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

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

[Christoppe]

update: I already figured how to do The roots are the following:

x1 = 1

x2 = 2

x3 =-1

Iteration = 9

Answer 1

[Star Strider]

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.)

(Previous post for reference: How to convert fprintf from single row to multiple rows?)

.

[Christoppe]

Hi it's a different code from what i posted a couple of days ago I was doing a secant back then. however, now I am doing a Jacobi method....

[Star Strider]

The same essential approach applies here:

format short
% Eqn=input('Enter the number of equations, n: ');
Eqn = 3;
% Error=input('Enter absolute errror, Ea: ')
Ea = 0.0001;
% A=input('Enter coefficients (DIAGONALLY DOMINANT!), Matrix A: ');
A = [4  -1 -1;-2 6 1;-1 1 7];
% B=input('Enter constants, Matrix B: ');
B = [3;9;-6];
% C=input('Enter the initial guess: x0,x1,x2: ');
C = [0;0;0];

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
    itervec(j,:) = [j x(:).'];
% fprintf('Iteration no %d \n',j)
% x
if abs(x-C)<=Ea
    break
end

C = x;
end

Results = array2table(itervec, 'VariableNames',{'Iteration', 'x1', 'x2', 'x3'})

Results = 9×4 table
    Iteration      x1        x2        x3   
    _________    ______    ______    _______

        1          0.75       1.5    -0.8571
        2        0.9107    1.8929    -0.9643
        3        0.9822    1.9643    -0.9975
        4        0.9917    1.9937    -0.9974
        5        0.9991    1.9968    -1.0003
        6        0.9991    1.9998    -0.9997
        7             1    1.9997    -1.0001
        8        0.9999         2         -1
        9             1         2         -1

The ‘Error’ variable is not defined, so I changed the if block test condition to use ‘Ea’ instead.

.

[Christoppe]

Hello sir, Thank you so much for your assistance, it means a lot to us who don't have enough knowledge in MATLAB. I hope I can return the favor someday...

[Star Strider]

As always, my pleasure!

The best way to return the favour is to use your growing knowledge to help others become proficient with MATLAB!