"Grad\"\>", ",", "\<\"HarmonicConjugate\"\>", ",", "\<\"HarmonicDecomposition\"\>", ",", "\<\"HilbertSchmidt\"\>", ",", "\<\"Homogeneous\"\>", ",", "\<\"Hyperplane\"\>", ",", "\<\"IdMatrix\"\>", ",", "\<\"IntegrateBall\"\>", ",", "\<\"IntegrateSphere\"\>", ",", "\<\"J\"\>", ",", "\<\"Kelvin\"\>", ",", "\<\"KelvinM\"\>", ",", "\<\"Laplacian\"\>", ",", "\<\"Multiple\"\>", ",", "\<\"Neumann\"\>", ",", "\<\"NormalD\"\>", ",", "\<\"Orthonormal\"\>", ",", "\<\"Partial\"\>", ",", "\<\"PoissonKernel\"\>", ",", "\<\"PoissonKernelH\"\>", ",", "\<\"Quadratic\"\>", ",", "\<\"Reflection\"\>", ",", "\<\"Region\"\>", ",", "\<\"S\"\>", ",", "\<\"Schwarz\"\>", ",", "\<\"SetDimension\"\>", ",", "\<\"Singularity\"\>", ",", "\<\"SurfaceArea\"\>", ",", "\<\"Taylor\"\>", ",", "\<\"Togetherness\"\>", ",", "\<\"TurnOff\"\>", ",", "\<\"TurnOn\"\>", ",", "\<\"Volume\"\>", ",", "\<\"ZeroToZero\"\>", ",", "\<\"ZonalHarmonic\"\>", ",", "\<\"\[Delta]\"\>", ",", "\<\"\[CapitalDelta]\"\>", ",", "\<\"\[CapitalPhi]\"\>"}], "}"}]], "Output", CellChangeTimes->{3.4384437326875*^9, 3.438806535109375*^9, 3.438888530734375*^9, 3.439174235828125*^9, 3.43920788225*^9, 3.439234350859375*^9, 3.439235054140625*^9}] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ "Calculus in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontSlant->"Plain"], "n"], TraditionalForm]]] }], "Section", CellTags->"Calculus in R^n"], Cell[CellGroupData[{ Cell["Norm", "Subsection", CellTags->"Norm"], Cell[TextData[{ StyleBox["Norm[x]", "InlineInput"], " is the Euclidean norm of a vector ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". 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Here ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " denotes a vector in Euclidean space, and ", Cell[BoxData[ FormBox[ SubscriptBox["x", "j"], TraditionalForm]]], " is the ", Cell[BoxData[ FormBox[ SuperscriptBox["j", "th"], TraditionalForm]]], " coordinate of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". The advantage of using ", StyleBox["Partial", "InlineInput"], " rather than the built-in ", Cell[BoxData[ FormBox["Mathematica", TraditionalForm]]], " function ", StyleBox["D", "InlineInput"], " or ", Cell[BoxData[ FormBox[ StyleBox["\[PartialD]", "InlineInput"], TraditionalForm]]], " is that ", StyleBox["Partial", "InlineInput"], " can deal with norms and dot products. 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Return to normal mode by pressing control-space. \ A suberscript or exponent can be typed by pressing control-6. Instead of \ entering ", Cell[BoxData[ StyleBox[ SubscriptBox["x", "j"], "InlineInput"]]], ", you can type ", StyleBox["x[j]", "InlineInput"], "." }], "Text"], Cell[TextData[{ "For taking multiple partial derivatives, ", StyleBox["Partial", "InlineInput"], " uses the same syntax as the built-in ", StyleBox["Mathematica", FontSlant->"Italic"], " function ", StyleBox["D", "InlineInput"], ". For example, the following command computes the partial derivative of |", Cell[BoxData[ FormBox["x", TraditionalForm]]], "| once with respect to the second coordinate, twice with respect to the \ first coordinate, and three times with respect to the fourth coordinate:" }], "Text", CellChangeTimes->{{3.4388713458125*^9, 3.43887134940625*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Partial", "[", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], ",", SubscriptBox["x", "2"], ",", RowBox[{"{", RowBox[{ SubscriptBox["x", "1"], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{ SubscriptBox["x", "4"], ",", "3"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"-", FractionBox[ RowBox[{"15", " ", RowBox[{"(", RowBox[{ RowBox[{"3", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubscriptBox["x", "2"], " ", SubscriptBox["x", "4"]}], "-", RowBox[{"21", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "2"], " ", SubscriptBox["x", "2"], " ", SubscriptBox["x", "4"]}], "-", RowBox[{"7", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubscriptBox["x", "2"], " ", SubsuperscriptBox["x", "4", "3"]}], "+", RowBox[{"63", " ", SubsuperscriptBox["x", "1", "2"], " ", SubscriptBox["x", "2"], " ", SubsuperscriptBox["x", "4", "3"]}]}], ")"}]}], SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "11"]]}]], "Output", CellChangeTimes->{3.438443767859375*^9, 3.4388065735*^9, 3.439234390609375*^9, 3.4392350698125*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["\[CapitalDelta] (Laplacian)", "Subsection", CellChangeTimes->{{3.4391696020625*^9, 3.439169614765625*^9}}, CellTags->"Laplacian"], Cell[TextData[{ Cell[BoxData[ FormBox[ SubscriptBox["\[CapitalDelta]", "x"], TraditionalForm]], FormatType->"TraditionalForm"], StyleBox["[f]", "InlineInput"], " is the Laplacian of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " with respect to ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.43916962475*^9, 3.439169633796875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", RowBox[{ SubscriptBox["x", "3"], " ", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}], "]"}]], "Input", CellChangeTimes->{{3.43916964028125*^9, 3.439169649546875*^9}}], Cell[BoxData[ FractionBox[ RowBox[{ SubscriptBox["x", "3"], "+", RowBox[{ RowBox[{"Dimension", "[", "x", "]"}], " ", SubscriptBox["x", "3"]}]}], RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]]], "Output", CellChangeTimes->{3.438806580140625*^9, 3.4388906460625*^9, 3.4388906844375*^9, 3.4391696511875*^9, 3.43923439415625*^9, 3.439235071875*^9}] }, Closed]], Cell[TextData[{ "To type ", StyleBox["\[CapitalDelta]", "InlineInput"], ", type escape-key D escape-key. Instead of ", StyleBox["\[CapitalDelta]", "InlineInput"], ", you can use ", StyleBox["Laplacian", "InlineInput"], "; thus ", Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox["Laplacian", "x"], "InlineInput"], TraditionalForm]], FormatType->"TraditionalForm"], StyleBox["[f]", "InlineInput"], " is the same as ", Cell[BoxData[ FormBox[ SubscriptBox["\[CapitalDelta]", "x"], TraditionalForm]], FormatType->"TraditionalForm"], StyleBox["[f]", "InlineInput"], "." }], "Text", CellChangeTimes->{{3.4391787251875*^9, 3.439178913875*^9}}], Cell[TextData[{ "Note that in the last result, the Laplacian depends upon ", StyleBox["Dimension[x]", "Output"], ", which is the dimension of the Euclidean space in which ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " lives. For work in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ", ", StyleBox["SetDimension[x, n]", "InlineInput"], " should be used to set the dimension of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " equal to ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "; here ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " can be a symbol or a specific integer, say 8. Be sure to use the ", StyleBox["SetDimension", "InlineInput"], " function rather than changing the value of ", StyleBox["Dimension[x]", "Output"], " directly. We illustrate this procedure by showing that for each ", Cell[BoxData[ FormBox["\[Zeta]", TraditionalForm]]], " with |", Cell[BoxData[ FormBox["\[Zeta]", TraditionalForm]]], "| = 1, the Poisson kernel ", Cell[BoxData[ FormBox["P", TraditionalForm]]], "( . , ", Cell[BoxData[ FormBox["\[Zeta]", TraditionalForm]]], ") is harmonic (in other words, what follows is a proof of Proposition 1.18 \ of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]):" }], "Text", CellChangeTimes->{{3.438890783890625*^9, 3.438890815796875*^9}, { 3.4388918445625*^9, 3.438891847171875*^9}, 3.4390858840625*^9, { 3.439176656015625*^9, 3.439176673796875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{ 3.438806583515625*^9, 3.439169683765625*^9, {3.4391765355*^9, 3.43917655925*^9}, 3.43917667925*^9, 3.439234401125*^9, 3.43923507375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", RowBox[{ RowBox[{"(", RowBox[{"1", "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], "/", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"x", "-", "\[Zeta]"}], "\[RightDoubleBracketingBar]"}], "n"]}], "]"}]], "Input", CellChangeTimes->{{3.439169691671875*^9, 3.439169700234375*^9}, { 3.439176637296875*^9, 3.4391766388125*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"4", " ", "n", " ", RowBox[{"(", RowBox[{ RowBox[{"-", RowBox[{"x", ".", "\[Zeta]"}]}], 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RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"x", "-", "\[Zeta]"}], "\[RightDoubleBracketingBar]"}], RowBox[{ RowBox[{"-", "4"}], "-", "n"}]], " ", RowBox[{"(", RowBox[{ RowBox[{"2", " ", RowBox[{"x", ".", "\[Zeta]"}]}], "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "\[Zeta]", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}]}]}], ")"}]}]}]], "Output", CellChangeTimes->{ 3.438806607171875*^9, 3.43916970278125*^9, {3.43917653746875*^9, 3.439176590734375*^9}, 3.439176681203125*^9, 3.439234403296875*^9, 3.43923507528125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"ExpandNorm", "[", "%", "]"}], " ", "/.", " ", RowBox[{ RowBox[{ "\[LeftDoubleBracketingBar]", "\[Zeta]", "\[RightDoubleBracketingBar]"}], "\[Rule]", "1"}]}]], "Input", CellChangeTimes->{{3.439176553015625*^9, 3.439176593515625*^9}, { 3.439176687828125*^9, 3.43917668875*^9}}], Cell[BoxData["0"], "Output", CellChangeTimes->{ 3.438806610515625*^9, {3.4391697071875*^9, 3.439169710671875*^9}, { 3.439176540328125*^9, 3.43917659409375*^9}, 3.439176690203125*^9, 3.43923440525*^9, 3.43923507725*^9}] }, Closed]], Cell[TextData[{ "Thus the Laplacian of ", Cell[BoxData[ FormBox["P", TraditionalForm]]], "( . , \[Zeta]) is 0, and hence ", Cell[BoxData[ FormBox["P", TraditionalForm]]], "( . , ", Cell[BoxData[ FormBox["\[Zeta]", TraditionalForm]]], ") is harmonic, as claimed." }], "Text", CellChangeTimes->{{3.439085874828125*^9, 3.439085876578125*^9}, { 3.439176696296875*^9, 3.439176708046875*^9}}], Cell[TextData[{ "If ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " and ", Cell[BoxData[ FormBox["y", TraditionalForm]]], " are symbols,then ", Cell[BoxData[ FormBox[ SubscriptBox["\[CapitalDelta]", RowBox[{"x", ",", "y"}]], TraditionalForm]], FormatType->"TraditionalForm"], StyleBox["[f]", "InlineInput"], " gives the Laplacian of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " with respect to the vector (", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], "), where ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " is thought of as a vector, ", Cell[BoxData[ FormBox["y", TraditionalForm]]], " is thought of as a real variable, and (", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], ") is thought of as a vector in a Euclidean space whose dimension is one \ more than the dimension of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". This format is often useful when working with functions defined on the \ upper half-space, as in Chapter 7 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]. Note that if we are thinking of (", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], ") as an element of ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ", then ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " should be defined to have dimension ", Cell[BoxData[ FormBox[ RowBox[{"n", "-", "1"}], TraditionalForm]]], ". We illustrate the use of this format of the ", StyleBox["\[CapitalDelta]", "InlineInput"], " function by showing that for each ", Cell[BoxData[ FormBox["t", TraditionalForm]]], " \[Epsilon] ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], RowBox[{"n", "-", "1"}]], TraditionalForm]]], ", the Poisson kernel ", Cell[BoxData[ FormBox[ SubscriptBox["P", "H"], TraditionalForm]]], "( ", Cell[BoxData[ FormBox[ RowBox[{".", " "}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["t", TraditionalForm]]], ") for