I'm trying to use the table function, How do I show all the outputs for each iteration?

clear all, format long 
xl=1.5;xu=3;error=abs(xu-xl);k=0;
number_of_significant_digits=3;pre=0.5*10^(2-number_of_significant_digits);
f=@(x)x^2-3*x+2;
xr=xu;
while abs(error)>pre
xrn=(xl+xu)/2;
if f(xrn)*f(xl)>0
    xl=xrn;
    
else 
    xu=xrn;
    
end
k=k+1;
error=(xrn-xr)*100/xrn;
xr=xrn;

end
k=k';
xr=xr';
error=error';
table(k,xr,error,f(xrn))

Answer 1

[Jon]

You need to assign your values in the loop to arrays so that they are saved. Could do something like this:

xl=1.5;
xu=3;
error=abs(xu-xl);
k=0;
number_of_significant_digits=3;
pre=0.5*10^(2-number_of_significant_digits);
f=@(x)x^2-3*x+2;
xr=xu;
% preallocate (we don't know size so just estimate)
numVals = 1000;
kvals = zeros(numVals,1);
errorVals = zeros(numVals,1);
xrVals = zeros(numVals,1);
fvals = zeros(numVals,1);

while abs(error)>pre
    xrn=(xl+xu)/2;
    if f(xrn)*f(xl)>0
        xl=xrn;

    else
        xu=xrn;

    end
    k=k+1;
    error = (xrn-xr)*100/xrn;
    xr = xrn;
    % save results in arrays
    kvals(k) = k;
    errorVals(k)=error;
    xrVals(k)=xr;
    fvals(k) = f(xrn);


end

% put in table, just use values that were actually assigned
table(kvals(1:k),xrVals(1:k),errorVals(1:k),fvals(1:k),'VariableNames',{'k','xr','error','f'})

ans = 11×4 table
    k       xr       error           f     
    __    ______    ________    ___________

     1      2.25     -33.333         0.3125
     2     1.875         -20       -0.10938
     3    2.0625      9.0909       0.066406
     4    1.9688     -4.7619      -0.030273
     5    2.0156      2.3256       0.015869
     6    1.9922     -1.1765     -0.0077515
     7    2.0039      0.5848      0.0039215
     8     1.998    -0.29326     -0.0019493
     9     2.001     0.14641     0.00097752
    10    1.9995    -0.07326    -0.00048804
    11    2.0002    0.036617      0.0002442

[Jon]

Alternatively, you could assign new rows to the table on each iteration like this which may be a little simpler, not sure about efficiency, but probably doesn't matter for this small size problem:

xl=1.5;
xu=3;
error=abs(xu-xl);
k=0;
number_of_significant_digits=3;
pre=0.5*10^(2-number_of_significant_digits);
f=@(x)x^2-3*x+2;
xr=xu;
% initialize table
% this first row will get overwritten anyhow just initialize with zeros
results = table(0,0,0,0,'VariableNames',{'k','xr','error','f'});
while abs(error)>pre
    xrn=(xl+xu)/2;
    if f(xrn)*f(xl)>0
        xl=xrn;

    else
        xu=xrn;

    end
    k=k+1;
    error = (xrn-xr)*100/xrn;
    xr = xrn;
    % save results in table
    results(k,:) = {k,xr,error,f(xrn)};


end
results

results = 11×4 table
    k       xr       error           f     
    __    ______    ________    ___________

     1      2.25     -33.333         0.3125
     2     1.875         -20       -0.10938
     3    2.0625      9.0909       0.066406
     4    1.9688     -4.7619      -0.030273
     5    2.0156      2.3256       0.015869
     6    1.9922     -1.1765     -0.0077515
     7    2.0039      0.5848      0.0039215
     8     1.998    -0.29326     -0.0019493
     9     2.001     0.14641     0.00097752
    10    1.9995    -0.07326    -0.00048804
    11    2.0002    0.036617      0.0002442

[Jon]

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