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Cell["\<\
M\[CapitalEAcute]TODOS DE RESOLUCI\[CapitalOAcute]N DE UNA ECUACI\
\[CapitalOAcute]N NO LINEAL.\
\>", "Title",
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Cell["M\[EAcute]todo de la bisecci\[OAcute]n", "Section",
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 "\nObtener, mediante el m\[EAcute]todo de la bisecci\[OAcute]n , una soluci\
\[OAcute]n aproximada de la ecuaci\[OAcute]n :\n",
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 " \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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Cell[TextData[{
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 "\n En primer lugar realizamos la representaci\[OAcute]n gr\[AAcute]fica \
para obtener un intervalo [a,b] en el que la funci\[OAcute]n cambie de signo"
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\[AAcute]s peque\[NTilde]a, con un error menor que 0.00001, mediante la \
iteraci\[OAcute]n de punto fijo, eligiendo una funci\[OAcute]n de iteraci\
\[OAcute]n g(x) que verifique las condiciones del teorema de punto fijo."
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para obtener el intervalo de trabajo y la aproximaci\[OAcute]n inicial."
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Cell["\<\
Interesa que \[VerticalSeparator]g'(x)\[RightBracketingBar] < 1. Una forma de \
obtener g(x) es la siguiente:
f(x) = 0 \[LongRightArrow] - \[Alpha] f(x) = 0 \[LongRightArrow] x-\[Alpha] \
f(x) = x \[LongRightArrow] g(x) = x - \[Alpha] f(x) \[LongRightArrow] 
g'(x) = 1-\[Alpha]f'(x)
\[VerticalSeparator]g'(x)\[RightBracketingBar] < 1 \[LongRightArrow] 0 < \
\[Alpha]f'(x) < 2.\
\>", "Text"],

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Cell["\<\
Probamos  con g(x) = x+ 0.2*f(x) ,  y comprobamos las condiciones del teorema.\
\>", "Text"],

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Se comprueba que \[ForAll] x \[Element] [-1,0] , g(x) (tambi\[EAcute]n)  \
\[ForAll] x \[Element] [-1,0]. A continuaci\[OAcute]n representamos g'(x) en \
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\>", "Text"],

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Cell["\<\
A la vista de la gr\[AAcute]fica, dos buenas aproximaciones son : 0.6 y 2. \
Utilizamos esta vez el comando While para realizar las iteraciones del m\
\[EAcute]todo de Newton-Raphson. 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.\
\>", "Text"],

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Cell[TextData[{
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 " una ecuaci\[OAcute]n.Podemos utilizar el comando Solve para obtener los \
valores de x que verifican la ecuaci\[OAcute]n:"
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Cell["\<\
El resultado es una lista de reglas en las que se asigna a x los valores de \
las raices. Para extraer de la soluci\[OAcute]n el resultado : \
\>", "Text"],

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Cell[TextData[{
 "El comando Solve siempre intenta dar como resultado f\[OAcute]rmulas expl\
\[IAcute]citas para la soluci\[OAcute]n de una ecuaci\[OAcute]n. Sin embargo, \
si la ecuaci\[OAcute]n es un poco compleja esto resultar\[AAcute] imposible. \
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 "tambi\[EAcute]n puede resolver ecuaciones con funciones trascendentes . En \
muchos de estos casos da un mensaje indicando que en el resultado no est\
\[AAcute]n todas las soluciones, "
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Cell["\<\
Se puede obtener una soluci\[OAcute]n num\[EAcute]rica de la ecuaci\[OAcute]n \
mediante el comando FindRoot. Este comando implementa una versi\[OAcute]n del \
m\[EAcute]todo de Newton-Raphson y necesita una aproximaci\[OAcute]n inicial. \
En la pr\[AAcute]ctica esta aproximaci\[OAcute]n se obtiene a partir de la \
representaci\[OAcute]n gr\[AAcute]fica.\
\>", "Text"],

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Cell["\<\
Resoluci\[OAcute]n de sistemas no lineales. El m\[EAcute]todo de Newton. El \
comando FindRoot.\
\>", "Section"],

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Cell["\<\
Como en el caso de una sola ec. no lineal, el comando FindRoot nos permite \
obtener una soluci\[OAcute]n num\[EAcute]rica (aproximada) de un sistema de \
ecuaciones no lineales partiendo de una aproximaci\[OAcute]n inicial. Una de \
sus ociones nos permite especificar el n\.ba m\[AAcute]ximo de iteraciones a \
realizar. Si se alcanza el n\.ba m\[AAcute]ximo de iteraciones sin haber \
obtenido la soluci\[OAcute]n el comando FindRoot nos devuelve la aproximaci\
\[OAcute]n calculada en la \[UAcute]ltima etapa que se puede utilizar como \
aproximaci\[OAcute]n inicial de un nuevo comando FindRoot.
Es conveniente realizar una representaci\[OAcute]n gr\[AAcute]fica, cuando \
sea posible, para obtener una buena aproximaci\[OAcute]n inicial.
El comando FindRoot puede resolver ec. no lineales en variable compleja. Para \
ello solo es preciso incluir la unidad imaginaria I en la \
aproximaci\[OAcute]n inicial. Por ejemplo:\
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Para resolver el sistema anterior utilizando el comando FindRoot:\
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Cell[TextData[{
 "1\.ba.- Comprobar gr\[AAcute]ficamente que la funci\[OAcute]n f(x) = x- 0.2 \
sen(x) - 0,5 tiene una \[UAcute]nica raiz real en el intervalo [0.5 , 1]. \
Aplicar los m\[EAcute]todos de la bisecci\[OAcute]n, punto fijo y \
Newton-Raphson, contando en cada caso el n\[UAcute]mero de iteraciones que se \
han realizado , para determinar la raiz con una precisi\[OAcute]n de ",
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 ". Establecer un n\.ba m\[AAcute]ximo de iteraciones a realizar igual a 50."
}], "Text"],

Cell["\ta) Bisecci\[OAcute]n", "Text"],

Cell["\tb) Punto fijo", "Text"],

Cell["\tc) Newton-Raphson", "Text"],

Cell[BoxData["\[IndentingNewLine]"], "Input",
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Cell[TextData[{
 "2\.ba.- Representar en una misma gr\[AAcute]fica las funciones f(x) = 2 \
cos(x) y g(x) = ",
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 " para obtener  aproximaciones  iniciales de las raices de 2cos(x) - ",
 Cell[BoxData[
  FormBox[
   SuperscriptBox["e", "x"], TraditionalForm]]],
 "= 0. \[DownQuestion]Cu\[AAcute]ntas ra\[IAcute]ces tiene?. \
\[DownQuestion]Cu\[AAcute]ntas son positivas?. Determinar las raices \
positivas con una precisi\[OAcute]n de ",
 Cell[BoxData[
  FormBox[
   SuperscriptBox["10", 
    RowBox[{"-", "8"}]], TraditionalForm]]],
 " utilizando el m\[EAcute]todo de Newton-Raphson ."
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Cell[BoxData["\[IndentingNewLine]"], "Input",
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Cell[TextData[{
 "3\.ba.- Comprobar de forma gr\[AAcute]fica que la ecuaci\[OAcute]n .\n\t\t4 \
sen(x) = 1+x\ntiene tres raices reales : ",
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 "<",
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 " . A continuaci\[OAcute]n determinar con una precisi\[OAcute]n de ",
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 " utilizando el mdo de la secante.\n\tb) ",
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 "       \"        \"   \"     \"   \" bisecci\[OAcute]n.\n\tc) ",
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 "       \"        \"   \"     \"   Newton."
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Cell[TextData[{
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 " primero por el m\[EAcute]todo de Newton y luego por el de la secante. \
Contar el n\.ba de iteraciones que se han realizado en ambos casos para \
comprobar cu\[AAcute]l de los dos m\[EAcute]todos tiene una convergencia m\
\[AAcute]s r\[AAcute]pida. "
}], "Text"],

Cell[BoxData["\[IndentingNewLine]"], "Input",
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Cell[TextData[{
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Justificar que la expresi\[OAcute]n :\n\t",
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la raiz de la ecuaci\[OAcute]n en el intervalo [0, \[Pi]/6]. Hallar dicha \
raiz con una precisi\[OAcute]n de ",
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 "."
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Cell[TextData[{
 "6\.ba.- Determinar la raiz cuadrada negativa de 0.5 con cuatro decimales y \
una precisi\[OAcute]n de ",
 Cell[BoxData[
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   SuperscriptBox["10", 
    RowBox[{"-", "3"}]], TraditionalForm]]],
 "considerando la funci\[OAcute]n f(x) = ",
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 " y resolviendo la ec. x = ",
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Cell[TextData[{
 "7\.ba.- Hacer una representaci\[OAcute]n gr\[AAcute]fica de las funciones \
f(x) = 2cosx y g(x) = ",
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 " para obtener estimaciones iniciales de las raices de la ec. 2cosx - ",
 Cell[BoxData[
  FormBox[
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 "=0. \[DownQuestion]Cu\[AAcute]ntas raices tiene ? \
\[DownQuestion]Cu\[AAcute]ntas son positivas?. Determinar las raices \
positivas con una precisi\[OAcute]n \[VerticalSeparator]f(r)\
\[VerticalSeparator] < ",
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iniciales ",
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redondeo a seis cifras decimales."
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\[EAcute]todo de Newton y a continuaci\[OAcute]n con el comando FindRoot:\n3 \
x - cos(y.z) - 1/2 = 0\n",
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