(* 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[ 118170, 2475] NotebookOptionsPosition[ 95564, 1983] NotebookOutlinePosition[ 112462, 2292] CellTagsIndexPosition[ 112375, 2287] WindowFrame->Normal*) (* Beginning of Notebook Content *) Notebook[{ Cell[CellGroupData[{ Cell["\<\ Tema : 8 Operadores L\[OAcute]gicos y Relacionales. Ciclos y Estructuras de Control. M\ \[EAcute]todos de resoluci\[OAcute]n de una Ecuaci\[OAcute]n No Lineal.\ \>", "Title", CellChangeTimes->{{3.42734278509206*^9, 3.4273428050291452`*^9}, 3.428126571703125*^9, {3.458980892578125*^9, 3.45898096840625*^9}, { 3.459666071703125*^9, 3.45966608446875*^9}, 3.4596661445*^9, { 3.45966645709375*^9, 3.459666512296875*^9}}, TextAlignment->Center, TextJustification->0], Cell[CellGroupData[{ Cell["Introduccion", "Section 1", CellChangeTimes->{{3.459240489828125*^9, 3.45924049259375*^9}}], Cell["\<\ En este tema se presentan los operadores relacionales (> , <, ==, ..), los \ operadores l\[OAcute]gicos (And, Or, ...), y los comandos que permiten \ implementar ciclos y estructuras de control, como If, Do, For y While. Como ejemplo de utilizaci\[OAcute]n de estos operadores y comandos se \ presentan los m\[EAcute]todos m\[AAcute]s sencillos para la \ resoluci\[OAcute]n de una ecuaci\[OAcute]n no lineal : el m\[EAcute]todo de \ la bisecci\[OAcute]n, el m\[EAcute]todo de Newton y el m\[EAcute]todo de la \ secante.\ \>", "Text", CellChangeTimes->{{3.459240504078125*^9, 3.45924068571875*^9}}] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Operadores l\[OAcute]gicos y relacionales", FontFamily->"Arial"]], "Section 1", TextAlignment->Left, TextJustification->1, Background->None], Cell["\<\ Los operadores l\[OAcute]gicos son: ! p : Contrario de p. Resulta verdadero si p es falso y al reves. p && q : ( And l\[OAcute]gico ) Es verdadero si p y q son verdaderos y falso \ en caso contrario. p \[Or]\[Or] q : (Or l\[OAcute]gico ) Es verdadero si p o q o ambos son \ verdaderos. Los operadores relacionales son : x == y ( x igual a y) x != y ( x distinto de y) x > y ; x < y; x >= y; x <= y. El resultado de una operaci\[OAcute]n l\[OAcute]gica o relacional es un valor \ l\[OAcute]gico verdadero o falso. \ \>", "Text", TextAlignment->Left, TextJustification->1, Background->None], Cell[CellGroupData[{ Cell["\<\ Ejemplo 1 \.ba. Obtener el valor l\[OAcute]gico, verdadero o falso, de las siguientes \ operaciones l\[OAcute]gicas: \t2>1\[And]\[Pi]>3 \t3>1\[And]4>2 \t1>2\[Or]\[Pi]>3 \t3+2 = 4 \t \t\ \>", "Subsection", CellChangeTimes->{{3.459239536578125*^9, 3.459239673671875*^9}, { 3.459239711390625*^9, 3.459239723984375*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"2", ">", "1"}], "&&", RowBox[{"Pi", ">", "3"}]}]], "Input", CellID->129342052], Cell[BoxData["True"], "Output", CellChangeTimes->{3.459239532828125*^9, 3.46156726703125*^9, 3.469795446546875*^9, 3.46979548921875*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"And", "[", RowBox[{ RowBox[{"3", ">", "1"}], ",", RowBox[{"4", ">", "2"}]}], "]"}]], "Input", CellChangeTimes->{{3.459239510109375*^9, 3.45923952809375*^9}}], Cell[BoxData["True"], "Output", CellChangeTimes->{3.459239529*^9, 3.461567267078125*^9, 3.469795446578125*^9, 