Question

4. (20 pts) In this problem, we combine the Steepest Descent method with Newton's method for solving the following nonlinear system. en +en-13 = 0, 12-2113 = 4. Use the Steepest Descent method with initial approximation x0,0,0 three iterations x(1), x(2), and x(3) to find the first ·Use x(3) fron the above the result as the initial approximation for Newton's iteration. Use the stopping criteria X(k)-s(k 1) < tol = 10 9 Display the results for each iteration in the above tabular manner

Answer 1

SMatlab code for Gradient Descent method and Newton method clear all close all n-0; eps-1; a-0.06 initialize iteration counter initialize error %set iteration parameter %set starting value fprintf( For Gradient Descent method.\n') fprintf("\tFor initial condition x1-%f, x2-% f , x3-Af functions for which root have to find syms x1 x2 x3 f.x2-@ ( x1 , x2 , x3) exp ( x1 ) +exp ( x2 )-x3; f x3- (xl, x2, x3) x2-2.*x1.*x3-4; %displaying the function fprintf ( Displaying the functions\n') disp(f_x1) disp(f_x2) disp(f_x3) fprintf( First three iterations using Steepest Descent\n') Computation loop for i= 1:3 gradf-double( [f_xl (xx (1) , xx (2), xx (3));f_x2 ( xx (1), xx(2), xx (3));f_x3 (xx (1), xx (2), xx yy-double ( xx-a*gradf ); n=n+1 ; eps norm (xx-YY) error (n) eps; xxYY fprint f ( "\tAfter %d iterations % iterate %counter+ 1 %update x x1=%f; x2=% f x3=%f. end fprintf( 'The solution for nonlinear equation using Steepest Descent is fprintf(nnFor Newton method.n') Matlab code for Newton method nonlinear function to be solved syms x1 x2 x3 f2(xl,x2, x3)-exp (xl)texp(x2)-x3; f3 (x1, x2,x3)x2-2.*x1.*x3-4; %finding the Jacobian matrix

f1 xl(xl, x2, x3)-diff(f1, x1); f1 x2(xl, x2, x3)-diff (fl, x2); f1 x3(xl, x2, x3)-diff(E1, x3); f2_x1(x1,x2, x3 )-diff (f2, x1); f2_x2 (xl, x2, x3)-diff (f2, x2); f2 x3 (xl, x2, x3)-diff (E2, x3); f3 x1(xl, x2, x3)-diff (E3, x1); f3_x2 (xl, x2, x3)-diff (f3, x2); f3 x3 (x1, x2, x3)-diff (E3, x3) f3x1 f3_x2 f3x3 1; fprintf The Jacobian matrix is\n') disp (jacl) All initial guess for x y z x11-1;x22-1 x33-1 fprint f ('For initial condition x1=%f, x2=%f, x3=%f In',x11,x22, x33) kmax-500; %maximum number of iterations %loop for Newton method fprintf( 'All iterations using Newton methodin') fprintfttxl\tx2\tx3\n') for i-1: kmax jac-jacl (x1l,x22, x33); ijac-inv (jac); xx-double( [x1l; x22;x33]- ijac [f1 (x11, x22,x33); £2 (x11,x22,x33); £3 (x11,x22, x33) ])i err-norm (xx- [x1l;x22 x11-double (xx(1)) x22-double (xx (2) x33-double (xx(3)); x33]); if err<101-9 break fprint f ('\t %f; %f %f.\n ' , xx ( 1 ) , xx ( 2 ) , xx ( 3 ) ) end fprintfnThe solution for nonlinear equation using \n\t x1一号f\n\t x2- %f\n\t x3= %f\n',x11,x22 , x33) Newton method is 응88888888 End of Code 888888 88888888 For Gradient Descent method

For initial condition x1=0.000000, Displaying the functions x2=0 .000000, x3=0.000000 C(x1,x2, x3)x1. *3+x1.*2. *x2-x1. *x3+6 @(X1 , x2,%)exp ( x1 ) +exp ( x2 )-x3 9(x1,x2,x3)x2-2. *x1. *x3-4 First three iterations using Steepest Descent After I iterations xì--0. 360000; x2--0. 120000 x3-0.240000. After 2 iterations x1=-0.72 1452; x2=-0.200676 x3=0.476832. After 3 iterations xl--1.073295; x2--0. 250319 x3-0. 687591 The solution for nonlinear equation using Steepest Descent is x1- -1.073295 x2= -0.250319 x3= 0.687591 For Newton method The Jacobian matrix is [ 3*x1"2 + 2*x2+x1 - x3, x1^2, exp(x2), -x1] exp(xI), -2 *x3, 1, -2 *x1] symbolic function inputs: xl, x2, x3 For initial condition x1=1.000000, x2=1.000000, x3=1.000000 All iterations using Newton method xl x2 x3 -0.906942; 1.069415 0.441649. -14. 799386; 0.244661-4.694808 -9.151190; -1.999543-1.589081. -5.180914; -4.617756-0.218574. -4.746648; 2.859214 0.091774. -3.320291; 1.872432 0.251691 -2.415314; 0.943168 0.528927. -2.040074; 0.231129 0.862376. -1.976524; -0.066344 1.023481 -1.970514; 0. 102868 1.041012. -1.970437; -0.103332 1.041224. -1.970437; -0.103332 1.041224. The solution for nonlinear equation using Newton method is x1- 1.970437 x2= -0.103332 x3- 1.041224 Published with MATLAB® R2018a

