MATLAB: Display Results as nicely formatted table

bisectionfunctionitteration

in this codes, how do I get the resalts as a nice table showing step by step iterations clarifying more collums would be

itteration, lower, upper, f(lower), f(upper), f(xc), f(c), xupper-xlower

function y = f(x) y = x.^3 – 2;

exists. Then:
>> format long
>> eps_abs = 1e-5;
>> eps_step = 1e-5;
>> a = 0.0;
>> b = 2.0;
>> while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    c = (a + b)/2;
    if ( f(c) == 0 )
       break;
    elseif ( f(a)*f(c) < 0 )
       b = c;
    else
       a = c;
    end
end
>> [a b]
ans = 1.259918212890625  1.259925842285156
>> abs(f(a))
ans =    0.0000135103601622
>> abs(f(b))
ans =    0.0000228224229404

Best Answer

I have no idea what f(xc) is so I didn't included it, don't use the following code to benchmark your function because it's slower (datasave isn't preallocated), use it only to get a nice table of values and have another function that doesn't make the table for benchmarks (tic toc or profiler).

Using the format short:

clc
fprintf('  itteration   lower    upper    f(lower)  f(upper)    f(c)  xupper-xlower\n')
clear
f=@(x) x.^3 - 2;
format short
eps_abs = 1e-5;
eps_step = 1e-5;
a = 0;
b = 2.0;
iter=0;
datasave=[];
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    iter=iter+1;
    c = (a + b)/2;
    datasave=[datasave; iter a b f(a) f(b) f(c) b-a];
    if ( f(c) == 0 )
       break;
    elseif ( f(a)*f(c) < 0 )
       b = c;
    else
       a = c;
    end
end
disp(datasave)

The output looks like this:

   itteration   lower    upper    f(lower)  f(upper)    f(c)  xupper-xlower
      1.0000         0    2.0000   -2.0000    6.0000       NaN    2.0000
      2.0000    1.0000    2.0000   -1.0000    6.0000   -1.0000    1.0000
      3.0000    1.0000    1.5000   -1.0000    1.3750    1.3750    0.5000
      4.0000    1.2500    1.5000   -0.0469    1.3750   -0.0469    0.2500
      5.0000    1.2500    1.3750   -0.0469    0.5996    0.5996    0.1250
      6.0000    1.2500    1.3125   -0.0469    0.2610    0.2610    0.0625
      7.0000    1.2500    1.2813   -0.0469    0.1033    0.1033    0.0313
      8.0000    1.2500    1.2656   -0.0469    0.0273    0.0273    0.0156
      9.0000    1.2578    1.2656   -0.0100    0.0273   -0.0100    0.0078
     10.0000    1.2578    1.2617   -0.0100    0.0086    0.0086    0.0039
     11.0000    1.2598    1.2617   -0.0007    0.0086   -0.0007    0.0020
     12.0000    1.2598    1.2607   -0.0007    0.0039    0.0039    0.0010
     13.0000    1.2598    1.2603   -0.0007    0.0016    0.0016    0.0005
     14.0000    1.2598    1.2600   -0.0007    0.0004    0.0004    0.0002
     15.0000    1.2599    1.2600   -0.0002    0.0004   -0.0002    0.0001
     16.0000    1.2599    1.2599   -0.0002    0.0001    0.0001    0.0001
     17.0000    1.2599    1.2599   -0.0000    0.0001   -0.0000    0.0000
     18.0000    1.2599    1.2599   -0.0000    0.0001    0.0001    0.0000
     19.0000    1.2599    1.2599   -0.0000    0.0000    0.0000    0.0000

Using the format long:

clc
fprintf('       itteration              lower             upper             f(lower)            f(upper)               f(c)           xupper-xlower\n')
clear
f=@(x) x.^3 - 2;
format long
eps_abs = 1e-5;
eps_step = 1e-5;
a = 0;
b = 2.0;
iter=0;
datasave=[];
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    iter=iter+1;
    c = (a + b)/2;
    datasave=[datasave; iter a b f(a) f(b) f(c) b-a];
    if ( f(c) == 0 )
       break;
    elseif ( f(a)*f(c) < 0 )
       b = c;
    else
       a = c;
    end
end
disp(datasave)