3.469795489265625*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"1", ">", "2"}], "||", RowBox[{"Pi", ">", "3"}]}]], "Input", CellID->17505479], Cell[BoxData["True"], "Output", CellChangeTimes->{3.459239659328125*^9, 3.46156726709375*^9, 3.469795446609375*^9, 3.469795489296875*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"3", "+", "2"}], "\[Equal]", "4"}]], "Input", CellChangeTimes->{{3.45923972746875*^9, 3.459239733265625*^9}}], Cell[BoxData["False"], "Output", CellChangeTimes->{{3.459239730984375*^9, 3.459239734421875*^9}, 3.461567267125*^9, 3.46979544665625*^9, 3.469795489328125*^9}] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["El comando If", "Section 1"], Cell["\<\ Mediante este comando se evalua una expresi\[OAcute]n u otra seg\[UAcute]n \ que una determinada condici\[OAcute]n sea verdadera o falsa. Su forma es : If[condici\[OAcute]n, t, f] Si la condici\[OAcute]n es verdadera se eval\[UAcute]a la expresi\[OAcute]n \ t, y si es falsa la f.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"x", "=", "7"}], ";", RowBox[{"y", "=", "10"}], ";", RowBox[{"If", "[", RowBox[{ RowBox[{"x", "\[GreaterEqual]", "y"}], ",", RowBox[{"x", "^", "2"}], ",", RowBox[{"y", "^", "3"}]}], "]"}]}]], "Input"], Cell[BoxData["1000"], "Output", CellChangeTimes->{3.461567267140625*^9, 3.4697954466875*^9, 3.469795489359375*^9}] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Ejemplo 2 \.ba.\nDefinir, utilizando el comando If, la siguiente funci\ \[OAcute]n:\n\t", Cell[BoxData[ FormBox[ TagBox[ RowBox[{ StyleBox[ RowBox[{"f", "(", "x", ")"}], ShowAutoStyles->False], StyleBox[" ", ShowAutoStyles->False], StyleBox["=", ShowAutoStyles->False], RowBox[{ StyleBox["{", ShowAutoStyles->False], StyleBox[GridBox[{ { RowBox[{ RowBox[{ RowBox[{"-", "1"}], " ", "x"}], " ", "<", " ", "0"}]}, { RowBox[{ RowBox[{"1", " ", "x"}], "\[GreaterEqual]", "0"}]} }], ShowAutoStyles->True]}]}], #& ], TraditionalForm]]] }], "Subsection", CellChangeTimes->{{3.459239767*^9, 3.45923989659375*^9}, 3.4592399379375*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"f", "[", "x_", "]"}], "=", RowBox[{"If", "[", RowBox[{ RowBox[{"x", "<", "0"}], ",", RowBox[{"-", "1"}], ",", "1"}], "]"}]}], ";", RowBox[{"Plot", "[", RowBox[{ RowBox[{"f", "[", "x", "]"}], ",", RowBox[{"{", RowBox[{"x", ",", RowBox[{"-", "2"}], ",", "2"}], "}"}]}], "]"}]}]], "Input", CellChangeTimes->{{3.45923990465625*^9, 3.4592399583125*^9}}], Cell[BoxData[ GraphicsBox[{{}, {}, {Hue[0.67, 0.6, 0.6], LineBox[CompressedData[" 1:eJxTTMoPSmViYGAwAWIQrWcv9ur////7GcDgg31Q+9nVxvEIfvn5ltz0ff/g /CtXuF8dTPwL58seyBcxl/kD52/NPsLKf/cXnJ/QY/jhZtNPOP/HpOqO/fY/ 4PzFfPckZzz8BufP8TIsutz5Fc5vykms0PT8AudXM68qL3r7Cc5PbDivcm3F Rzh/6t6nM+ZkfYDzd93X4Fmj9A7OVxZwcAqZ9xrO36TVkfXo8Qs4//7RjQpm Zs/g/Kc6U432b3sM5wvWrTTTY3wI5+uYvddZ+vkOnN/s8vbCzorrcL5oyLoJ keKXEPZ9O8p/QuUUnL/09tPely374PzYQJktt5OXwPk7uaqvHipdbA/jmz5i lGuM2A/nK3/Y/dPq0Ek433PxzbN+EZfgfFdfIeavW6/D+f3x1xsMtt+B84/7 yfIGCD2E898qitZdmvUYzi8t0U7sF3sG51+/IxCUduIFnJ9QafOc3ec1nH9Y 27zrMOs7OF+Nc8+LJR4f4HzZPdmurB0f4XyGX0wrix59gvO/B+6NO6r6Bc6/ bRIvrdryFc4vCBcQVLrzDc6XnPIkgkX2B5y/LnrB8s8VP+H8cNt6jxN7f8H5 G+Nylaz+/Ibzf3A5M+8M/QvnB120NpOd8Q/Od+DQnMDn8x/Oh+YXOB8AAVob og== "]]}}, AspectRatio->NCache[GoldenRatio^(-1), 0.6180339887498948], Axes->True, AxesOrigin->{0, 0}, PlotRange->{{-2, 2}, {0., 2.