the upper half-space is harmonic (see page 145 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "] for another proof of this):" }], "Text", CellChangeTimes->{{3.4390855303125*^9, 3.439085532765625*^9}, { 3.43908584396875*^9, 3.439085844796875*^9}, {3.439170456109375*^9, 3.43917048028125*^9}, {3.4391725664375*^9, 3.43917256825*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", RowBox[{"n", "-", "1"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(\\(\\(-1\\)\\) \ + n\\), \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional \ real Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.4388066260625*^9, 3.439172587875*^9, 3.43923440765625*^9, 3.4392350815625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", RowBox[{"x", ",", "y"}]], "[", RowBox[{"y", "/", SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"x", "-", "t"}], "\[RightDoubleBracketingBar]"}], "2"], "+", SuperscriptBox["y", "2"]}], ")"}], RowBox[{"n", "/", "2"}]]}], "]"}]], "Input", CellChangeTimes->{{3.439172574625*^9, 3.43917259375*^9}}], Cell[BoxData[ RowBox[{"-", FractionBox[ RowBox[{"n", " ", "y", " ", SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox["y", "2"], "+", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "t"}], "+", "x"}], "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], RowBox[{ RowBox[{"-", "2"}], "-", FractionBox["n", "2"]}]], " ", RowBox[{"(", RowBox[{ RowBox[{"2", " ", RowBox[{"t", ".", "x"}]}], "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "t", "\[RightDoubleBracketingBar]"}], "2"], "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], "+", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "t"}], "+", "x"}], "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], " ", RowBox[{"(", RowBox[{ RowBox[{"-", SuperscriptBox["y", "2"]}], "+", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "t"}], "+", "x"}], "\[RightDoubleBracketingBar]"}], "2"], "+", RowBox[{"n", " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "t"}], "+", "x"}], "\[RightDoubleBracketingBar]"}], "2"]}]}], ")"}]}], SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "t"}], "+", "x"}], "\[RightDoubleBracketingBar]"}], "2"]]}]], "Output", CellChangeTimes->{3.4388066355625*^9, 3.439172596109375*^9, 3.439234409609375*^9, 3.439235083484375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"ExpandNorm", "[", "%", "]"}]], "Input"], Cell[BoxData["0"], "Output", CellChangeTimes->{3.43880663840625*^9, 3.439172602875*^9, 3.43923441165625*^9, 3.43923508525*^9}] }, Closed]], Cell[TextData[{ "Thus the Laplacian of ", Cell[BoxData[ FormBox[ SubscriptBox["P", "H"], TraditionalForm]]], "( ", Cell[BoxData[ FormBox[ RowBox[{".", " "}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["t", TraditionalForm]]], ") is 0, and hence ", Cell[BoxData[ FormBox[ SubscriptBox["P", "H"], TraditionalForm]]], "( ", Cell[BoxData[ FormBox[ RowBox[{".", " "}], TraditionalForm]]], ", ", Cell[BoxData[ FormBox["t", TraditionalForm]]], ") is harmonic, as claimed." }], "Text", CellChangeTimes->{{3.4390858521875*^9, 3.439085858453125*^9}}], Cell[TextData[{ "The expression ", Cell[BoxData[ FormBox[ SubscriptBox["\[CapitalDelta]", "v"], TraditionalForm]], FormatType->"TraditionalForm"], StyleBox["[f]", "InlineInput"], ", where ", StyleBox["v", FontSlant->"Italic"], " is a list of explicit coordinates, gives the Laplacian of ", StyleBox["f", FontSlant->"Italic"], " with respect to the coordinates in the list ", StyleBox["v", FontSlant->"Italic"], ". For example, below we compute the Laplacian of ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "2"], SuperscriptBox["y", "3"], SuperscriptBox["z", "4"]}], TraditionalForm]], FormatType->"TraditionalForm"], " with respect to the usual coordinates ", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], ", and ", StyleBox["z", FontSlant->"Italic"], " of ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.439173514046875*^9, 3.43917378090625*^9}, { 3.439173826296875*^9, 3.43917384528125*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]], "[", RowBox[{ SuperscriptBox["x", "2"], SuperscriptBox["y", "3"], SuperscriptBox["z", "4"]}], "]"}]], "Input", CellChangeTimes->{{3.43917322184375*^9, 3.43917327396875*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"12", " ", SuperscriptBox["x", "2"], " ", SuperscriptBox["y", "3"], " ", SuperscriptBox["z", "2"]}], "+", RowBox[{"6", " ", SuperscriptBox["x", "2"], " ", "y", " ", SuperscriptBox["z", "4"]}], "+", RowBox[{"2", " ", SuperscriptBox["y", "3"], " ", SuperscriptBox["z", "4"]}]}]], "Output", CellChangeTimes->{3.4391732755*^9, 3.439234416265625*^9, 3.439235087078125*^9}] }, Closed]], Cell[TextData[{ "Note that in the example above, ", StyleBox["x", FontSlant->"Italic"], " is no longer being used as a vector but as the first coordinate in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], ". Because ", StyleBox["x", FontSlant->"Italic"], " has been used in a list of coordinates, the software no longer treats ", StyleBox["x", FontSlant->"Italic"], " as a vector." }], "Text", CellChangeTimes->{{3.4391739675*^9, 3.43917401546875*^9}, { 3.439174146140625*^9, 3.43917418484375*^9}}], Cell[TextData[{ "If ", StyleBox["m", FontSlant->"Italic"], " is a postive integer, then ", Cell[BoxData[ FormBox[ StyleBox[ SuperscriptBox[ SubscriptBox["\[CapitalDelta]", "x"], StyleBox["m", FontSlant->"Plain"]], "InlineInput"], TraditionalForm]]], " can be used, with the same syntax as ", Cell[BoxData[ FormBox[ SubscriptBox["\[CapitalDelta]", "x"], TraditionalForm]], FormatType->"TraditionalForm"], ", to evaluate the ", Cell[BoxData[ FormBox[ SuperscriptBox["m", "th"], TraditionalForm]]], " power of the Laplacian. For example, here we find the biLaplacian (the \ Laplacian of the Laplacian) of 1/|", Cell[BoxData[ FormBox["x", TraditionalForm]]], "| on ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "8"], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.43917006415625*^9, 3.43917009059375*^9}, { 3.43917017609375*^9, 3.439170180875*^9}, {3.4391738975625*^9, 3.4391738988125*^9}, {3.439174061203125*^9, 3.439174069671875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", " ", "8"}], "]"}]], "Input", CellChangeTimes->{{3.43917390678125*^9, 3.439173914265625*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(8\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.439173914984375*^9, 3.43923441840625*^9, 3.4392350891875*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SuperscriptBox[ SubscriptBox["\[CapitalDelta]", "x"], "2"], "[", RowBox[{"1", "/", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}], "]"}]], "Input", CellChangeTimes->{{3.4391700995625*^9, 3.439170112265625*^9}}], Cell[BoxData[ FractionBox["45", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "5"]]], "Output", CellChangeTimes->{3.438806617390625*^9, 3.439170118921875*^9, 3.4391739175625*^9, 3.439234421125*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Grad", "Subsection", CellTags->"Grad"], Cell[TextData[{ StyleBox["Grad[f, x]", "InlineInput"], " is the gradient of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " with respect to the vector ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " (", StyleBox["Mathematica", FontSlant->"Italic"], "'s convention of spelling out names in full would imply that this function \ should be called ", StyleBox["Gradient", "InlineInput"], " instead of ", StyleBox["Grad", "InlineInput"], ", but ", StyleBox["Gradient", "InlineInput"], " is already used by ", StyleBox["Mathematica", FontSlant->"Italic"], " for an unrelated purpose):" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Grad", "[", RowBox[{ RowBox[{ RowBox[{"a", ".", "x"}], "+", SubscriptBox["x", "k"], "+", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"a", "+", FractionBox["x", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]], "+", SubscriptBox["\[Delta]", "k"]}]], "Output", CellChangeTimes->{3.438806645328125*^9, 3.43920789825*^9, 3.439234427875*^9, 3.439235093765625*^9}] }, Closed]], Cell[TextData[{ "In the last output, ", Cell[BoxData[ FormBox[ StyleBox[ SubscriptBox["\[Delta]", "k"], "Output"], TraditionalForm]], FormatType->"TraditionalForm"], " denotes the vector that equals 1 in the ", Cell[BoxData[ FormBox[ SuperscriptBox["k", "th"], TraditionalForm]]], " coordinate and 0 in the other coordinates. To type \[Delta], type \ escape-key d escape-key." }], "Text", CellChangeTimes->{{3.4392079085625*^9, 3.439207912765625*^9}, { 3.43920803740625*^9, 3.439208084046875*^9}, {3.43920851221875*^9, 3.43920853153125*^9}}, CellTags->"Delta"] }, Closed]], Cell[CellGroupData[{ Cell["NormalD", "Subsection", CellTags->"NormalD"], Cell[TextData[{ StyleBox["NormalD[f, z]", "InlineInput"], " equals the outward normal derivative of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " at a point ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " on the unit sphere." }], "Text"], Cell[TextData[{ "In the section of Chapter 1 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "] titled ", StyleBox["The Poisson Kernel for the Ball", FontSlant->"Italic"], ", the formula for the Poisson kernel was derived by showing that ", Cell[BoxData[ FormBox["P", TraditionalForm]]], "(", Cell[BoxData[ FormBox["x", TraditionalForm]]], ",\[NonBreakingSpace]", Cell[BoxData[ FormBox["z", TraditionalForm]]], ") equals the normal derivative (with respect to ", Cell[BoxData[ FormBox["z", TraditionalForm]]], ") of" }], "Text"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", "x"}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], "-", RowBox[{ SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", FractionBox["x", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]]}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]]}]}], ")"}], "/", RowBox[{ RowBox[{"(", RowBox[{"2", "-", "n"}], ")"}], "."}]}], TraditionalForm]], "Text", TextAlignment->Center], Cell["\<\ The computation of that normal derivative is a bit complicated and was left \ to the reader. Now we show how this computation can be done easily by a \ computer. We begin by taking the normal derivative of the function above:\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"NormalD", "[", RowBox[{ RowBox[{ RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", "x"}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], "-", RowBox[{ SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", FractionBox["x", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]]}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]]}]}], ")"}], "/", RowBox[{"(", RowBox[{"2", "-", "n"}], ")"}]}], ",", "z"}], "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"2", "-", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"1", "-", RowBox[{"x", ".", "z"}]}], ")"}], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "x"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]]}], "-", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"-", "2"}], "+", "n"}], ")"}], " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]], " ", RowBox[{"(", RowBox[{ RowBox[{"x", ".", "z"}], "-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", FractionBox["x", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]]}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]]}]}], RowBox[{"2", "-", "n"}]]], "Output", CellChangeTimes->{ 3.4388066765625*^9, 3.439234429953125*^9, {3.43923509625*^9, 3.4392351039375*^9}}] }, Closed]], Cell[TextData[{ "Now we ask the computer to replace ", Cell[BoxData[ FormBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", FractionBox["x", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]]}], "\[RightDoubleBracketingBar]"}], TraditionalForm]]], " in the last output with ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", "x"}], "\[RightDoubleBracketingBar]"}], "/", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}], TraditionalForm]]], "; the Symmetry Lemma (1.13 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]) implies that the two expressions are equal:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"%", " ", "/.", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", FractionBox["x", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]]}], "\[RightDoubleBracketingBar]"}], "\[Rule]", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", "x"}], "\[RightDoubleBracketingBar]"}], "/", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}]}]}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{ RowBox[{ RowBox[{"(", RowBox[{"2", "-", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"1", "-", RowBox[{"x", ".", "z"}]}], ")"}], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "x"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]]}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"-", "2"}], "+", "n"}], ")"}], " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]], " ", RowBox[{"(", RowBox[{ RowBox[{"-", RowBox[{"x", ".", "z"}]}], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], " ", SuperscriptBox[ RowBox[{"(", FractionBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "x"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]], ")"}], RowBox[{"-", "n"}]]}]}], RowBox[{"2", "-", "n"}]]], "Output", CellChangeTimes->{ 3.438806685203125*^9, 3.439234432375*^9, {3.439235098265625*^9, 3.43923510575*^9}}] }, Closed]], Cell[TextData[{ "Applying the ", StyleBox["Mathematica", FontSlant->"Italic"], " function ", StyleBox["PowerExpand", "InlineInput"], " to the last output gives the desired result:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"PowerExpand", "[", "%", "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"-", RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}]}], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "x"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]]}]], "Output", CellChangeTimes->{3.438806688375*^9, 3.4392344345625*^9, 3.43923510765625*^9} ] }, Closed]], Cell["\<\ The last output completes the derivation of the formula for the Poisson \ kernel for the ball.\ \>", "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Matrices", "Subsection", CellTags->"Matrices"], Cell[CellGroupData[{ Cell["Matrix dimensions", "Subsubsection"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " uses ", StyleBox[".", "InlineInput"], " to represent matrix multiplication as well as dot product. When using this \ package, you must use the ", StyleBox["SetDimension", "InlineInput"], " command to tell ", StyleBox["Mathematica ", FontSlant->"Italic"], "whenever a symbol is being thought of as a matrix. To think of ", Cell[BoxData[ FormBox["A", TraditionalForm]]], " as an ", Cell[BoxData[ FormBox["m", TraditionalForm]]], "-by-", Cell[BoxData[ FormBox["n", TraditionalForm]]], " matrix, enter the command ", StyleBox["SetDimension[A, {m, n}]", "InlineInput"], "; here ", StyleBox["m", FontSlant->"Italic"], " and ", StyleBox["n", FontSlant->"Italic"], " can be symbols or specific positive integers such as 7 and 3. For the \ examples in this section, we want to think of ", Cell[BoxData[ FormBox["A", TraditionalForm]]], " as a square matrix, so we make the two dimensions of ", Cell[BoxData[ FormBox["A", TraditionalForm]]], " equal:" }], "Text", CellChangeTimes->{{3.4388909855625*^9, 3.438891033890625*^9}, 3.438891140125*^9, {3.438891817703125*^9, 3.4388918231875*^9}, 3.438891953171875*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"A", ",", " ", RowBox[{"{", RowBox[{"n", ",", "n"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Matrix\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(A\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a matrix of \ size \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-by-\\\\\\\"\\\", \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\".\\\\\\\"\\\", \\\"MT\\\ \"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.4388066990625*^9, 3.438891077796875*^9, 3.439174322609375*^9, 3.439174521953125*^9, 3.439234440953125*^9, 3.4392351119375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Grad", "[", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"A", ".", "x"}], "\[RightDoubleBracketingBar]"}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{ RowBox[{"(", RowBox[{"A", ".", "x"}], ")"}], ".", "A"}], RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"A", ".", "x"}], "\[RightDoubleBracketingBar]"}]]], "Output", CellChangeTimes->{3.438806702015625*^9, 3.438891084453125*^9, 3.439234443015625*^9, 3.439235114125*^9}] }, Closed]], Cell[TextData[{ "To interpret input or output involving products of matrices and vectors, \ think of vectors as either column vectors or row vectors, whichever makes \ sense in context. Thus in the input above, ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " is obviously a vector, because we are taking the gradient of some function \ with respect to ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". For ", Cell[BoxData[ FormBox[ RowBox[{"A", ".", "x"}], TraditionalForm]]], " to make sense as a vector in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ", we must think of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " as a column vector (an ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-by-1 matrix). To interpret ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"A", ".", "x"}], ")"}], ".", "A"}], TraditionalForm]]], " in the last output, we again think of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " as a column vector (an ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-by-1 matrix), so that ", Cell[BoxData[ FormBox[ RowBox[{"A", ".", "x"}], TraditionalForm]]], " makes sense as an element of ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ". Because ", Cell[BoxData[ FormBox[ RowBox[{"A", ".", "x"}], TraditionalForm]]], " is then multiplied by the ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "-by-", Cell[BoxData[ FormBox["n", TraditionalForm]]], " matrix ", Cell[BoxData[ FormBox["A", TraditionalForm]]], ", we must think of ", Cell[BoxData[ FormBox[ RowBox[{"A", ".", "x"}], TraditionalForm]]], " as a row vector (an 1-by-", Cell[BoxData[ FormBox["n", TraditionalForm]]], " matrix), so that ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"A", ".", "x"}], ")"}], ".", "A"}], TraditionalForm]]], " is a vector, as expected here (because it is the gradient of a function)." }], "Text", CellChangeTimes->{{3.4388910423125*^9, 3.438891043421875*^9}}], Cell[TextData[{ "In the example above we did not enter the command ", StyleBox["SetDimension[x, n]", "InlineInput"], " to tell ", StyleBox["Mathematica", FontSlant->"Italic"], " that ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " is a vector (although it would not have hurt to do so). The HFT7.m package \ can almost always use the context to distinguish scalars from vectors (if \ both interpretations make sense, the package assumes that a symbol represents \ a scalar). However, you must always explicitly use the ", StyleBox["SetDimension", "InlineInput"], " command to tell ", StyleBox["Mathematica", FontSlant->"Italic"], " which objects are matrices." }], "Text", CellChangeTimes->{3.438803381234375*^9, 3.43886989175*^9}], Cell[TextData[{ "The expression ", StyleBox["Dimension[A]", "InlineInput"], " gives the dimensions of a matrix ", StyleBox["A", FontSlant->"Italic"], ". However, be sure to use the ", StyleBox["SetDimension", "InlineInput"], " function rather than changing the value of ", StyleBox["Dimension[A]", "InlineInput"], " directly. 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To interpret ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"(", RowBox[{"A", ".", "x"}], ")"}], ".", SubscriptBox["A", RowBox[{".", RowBox[{",", "3"}]}]]}], TraditionalForm]]], " in the last output, we think of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " as a column vector, so that ", Cell[BoxData[ FormBox[ RowBox[{"A", ".", "x"}], TraditionalForm]]], " is a vector. 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Let's find the Laplacian of the same function:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", RowBox[{"f", "[", SuperscriptBox[ RowBox[{"g", "[", RowBox[{"3", "x"}], "]"}], "2"], "]"}], "]"}]], "Input", CellChangeTimes->{{3.439174651859375*^9, 3.43917465940625*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"18", " ", RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"Grad", "[", "g", "]"}], "[", RowBox[{"3", " ", "x"}], "]"}], "\[RightDoubleBracketingBar]"}], "2"], "+", RowBox[{ RowBox[{"g", "[", RowBox[{"3", " ", "x"}], "]"}], " ", RowBox[{ RowBox[{"\[CapitalDelta]", "[", "g", "]"}], "[", RowBox[{"3", " ", "x"}], "]"}]}]}], ")"}], " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", SuperscriptBox[ RowBox[{"g", "[", RowBox[{"3", " ", "x"}], "]"}], "2"], "]"}]}], "+", RowBox[{"36", " ", SuperscriptBox[ RowBox[{"g", "[", RowBox[{"3", " ", "x"}], "]"}], "2"], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"Grad", "[", "g", "]"}], "[", RowBox[{"3", " ", "x"}], "]"}], "\[RightDoubleBracketingBar]"}], "2"], " ", RowBox[{ SuperscriptBox["f", "\[Prime]\[Prime]", MultilineFunction->None], "[", SuperscriptBox[ RowBox[{"g", "[", RowBox[{"3", " ", "x"}], "]"}], "2"], "]"}]}]}]], "Output", CellChangeTimes->{3.438806810046875*^9, 3.439174660671875*^9, 3.439234597859375*^9, 3.43923516446875*^9}] }, Closed]], Cell[TextData[{ StyleBox["\[CapitalDelta][g][3x]", "InlineInput"], " denotes, of course, the Laplacian of ", Cell[BoxData[ FormBox["g", TraditionalForm]]], ", evaluated at ", Cell[BoxData[ FormBox[ RowBox[{"3", " ", "x"}], TraditionalForm]]], "." }], "Text", CellChangeTimes->{{3.439174679375*^9, 3.439174680734375*^9}}], Cell[TextData[{ "All our differentiation commands (", StyleBox["Partial", "InlineInput"], ", ", StyleBox["\[CapitalDelta]", "InlineInput"], ", ", StyleBox["Gradient", "InlineInput"], ", ", StyleBox["NormalD", "InlineInput"], ", ", StyleBox["Divergence", "InlineInput"], ", ", StyleBox["Jacobian", "InlineInput"], ") and ", StyleBox["Homogeneous", "InlineInput"], " and ", StyleBox["Taylor", "InlineInput"], " can be used with non-explicit functions. For example, if working in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "2"], TraditionalForm]]], ", to find the term of degree 3 in the homogeneous expansion of ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", SuperscriptBox[ RowBox[{"(", RowBox[{"g", "(", RowBox[{"3", "x"}], ")"}], ")"}], "2"], ")"}], TraditionalForm]]], " about the origin, you would enter the command ", StyleBox["SetDimension[x, 2]", "InlineInput"], " and then the command ", StyleBox["Homogeneous[", "InlineInput"], StyleBox[Cell[BoxData[ RowBox[{"f", "[", SuperscriptBox[ RowBox[{"g", "[", RowBox[{"3", "x"}], "]"}], "2"], "]"}]], "InlineInput"], "InlineInput"], StyleBox[", 3, x].", "InlineInput"], " Try it! In the answer you will see terms such as ", Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{ SubscriptBox["Partial", RowBox[{"{", RowBox[{"1", ",", "1", ",", "2"}], "}"}]], "[", "g", "]"}], "[", "0", "]"}], "Output"]]], ", which denotes the partial derivative of ", Cell[BoxData[ FormBox["g", TraditionalForm]]], ", twice with respect to the first variable and once with respect to the \ second variable, evaluated at 0." }], "Text", CellChangeTimes->{{3.439174699828125*^9, 3.439174701203125*^9}}] }, Closed]], Cell[CellGroupData[{ Cell["Vector-valued functions", "Subsection", CellTags->"Vector-valued