Issue finding numerical error (problem with MATLAB's precision?)
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I'm trying to recreate some code that a TA showed me the other day. Suppose we have the system Ax=b. Let's say A = [2,1,1,0;4,3,3,1;8,7,9,5;6,7,9,8]. And let's say b = [1;1;-1;-3]. I saw a trick in MATLAB that can solve this equation using the "\" symbol. So A \ b should yield the following:
ans =
1.0000
0.0000
-1.0000
-0.0000
If you subtract this from the true solution:
(A \ b) - [1;0;-1;0]
ans =
1.0e-15 *
-0.2220
0.2538
0
-0.0912
Then you can actually see the error in MATLAB's solution versus the actual solution. My issue is that I cannot recreate this. When I do A \ b I get this:
ans =
1
0
-1
0
>> (A \ b) - [1;0;-1;0]
ans =
0
0
0
0
Therefore I cannot see the error, or this answer is just very imprecise-I'm not sure. How can I recreate the correct output? Is there another way I can see numerical error from an approximated solution versus the actual solution in MATLAB?
0 Comments
Answers (2)
Andrew Newell
on 16 Mar 2015
Edited: Andrew Newell
on 16 Mar 2015
That pesky MATLAB - getting more accurate all the time! Here is an example that I got from the blog article Floating Point Comparisons in Matlab:
x = 0.8-0.7;
x-0.1
On my computer, that returns 8.3267e-17 (which is still pretty good!).
2 Comments
Andrew Newell
on 17 Mar 2015
Edited: Andrew Newell
on 17 Mar 2015
Did you try this example from the same article?
K = 7;
s = zeros(K,1);
for k = 1:K
s(k) = sum((10^-k)*ones(10^k,1));
end
s
e = abs(s - 1)
Jan
on 17 Mar 2015
Edited: Jan
on 17 Mar 2015
Which format is active in the command window? What do you get with this:
format long g
(A \ b) - [1;0;-1;0]
Are A and B arrays of type double?
3 Comments
Andrew Newell
on 17 Mar 2015
I fixed that for you. You should be able to click on "Edit" and fix it yourself, though.
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