}}, PlotRangeClipping->True, PlotRangePadding->{ Scaled[0.02], Scaled[0.02]}]], "Output", CellChangeTimes->{3.459239959203125*^9, 3.459239999453125*^9, 3.4615672671875*^9, 3.469795446734375*^9, 3.469795489421875*^9}] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Ciclos y estructuras de Control ", "Section 1", Background->None], Cell[TextData[{ "La ejecuci\[OAcute]n de un programas de ", StyleBox["Mathematica", FontSlant->"Italic"], " implica la evaluaci\[OAcute]n de una sucesi\[OAcute]n de expresiones de ", StyleBox["Mathematica", FontSlant->"Italic"], ". En programas sencillos las expresiones a evaluar se suelen encontrar \ separadas por ; y se eval\[UAcute]an una tras otra. A menudo, sin embargo, \ puede ser necesario evaluar un mismo grupo de expresiones varias veces, en \ una especie de \"ciclo\". Esto es lo que ocurre en muchos m\[EAcute]todos \ iterativos.\n\nEntre otros, ", StyleBox["Mathematica", FontSlant->"Italic"], " dispone de los siguientes comandos para implementar ciclos: Do, For y \ While. El comando m\[AAcute]s sencillo es el Do que se parece mucho al \ comando Table.\n\nLa sintaxis de estos comandos es la siguiente: \n" }], "Text", PageBreakWithin->Automatic, PageBreakBelow->Automatic, GroupPageBreakWithin->Automatic, CellChangeTimes->{{3.4275417334648533`*^9, 3.4275417915348682`*^9}}, TextJustification->1, ImageSize->{150, 150}, RenderingOptions->{"ImageCacheDepth"->"DeepestScreen"}, Background->None] }, Open ]], Cell[CellGroupData[{ Cell["1 \.ba.- Do", "Section"], Cell[TextData[{ "\n\t", StyleBox["Do[expr, {i, imax}]\n\tDo[expr, {i, imin, imax, di}]\n\tDo[expr, \ {n}]", FontWeight->"Bold"], ".\n \tEste comando es muy parecido al comando Table con la diferencia de \ que no se genera una lista, simplemente se eval\[UAcute]a expr tantas veces \ como indique el contador i." }], "Text", CellChangeTimes->{3.4275416552720547`*^9, 3.4592401459375*^9}], Cell[CellGroupData[{ Cell["\<\ Ejemplo 3 Presentar en pantalla utilizando el comando Print los cubos de los 10 \ primeros n\[UAcute]meros pares\ \>", "Subsection", CellChangeTimes->{{3.459240152953125*^9, 3.459240205640625*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Do", "[", RowBox[{ RowBox[{"Print", "[", RowBox[{"n", "^", "3"}], "]"}], ",", RowBox[{"{", RowBox[{"n", ",", "2", ",", "20", ",", "2"}], "}"}]}], "]"}]], "Input", CellChangeTimes->{{3.45924021078125*^9, 3.45924023834375*^9}}], Cell[BoxData["8"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.469795489453125*^9}], Cell[BoxData["64"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.469795489453125*^9}], Cell[BoxData["216"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["512"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["1000"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["1728"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["2744"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["4096"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["5832"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}], Cell[BoxData["8000"], "Print", CellChangeTimes->{3.459240239265625*^9, 3.461567267203125*^9, 3.469795446765625*^9, 3.46979548946875*^9}] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["2 \.ba.