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" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(u = {{1, 1}, {1, 0}}; GramSchmidt[u]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \(GramSchmidt[{{1, 1}, {1, 0}}]\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \("In[2]:=", FontSize->14], Cell[BoxData[ RowBox[{\(GramSchmidt::"shdw"\), \(\(:\)\(\ \)\), "\\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"General::shdw\\\"]\\)\"\>"}]], "Message", CellLabel->"From In[2]:="] }, Open ]], Cell["En este caso lo que debemos hacer es ", "Text"], Cell[BoxData[ \(Remove[GramSchmidt]\)], "Input", CellLabel->"In[3]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(GramSchmidt[u]\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \({{1\/\@2, 1\/\@2}, {1\/\@2, \(-\(1\/\@2\)\)}}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell["\", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Names["\"]\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \({"GramSchmidt", "Householder", "InnerProduct", "Normalize", "Normalized", "Projection"}\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[TextData[{ "Tambi\[EAcute]n debemos tener cuidado y no cometer errores \ tipogr\[AAcute]ficos cuando cargamos un paquete pues podemos tener problemas, \ por lo que resulta conveniente ir a la ayuda de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " y copiar y pegar en nuestro documento la funci\[OAcute]n que nos carga \ cada paquete que necesitemos.\n\nMientras no cerremos la sesi\[OAcute]n de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " los paquetes que hallamos cargado los tendremos disponibles para utilizar \ cualquiera de las funciones que est\[EAcute]n contenidas en ellos. A veces al \ cargar un paquete se cargan de forma autom\[AAcute]tica otros paquetes. 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Hallar una base ortogonal de", FontColor->RGBColor[0, 0, 1]], " S con el producto escalar del problema." }], "Subsubsection", CellDingbat->None, TextJustification->1, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(\(us = {{2, 1, 1}, {1, 0, \(-1\)}};\)\)], "Input", CellLabel->"In[14]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"es", "=", RowBox[{"GramSchmidt", "[", RowBox[{"us", ",", \(InnerProduct \[Rule] pe\), ",", StyleBox[\(Normalized \[Rule] False\), FontColor->RGBColor[1, 0, 0]]}], "]"}]}]], "Input", CellLabel->"In[15]:=", FontSize->14], Cell[BoxData[ \({{2, 1, 1}, {7\/11, \(-\(2\/11\)\), \(-\(13\/11\)\)}}\)], "Output", CellLabel->"Out[15]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(os = GramSchmidt[us, InnerProduct \[Rule] pe]\)], "Input", CellLabel->"In[16]:=", FontSize->14], Cell[BoxData[ \({{2\/\@11, 1\/\@11, 1\/\@11}, {7\/\(3\ \@22\), \(-\(\@\(2\/11\)\/3\)\), \(-\(13\/\(3\ \ \@22\)\)\)}}\)], "Output", 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FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ "Podemos calcular el producto escalar usual entre vectores de ", Cell[BoxData[ FormBox[ SuperscriptBox["\[DoubleStruckCapitalR]", StyleBox["n", FontSlant->"Plain"]], TraditionalForm]]], "de dos maneras:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({1, 2, 3, \(-1\)} . {\(-1\), 3, 5, 0}\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \(20\)], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Dot[{1, 2, 3, \(-1\)}, {\(-1\), 3, 5, 0}]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \(20\)], "Output", CellLabel->"Out[2]="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["b.