Question

Problem 12.12

Pls show the m-file

Develop your own M-file function for the Gauss-Seidel method without relaxation based on Fig. 12.2, but change the first line so that it returns the approximate error and the number of iterations: function [x, ea, iter] = ... GaussSeidel (, b, es, maxit) Test it by duplicating Example 12.1 and then use it to solve Prob. 12.2a. Develop your own M-file function for Gauss-Seidel with relaxation. Here is the function's first line: function [x, ea, iter] = GaussSeidelR (A, b, 1 ambda, es, maxit) In the event that the user docs not enter a value for lambda, set the default value as lambda = 1. Test it by duplicating Example 12.2 and then use it to solve Prob. 12.2b. Develop your own M-file function for the Newton-Raphson method for nonlinear systems of equations based on Fie. 12.4. Test it by solving Example 12.4 and then use it

Answer 1

function [x1 , ea,iterval] //if conditiorn -ĢaussseidelR (Arba omg, esvaïr max..it ) if nargin<2 errer('2 arguments'); end //check the max iteration if nargin<5| lisempty (maxit) max it-50 end if nargin<4 1 lisempt (esval) esval=0.00001; end if nargin<31 113emptỵ (omg) end [ml, n1] size(A) ; if ml~-n1, error ('Matrix A 'end C1 A;

for 1:n1 = end x1=x1'; %Transpose matrix for = 1:n1 end ER五五= 1:n! end While iterval max ït xoldval = x1;

eal (1) = abs ( (x1 (2) -xoldval (2) ) /x1 () ) * 100; end end iterval = iterva1+1; %while loop if max (eal) <=e s val break end ea end Smain.m AI8-0.4 0:-0.4 0.8-0.4:0 -0.4 0.8] b[41;25;105] ,,Mal 0.00001 Gauss SeideR(A,b,1,esval, 50)

Output: 0.8000 0.4000 -0.4000 0.8000 -0.4000 0 0.4000 0.8000 41 25 105 esvai 1.0000e-05 ans = 173.7500 245.0000 253.7500

copyable code:

% GaussSeidelR.m

function [x1,ea,iterval] = GaussSeidelR(A,b,omg,esval,max_it)

//if condition

if nargin<2

     error('2 arguments');

end

//check the max iteration

if nargin<5||isempty(max_it)

     max_it=50;

end

if nargin<4||isempty(esval)

     esval=0.00001;

end

if nargin<3||isempty(omg)

     omg=1;

end

[m1,n1] = size(A);

if m1~=n1, error('Matrix A '); end

C1 = A;

for i = 1:n1

     C1(i,i) = 0; %diagonal

     x1(i) = 0;

end

x1 = x1'; %Transpose matrix

for i = 1:n1

     C1(i,1:n1) = C1(i,1:n1)/A(i,i);

end

for i = 1:n1

     d1(i) = b(i)/A(i,i);

end

iterval = 0;

while iterval<=max_it

   

     xoldval = x1;

     for i = 1:n1

         x1(i) = d1(i)-C1(i,:)*x1;

         x1(i) = omg*x1(i) + (1-omg)*xoldval(i);

       

         if x1(i) ~= 0

             ea1(i) = abs((x1(i) - xoldval(i))/x1(i)) * 100;

         end

     end

     iterval = iterval+1;

  

     %while loop

       if max(ea1)<=esval

         break

       end

%     iterval

%     x1

%     ea

end

%main.m

A=[0.8 -0.4 0;-0.4 0.8 -0.4;0 -0.4 0.8]

b=[41;25;105]

esval=0.00001

GaussSeidelR(A,b,1,esval,50)