- For", "Section"], Cell["\<\ \tFor[inicio, test, incr, expr] \tEl comando For tiene cuatro partes : \t- inicio : se inicializan con el valor deseado las variables que \ intervienen en el ciclo. Ir\[AAcute]n separadas por ;. \t- test : condici\[OAcute]n l\[OAcute]gica que ser\[AAcute] evaluada cada \ vez que se recorra el ciclo. Cuando test es falso se sale del ciclo. \t- incr : sirve para incrementar el contador utilizado. - expr : est\[AAcute] constitu\[IAcute]do por el conjunto de \ acciones, separadas por ; , que se ejecutan cada vez que se recorre el ciclo.\ \>", "Text", CellChangeTimes->{ 3.427541713120646*^9, {3.459240024875*^9, 3.45924004434375*^9}}], Cell[CellGroupData[{ Cell["\<\ Ejemplo 4 Repetir el ejercicio anterior utilizando el comando For.\ \>", "Subsection", CellChangeTimes->{{3.459240152953125*^9, 3.459240205640625*^9}, { 3.459240262203125*^9, 3.459240283125*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"For", "[", RowBox[{ RowBox[{"k", "=", "1"}], ",", RowBox[{"k", "\[LessEqual]", "10"}], ",", RowBox[{"k", "++"}], ",", RowBox[{"Print", "[", RowBox[{ RowBox[{"(", RowBox[{"2", "k"}], ")"}], "^", "3"}], "]"}]}], "]"}]], "Input", CellChangeTimes->{{3.45924028659375*^9, 3.45924035459375*^9}}], Cell[BoxData["8"], "Print", CellChangeTimes->{3.459240355828125*^9, 3.461567267234375*^9, 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3.45924044665625*^9}, 3.461567267265625*^9, 3.469795446859375*^9, 3.469795489703125*^9}] }, Open ]] }, Open ]] }, Open ]], Cell["\<\ M\[CapitalEAcute]TODOS NUM\[CapitalEAcute]RICOS DE RESOLUCI\[CapitalOAcute]N \ DE UNA ECUACI\[CapitalOAcute]N NO LINEAL.\ \>", "Section 1", CellChangeTimes->{{3.42812980925*^9, 3.428129812421875*^9}}], Cell[CellGroupData[{ Cell["M\[EAcute]todo de la bisecci\[OAcute]n", "Section", Background->None], Cell["\<\ El m\[EAcute]todo de la bisecci\[OAcute]n es un m\[EAcute]todo muy sencillo \ para resolver una ecuaci\[OAcute]n no lineal : \tf (x) = 0 B\[AAcute]sicamente consiste en localizar un intervalo [a, b] en el que la \ funci\[OAcute]n cambie de signo. Suponiendo que f sea continua en [a, b], \ entonces existe al menos un punto c \[Epsilon] [a, b] en el que f (c) = 0. Para localizar c , en cada iteraci\[OAcute]n, se divide el intervalo en dos \ partes iguales eligiendo el subintervalo en el que se mantiene el cambio de \ signo de la funci\[OAcute]n. Este m\[EAcute]todo tiene el inconveniente de que su convergencia es muy \ lenta.\ \>", "Text", CellChangeTimes->{{3.4592407109375*^9, 3.459240957671875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Ejemplo", FontColor->GrayLevel[0]], " 6. \nObtener, mediante el m\[EAcute]todo de la bisecci\[OAcute]n , una \ soluci\[OAcute]n aproximada de la ecuaci\[OAcute]n :\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SqrtBox[ RowBox[{ SuperscriptBox["x", "2"], "+", "1"}]], " ", "=", " ", RowBox[{"tan", "(", "x", ")"}]}], ",", " ", RowBox[{"0", " ", "<", " ", "x", " ", "<", " ", RowBox[{"\[Pi]", "/", "2"}]}]}], TraditionalForm]]], " \nEn cada iteraci\[OAcute]n se calcula tambi\[EAcute]n la longitud del \ intervalo. El proceso finalizar\[AAcute] cuando la longitud del intervalo sea \ menor que 0.001 o el valor de la funci\[OAcute]n en el resultado de la \ \[UAcute]ltima iteraci\[OAcute]n sea menor que 0.001. 