Functions"], Cell[TextData[{ "Suppose we compute the partial derivative with respect to ", Cell[BoxData[ FormBox[ SubscriptBox["x", "1"], TraditionalForm]]], " of the function that takes ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " to ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", RowBox[{"g", "(", SuperscriptBox[ RowBox[{"h", "(", "x", ")"}], "2"], ")"}], ")"}], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Partial", "[", RowBox[{ RowBox[{"f", "[", RowBox[{"g", "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], "]"}], ",", SubscriptBox["x", "1"]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"2", " ", RowBox[{"h", "[", "x", "]"}], " ", RowBox[{ SuperscriptBox["f", "\[Prime]", MultilineFunction->None], "[", RowBox[{"g", "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], "]"}], " ", RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], " ", RowBox[{ RowBox[{ SubscriptBox["Partial", RowBox[{"{", "1", "}"}]], "[", "h", "]"}], "[", "x", "]"}]}]], "Output", CellChangeTimes->{3.438806820515625*^9, 3.43923460125*^9, 3.439235168140625*^9}] }, Closed]], Cell[TextData[{ "The example above illustrates a general principle: This package assumes \ that all functions are real valued, unless told otherwise. Thus the last \ output is correct if we are thinking of ", Cell[BoxData[ FormBox["g", TraditionalForm]]], " as a real-valued function. If we want to think of ", Cell[BoxData[ FormBox["g", TraditionalForm]]], " as taking values in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ", we must first enter the command ", StyleBox["SetDimension[g[_], n]", "InlineInput"], ", which instructs the computer that ", Cell[BoxData[ FormBox["g", TraditionalForm]]], " is a function with range in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "n"], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{3.438888778234375*^9}], Cell[CellGroupData[{ Cell[BoxData[ StyleBox[ RowBox[{"SetDimension", "[", RowBox[{ RowBox[{"g", "[", "_", "]"}], ",", " ", "n"}], "]"}], "InlineInput"]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"VectorValuedFunction\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(g\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a function \ taking values in \\\\\\\"\\\", \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional Euclidean \ space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.43880682425*^9, 3.439234607234375*^9, 3.439235187328125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Partial", "[", RowBox[{ RowBox[{"f", "[", RowBox[{"g", "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], "]"}], ",", SubscriptBox["x", "1"]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"2", " ", RowBox[{ RowBox[{ RowBox[{"Grad", "[", "f", "]"}], "[", RowBox[{"g", "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], "]"}], ".", RowBox[{"(", RowBox[{ RowBox[{"h", "[", "x", "]"}], " ", RowBox[{ SuperscriptBox["g", "\[Prime]", MultilineFunction->None], "[", SuperscriptBox[ RowBox[{"h", "[", "x", "]"}], "2"], "]"}], " ", RowBox[{ RowBox[{ SubscriptBox["Partial", RowBox[{"{", "1", "}"}]], "[", "h", "]"}], "[", "x", "]"}]}], ")"}]}]}]], "Output", CellChangeTimes->{3.43880682721875*^9, 3.4392346164375*^9, 3.4392351895625*^9}] }, Closed]], Cell[TextData[{ "Note that the last output involves the dot product of two vectors: the \ gradient of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], ", evaluated at ", Cell[BoxData[ FormBox[ RowBox[{"g", "(", SuperscriptBox[ RowBox[{"h", "(", "x", ")"}], "2"], ")"}], TraditionalForm]]], ", and the derivative of ", Cell[BoxData[ FormBox["g", TraditionalForm]]], ", evaluated at ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{"h", "(", "x", ")"}], "2"], TraditionalForm]]], ". The last two outputs should be compared." }], "Text", CellChangeTimes->{3.439174731515625*^9}] }, Closed]], Cell[CellGroupData[{ Cell["Volume", "Subsection", CellTags->"Volume"], Cell[TextData[{ StyleBox["Volume[n]", "InlineInput"], " is the volume (unnormalized) of the unit ball in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], StyleBox["n", FontSize->10]], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Volume", "[", "4", "]"}]], "Input"], Cell[BoxData[ FractionBox[ SuperscriptBox["\[Pi]", "2"], "2"]], "Output", CellChangeTimes->{3.438806832796875*^9, 3.439235194484375*^9}] }, Closed]], Cell[TextData[{ StyleBox["Volume", "InlineInput"], " is computed using the formula given by Exercise 6 in Appendix A of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["SurfaceArea", "Subsection", CellTags->"SurfaceArea"], Cell[TextData[{ StyleBox["SurfaceArea[n]", "InlineInput"], " is the surface area (unnormalized) of the unit sphere in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], StyleBox["n", FontSize->10]], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SurfaceArea", "[", "57", "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{"536870912", " ", SuperscriptBox["\[Pi]", "28"]}], "8687364368561751199826958100282265625"]], "Output", CellChangeTimes->{3.43880684415625*^9, 3.43923519678125*^9}] }, Closed]], Cell[TextData[{ StyleBox["SurfaceArea[n]", "InlineInput"], " is computed by multiplying the formula for ", StyleBox["Volume[n]", "InlineInput"], " by ", Cell[BoxData[ FormBox["n", TraditionalForm]]], "; see A.2 in Appendix A of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["IntegrateSphere", "Subsection", CellTags->"IntegrateSphere"], Cell[TextData[{ StyleBox["IntegrateSphere[f, x]", "InlineInput"], " equals the integral of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], ", with respect to normalized surface area measure, over the unit sphere in \ the Euclidean space defined by ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". Here ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " should be a polynomial function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438806853390625*^9, 3.439235203375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"IntegrateSphere", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "2"], " ", SuperscriptBox[ SubscriptBox["x", "2"], "4"], " ", SuperscriptBox[ SubscriptBox["x", "3"], "6"]}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ FractionBox["45", RowBox[{"n", " ", RowBox[{"(", RowBox[{"2", "+", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"4", "+", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"6", "+", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"8", "+", "n"}], ")"}], " ", RowBox[{"(", RowBox[{"10", "+", "n"}], ")"}]}]]], "Output", CellChangeTimes->{3.438806856625*^9, 3.439235205*^9}] }, Closed]], Cell[TextData[{ StyleBox["IntegrateSphere", "InlineInput"], " is computed by using the results in Section 3 of [", ButtonBox["W", BaseStyle->"Hyperlink", ButtonData:>"W"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["IntegrateBall", "Subsection", CellTags->"IntegrateBall"], Cell[TextData[{ StyleBox["IntegrateBall[f, x]", "InlineInput"], " equals the integral of ", Cell[BoxData[ FormBox["f", TraditionalForm]]], ", with respect to (unnormalized) volume measure, over the unit ball in the \ Euclidean space defined by ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". Here ", Cell[BoxData[ FormBox["f", TraditionalForm]]], " should be a function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " and ||", Cell[BoxData[ FormBox[ RowBox[{"x", "||"}], TraditionalForm]]], " that is a polynomial in ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " and a function in ||", Cell[BoxData[ FormBox[ RowBox[{"x", "||"}], TraditionalForm]]], " for which ", StyleBox["Mathematica", FontSlant->"Italic"], " can find an explicit antiderivative:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "7"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(7\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.43880686578125*^9, 3.4392352075625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"IntegrateBall", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "2"], " ", RowBox[{ SuperscriptBox[ SubscriptBox["x", "2"], "4"], "/", RowBox[{"(", RowBox[{"1", "+", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}], ")"}]}]}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{"16", " ", SuperscriptBox["\[Pi]", "3"], " ", RowBox[{"(", RowBox[{ RowBox[{"-", FractionBox["18107", "27720"]}], "+", RowBox[{"Log", "[", "2", "]"}]}], ")"}]}], "3465"]], "Output", CellChangeTimes->{3.43880688178125*^9, 3.43923521146875*^9}] }, Closed]], Cell[TextData[{ StyleBox["IntegrateBall", "InlineInput"], " is computed by converting to polar coordinates and then using the function \ ", StyleBox["IntegrateSphere ", "InlineInput"], "(see [", ButtonBox["R", BaseStyle->"Hyperlink", ButtonData:>"R"], "], Chapter 8, Exercise 6)." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Boundary Value Problems", "Section", CellTags->"Boundary Value Problems"], Cell[CellGroupData[{ Cell["Dirichlet problems", "Subsection", CellTags->"Dirichlet"], Cell[CellGroupData[{ Cell["The Dirichlet problem on the sphere", "Subsubsection"], Cell[TextData[{ StyleBox["Dirichlet[p, x]", "InlineInput"], " is the solution to the standard Dirichlet problem: find the harmonic \ function on the unit ball in the Euclidean space defined by ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " that equals ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " on the unit sphere. Thus ", StyleBox["Dirichlet[p, x]", "InlineInput"], " is the Poisson integral of ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " as a function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ". Here ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " must be a polynomial function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "5"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(5\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438806903*^9, 3.439174750234375*^9, 3.43923521990625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], " ", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ FractionBox["1", "15015"], RowBox[{"(", RowBox[{"143", "-", RowBox[{"273", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], "+", RowBox[{"165", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"]}], "-", RowBox[{"35", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "6"]}], "+", RowBox[{"910", " ", SubsuperscriptBox["x", "1", "2"]}], "-", RowBox[{"1540", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "2"]}], "+", RowBox[{"630", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "1", "2"]}], "+", RowBox[{"1155", " ", SubsuperscriptBox["x", "1", "4"]}], "-", RowBox[{"1155", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "4"]}], "+", RowBox[{"455", " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"770", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"315", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"6930", " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"6930", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"15015", " ", SubsuperscriptBox["x", "1", "4"], " ", SubsuperscriptBox["x", "2", "2"]}]}], ")"}]}]], "Output", CellChangeTimes->{ 3.43880690565625*^9, {3.439174752578125*^9, 3.43917477425*^9}, 3.439235221953125*^9}] }, Closed]], Cell[TextData[{ "The last output is a harmonic function on ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "5"], TraditionalForm]]], "that equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], TraditionalForm]]], " on the unit sphere. To check this, first we take the Laplacian of the last \ output, and then we evaluate the last output on the unit sphere:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.43917476346875*^9, 3.43917477115625*^9}}], Cell[BoxData["0"], "Output", CellChangeTimes->{ 3.438806908484375*^9, {3.4391747578125*^9, 3.43917477675*^9}, 3.439235225140625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"%%", " ", "/.", " ", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "\[Rule]", "1"}]}]], "Input"], Cell[BoxData[ RowBox[{ SubsuperscriptBox["x", "1", "4"], " ", SubsuperscriptBox["x", "2", "2"]}]], "Output", CellChangeTimes->{3.438806910609375*^9, 3.43923522703125*^9}] }, Closed]], Cell["Thus we have indeed solved the specified Dirichlet problem.", "Text"], Cell[TextData[{ "The solution to the Dirichlet problem on the sphere is computed by using \ the algorithm described in [", ButtonBox["AR", BaseStyle->"Hyperlink", ButtonData:>"AR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["The Dirichlet problem on the exterior of the sphere", "Subsubsection"], Cell[TextData[{ "The ", StyleBox["Region", "InlineInput"], " option allows the user to solve Dirichlet problems on regions other than \ the sphere. Currently supported values for ", StyleBox["Region", "InlineInput"], " are ", StyleBox["Sphere", "InlineInput"], ", ", StyleBox["ExteriorSphere", "InlineInput"], ", ", StyleBox["Annulus", "InlineInput"], ", and ", StyleBox["Quadratic", "InlineInput"], ". The default value of ", StyleBox["Region", "InlineInput"], " is ", StyleBox["Sphere", "InlineInput"], "; thus ", StyleBox["Dirichlet[p, x, Region\[Rule]Sphere]", "InlineInput"], " is the same as ", StyleBox["Dirichlet[p, x]", "InlineInput"], "." }], "Text"], Cell[TextData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{"p", ",", "x", ",", " ", RowBox[{"Region", "\[Rule]", "ExteriorSphere"}]}], "]"}]], "InlineInput"], " gives the solution to the standard exterior Dirichlet problem: find the \ harmonic function on the exterior (including \[Infinity]) of the unit ball in \ the Euclidean space defined by ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " that equals ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " on the unit sphere. This function is the exterior Poisson integral of ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " as a function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " (see Theorem 4.11 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]). Here ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " must be a polynomial function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], " ", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], ",", "x", ",", " ", RowBox[{"Region", "\[Rule]", "ExteriorSphere"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ FractionBox["1", RowBox[{"15015", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "15"]}]], RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", "35"}], " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "6"]}], "+", RowBox[{"165", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "8"]}], "-", RowBox[{"273", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "10"]}], "+", RowBox[{"143", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "12"]}], "+", RowBox[{"630", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "1", "2"]}], "-", RowBox[{"1540", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "6"], " ", SubsuperscriptBox["x", "1", "2"]}], "+", RowBox[{"910", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "8"], " ", SubsuperscriptBox["x", "1", "2"]}], "-", RowBox[{"1155", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "4"]}], "+", RowBox[{"1155", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "1", "4"]}], "+", RowBox[{"315", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"770", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "6"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"455", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "8"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"6930", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"6930", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"15015", " ", SubsuperscriptBox["x", "1", "4"], " ", SubsuperscriptBox["x", "2", "2"]}]}], ")"}]}]], "Output", CellChangeTimes->{3.4388069183125*^9, 3.43923523446875*^9}] }, Closed]], Cell[TextData[{ "The last output is a harmonic function on the exterior of the unit ball in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "5"], TraditionalForm]]], " (because the dimension of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " was set to 5 earlier) that equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], TraditionalForm]]], " on the unit sphere. This function is the Kelvin transform of the solution \ that would have been obtained without the ", StyleBox["Region\[Rule]Sphere", "InlineInput"], " option; see Chapter 4 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["The Dirichlet problem on annular regions", "Subsubsection"], Cell[TextData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{"{", RowBox[{"p", ",", "q"}], "}"}], ",", "x", ",", " ", RowBox[{"Region", "\[Rule]", RowBox[{"Annulus", "[", RowBox[{"r", ",", "s"}], "]"}]}]}], " ", "]"}]], "InlineInput"], " gives the solution to the annular Dirichlet problem: find the harmonic \ function on the annular region with inner radius ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " and outer radius ", Cell[BoxData[ FormBox["s", TraditionalForm]]], " that equals ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " on the sphere of radius ", Cell[BoxData[ FormBox["r", TraditionalForm]]], " and equals ", Cell[BoxData[ FormBox["q", TraditionalForm]]], " on the sphere of radius ", Cell[BoxData[ FormBox["s", TraditionalForm]]], ". 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Thus the last output obviously equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "3"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], TraditionalForm]]], " on the elliptic parabolic given by the equation ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox[ SubscriptBox["x", "1"], "2"]}], "+", RowBox[{"3", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}]}], " ", "=", " ", RowBox[{ RowBox[{"4", SubscriptBox["x", "3"]}], "+", " ", "1"}]}], TraditionalForm]]], ". Thus to verify that the last output is correct, we only need check that \ it is harmonic:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.439174827765625*^9, 3.4391748341875*^9}}], Cell[BoxData["0"], "Output", CellChangeTimes->{3.438806950125*^9, 3.439174834859375*^9, 3.439175144109375*^9, 3.439235323953125*^9}] }, Closed]], Cell[TextData[{ "The solution to the Dirichlet problem on a quadratic surface is computed by \ using the algorithm developed in [", ButtonBox["AGV", BaseStyle->"Hyperlink", ButtonData:>"AGV"], "]. That paper also shows that if each coordinate of ", Cell[BoxData[ FormBox["b", TraditionalForm]]], " is nonnegative and at least one coordinate of ", Cell[BoxData[ FormBox["b", TraditionalForm]]], " is nonzero, and if the quadratic surface in question is nontrivial in the \ sense that there exits ", Cell[BoxData[ FormBox[ RowBox[{"x", " ", "\[Epsilon]", " ", SuperscriptBox[ StyleBox["R", FontWeight->"Bold"], "n"]}], TraditionalForm]]], " such that" }], "Text"], Cell[TextData[{ Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"b", ".", SuperscriptBox["x", "2"]}], "+", " ", RowBox[{"c", ".", "x"}], " ", "+", " ", "d"}], " ", "<", " ", "0"}], TraditionalForm]]], "," }], "Text", TextAlignment->Center], Cell[TextData[{ "then the polynomial produced by this algorithm is the unique harmonic \ polynomial that equals ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " on the quadratic surface in question." }], "Text"], Cell[TextData[{ "Using the notation above, if ", Cell[BoxData[ FormBox["c", TraditionalForm]]], " consists of all 0's, then it may be omitted. For example, here we find the \ harmonic polynomial that equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], TraditionalForm]]], "on the ellipsoid given by the equation ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox[ SubscriptBox["x", "1"], "2"]}], "+", RowBox[{"3", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], "+", RowBox[{"4", SuperscriptBox[ SubscriptBox["x", "3"], "2"]}]}], "=", "1"}], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], ",", "x", ",", RowBox[{"Region", "\[Rule]", RowBox[{"Quadratic", "[", RowBox[{ RowBox[{"{", RowBox[{"2", ",", "3", ",", "4"}], "}"}], ",", RowBox[{"-", "1"}]}], "]"}]}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{ SubsuperscriptBox["x", "1", "4"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{ FractionBox["1", "39309793479"], RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "+", RowBox[{"2", " ", SubsuperscriptBox["x", "1", "2"]}], "+", RowBox[{"3", " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"4", " ", SubsuperscriptBox["x", "3", "2"]}]}], ")"}], " ", RowBox[{"(", RowBox[{ RowBox[{"-", "52529359"}], "-", RowBox[{"511000770", " ", SubsuperscriptBox["x", "1", "2"]}], "-", RowBox[{"801788415", " ", SubsuperscriptBox["x", "1", "4"]}], "+", RowBox[{"102950118", " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"945182934", " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"67605435", " ", SubsuperscriptBox["x", "2", "4"]}], "-", RowBox[{"64713579", " ", SubsuperscriptBox["x", "3", "2"]}], "-", RowBox[{"5766993996", " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{"204109461", " ", SubsuperscriptBox["x", "2", "2"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{"342331356", " ", SubsuperscriptBox["x", "3", "4"]}]}], ")"}]}]}]}]], "Output", CellChangeTimes->{3.438806964796875*^9, 3.439235340203125*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Generalized Dirichlet problems", "Subsubsection"], Cell[TextData[{ "The generalized Dirichlet problem asks for a function with specified \ boundary values on some surface and with specified Laplacian inside that \ surface. Thus if the Laplacian is required to be 0, then this is the same as \ the usual Dirichlet problem. The generalized Dirichlet problem with specified \ Laplacian equal to ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " is solved by adding the option ", Cell[BoxData[ StyleBox[ RowBox[{"\[CapitalDelta]", "\[Rule]", " ", "p"}], "InlineInput"]]], " to the ", StyleBox["Dirichlet", "InlineInput"], " function; here ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " must be a polynomial. For example, we can find the function that equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "3"], SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], TraditionalForm]]], "on the unit sphere in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], " and whose Laplacian equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "2"], "2"], SubscriptBox["x", "3"]}], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.439175402171875*^9, 3.43917540346875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "3"], SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], ",", " ", "x", ",", " ", RowBox[{"\[CapitalDelta]", "\[Rule]", " ", RowBox[{ SuperscriptBox[ SubscriptBox["x", "2"], "2"], SubscriptBox["x", "3"]}]}]}], "]"}]], "Input", CellChangeTimes->{{3.43917541034375*^9, 3.439175412109375*^9}}], Cell[BoxData[ RowBox[{ FractionBox["1", "1260"], RowBox[{"(", RowBox[{ RowBox[{"108", " ", SubscriptBox["x", "1"]}], "-", RowBox[{"168", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubscriptBox["x", "1"]}], "+", RowBox[{"60", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubscriptBox["x", "1"]}], "+", RowBox[{"140", " ", SubsuperscriptBox["x", "1", "3"]}], "-", RowBox[{"140", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "3"]}], "+", RowBox[{"420", " ", SubscriptBox["x", "1"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"420", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubscriptBox["x", "1"], " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"1260", " ", SubsuperscriptBox["x", "1", "3"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"9", " ", SubscriptBox["x", "3"]}], "+", RowBox[{"14", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubscriptBox["x", "3"]}], "-", RowBox[{"5", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SubscriptBox["x", "3"]}], "-", RowBox[{"70", " ", SubsuperscriptBox["x", "2", "2"], " ", SubscriptBox["x", "3"]}], "+", RowBox[{"70", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "2", "2"], " ", SubscriptBox["x", "3"]}]}], ")"}]}]], "Output", CellChangeTimes->{ 3.438806970015625*^9, {3.4391754245625*^9, 3.43917543909375*^9}, 3.43923534453125*^9}] }, Closed]], Cell[TextData[{ "We can check that output above correctly solves the generalized Dirichlet \ problem by verifying that the Laplacian of the last output indeed equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "2"], "2"], SubscriptBox["x", "3"]}], TraditionalForm]]], " and that the last output restricted to the unit sphere indeed equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "3"], SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.439175450296875*^9, 3.43917545715625*^9}}], Cell[BoxData[ RowBox[{ SubsuperscriptBox["x", "2", "2"], " ", SubscriptBox["x", "3"]}]], "Output", CellChangeTimes->{3.438806972765625*^9, 3.439175457890625*^9, 3.439235347296875*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"%%", " ", "/.", " ", RowBox[{ RowBox[{"\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "\[Rule]", "1"}]}]], "Input"], Cell[BoxData[ RowBox[{ SubsuperscriptBox["x", "1", "3"], " ", SubsuperscriptBox["x", "2", "2"]}]], "Output", CellChangeTimes->{3.438806974921875*^9, 3.439175461828125*^9, 3.43923535553125*^9}] }, Closed]], Cell[TextData[{ "The ", StyleBox["\[CapitalDelta]", "InlineInput"], " option can be used combined with the ", StyleBox["Region", "InlineInput"], " option to solve the generalized Dirichlet problem on any region for which \ this package can solve the usual Dirichlet problem. The order of the options \ does not matter. Here we find the function on ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], " whose Laplacian equals ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["x", "2"], "2"], TraditionalForm]]], "and that equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], TraditionalForm]]], "on the ellipsoid ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox[ SubscriptBox["x", "1"], "2"]}], "+", RowBox[{"3", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], "+", RowBox[{"4", SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], "-", " ", "1"}], " ", "=", " ", "0"}], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.439175474625*^9, 3.439175475796875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], ",", "x", ",", RowBox[{"\[CapitalDelta]", "\[Rule]", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], ",", RowBox[{"Region", "\[Rule]", RowBox[{"Quadratic", "[", RowBox[{ RowBox[{"{", RowBox[{"2", ",", "3", ",", "4"}], "}"}], ",", RowBox[{"-", "1"}]}], "]"}]}]}], "]"}]], "Input", CellChangeTimes->{{3.439175485265625*^9, 3.4391754865*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ SubsuperscriptBox["x", "1", "4"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{ FractionBox["1", "157239173916"], RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "+", RowBox[{"2", " ", SubsuperscriptBox["x", "1", "2"]}], "+", RowBox[{"3", " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"4", " ", SubsuperscriptBox["x", "3", "2"]}]}], ")"}], " ", RowBox[{"(", RowBox[{"82719889", "-", RowBox[{"2354066130", " ", SubsuperscriptBox["x", "1", "2"]}], "-", RowBox[{"3207153660", " ", SubsuperscriptBox["x", "1", "4"]}], "+", RowBox[{"3779085195", " ", SubsuperscriptBox["x", "2", "2"]}], "+", RowBox[{"3780731736", " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "2", "2"]}], "-", RowBox[{"270421740", " ", SubsuperscriptBox["x", "2", "4"]}], "-", RowBox[{"680540064", " ", SubsuperscriptBox["x", "3", "2"]}], "-", RowBox[{"23067975984", " ", SubsuperscriptBox["x", "1", "2"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{"816437844", " ", SubsuperscriptBox["x", "2", "2"], " ", SubsuperscriptBox["x", "3", "2"]}], "+", RowBox[{"1369325424", " ", SubsuperscriptBox["x", "3", "4"]}]}], ")"}]}]}]}]], "Output", CellChangeTimes->{3.43880697771875*^9, 3.439175498796875*^9, 3.439235366828125*^9}] }, Closed]], Cell[TextData[{ "The form of the last output shows, without calculation, that it obviously \ equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], TraditionalForm]]], "on the ellipsoid ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox[ SubscriptBox["x", "1"], "2"]}], "+", RowBox[{"3", SuperscriptBox[ SubscriptBox["x", "2"], "2"]}], "+", RowBox[{"4", SuperscriptBox[ SubscriptBox["x", "3"], "2"]}], "-", " ", "1"}], " ", "=", " ", "0"}], TraditionalForm]]], ". Thus to verify that we have correctly solved this generalized Dirichlet \ problem, we only need to check that the Laplacian equals ", Cell[BoxData[ FormBox[ SuperscriptBox[ SubscriptBox["x", "2"], "2"], TraditionalForm]]], ":" }], "Text", CellChangeTimes->{{3.439175516359375*^9, 3.439175526421875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", "x"], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.439175554140625*^9, 3.439175560609375*^9}}], Cell[BoxData[ SubsuperscriptBox["x", "2", "2"]], "Output", CellChangeTimes->{3.438806982484375*^9, 3.43917556134375*^9, 3.439235373359375*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Dirichlet problems with explicit coordinates", "Subsubsection"], Cell[TextData[{ "So far we have worked with symbolic vectors, which we have usually called ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " or ", Cell[BoxData[ FormBox["z", TraditionalForm]]], ". Vectors can also be described by giving explicit coordinates in the form \ of a list. Both formats (symbolic vectors and explicit lists) work with all \ functions in this package. For example, suppose we are working in ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], " and want to use ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], ", and ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " instead of ", Cell[BoxData[ FormBox[ SubscriptBox["x", "1"], TraditionalForm]]], ", ", Cell[BoxData[ FormBox[ SubscriptBox["x", "2"], TraditionalForm]]], ", and ", Cell[BoxData[ FormBox[ SubscriptBox["x", "3"], TraditionalForm]]], ". Here is how to find the Poisson integral of ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ SuperscriptBox["x", "3"], "y"}], TraditionalForm], SuperscriptBox["z", "2"]}], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox["x", "3"], "y", " ", SuperscriptBox["z", "2"]}], ",", " ", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ FractionBox["1", "231"], " ", RowBox[{"(", RowBox[{ RowBox[{"11", " ", "x", " ", "y"}], "+", RowBox[{"3", " ", SuperscriptBox["x", "3"], " ", "y"}], "-", RowBox[{"14", " ", SuperscriptBox["x", "5"], " ", "y"}], "-", RowBox[{"18", " ", "x", " ", SuperscriptBox["y", "3"]}], "-", RowBox[{"7", " ", SuperscriptBox["x", "3"], " ", SuperscriptBox["y", "3"]}], "+", RowBox[{"7", " ", "x", " ", SuperscriptBox["y", "5"]}], "+", RowBox[{"45", " ", "x", " ", "y", " ", SuperscriptBox["z", "2"]}], "+", RowBox[{"161", " ", SuperscriptBox["x", "3"], " ", "y", " ", SuperscriptBox["z", "2"]}], "-", RowBox[{"49", " ", "x", " ", SuperscriptBox["y", "3"], " ", SuperscriptBox["z", "2"]}], "-", RowBox[{"56", " ", "x", " ", "y", " ", SuperscriptBox["z", "4"]}]}], ")"}]}]], "Output", CellChangeTimes->{3.438806989703125*^9, 3.439175688765625*^9, 3.439235376296875*^9}] }, Closed]], Cell["\<\ To check that the last output is correct, first we verify that it is harmonic \ by checking that its Laplacian equals 0:\ \>", "Text", CellChangeTimes->{{3.43917570103125*^9, 3.43917570146875*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.43917572153125*^9, 3.439175738171875*^9}}], Cell[BoxData["0"], "Output", CellChangeTimes->{3.4388069925*^9, 3.43917573890625*^9, 3.439235378421875*^9} ] }, Closed]], Cell[TextData[{ "Now we check that the difference between the claimed solution to the \ Dirichlet problem equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["x", "3"], SuperscriptBox["yz", "2"]}], TraditionalForm]]], " on the unit sphere by verifying that the difference between these two \ functions, when restricted to the unit sphere, equals 0:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{ SuperscriptBox["x", "3"], "y", " ", SuperscriptBox["z", "2"]}], "-", " ", "%%"}], " ", "/.", " ", RowBox[{"z", "\[Rule]", SqrtBox[ RowBox[{"1", "-", SuperscriptBox["x", "2"], "-", SuperscriptBox["y", "2"]}]]}]}]], "Input"], Cell[BoxData["0"], "Output", CellChangeTimes->{3.438806998109375*^9, 3.439235383859375*^9}] }, Closed]], Cell[TextData[{ "Here is another example using explicit coordinates. In this example, we \ find the function on ", Cell[BoxData[ FormBox[ SuperscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "3"], TraditionalForm]]], " whose Laplacian equals ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["y", "2"], "z"}], TraditionalForm]]], " and that equals ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ SuperscriptBox["x", "3"], "y"}], TraditionalForm], SuperscriptBox["z", "2"]}], TraditionalForm]]], " on the ellipsoind ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox["x", "2"]}], "+", RowBox[{"3", SuperscriptBox["y", "2"]}], "+", RowBox[{"4", SuperscriptBox["z", "2"]}], "-", " ", "1"}], " ", "=", " ", "0"}], TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Dirichlet", "[", RowBox[{ RowBox[{ SuperscriptBox["x", "3"], "y", " ", SuperscriptBox["z", "2"]}], ",", " ", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}], ",", " ", RowBox[{"Region", "\[Rule]", RowBox[{"Quadratic", "[", RowBox[{ RowBox[{"{", RowBox[{"2", ",", "3", ",", "4"}], "}"}], ",", RowBox[{"-", "1"}]}], "]"}]}], ",", " ", RowBox[{"\[CapitalDelta]", "\[Rule]", RowBox[{ SuperscriptBox["y", "2"], SuperscriptBox["z", "3"]}]}]}], "]"}]], "Input", CellChangeTimes->{{3.439175832515625*^9, 3.439175833921875*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox["x", "3"], " ", "y", " ", SuperscriptBox["z", "2"]}], "+", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"-", "1"}], "+", RowBox[{"2", " ", SuperscriptBox["x", "2"]}], "+", RowBox[{"3", " ", SuperscriptBox["y", "2"]}], "+", RowBox[{"4", " ", SuperscriptBox["z", "2"]}]}], ")"}], " ", RowBox[{"(", RowBox[{ RowBox[{"-", FractionBox[ RowBox[{"961", " ", "x", " ", "y"}], "150917"]}], "-", FractionBox[ RowBox[{"223", " ", SuperscriptBox["x", "3"], " ", "y"}], "7943"], "+", FractionBox[ RowBox[{"93", " ", "x", " ", SuperscriptBox["y", "3"]}], "7943"], "+", FractionBox[ RowBox[{"6128862523", " ", "z"}], "154091469268302"], "-", FractionBox[ RowBox[{"424717558", " ", SuperscriptBox["x", "2"], " ", "z"}], "4532102037303"], "+", FractionBox[ RowBox[{"44102", " ", SuperscriptBox["x", "4"], " ", "z"}], "734657487"], "+", FractionBox[ RowBox[{"3416529117", " ", SuperscriptBox["y", "2"], " ", "z"}], "6042802716404"], "-", FractionBox[ RowBox[{"448445", " ", SuperscriptBox["x", "2"], " ", SuperscriptBox["y", "2"], " ", "z"}], "489771658"], "-", FractionBox[ RowBox[{"1320573", " ", SuperscriptBox["y", "4"], " ", "z"}], "979543316"], "-", FractionBox[ RowBox[{"571", " ", "x", " ", "y", " ", SuperscriptBox["z", "2"]}], "7943"], "+", FractionBox[ RowBox[{"1235669309", " ", SuperscriptBox["z", "3"]}], "18128408149212"], "-", FractionBox[ RowBox[{"213085", " ", SuperscriptBox["x", "2"], " ", SuperscriptBox["z", "3"]}], "1469314974"], "+", FractionBox[ RowBox[{"8956053", " ", SuperscriptBox["y", "2"], " ", SuperscriptBox["z", "3"]}], "979543316"], "-", FractionBox[ RowBox[{"297101", " ", SuperscriptBox["z", "5"]}], "734657487"]}], ")"}]}]}]], "Output", CellChangeTimes->{3.438807008859375*^9, 3.43917584003125*^9, 3.439235392015625*^9}] }, Closed]], Cell[TextData[{ "Obviously the last output equals ", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ SuperscriptBox["x", "3"], "y"}], TraditionalForm], SuperscriptBox["z", "2"]}], TraditionalForm]]], " on the ellipsoind ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"2", SuperscriptBox["x", "2"]}], "+", RowBox[{"3", SuperscriptBox["y", "2"]}], "+", RowBox[{"4", SuperscriptBox["z", "2"]}], "-", " ", "1"}], " ", "=", " ", "0"}], TraditionalForm]]], ". We now check that its Laplacian is the desired function:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ SubscriptBox["\[CapitalDelta]", RowBox[{"{", RowBox[{"x", ",", "y", ",", "z"}], "}"}]], "[", "%", "]"}]], "Input", CellChangeTimes->{{3.439175845546875*^9, 3.439175860828125*^9}}], Cell[BoxData[ RowBox[{ SuperscriptBox["y", "2"], " ", SuperscriptBox["z", "3"]}]], "Output", CellChangeTimes->{3.43880701209375*^9, 3.4391758678125*^9, 3.439235394953125*^9}] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["HarmonicDecomposition", "Subsection", CellTags->"HarmonicDecomposition"], Cell[TextData[{ StyleBox["HarmonicDecomposition[p, x]", "InlineInput"], " gives the decomposition of ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " into a sum of harmonic polynomials (on the Euclidean space defined by ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ") times powers of ||", Cell[BoxData[ FormBox["x", TraditionalForm]]], "||. Here ", Cell[BoxData[ FormBox["p", TraditionalForm]]], " must be a polynomial function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438443779859375*^9, 3.43880701815625*^9, 3.439235397515625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"HarmonicDecomposition", "[", RowBox[{ SuperscriptBox[ SubscriptBox["x", "1"], "4"], ",", "x"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ FractionBox[ RowBox[{"3", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"]}], RowBox[{"8", "+", RowBox[{"6", " ", "n"}], "+", SuperscriptBox["n", "2"]}]], "-", FractionBox[ RowBox[{"6", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SubsuperscriptBox["x", "1", "2"]}], RowBox[{"4", "+", "n"}]], "+", SubsuperscriptBox["x", "1", "4"]}], ",", "0"}], "}"}], ",", RowBox[{"{", RowBox[{ FractionBox[ RowBox[{"6", " ", RowBox[{"(", RowBox[{ RowBox[{"-", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], "+", RowBox[{"n", " ", SubsuperscriptBox["x", "1", "2"]}]}], ")"}]}], RowBox[{"n", " ", RowBox[{"(", RowBox[{"4", "+", "n"}], ")"}]}]], ",", "2"}], "}"}], ",", RowBox[{"{", RowBox[{ FractionBox["3", RowBox[{"n", " ", RowBox[{"(", RowBox[{"2", "+", "n"}], ")"}]}]], ",", "4"}], "}"}]}], "}"}]], "Output", CellChangeTimes->{3.4384437831875*^9, 3.43880702003125*^9, 3.43923539959375*^9}] }, Closed]], Cell[TextData[{ "The output given by ", StyleBox["HarmonicDecomposition[p, x]", "InlineInput"], " consists of a list of pairs. 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Here each of ", StyleBox["x", "InlineInput"], " and ", StyleBox["c", "InlineInput"], " can be either an expression representing a vector or a list of coordinates \ for a vector:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Reflection", "[", RowBox[{ RowBox[{"{", RowBox[{"2", ",", "4", ",", " ", "5"}], "}"}], ",", " ", RowBox[{"Sphere", "[", RowBox[{"7", ",", " ", RowBox[{"{", RowBox[{"3", ",", " ", "1", ",", " ", "6"}], "}"}]}], "]"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{"-", FractionBox["16", "11"]}], ",", FractionBox["158", "11"], ",", FractionBox["17", "11"]}], "}"}]], "Output", CellChangeTimes->{3.43880717221875*^9, 3.439235541265625*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Reflection", "[", RowBox[{ RowBox[{ RowBox[{"2", "x"}], "+", "y"}], ",", " ", RowBox[{"Sphere", "[", RowBox[{"3", ",", " ", "z"}], "]"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"z", "+", FractionBox[ RowBox[{"9", " ", RowBox[{"(", RowBox[{ RowBox[{"2", " ", "x"}], "+", "y", "-", "z"}], ")"}]}], SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"2", " ", "x"}], "+", "y", "-", "z"}], "\[RightDoubleBracketingBar]"}], "2"]]}]], "Output", CellChangeTimes->{3.438807174328125*^9, 3.43923554821875*^9}] }, Closed]], Cell[TextData[{ StyleBox["Reflection[x, Sphere[r, c]]", "InlineInput"], " is computed using equation 4.13 in [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"], Cell[TextData[{ StyleBox["Reflection[x, Hyperplane[b, t]]", "InlineInput"], " is the reflection of a vector ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " with respect to the hyperplane consisting of those vectors whose inner \ product with ", Cell[BoxData[ FormBox["b", TraditionalForm]]], " equals ", Cell[BoxData[ FormBox["t", TraditionalForm]]], ". 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Here each of ", StyleBox["x", "InlineInput"], " and ", StyleBox["b", "InlineInput"], " can be either an expression representing a vector or a list of coordinates \ for a vector:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Reflection", "[", RowBox[{"x", " ", ",", " ", RowBox[{"Hyperplane", "[", RowBox[{ RowBox[{"{", RowBox[{"1", ",", " ", "4", ",", "5"}], "}"}], ",", " ", "7"}], "]"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{ FractionBox["1", "21"], " ", RowBox[{"(", RowBox[{"7", "+", RowBox[{"20", " ", SubscriptBox["x", "1"]}], "-", RowBox[{"4", " ", SubscriptBox["x", "2"]}], "-", RowBox[{"5", " ", SubscriptBox["x", "3"]}]}], ")"}]}], ",", RowBox[{ FractionBox["1", "21"], " ", RowBox[{"(", RowBox[{"28", "-", RowBox[{"4", " ", SubscriptBox["x", "1"]}], "+", RowBox[{"5", " ", SubscriptBox["x", "2"]}], "-", RowBox[{"20", " ", SubscriptBox["x", "3"]}]}], ")"}]}], ",", RowBox[{ FractionBox["1", "21"], " ", RowBox[{"(", RowBox[{"35", "-", RowBox[{"5", " ", SubscriptBox["x", "1"]}], "-", RowBox[{"20", " ", SubscriptBox["x", "2"]}], "-", RowBox[{"4", " ", SubscriptBox["x", "3"]}]}], ")"}]}]}], "}"}]], "Output", CellChangeTimes->{3.43880718253125*^9, 3.439235551984375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Reflection", "[", RowBox[{ RowBox[{"{", RowBox[{ SubscriptBox["x", "1"], ",", SubscriptBox["x", "2"]}], "}"}], ",", " ", RowBox[{"Hyperplane", "[", RowBox[{ RowBox[{"{", RowBox[{"4", ",", "5"}], "}"}], ",", " ", "7"}], "]"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{ RowBox[{ FractionBox["1", "41"], " ", RowBox[{"(", RowBox[{"56", "+", RowBox[{"9", " ", SubscriptBox["x", "1"]}], "-", RowBox[{"40", " ", SubscriptBox["x", "2"]}]}], ")"}]}], ",", RowBox[{ FractionBox["1", "41"], " ", RowBox[{"(", RowBox[{"70", "-", RowBox[{"40", " ", SubscriptBox["x", "1"]}], "-", RowBox[{"9", " ", SubscriptBox["x", "2"]}]}], ")"}]}]}], "}"}]], "Output", CellChangeTimes->{3.4388071849375*^9, 3.439235554203125*^9}] }, Closed]], Cell[TextData[{ StyleBox["Reflection[x, Hyperplane[b, t]]", "InlineInput"], " is computed using Exercise 9 on page 71 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Kelvin", "Subsection", CellTags->"Kelvin"], Cell[TextData[{ StyleBox["Kelvin[u, x]", "InlineInput"], " is the Kelvin transform of ", Cell[BoxData[ FormBox["u", TraditionalForm]]], ", thought of as a function of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438807213109375*^9, 3.439176795484375*^9, 3.4392355566875*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Kelvin", "[", RowBox[{ RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"x", ".", "y"}], ")"}], "2"], "+", RowBox[{ SuperscriptBox[ SubscriptBox["x", "3"], "5"], " ", RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}]}]}], ",", "x"}], "]"}]], "Input", CellChangeTimes->{3.439176808375*^9, 3.439176842484375*^9}], Cell[BoxData[ RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], RowBox[{ RowBox[{"-", "9"}], "-", "n"}]], " ", RowBox[{"(", RowBox[{ RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"x", ".", "y"}], ")"}], "2"], " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "7"]}], "+", SubsuperscriptBox["x", "3", "5"]}], ")"}]}]], "Output", CellChangeTimes->{3.43917684346875*^9, 3.43923555875*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Phi", "Subsection", CellTags->"Phi"], Cell[TextData[{ StyleBox["\[CapitalPhi][z]", "InlineInput"], " is is the modified inversion introduced on page 154 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]. The modified inversion \[CapitalPhi] is defined by" }], "Text"], Cell[TextData[{ "\[CapitalPhi](z) = 2", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ FractionBox[ RowBox[{"z", "-", StyleBox["S", FontWeight->"Bold", FontSlant->"Plain"]}], SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", "-", StyleBox["S", FontWeight->"Bold", FontSlant->"Plain"]}], "\[RightDoubleBracketingBar]"}], "2"]], FontSize->16], " ", "+", " "}], TraditionalForm]]], StyleBox["S", FontWeight->"Bold"], "," }], "Text", TextAlignment->Center], Cell[TextData[{ "where ", StyleBox["S", FontWeight->"Bold"], " is the south pole (0, 0, ..., 0, -1). The modified inversion \[CapitalPhi] \ is useful for translating questions about harmonic functions on balls to \ questions about harmonic functions on half-spaces, and vice versa." }], "Text"], Cell[TextData[{ "Here is a proof of that \[CapitalPhi](\[CapitalPhi](", Cell[BoxData[ FormBox["z", TraditionalForm]]], ")) = ", Cell[BoxData[ FormBox["z", TraditionalForm]]], ", as claimed in Proposition 7.18 (a) of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalPhi]", "[", "z", "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(S\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(Dimension[z]\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438807221703125*^9, 3.439235563046875*^9}], Cell[BoxData[ RowBox[{"S", "+", FractionBox[ RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], ")"}]}], SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], "2"]]}]], "Output", CellChangeTimes->{3.43880722171875*^9, 3.43923556309375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalPhi]", "[", "%", "]"}]], "Input"], Cell[BoxData["z"], "Output", CellChangeTimes->{3.43880722440625*^9, 3.43923556715625*^9}] }, Closed]], Cell[TextData[{ "Another, and often more convenient, form of ", StyleBox["\[CapitalPhi]", "InlineInput"], " is given by ", StyleBox["\[CapitalPhi][x, y]", "InlineInput"], ". Here ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " is a vector, ", Cell[BoxData[ FormBox["y", TraditionalForm]]], " is a real number, and thus (", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], ") is a vector in a Euclidean space whose dimension is one more than the \ dimension of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], "." }], "Text"], Cell[TextData[{ "Here is a proof of the assertion of Exercise 6 in Chapter 7 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "], which asserts that" }], "Text"], Cell[TextData[{ "1 - ", Cell[BoxData[ FormBox[ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"\[CapitalPhi]", "(", RowBox[{"x", ",", " ", "y"}], ")"}], "\[RightDoubleBracketingBar]"}], "2"], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ StyleBox[ FractionBox[ RowBox[{"4", "y"}], RowBox[{ SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], "+", " ", SuperscriptBox[ RowBox[{"(", RowBox[{"y", " ", "+", " ", "1"}], ")"}], "2"]}]], FontSize->16], TraditionalForm]]], ":" }], "Text", TextAlignment->Center], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[CapitalPhi]", "[", RowBox[{"x", ",", "y"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{"{", RowBox[{ FractionBox[ RowBox[{"2", " ", "x"}], RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", "y"}], ")"}], "2"], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]], ",", RowBox[{ RowBox[{"-", "1"}], "+", FractionBox[ RowBox[{"2", " ", RowBox[{"(", RowBox[{"1", "+", "y"}], ")"}]}], RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", "y"}], ")"}], "2"], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]}]}], "}"}]], "Output", CellChangeTimes->{3.438807236921875*^9, 3.439235569296875*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"1", "-", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", "%", "\[RightDoubleBracketingBar]"}], "2"]}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{"4", " ", "y"}], RowBox[{"1", "+", RowBox[{"2", " ", "y"}], "+", SuperscriptBox["y", "2"], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]], "Output", CellChangeTimes->{3.4388072394375*^9, 3.439235571390625*^9}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["KelvinM", "Subsection", CellTags->"KelvinM"], Cell[TextData[{ StyleBox["KelvinM[u, z]", "InlineInput"], " is the modified Kelvin transform defined to be equal to" }], "Text"], Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["2", RowBox[{ RowBox[{"(", RowBox[{"n", "-", "2"}], ")"}], "/", "2"}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{"z", " ", "-", " ", StyleBox["S", FontWeight->"Bold", FontSlant->"Plain"]}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], " ", RowBox[{ RowBox[{"u", "(", RowBox[{"\[CapitalPhi]", "(", "z", ")"}], ")"}], "."