The output looks like this:

     itteration              lower             upper             f(lower)            f(upper)               f(c)           xupper-xlower
     1.000000000000000                   0   2.000000000000000  -2.000000000000000   6.000000000000000                 NaN   2.000000000000000
     2.000000000000000   1.000000000000000   2.000000000000000  -1.000000000000000   6.000000000000000  -1.000000000000000   1.000000000000000
     3.000000000000000   1.000000000000000   1.500000000000000  -1.000000000000000   1.375000000000000   1.375000000000000   0.500000000000000
     4.000000000000000   1.250000000000000   1.500000000000000  -0.046875000000000   1.375000000000000  -0.046875000000000   0.250000000000000
     5.000000000000000   1.250000000000000   1.375000000000000  -0.046875000000000   0.599609375000000   0.599609375000000   0.125000000000000
     6.000000000000000   1.250000000000000   1.312500000000000  -0.046875000000000   0.260986328125000   0.260986328125000   0.062500000000000
     7.000000000000000   1.250000000000000   1.281250000000000  -0.046875000000000   0.103302001953125   0.103302001953125   0.031250000000000
     8.000000000000000   1.250000000000000   1.265625000000000  -0.046875000000000   0.027286529541016   0.027286529541016   0.015625000000000
     9.000000000000000   1.257812500000000   1.265625000000000  -0.010024547576904   0.027286529541016  -0.010024547576904   0.007812500000000
    10.000000000000000   1.257812500000000   1.261718750000000  -0.010024547576904   0.008573234081268   0.008573234081268   0.003906250000000
    11.000000000000000   1.259765625000000   1.261718750000000  -0.000740073621273   0.008573234081268  -0.000740073621273   0.001953125000000
    12.000000000000000   1.259765625000000   1.260742187500000  -0.000740073621273   0.003912973217666   0.003912973217666   0.000976562500000
    13.000000000000000   1.259765625000000   1.260253906250000  -0.000740073621273   0.001585548394360   0.001585548394360   0.000488281250000
    14.000000000000000   1.259765625000000   1.260009765625000  -0.000740073621273   0.000422512079240   0.000422512079240   0.000244140625000
    15.000000000000000   1.259887695312500   1.260009765625000  -0.000158837092386   0.000422512079240  -0.000158837092386   0.000122070312500
    16.000000000000000   1.259887695312500   1.259948730468750  -0.000158837092386   0.000131823412403   0.000131823412403   0.000061035156250
    17.000000000000000   1.259918212890625   1.259948730468750  -0.000013510360162   0.000131823412403  -0.000013510360162   0.000030517578125
    18.000000000000000   1.259918212890625   1.259933471679688  -0.000013510360162   0.000059155646067   0.000059155646067   0.000015258789063
    19.000000000000000   1.259918212890625   1.259925842285156  -0.000013510360162   0.000022822422940   0.000022822422940   0.000007629394531

Related Solutions

MATLAB: How to extract specific rows from one array to another array

The problem is that you test for equivalence of floating point values. This is bound to cause errors, as has been discussed hundreds of times on this forum:

https://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

https://www.mathworks.com/matlabcentral/answers/350303-why-0-6-is-not-equal-to-0-2-3

http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

etc., etc.

The solution is to use ismembertol, or compare the absolute difference against some tolerance value:

idx = false(size(NA2,1),1);
tol = 0.0001;
for k = 1:size(CapacityOut,1)
   idx = idx | abs(NA2(:,1)-CapacityOut(k))<tol;
end
out = NA2(idx,:);

giving:

>> out
out =
   0.00000   0.98000   1.00000   0.02000  -3.10000   1.00000   1.00000
   1.10000   0.98000   0.05910   0.02000  -2.00000   1.00000   0.07790
   2.00000   0.98000   0.03000   0.02000  -1.10000   1.00000   0.04940
   3.10000   0.98000   0.00000   0.02000  -0.00000   1.00000   0.02000
   4.20000   0.98000   0.00000   0.02000   1.10000   0.05910   0.00120
   5.10000   0.98000   0.00000   0.02000   2.00000   0.00000   0.00000
   6.20000   0.98000   0.00000   0.02000   3.10000   0.00000   0.00000
>>

MATLAB: Error: “Not enough input arguments. Error in f (line 2) y = (x.^3) – 2.” I also want to verify that this code is using the bisection method to estimate a zero until two successive estimates have a diff. less than 10-6.

You need to have two files for this. One for your function that you intend to find the root of. And another for the script to do the bisection by calling that function. So, first create a file called f.m and have it contain your function:

% File f.m
function y = f(x)
   y = (x^3) - 2;
end

Then delete those lines from the top of your script file. E.g., have another file called bisection.m that contains your other code. E.g.,

% File bisection.m
format long
eps_abs = 1e-6;
eps_step = 1e-6;
a = 1.0;
b = 2.0;
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    c = (a + b)/2;
    if ( f(c) == 0 )
           break;
    elseif ( f(a)*f(c) < 0 )
           b = c;
    else
           a = c;
    end
end
[a b]
abs(f(a))
abs(f(b))

Then type the following at the command line to run your script (which will call your function f):

bisection

Best Answer

Related Solutions

MATLAB: How to extract specific rows from one array to another array

MATLAB: Error: “Not enough input arguments. Error in f (line 2) y = (x.^3) – 2.” I also want to verify that this code is using the bisection method to estimate a zero until two successive estimates have a diff. less than 10-6.

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