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Se necesita partir de una aproximaci\[OAcute]n inicial de \ la raiz, ", Cell[BoxData[ FormBox[ SubscriptBox["p", "0"], TraditionalForm]]], ". \nEn la siguiente figura se explica de forma sencilla en qu\[EAcute] \ consiste gr\[AAcute]ficamente el m\[EAcute]todo y c\[OAcute]mo se obtiene la \ f\[OAcute]rmula de recurrencia que permite construir la sucesi\[OAcute]n.\n\ Partiendo de la aproximaci\[OAcute]n inicial ", Cell[BoxData[ FormBox[ SubscriptBox["p", "0"], TraditionalForm]]], ", trazamos la tangente a la curva en el punto ", Cell[BoxData[ FormBox[ RowBox[{"(", SubscriptBox["p", "0"]}], TraditionalForm]]], ", f(", Cell[BoxData[ FormBox[ SubscriptBox["p", "0"], TraditionalForm]]], ")). Esta tangente corta al eje horizontal en el punto ", Cell[BoxData[ FormBox[ SubscriptBox["p", "1"], TraditionalForm]]], " que es el siguiente t\[EAcute]rmino de la sucesi\[OAcute]n. Calculando la \ tangente y hallando el corte con el eje horizontal se tiene:\n\t", Cell[BoxData[ FormBox[ SubscriptBox["p", "1"], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SubscriptBox["p", "0"], TraditionalForm]]], " - ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", SubscriptBox["p", "0"]}], TraditionalForm]]], ") / f ' ", Cell[BoxData[ FormBox[ RowBox[{"(", SubscriptBox["p", "0"]}], TraditionalForm]]], ")\nDel mismo modo se obtiene ", Cell[BoxData[ FormBox[ SubscriptBox["p", "2"], TraditionalForm]]], ", ..." }], "Text", CellChangeTimes->{ 3.427541848634506*^9, {3.4275420185221167`*^9, 3.427542025305037*^9}, { 3.45924103440625*^9, 3.459241050640625*^9}, {3.45924136159375*^9, 3.45924163146875*^9}}], Cell[GraphicsData["Metafile", "\<\ CF5dJ6E]HGAYHf4PEfU^I6mgLb15CDHPAVmbKF5d0@0008DX0@0006`000310@00i0D00=`>001D3@00 K`L00>0H003D?P00>cP00215CDH00040:8D008H2000900000000000000000000`1800<0J003;0000 8@4000000000000000000?PH0`3XJ0@0AP0002`0000P0000ADe6:`500@0L00004000008@`=/00000 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"-", "1"}]], ")"}]}]}], ",", " ", RowBox[{"k", "=", "1"}], ",", "2", ",", RowBox[{"...", "."}]}], TraditionalForm]]], "\nEn el l\[IAcute]mite, k\[Rule]\[Infinity], se obtendr\[IAcute]a la soluci\ \[OAcute]n exacta. En la pr\[AAcute]ctica se va construyendo la sucesi\ \[OAcute]n hasta que la distancia entre las dos \[UAcute]ltimas \ aproximaciones sea menor que una precisi\[OAcute]n establecida de antemano.\n" }], "Text", CellChangeTimes->{ 3.427541848634506*^9, {3.4275420185221167`*^9, 3.427542025305037*^9}, { 3.45924103440625*^9, 3.459241050640625*^9}, {3.459241353953125*^9, 3.459241355328125*^9}, {3.4592416440625*^9, 3.459241733890625*^9}, { 3.45924186203125*^9, 3.4592418756875*^9}}], Cell[CellGroupData[{ Cell[TextData[{ "Ejemplo 7\nHallar mediante el m\[EAcute]todo de Newton y con un error menor \ que ", Cell[BoxData[ FormBox[ SuperscriptBox["10", RowBox[{"-", "4"}]], TraditionalForm]]], " la primera raiz positiva de la ecuaci\[OAcute]n:\n\t0.5 ", Cell[BoxData[ FormBox[ SuperscriptBox["e", RowBox[{"x", "/", "3"}]], TraditionalForm]]], "-sin(x) = 0" }], "Subsection", CellChangeTimes->{{3.427541875397113*^9, 3.427541986163279*^9}, { 3.4281298554375*^9, 3.42812985575*^9}, 3.4592417590625*^9, { 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Se utiliza como condici\[OAcute]n que la \ diferencia en valor absoluto entre las dos \[UAcute]ltimas iteraciones sea \ menor que eps y que el n\.ba de iteraciones sea inferior a nmax. No es necesario guardar todas las aproximaciones de la soluci\[OAcute]n que \ se van calculando en cada una de las iteraciones. Solamente se guardan las \ dos \[UAcute]ltimas en las variables xv y xn, y esto nos obliga a hacer una \ actualizaci\[OAcute]n al terminar el ciclo, xv = xn.\ \>", "Text", CellChangeTimes->{{3.45966847325*^9, 3.45966860425*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"xv", "=", "0.5"}], ";", RowBox[{"error", "=", "1"}], ";", RowBox[{"eps", "=", SuperscriptBox["10", RowBox[{"-", "4"}]]}], ";", RowBox[{"nmax", "=", "50"}], ";", RowBox[{"n", "=", "1"}], ";"}]], "Input", CellChangeTimes->{{3.459241883828125*^9, 3.459241885453125*^9}, 3.469795474953125*^9}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"While", "[", RowBox[{ RowBox[{ RowBox[{"error", ">", "eps"}], "&&", RowBox[{"n", "<", "nmax"}]}], ",", RowBox[{ RowBox[{"xn", "=", RowBox[{"xv", "-", RowBox[{ RowBox[{"f", "[", "xv", "]"}], "/", RowBox[{ RowBox[{"f", "'"}], "[", "xv", "]"}]}]}]}], ";", RowBox[{"error", "=", RowBox[{"Abs", "[", RowBox[{"xn", "-", "xv"}], "]"}]}], ";", RowBox[{"Print", "[", RowBox[{ "\"\\"", ",", "n", ",", "\"\< la sol.aprox.es \>\"", ",", "xn"}], "]"}], ";", RowBox[{"xv", "=", "xn"}], ";", RowBox[{"n", "++"}]}]}], "]"}]], "Input", CellChangeTimes->{{3.459241888328125*^9, 3.45924189265625*^9}}], Cell[BoxData[ InterpretationBox[ RowBox[{"\<\"En la iter. \"\>", "\[InvisibleSpace]", "1", "\[InvisibleSpace]", "\<\" la sol.aprox.es \"\>", "\[InvisibleSpace]", "0.6634441598425319`"}], SequenceForm["En la iter. 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Una forma de \ evitar la evaluaci\[OAcute]n de esta derivada es sustituirla por una \ aproximaci\[OAcute]n:\n\t", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{"f", "'"}], RowBox[{"(", SubscriptBox["x", RowBox[{"k", "-", "1"}]], ")"}]}], " ", "="}], TraditionalForm]]], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ RowBox[{"f", "(", SubscriptBox["x", RowBox[{"k", "-", "1"}]], ")"}], "-", RowBox[{"f", "(", SubscriptBox["x", RowBox[{"k", "-", "2"}]], ")"}]}], RowBox[{ SubscriptBox["x", RowBox[{"k", "-", "1"}]], "-", SubscriptBox["x", RowBox[{"k", "-", "2"}]]}]], TraditionalForm]]], "\nSustituyendo esta aproximaci\[OAcute]n en la expresi\[OAcute]n anterior \ se obtiene:\n\t", Cell[BoxData[ FormBox[ SubscriptBox["x", "k"], TraditionalForm]]], "= ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SubscriptBox["x", RowBox[{"k", "-", "1"}]], "-", " ", RowBox[{ RowBox[{"f", "(", SubscriptBox["x", RowBox[{"k", "-", "1"}]], ")"}], "*", RowBox[{ RowBox[{"(", RowBox[{ SubscriptBox["x", RowBox[{"k", "-", "1"}]], "-", SubscriptBox["x", RowBox[{"k", "-", "2"}]]}], ")"}], "/", RowBox[{"(", RowBox[{ RowBox[{"f", "(", SubscriptBox["x", RowBox[{"k", "-", "1"}]], ")"}], "-", RowBox[{"f", "(", SubscriptBox["x", RowBox[{"k", "-", "2"}]], ")"}]}], ")"}]}]}]}], ",", " ", RowBox[{"k", "=", "1"}], ",", "2", ",", RowBox[{"...", "."