How should my set my obtained answers into a table form?

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Christoppe
Christoppe on 27 Jun 2022
Commented: Star Strider on 27 Jun 2022
Hi! May I kindly ask for your help I have been working on Jacobi Method since yesterday and I haven’t figured it out yet I don’t have someone to ask. I would appreciate your help…
I wanted to show the results of my codes like this.....
example values only....
Iteration x1 x2 x3
1 3.00000 4.00000 4.60000
2 4.00000 4.60000 4.27076
3 4.60000 4.27076 4.30295
4 4.27076 4.30295 4.30697
5 4.30295 4.30697 4.30691
The roots are the following:
x1 = 1
x2 = 2
x3 =-1
Iteration = 9
Inputs are....
Enter the number of equations, n:
3
Enter absolute errror, Ea:
0.0001
Enter coefficient (Diagaonally Dominant), Matrix A:
[4 -1 -1;-2 6 1;-1 1 7]
Enter constants, Matrix B:
[3;9;-6]
Enter the initial guess, x0,x1,x2:
[0;0;0]
CODE
format short
Eqn=input('Enter the number of equations, n: ');
Error=input('Enter absolute errror, Ea: ')
A=input('Enter coefficients (DIAGONALLY DOMINANT!), Matrix A: ');
B=input('Enter constants, Matrix B: ');
C=input('Enter the initial guess: x0,x1,x2: ');
D=[A B]
N=length(B);
x=zeros(N,1);
for j=1:100
for i=1:N
x(i) = round((B(i)/A(i,i)) - (A(i,[1:i-1,i+1:N])*C([1:i-1,i+1:N]))/A(i,i),4);
end
fprintf('Iteration no %d \n',j)
x
if abs(x-C)<=Error
break
end
C = x;
end
  1 Comment
Christoppe
Christoppe on 27 Jun 2022
update: I already figured how to do The roots are the following:
x1 = 1
x2 = 2
x3 =-1
Iteration = 9

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Accepted Answer

Star Strider
Star Strider on 27 Jun 2022
Copying from my answer to your previous post:
% Secant Method in MATLAB
% a=input('Enter function:','s');
% f=inline(a)
a = 'x^3 - 4*x^2 + x - 10'
a = 'x^3 - 4*x^2 + x - 10'
f = str2func(['@(x)' a])
f = function_handle with value:
@(x)x^3-4*x^2+x-10
% x0 = input ('Enter the first initial guess: ')
% x1 = input ('Enter the second initial guess: ')
% Ea = input ('Enter absolute error: ')
x0 = 3;
x1 = 4;
Ea = 0.0001;
iteration=0;
% disp('Iteration x0 x1 x2 f(x0) f(x1) f(x2) Ea')
for i=1:1000
x2 = x0 - ((f(x0))*((x1-x0)/(f(x1)-f(x0))));
Error = round(abs(x2-x1),4);
% fprintf(['%i %11.5f %11.5f %11.5f %11.5f %11.5f %11.5f %11.5f\n'],i,x0,x1,x2,f(x0),f(x1),f(x2),Error)
itervec(i,:) = [i,x0,x1,x2,f(x0),f(x1),f(x2),Error];
iteration=iteration+1;
if (Error<=Ea)
Root=x2
Error=Ea;
Iteration=iteration
break
end
x0=x1;
x1=x2;
end
Root = 4.3069
Iteration = 5
Results = array2table(itervec, 'VariableNames',{'Iteration', 'x0', 'x1', 'x2', 'f(x0)', 'f(x1)', 'f(x2)', 'Ea'})
Results = 5×8 table
Iteration x0 x1 x2 f(x0) f(x1) f(x2) Ea _________ ______ ______ ______ _________ _________ ___________ ______ 1 3 4 4.6 -16 -6 7.296 0.6 2 4 4.6 4.2708 -6 7.296 -0.79078 0.3292 3 4.6 4.2708 4.303 7.296 -0.79078 -0.087734 0.0322 4 4.2708 4.303 4.307 -0.79078 -0.087734 0.0012921 0.004 5 4.303 4.307 4.3069 -0.087734 0.0012921 -2.0587e-06 0.0001
Creating the table requires saving the intermediate results to a vector and then using array2table to load them into a table. (I am not certain what to do with the ‘Root’ variable, since it was not part of the original output array.)
.
  4 Comments
Star Strider
Star Strider on 27 Jun 2022
As always, my pleasure!
The best way to return the favour is to use your growing knowledge to help others become proficient with MATLAB!

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