%%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 %%%%%%%%%%%%%%%
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Problem 1 (Matlab): One of the most fundamental root finding algorithms is Newton's Method. Given a real-valued, differentiable function f, Newton's method is given by 1. Initialization: Pick a point xo which is near the root of f Iteratively define points rn+1 for n = 0,1,2,..., by 2. Iteration: f(xn) nt1 In 3. Termination: Stop when some stopping criterion occurs said in the literature). For the purposes of this problem, the stopping criterion will be 100 iterations (This sounds vague,...
This is Matlab Problem and I'll attach problem1 and its answer
for reference.
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