}]}], TraditionalForm]]], "\nA diferencia del m\[EAcute]todo anterior \[EAcute]ste, para comenzar a \ construir la sucesi\[OAcute]n de valores aproximados, requiere partir de dos \ valores, ", Cell[BoxData[ FormBox[ SubscriptBox["x", "0"], TraditionalForm]]], " y ", Cell[BoxData[ FormBox[ SubscriptBox["x", "1"], TraditionalForm]]], ". Igual que en el caso anterior, no es necesario guardar todas las \ aproximaciones. En este caso ser\[AAcute]n necesarias en cada \ iteraci\[OAcute]n tres variables, y tambi\[EAcute]n ser\[AAcute] precisa, al \ terminar el ciclo, una actualizaci\[OAcute]n de las variables." }], "Text", CellChangeTimes->{ 3.427541848634506*^9, {3.4275420185221167`*^9, 3.427542025305037*^9}, { 3.459241793609375*^9, 3.459241839328125*^9}, {3.4596683583125*^9, 3.45966843865625*^9}, {3.459668615328125*^9, 3.459668730984375*^9}}] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Ejemplo 8 Repetir el ejercicio anterior utilizando el m\[EAcute]todo de la secante para \ calcular la segunda raiz positiva de la ecuaci\[OAcute]n.\ \>", "Subsection", CellChangeTimes->{{3.427541875397113*^9, 3.427541986163279*^9}, { 3.4281298554375*^9, 3.42812985575*^9}, 3.4592417590625*^9, { 3.45966831275*^9, 3.459668339*^9}, {3.45966880346875*^9, 3.459668829640625*^9}}], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"f", "[", "x_", "]"}], "=", RowBox[{ RowBox[{"0.5", "*", RowBox[{"Exp", "[", RowBox[{"x", "/", "3"}], "]"}]}], "-", RowBox[{"Sin", "[", "x", "]"}]}]}], ";"}]], "Input", CellChangeTimes->{3.459668756828125*^9}], Cell[TextData[{ "A la vista de la gr\[AAcute]fica anterior elegimos como valores de partida ", Cell[BoxData[ FormBox[ SubscriptBox["x", "0"], TraditionalForm]]], "= 1.4 y ", Cell[BoxData[ FormBox[ SubscriptBox["x", "1"], TraditionalForm]]], "= 2.2" }], "Text", CellChangeTimes->{{3.459668853875*^9, 3.459668936453125*^9}}], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"For", "[", RowBox[{ RowBox[{ RowBox[{"x0", "=", "1.4"}], ";", RowBox[{"x1", "=", "2.2"}], ";", RowBox[{"error", "=", "1"}], ";", RowBox[{"n", "=", "1"}]}], ",", RowBox[{"error", ">", "0.0001"}], ",", RowBox[{"n", "++"}], ",", RowBox[{ RowBox[{"x2", "=", RowBox[{"x1", "-", RowBox[{ RowBox[{"f", "[", "x1", "]"}], "*", RowBox[{ RowBox[{"(", RowBox[{"x1", "-", "x0"}], ")"}], "/", RowBox[{"(", RowBox[{ RowBox[{"f", "[", "x1", "]"}], "-", RowBox[{"f", "[", "x0", "]"}]}], ")"}]}]}]}]}], ";", RowBox[{"error", "=", RowBox[{"Abs", "[", RowBox[{"x2", "-", "x1"}], "]"}]}], ";", RowBox[{"Print", "[", RowBox[{ "\"\\"", ",", "n", ",", "\"\< la sol.aprox.es \>\"", ",", "x2"}], "]"}], ";", RowBox[{"x0", "=", "x1"}], ";", RowBox[{"x1", "=", "x2"}]}]}], "]"}]], "Input", CellChangeTimes->{{3.459241888328125*^9, 3.45924189265625*^9}, { 3.459668979046875*^9, 3.459669117375*^9}}], Cell[BoxData[ InterpretationBox[ RowBox[{"\<\"En la iter. \"\>", "\[InvisibleSpace]", "1", "\[InvisibleSpace]", "\<\" la sol.aprox.es \"\>", "\[InvisibleSpace]", "1.7577832990622198`"}], SequenceForm["En la iter. 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