1. Considere el siguiente sistema de ecuaciones no lineales de 3 incgnitas 3 ecuaciones.

Pruebe con un punto inicial Itere hasta que x (k) - x (k-1) preciter = iter+1x = x-inv(Jn(x))*Fn(x)if iter > 1000error('parece que newton no converge');endend

function z=Fn(x)z=[7*x(1)+5*x(2)-x(3).^2*sin(x(1))-12 -x(1).^4+(cos(x(2))).^2+2*x(3).^3-8 6*x(1)+2*x(2)-x(3)+34];

function z=Jn(x)z=[7*x(2)-x(3)^2*cos(x(1)) 7*x(1)+5 2*x(3)*sin(x(1)) -4*x(1)^3 -2*sin(x(2)) 6*x(3)^2 6 2 -1]; >> [x,iter]= NEWTON_NL([10 20 -50]',0.000001)

iter = 1x = 11.8604 -68.6715 -32.1808 iter = 2x = 12.9660 -64.4381 -17.0802 iter = 3x = 13.4431 -53.7647 7.1289 iter = 4x = 8.4431 -66.0748 -47.4912 iter = 5x = 13.3909 -72.3751 -30.4046 iter = 6x = 12.7696 -63.0822 -15.5469 iter = 7x = 13.6994 -51.4428 13.3110 iter = 8x = 1.0e+003 * 0.1207 0.1593 1.0770 iter = 9x = 113.1501 13.9779 740.8566 iter = 10x = 112.9279 -84.1402 543.2871 iter = 11x = 113.0336 -128.9170 454.3673 iter = 12x = 113.0653 -138.7739 434.8442 iter = 13x = 113.0958 -139.2390 434.0967 iter = 14x = 113.0978 -139.2405 434.1056 iter = 15x = 113.0978 -139.2405 434.1057 iter= 16x = 113.0978 -139.2405 434.1057 iter = 17x = 113.0978 -139.2405 434.1057 iter = 18x = 113.0978 -139.2405 434.105

2. Resolver el sistema de ecuaciones no lineales

sin(xy) + z = 12 sin(xz) + y = 23 sin(yz) + x = 3

pruebe con un punto inicial [0 0 0] y con parmetro de precisin "e= 10-6:

SOLUCION

function z=Jn(x)z=[x(2)*cos(x(1)*x(2)) x(1)*cos(x(1)*x(2)) 1 2*x(3)*cos(x(1)*x(3)) 1 2*x(1)*cos(x(1)*x(3)) 1 3*x(3)*cos(x(2)*x(3)) 3*x(2)*cos(x(2)*x(3))];

function z=Fn(x)z=[sin(x(1)*x(2))+x(3)-1 2*sin(x(1)*x(3))+x(2)-2 3*sin(x(2)*x(3))+x(1)-3];

function z=Jn(x)z=[x(2)*cos(x(1)*x(2)) x(1)*cos(x(1)*x(2)) 1 2*x(3)*cos(x(1)*x(3)) 1 2*x(1)*cos(x(1)*x(3)) 1 3*x(3)*cos(x(2)*x(3)) 3*x(2)*cos(x(2)*x(3))]; >> [x,iter]= NEWTON_NL([0 0 0]',0.000001)

iter =

1

x =

3 2 1

iter =

2

x =

2.0809 2.5545 1.4472

iter =

3

x =

2.6155 2.3971 1.2345

iter =

4

x =

2.5734 2.3320 1.2846

iter =

5

x =

2.5713 2.3307 1.2864

iter =

6

x =

2.5713 2.3307 1.2864

x =

2.5713 2.3307 1.2864

iter =

6

3. Resolver el sistema de ecuaciones no lineales

3x cos(yz) = 1=24x2 625y2 + 2y = 1exy + 20z + 10_33 = 0pruebe con un punto inicial [1 2 3] y con parmetro de precisin e= 10-6:

SOLUCION

function [x,iter]= NEWTON_NL(x,prec)iter =0;while norm(Fn(x))>preciter = iter+1x = x-inv(Jn(x))*Fn(x)if iter > 1000error('parece que newton no converge');endend

function z=Fn(x)z=[3*x(1)-cos(x(2)*x(3))-1/2 4*x(1)^2-625*x(2)^2+2*x(2)-1 exp(-x(1)*x(2))+20*x(3)+((10*pi-3)/3)];

function z=Jn(x)z=[3 x(3)*sin(x(2)*x(3)) x(2)*sin(x(2)*x(3)) 8*x(1) -1250*x(2)+2 0 -x(2)*exp(-x(1)*x(2)) -x(1)*exp(-x(1)*x(2)) 20];

>> [x,iter]= NEWTON_NL([1 2 3]',0.000001)

iter =

1

x =

-0.4467 0.9974 -0.5067

iter =

2

x =

0.5068 0.4966 -0.4600

iter =

3

x =

0.4978 0.2491 -0.5175

iter =

4

x =

0.4999 0.1253 -0.5205

iter =

5

x =

0.5000 0.0635 -0.5220

iter =

6

x =

0.5000 0.0326 -0.5228

iter =

7

x =

0.5000 0.0171 -0.5232

iter =

8

x =

0.5000 0.0094 -0.5234

iter =

9

x =

0.5000 0.0057 -0.5235

iter =

10

x =

0.5000 0.0040 -0.5235

iter =

11

x =

0.5000 0.0033 -0.5235

iter =

12

x =

0.5000 0.0032 -0.5235

iter =

13

x =

0.5000 0.0032 -0.5235

x =

0.5000 0.0032 -0.5235

iter =

13

SOLUCION