- ", FontColor->RGBColor[1, 0, 0]], StyleBox["Calcular la norma del vector ", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ FormBox[ RowBox[{ OverscriptBox[ StyleBox["z", FontSlant->"Plain"], "_"], "=", " ", \((1, 0, \(-2\), 4)\)}], TraditionalForm]], 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\ Justificar la respuesta.", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]] }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 4.3 (El espacio vectorial de las funciones \ continuas sobre un intervalo cerrado y acotado con el producto escalar \ usual)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", TextJustification->1, FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ "En ", Cell[BoxData[ RowBox[{ StyleBox[" ", FontFamily->"Times"], RowBox[{ StyleBox["(", FontFamily->"Times"], StyleBox[" ", FontFamily-> "Times"], \(\[ScriptCapitalC]\_\([0, 1]\)\ , \ \ \[LeftAngleBracket]\ \ , \ \[RightAngleBracket]\), ")"}]}]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], " con el producto escalar usual, se pide: ", Cell[BoxData[ StyleBox[" ", FontColor->GrayLevel[0]]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], "\n", StyleBox["a.- ", FontColor->RGBColor[1, 0, 0]], " Hallar una base ortonormal ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["B", FontSlant->"Plain"], "ON"], TraditionalForm]]], " del subespacio vectorial S", Cell[BoxData[ \(\(\(=\)\(\[ScriptCapitalL] \(({f\_1, f\_2, f\_3})\)\)\)\)], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], ", con\n\t\t\t", Cell[BoxData[ \(\(f\_1\) \((x)\) = 1\ \ ; \ \ \(f\_2\) \((x)\) = x\ \ ; \ \ \(f\_3\) \((x)\) = x\^2\)], TextAlignment->Left, TextJustification->1, FontFamily->"Times", FontSize->14], "\t\t\t" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[TextData[{ " Recordemos que el producto escalar usual en el espacio vectorial ", Cell[BoxData[ \(\[ScriptCapitalC]\_\([0, 1]\)\)], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], " es:\n\n\t\t\t\t\t\t\t\t", Cell[BoxData[ StyleBox[\(\[LeftAngleBracket]f, g\[RightAngleBracket] = \[Integral]\_0\%1 f \((x)\) g \((x)\) dx\), FontFamily->"Times", FontWeight->"Bold"]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14] }], "Text"], Cell[BoxData[ \(pe[f_, g_] := Integrate[f\ g, {x, 0, 1}]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \(f1[x_] := 1; f2[x_] := x; f3[x_] := x^2; u = {f1[x], f2[x], f3[x]};\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[TextData[{ "Comprobar que la familia de funciones", Cell[BoxData[ \(\(\(\ \)\({f\_1, f\_2, f\_3}\)\)\)], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], " es libre no es sencillo si utilizamos exclusivamente la \ definici\[OAcute]n de sistema libre de vectores. Hemos visto que no es \ necesario comprobar que el sistema sea libre pues el algoritmo de \ Gram-Schmidt va a detectar si el sistema es ligado. Por lo tanto aplicamos \ directamente la funci\[OAcute]n", StyleBox[" GramSchmidt ", FontFamily->"Courier", FontWeight->"Bold"], "a esta familia de funciones." }], "Text"], Cell[BoxData[ \("In[3]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(w = GramSchmidt[u, InnerProduct -> pe] // Simplify\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \({1, \@3\ \((\(-1\) + 2\ x)\), \@5\ \((1 - 6\ x + 6\ x\^2)\)}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[TextData[{ "Como hemos obtenido una lista de tres funciones no nulas esto significa \ que el sistema ", Cell[BoxData[ StyleBox[\({f\_1, f\_2, f\_3}\), FontFamily->"Helvetica"]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], "es libre y que la colecci\[OAcute]n que aparece en ", StyleBox["Out[4]", FontFamily->"Courier", FontWeight->"Bold"], " es una base ortonormal del subespacio S engendrado por ", Cell[BoxData[ StyleBox[\({f\_1, f\_2, f\_3}\), 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Cell[TextData[{ StyleBox["Sugerencia:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontSize->14, FontColor->RGBColor[1, 0, 1]], "Escribir correctamente la soluci\[OAcute]n justificando la respuesta e \ indicando que entradas/salidas del segmento de ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " adjunto se han utilizado." }], "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 4.4 (Soluci\[OAcute]n aproximada de un \ sistema de ecuaciones lineales incompatible)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", TextJustification->1, FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Consideremos el sistema\n\t\t\t", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ StyleBox[\(\(x\_1\) a\&_\_1 + \(x\_2\) a\&_\_2 + \(x\_3\) a\&_\_3 = b\&_\), FontFamily->"Times"]], TextAlignment->Left, TextJustification->1, 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incompatible." }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(u = {{1, 0, 1, \(-1\)}, {0, 2, 1, 2}, {\(-1\), 1, 1, 1}}; b = {1, 1, 1, 1};\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(s = Solve[x1\ u[\([1]\)] + x2\ u[\([2]\)] + x3\ u[\([3]\)] \[Equal] b, {x1, x2, x3}]\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \({}\)], "Output", CellLabel->"Out[2]="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["b.- ", FontColor->RGBColor[1, 0, 0]], "Calcular la soluci\[OAcute]n aproximada utilizando el m\[EAcute]todo de m\ \[IAcute]nimos cuadrados.\t\t\t" }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \("In[3]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(w = GramSchmidt[u] // Simplify\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \({{1\/\@3, 0, 1\/\@3, \(-\(1\/\@3\)\)}, {1\/\@78, \@\(6\/13\), 2\ \@\(2\/39\), 5\/\@78}, {\(-\(11\/\@195\)\), \(-\(1\/\@195\)\), 8\/\@195, \(-\@\(3\/65\)\)}}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[TextData[{ StyleBox["Observaci\[OAcute]n:", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], StyleBox[" ", FontWeight->"Bold", FontColor->RGBColor[1, 0, 1]], "En general nos resulta m\[AAcute]s c\[OAcute]modo con ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " ortonormalizar una familia de vectores que pedirle que solamente nos \ palique el proceso de Gram-Schmidt sin normalizar los vectores resultantes." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(bp = Sum[Projection[b, w[\([i]\)]], {i, 3}] // Simplify\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \({14\/15, 19\/15, 13\/15, 4\/5}\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(s = Solve[x1\ u[\([1]\)] + x2\ u[\([2]\)] + x3\ u[\([3]\)] \[Equal] bp, {x1, x2, x3}]\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[BoxData[ \({{x1 \[Rule] 7\/15, x2 \[Rule] 13\/15, x3 \[Rule] \(-\(7\/15\)\)}}\)], "Output", CellLabel->"Out[6]="] }, Open ]], Cell[BoxData[ \(Clear["\", $Line]\)], "Input", CellLabel->"In[7]:=", FontSize->14] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["EJERCICIO 4.5 (Otro problema sobre el espacio \ vectorial de las funciones continuas sobre un intervlo cerrado y acotado con \ el producto escalar usual)", FontColor->RGBColor[0, 0, 1]]], "SectionFirst", FontColor->RGBColor[1, 0, 1], Background->RGBColor[0.8, 1, 0.4]], Cell[CellGroupData[{ Cell[TextData[{ "Sea el espacio vectorial eucl\[IAcute]deo ", Cell[BoxData[ RowBox[{" ", RowBox[{ StyleBox["(", FontFamily->"Helvetica"], StyleBox[" ", FontFamily->"Helvetica", FontColor->GrayLevel[0]], StyleBox[\(\[ScriptCapitalC]\_\([0, 2 \[Pi]]\)\ , 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{q,f,g} forman un sistema ortonormal\t\t\t", FontColor->RGBColor[0, 0, 1]] }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell["Podemos proceder de distintas formas:", "Text", CellDingbat->None], Cell[TextData[StyleBox["Calcular la matriz de Gram del producto escalar usual \ en la base B={q,f,g} de un cierto subespacio S y ver que se trata de la \ matriz unidad.", FontColor->GrayLevel[0]]], "Text", CellDingbat->"\[FilledSmallCircle]", FontColor->RGBColor[0, 1, 0]], Cell[TextData[{ StyleBox["Aplicar la funci\[OAcute]n de", FontColor->GrayLevel[0]], " ", StyleBox["Mathematica", FontWeight->"Bold", FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], " ", StyleBox["GramSchmidt", FontFamily->"Courier", FontWeight->"Bold", FontColor->GrayLevel[0]], StyleBox[" a la familia de funciones B y ver que obtenemos la misma familia \ de funciones.", FontColor->GrayLevel[0]] }], "Text", CellDingbat->"\[FilledSmallCircle]", FontColor->RGBColor[0, 