Plot data from while loop

I've written an algorithm that can approximate the root of a multivariable function using the Newton-Raphon method. The algorithm, given 2 multivariable functions and 2 guess values, runs through an equation multiple times, returning the approximate root and error each iteration. I am trying to plot the error against the iteration. However, when i do so, i get a blank plot. How would I approach plotting this error data (eax and eay [found in part 3]) from the while loop against the iteration number?

Thank you

Below is the code:

disp(['<strong>Method: </strong>' 'Newton-Raphson for Simultaneous  Equations'])
n = input('Desired Accuracy: ');
e = 0.5*(10^(2-n));
syms x y
Data_Vector = [f1 f2 diff(f1,x) diff(f1,y) diff(f2,x) diff(f2,y)];     
Iteration = 0;
x=input('x_guess: ');
y=input('y_guess: ');

  
while (1)
disp('------------')
    Iteration = Iteration + 1;
disp(['<strong>Iteration: </strong>' num2str(Iteration) ])
%Part 1: Initializing
    x;
    y;
    i = Iteration - 1;
    disp([ '(x_' num2str(i) ',' 'y_' num2str(i) ')' ' = ' '(' num2str(x) ',' num2str(y) ')'])
    Data_Vector_Num1 = subs(Data_Vector);
    Data_Vector_Num2 = double(Data_Vector_Num1);
    
%Part 2: Evaluation
    %Part 2A: X - Root
     xr = x - (Data_Vector_Num2(1)*Data_Vector_Num2(6)-Data_Vector_Num2(2)*Data_Vector_Num2(4))/(Data_Vector_Num2(3)*Data_Vector_Num2(6)-Data_Vector_Num2(4)*Data_Vector_Num2(5));
     
    %Part 2B: Y - Root
     yr = y  - (Data_Vector_Num2(2)*Data_Vector_Num2(3)-Data_Vector_Num2(1)*Data_Vector_Num2(5))/(Data_Vector_Num2(3)*Data_Vector_Num2(6)-Data_Vector_Num2(4)*Data_Vector_Num2(5));

     
%Part 3: Approximate Error
    
        eax = abs((xr-x)/xr)*100;
        eay = abs((yr-y)/yr)*100;
        
        

%Part 4: Resetting X & Y
        
    x=xr;
    y=yr;
    
%Part 5: Root Check
    f1_rc_a = subs(f1);
    f1_rc_b = double(f1_rc_a);
    f2_rc_a = subs(f2);
    f2_rc_b = double(f2_rc_a);
    disp([ '(x_' num2str(Iteration) ',' 'y_' num2str(Iteration) ')' ' = ' '(' num2str(x) ',' num2str(y) ')'])
    disp(['Value of Root: ' 'f1(' num2str(x) ',' num2str(y) ')' ' = ' num2str(f1_rc_b) '  ||  ' 'f2(' num2str(x) ',' num2str(y) ')' ' = ' num2str(f2_rc_b)])
    disp(['Approximate Percent Error: ' 'x:' num2str(eax) '%' ' y:' num2str(eay) '%'])
    

    if eax < e && eay < e
        break,end

end

%Part 6: Summary
    disp('=======================')
    disp('<strong>Summary of Results: </strong>')
    disp(['<strong>Method: </strong>' 'Newton-Raphson for Simultaneous Equations'])
    disp(['<strong>Number of Iterations:</strong> ' num2str(Iteration)])
    disp('<strong>Approximate Root: </strong>')
    [round(x,n,'significant') round(y,n,'significant')]
            
    disp(['<strong>Value of Root: </strong>' 'f1(' num2str(x) ',' num2str(y) ')' ' = ' num2str(f1_rc_b) '      ' 'f2(' num2str(x) ',' num2str(y) ')' ' = ' num2str(f2_rc_b)])
    disp([ '<strong>Accuracy: </strong>' num2str((n)) ' Significant Figures'])

Here, I'll attatch a sample output:

    Method: Newton-Raphson for Simultaneous  Equations
Desired Accuracy: 4
x_guess: 1.5
y_guess: 3.5
------------
Iteration: 1
(x_0,y_0) = (1.5,3.5)
(x_1,y_1) = (2.036,2.8439)
Value of Root: f1(2.036,2.8439) = -0.064375  ||  f2(2.036,2.8439) = -4.7562
Approximate Percent Error: x:26.3272% y:23.0715%
------------
Iteration: 2
(x_1,y_1) = (2.036,2.8439)
(x_2,y_2) = (1.9987,3.0023)
Value of Root: f1(1.9987,3.0023) = -0.0045199  ||  f2(1.9987,3.0023) = 0.049571
Approximate Percent Error: x:1.8676% y:5.2764%
------------
Iteration: 3
(x_2,y_2) = (1.9987,3.0023)
(x_3,y_3) = (2,3)
Value of Root: f1(2,3) = -1.2861e-06  ||  f2(2,3) = -2.214e-05
Approximate Percent Error: x:0.064969% y:0.076305%
------------
Iteration: 4
(x_3,y_3) = (2,3)
(x_4,y_4) = (2,3)
Value of Root: f1(2,3) = 1.501e-13  ||  f2(2,3) = 2.7769e-12
Approximate Percent Error: x:8.0619e-07% y:1.9554e-05%
=======================
Summary of Results: 
Method: Newton-Raphson for Simultaneous Equations
Number of Iterations: 4
Approximate Root: 

ans =

     2     3

Value of Root: f1(2,3) = 1.501e-13      f2(2,3) = 2.7769e-12
Accuracy: 4 Significant Figures

Answer 1

[Mathieu NOE]

hello

you have to index the error variables with Iteration

otherwise you're overwritting each time (and you get only a scalar at the end - that's why the plot is blank)

eax(Iteration) = abs((xr-x)/xr)*100;
eay(Iteration) = abs((yr-y)/yr)*100;

[Michael Jacobson]

Ah ok, I did that and now the plot is coming out correctly. Thank you for the help

[Mathieu NOE]

you're welcome