{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Analisis-practica4

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Ejercicio 2.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern