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Analisis-practica4 - Anlisis Numrico a e Mtodo del punto jo...

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An´ alisis Num´ erico etodo del punto fijo y M´ etodo de Newton 26 de marzo de 2011 Ejercicio 1. Implemente el m´ etodo del punto fjo para el siguiente sistema de ecuaciones lin- eales: ( x = 2 x 2 - 2 y 2 +1 4 y = - 4+ x 4 +4 y 4 +8 y +4 12 Utilicemos el programa siguiente: function x=pf(x0,y0,tol,fun1,fun2) x=X0(1); y=X0(2); k=1; while (eval(fun1)¿x — eval(fun2)¿y) F=[eval(fun1),eval(fun2)]’ X=[x,y]’; Q=Jacobiana(X,0;001,fun1,fun2); Y=QF; k=k+1; x=x+Y(1); y=y+Y(2); end x=[x,y]; function J=Jacobiana(X,h,fun1,fun2) x=X(1); y=X(2); df1dx=(evaluar(x+h,y,fun1)- evaluar(x,y,fun1))/h; df1dy=(evaluar(x,y+h,fun1)- evaluar(x,y,fun1))/h; df2dx=(evaluar(x+h,y,fun2)- evaluar(x,y,fun2))/h; df2dy=(evaluar(x,y+h,fun2)- evaluar(x,y,fun2))/h; J=[df1dx df1dy; df2dx df2dy]; function f=evaluar(x,y,fun) f=eval(fun); Para los valores iniciales x 0 = 1 y y 0 = 1 con una tolerancia tol = 0; 1 se obtienen los siguientes resultados: k x k y k f ( x k ) g ( y k ) 1 0.2500 1.08331 1 1 2 0.1383 0.4178 0.03220 0.5719 3 0.6369 0.1395 -0.8894 0.1309 . . . . . . . . . . . . . . . 17 0.2436 0.0399 0.2789 0.0269 Con la aproximaxion dada por tol se alcanza la solucion en 17 iteraciones 1
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Ejercicio 2.
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