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

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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
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

Star Strider on 27 Jun 2022
% 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.)
.
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!