}]}], TraditionalForm]]]], "Text", TextAlignment->Center], Cell[TextData[{ "This modified Kelvin transform is defined on page 155 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]. Here is a proof of the assertion made there that the modified Kelvin \ transform is its own inverse:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"z", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(z\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.43880724828125*^9, 3.439235579359375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"KelvinM", "[", RowBox[{ RowBox[{"u", "[", "z", "]"}], ",", "z"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ SuperscriptBox["2", RowBox[{ RowBox[{"-", "1"}], "+", FractionBox["n", "2"]}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"2", "-", "n"}]], " ", RowBox[{"u", "[", RowBox[{"S", "+", FractionBox[ RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], ")"}]}], SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], "2"]]}], "]"}]}]], "Output", CellChangeTimes->{3.438807251796875*^9, 3.4392355831875*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"KelvinM", "[", RowBox[{"%", ",", " ", "z"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ SuperscriptBox[ RowBox[{"(", FractionBox["1", RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}]], ")"}], RowBox[{"-", "n"}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]], " ", RowBox[{"u", "[", "z", "]"}]}]], "Output", CellChangeTimes->{3.438807254109375*^9, 3.439235585484375*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"PowerExpand", "[", "%", "]"}]], "Input"], Cell[BoxData[ RowBox[{"u", "[", "z", "]"}]], "Output", CellChangeTimes->{3.438807256203125*^9, 3.43923558753125*^9}] }, Closed]], Cell[TextData[{ "Another form of ", StyleBox["KelvinM", "InlineInput"], " is given by ", StyleBox["KelvinM[u, x, y]", "InlineInput"], ". Here ", Cell[BoxData[ FormBox["x", TraditionalForm]]], " is a vector, ", Cell[BoxData[ FormBox["y", TraditionalForm]]], " is a real variable, and ", Cell[BoxData[ FormBox["u", TraditionalForm]]], " is a function of (", Cell[BoxData[ FormBox["x", TraditionalForm]]], ", ", Cell[BoxData[ FormBox["y", TraditionalForm]]], "), which is a vector in a Euclidean space whose dimension is one more than \ the dimension of ", Cell[BoxData[ FormBox["x", TraditionalForm]]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", RowBox[{"n", "-", "1"}]}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(\\(\\(-1\\)\\) \ + n\\), \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional \ real Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.438807258796875*^9, 3.439235590203125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"KelvinM", "[", RowBox[{ RowBox[{"x", "[", "4", "]"}], ",", "x", ",", "y"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ SuperscriptBox["2", RowBox[{"n", "/", "2"}]], " ", SuperscriptBox[ RowBox[{"(", RowBox[{ SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", "y"}], ")"}], "2"], "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], RowBox[{ RowBox[{"-", "n"}], "/", "2"}]], " ", SubscriptBox["x", "4"]}]], "Output", CellChangeTimes->{3.438807262078125*^9, 3.439235595328125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"KelvinM", "[", RowBox[{ RowBox[{"z", "[", "4", "]"}], ",", "z"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ SuperscriptBox["2", RowBox[{ RowBox[{"-", "1"}], "+", FractionBox["n", "2"]}]], " ", SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], RowBox[{"-", "n"}]], " ", RowBox[{"(", RowBox[{ RowBox[{ RowBox[{"-", "2"}], " ", SubscriptBox["S", "4"]}], "+", RowBox[{ SuperscriptBox[ RowBox[{"\[LeftDoubleBracketingBar]", RowBox[{ RowBox[{"-", "S"}], "+", "z"}], "\[RightDoubleBracketingBar]"}], "2"], " ", SubscriptBox["S", "4"]}], "+", RowBox[{"2", " ", SubscriptBox["z", "4"]}]}], ")"}]}]], "Output", CellChangeTimes->{3.438807264375*^9, 3.439235599265625*^9}] }, Closed]], Cell[TextData[{ "If we had used the input ", StyleBox["KelvinM[ z[4], z ]", "InlineInput"], ", the result would have included terms of the form ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["S", FontWeight->"Bold", FontSlant->"Plain"], "4"], TraditionalForm]]], ", which denotes the fourth coordinate of the south pole ", StyleBox["S", FontWeight->"Bold"], ". Of course ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["S", FontWeight->"Bold", FontSlant->"Plain"], "4"], TraditionalForm]]], " equals either 0 or -1, but since the dimension of ", Cell[BoxData[ FormBox["z", TraditionalForm]]], " has been set to equal the symbol ", Cell[BoxData[ FormBox["n", TraditionalForm]]], ", the computer has no way of determining whether or not ", Cell[BoxData[ FormBox["n", TraditionalForm]]], " equals 4. With the input ", StyleBox["KelvinM[ z[4], z ]", "InlineInput"], " there is no ambiguity; ", Cell[BoxData[ FormBox[ SubscriptBox["x", "4"], TraditionalForm]]], " cannot be the last coordinate, because the last coordinate is denoted by \ ", Cell[BoxData[ FormBox["y", TraditionalForm]]], "." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Kernels", "Section", CellTags->"Kernels"], Cell[CellGroupData[{ Cell["PoissonKernel", "Subsection", CellTags->"PoissonKernel"], Cell[TextData[{ StyleBox["PoissonKernel[x, y]", "InlineInput"], " is the extended Poisson kernel ", Cell[BoxData[ FormBox["P", TraditionalForm]]], "(", Cell[BoxData[ FormBox["x", TraditionalForm]]], ",", Cell[BoxData[ FormBox["y", TraditionalForm]]], ") for the unit ball as defined by Equation 6.21 of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "n"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(n\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.43880729778125*^9, 3.43923560903125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"PoissonKernel", "[", RowBox[{"x", ",", "y"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"(", RowBox[{"1", "-", RowBox[{ SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "y", "\[RightDoubleBracketingBar]"}], "2"]}]}], ")"}], " ", SuperscriptBox[ RowBox[{"(", RowBox[{"1", "-", RowBox[{"2", " ", RowBox[{"x", ".", "y"}]}], "+", RowBox[{ 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The user must first set the dimension of the Euclidean space to a \ positive integer value:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"SetDimension", "[", RowBox[{"x", ",", "5"}], "]"}]], "Input"], Cell[BoxData[ RowBox[{ RowBox[{"SetDimension", "::", "\<\"Vector\"\>"}], RowBox[{ ":", " "}], "\<\"\[NoBreak]\\!\\(\\*StyleBox[\\!\\(x\\), \\\"MT\\\"]\\)\ \[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\" will be considered to be a vector in \ \\\\\\\"\\\", \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\!\\(5\\), \ \\\"MT\\\"]\\)\[NoBreak]\\!\\(\\*StyleBox[\\\"\\\\\\\"-dimensional real \ Euclidean space.\\\\\\\"\\\", \\\"MT\\\"]\\)\"\>"}]], "Message", "MSG", CellChangeTimes->{3.43880739046875*^9, 3.439235715875*^9, 3.43923602434375*^9, 3.439237799203125*^9}] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Schwarz", "[", "x", "]"}]], "Input"], Cell[BoxData[ FractionBox[ RowBox[{"2", "+", RowBox[{"4", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], "+", RowBox[{"2", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"]}], "-", RowBox[{"2", " ", SqrtBox[ RowBox[{"1", "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]}], "-", RowBox[{"3", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"], " ", SqrtBox[ RowBox[{"1", "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]}], "+", RowBox[{"3", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "4"], " ", SqrtBox[ RowBox[{"1", "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]}], "+", RowBox[{"2", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "6"], " ", SqrtBox[ RowBox[{"1", "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}]]}]}], RowBox[{"2", " ", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "3"], " ", SuperscriptBox[ RowBox[{"(", RowBox[{"1", "+", SuperscriptBox[ RowBox[{ "\[LeftDoubleBracketingBar]", "x", "\[RightDoubleBracketingBar]"}], "2"]}], ")"}], "2"]}]]], "Output", CellChangeTimes->{3.43880742334375*^9, 3.43923575425*^9, 3.43923605565625*^9, 3.43923783025*^9}] }, Closed]], Cell[TextData[{ StyleBox["Schwarz[x]", "InlineInput"], " is computed by evaluating the Poisson integral that defines ", Cell[BoxData[ FormBox["U", TraditionalForm]]], "(\[LeftDoubleBracketingBar]x\[RightDoubleBracketingBar] ", StyleBox["N", FontWeight->"Bold"], "), iterating an appropriate formula from Appendix A of [", ButtonBox["ABR", BaseStyle->"Hyperlink", ButtonData:>"ABR"], "]." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Togetherness", "Subsection", CellTags->"TurnOff and TurnOn"], Cell[TextData[{ "When not using the HFT7.m software package, ", StyleBox["Mathematica", FontSlant->"Italic"], " might tell the user that the result of a calculation is ", Cell[BoxData[ FormBox[ RowBox[{ SuperscriptBox["t", "2"], "+", RowBox[{ RowBox[{"(", RowBox[{"1", "-", "t"}], ")"}], " ", RowBox[{"(", RowBox[{"1", " ", "+", " ", "t"}], ")"}]}]}], TraditionalForm]]], ". 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Thus \ the HFT7.m package redefines the ", StyleBox["Mathematica", FontSlant->"Italic"], " variable ", StyleBox["$Post", "InlineInput"], " so that for each output ", Cell[BoxData[ FormBox["X", TraditionalForm]]], ", ", StyleBox["Mathematica", FontSlant->"Italic"], " will display either ", Cell[BoxData[ FormBox["X", TraditionalForm]]], " or the result of applying ", StyleBox["Together", "InlineInput"], " to ", Cell[BoxData[ FormBox["X", TraditionalForm]]], " and all its subexpressions, whichever is shorter (if the user had defined ", StyleBox["$Post", "InlineInput"], " before starting the HFT7.m package, the HFT7.m version of ", StyleBox["$Post", "InlineInput"], " is composed with the user's). This procedure is invisible to the user and \ usually requires only a short amount of computer time. 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