1, 0]], Cell["\", "Text", CellDingbat->None], Cell[BoxData[ \(pe[f_, g_] := Integrate[f\ g, {x, 0, 2 Pi}]\)], "Input", CellLabel->"In[1]:=", FontSize->14], Cell[BoxData[ \(q[x_] := 1/Sqrt[2\ Pi]; f[x_] := \((1/Sqrt[\ Pi])\) Sin[x]; g[x_] := \((1/Sqrt[\ Pi])\) Cos[x]; u = {q[x], f[x], g[x]};\)], "Input", CellLabel->"In[2]:=", FontSize->14], Cell[BoxData[ \("In[3]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(w = GramSchmidt[u, InnerProduct -> pe] // Simplify\)], "Input", CellLabel->"In[4]:=", FontSize->14], Cell[BoxData[ \({1\/\@\(2\ \[Pi]\), Sin[x]\/\@\[Pi], Cos[x]\/\@\[Pi]}\)], "Output", CellLabel->"Out[4]="] }, Open ]], Cell[TextData[{ "Por lo tanto el sistema de funciones ", StyleBox["B={q,f,g}", FontColor->GrayLevel[0]], " es ortonormal.\nVamos a calcular tambi\[EAcute]n la matriz de Gram en la \ base B del subespacio engendrado por B y comprobaremos que se trata de la \ matriz unidad." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Table[pe[u[\([i]\)], u[\([j]\)]], {i, 3}, {j, 3}] \[Equal] IdentityMatrix[3]\)], "Input", CellLabel->"In[5]:=", FontSize->14], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[5]="] }, Open ]], Cell["O bien:", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(gu = Table[pe[u[\([i]\)], u[\([j]\)]], {i, 3}, {j, 3}]\)], "Input", CellLabel->"In[6]:=", FontSize->14], Cell[BoxData[ \({{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}\)], "Output", CellLabel->"Out[6]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(MatrixForm[gu]\)], "Input", CellLabel->"In[7]:=", FontSize->14], Cell[BoxData[ TagBox[ RowBox[{"(", "\[NoBreak]", GridBox[{ {"1", "0", "0"}, {"0", "1", "0"}, {"0", "0", "1"} }, RowSpacings->1, ColumnSpacings->1, ColumnAlignments->{Left}], "\[NoBreak]", ")"}], Function[ BoxForm`e$, MatrixForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[7]//MatrixForm="] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["b.- ", FontColor->RGBColor[1, 0, 0]], "Hallar la mejor aproximaci\[OAcute]n vp de la funci\[OAcute]n ", Cell[BoxData[ StyleBox[\(v \((x)\) = x\^2 - 1\), FontFamily->"Helvetica"]], TextAlignment->Left, TextJustification->1, FontFamily->"Times New Roman", FontSize->14], "en el subespacio S=\[ScriptCapitalL]({q,f,g}) y la norma de la funci\ \[OAcute]n error.\nRepresentar en una \[UAcute]nica gr\[AAcute]fica las \ funciones vp(x) y v(x)." }], "Subsubsection", CellDingbat->None, TextAlignment->AlignmentMarker, FontSize->14, FontWeight->"Bold", FontColor->RGBColor[0, 0, 1]], Cell[BoxData[ \(\(v[x_] := x^2 - 1;\)\)], "Input", CellLabel->"In[8]:=", FontSize->14], Cell[CellGroupData[{ Cell[BoxData[ \(vp = Sum[Projection[v[x], u[\([i]\)], InnerProduct \[Rule] pe], {i, 3}] // Simplify\)], "Input", CellLabel->"In[9]:=", FontSize->14], Cell[BoxData[ \(\(-1\) + \(4\ \[Pi]\^2\)\/3 + 4\ Cos[x] - 4\ \[Pi]\ Sin[x]\)], "Output",\ CellLabel->"Out[9]="] }, Open ]], Cell[TextData[{ "Podemos representar gr\[AAcute]ficamente las dos funciones con la funci\ \[OAcute]n ", StyleBox["Plot", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", FontWeight->"Bold"], "y utilizar estilos diferentes para cada una de las dos fucniones con \ objeto de identificarlas:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Plot", "[", RowBox[{ RowBox[{"{", RowBox[{ StyleBox[\(v[x]\), FontColor->RGBColor[0, 1, 0]], ",", "vp"}], "}"}], ",", \({x, 0, 2 Pi}\), ",", RowBox[{"PlotStyle", "\[Rule]", RowBox[{"{", RowBox[{ StyleBox[\({Dashing[{0.04}], Thickness[0.01]}\), FontColor->RGBColor[0, 1, 0]], ",", \(Thickness[0.005]\)}], "}"}]}]}], "]"}], ";"}]], "Input", CellLabel->"In[10]:=", FontSize->14], Cell[GraphicsData["PostScript", "\"], "Graphics", CellLabel->"From In[10]:=", ImageSize->{288, 177.938}, ImageMargins->{{35, 0}, {0, 0}}, ImageRegion->{{0, 1}, {0, 1}}, ImageCache->GraphicsData["Bitmap", "\omWI f@003ooIfMT00ol0003ofMWIomWIf@0JomWIf@03o`000?oIfMWofMWI063ofMWI00?o0000omWIfOoI 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