%%Matlab code for Gradient Descent method and Newton method
clear all
close all
n=0;            %initialize iteration counter
eps=1;          %initialize error
a=0.06;         %set iteration parameter
xx=[0;0;0];     %set starting value
fprintf('For Gradient Descent method.\n')
fprintf('\tFor initial condition x1=%f, x2=%f, x3=%f \n',xx(1),xx(2),xx(3))
%functions for which root have to find

syms x1 x2 x3
f_x1=@(x1,x2,x3) x1.^3+x1.^2.*x2-x1.*x3+6;
f_x2=@(x1,x2,x3) exp(x1)+exp(x2)-x3;
f_x3=@(x1,x2,x3) x2-2.*x1.*x3-4;

%displaying the function

fprintf('Displaying the functions\n')
disp(f_x1)
disp(f_x2)
disp(f_x3)

fprintf('First three iterations using Steepest Descent\n')
%Computation loop
for i=1:3
    gradf=double([f_x1(xx(1),xx(2),xx(3));f_x2(xx(1),xx(2),xx(3));f_x3(xx(1),xx(2),xx(3))]); %gradf(x)
    yy=double(xx-a*gradf);                             %iterate
    n=n+1;
    eps=norm(xx-yy);                                   %counter+1
    error(n)=eps;
    xx=yy;                                             %update x
    fprintf('\tAfter %d iterations x1=%f; x2=%f x3=%f.\n',n,xx(1),xx(2),xx(3))
end

fprintf('The solution for nonlinear equation using Steepest Descent is\n\t x1= %f\n\t x2= %f\n\t x3= %f\n',xx(1),xx(2),xx(3));

fprintf('\n\nFor Newton method.\n')

%Matlab code for Newton method
%nonlinear function to be solved

syms x1 x2 x3
f1(x1,x2,x3)=x1.^3+x1.^2.*x2-x1.*x3+6;
f2(x1,x2,x3)=exp(x1)+exp(x2)-x3;
f3(x1,x2,x3)=x2-2.*x1.*x3-4;

%finding the Jacobian matrix
f1_x1(x1,x2,x3)=diff(f1,x1);
f1_x2(x1,x2,x3)=diff(f1,x2);
f1_x3(x1,x2,x3)=diff(f1,x3);

f2_x1(x1,x2,x3)=diff(f2,x1);
f2_x2(x1,x2,x3)=diff(f2,x2);
f2_x3(x1,x2,x3)=diff(f2,x3);

f3_x1(x1,x2,x3)=diff(f3,x1);
f3_x2(x1,x2,x3)=diff(f3,x2);
f3_x3(x1,x2,x3)=diff(f3,x3);

jac1=[f1_x1 f1_x2 f1_x3 ;...
            f2_x1 f2_x2 f2_x3 ;...
            f3_x1 f3_x2 f3_x3 ];
fprintf('The Jacobian matrix is\n')          
disp(jac1)

%All initial guess for x y z  
x11=1;x22=1;x33=1;

    fprintf('For initial condition x1=%f, x2=%f, x3=%f \n',x11,x22,x33)
    kmax=500; %maximum number of iterations
    %loop for Newton method
  
    fprintf('All iterations using Newton method\n')
    fprintf('\t\tx1\tx2\tx3\n')
    for i=1:kmax

        jac=jac1(x11,x22,x33);
        ijac=inv(jac);
        xx=double([x11;x22;x33]-ijac*[f1(x11,x22,x33);f2(x11,x22,x33);f3(x11,x22,x33)]);
        err=norm(xx-[x11;x22;x33]);
        x11=double(xx(1));
        x22=double(xx(2));
        x33=double(xx(3));

        if err<10^-9
            break
        end
        fprintf('\t %f; %f %f.\n',xx(1),xx(2),xx(3))
    end

fprintf('\nThe solution for nonlinear equation using Newton method is\n\t x1= %f\n\t x2= %f\n\t x3= %f\n',x11,x22,x33)

%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%