I am trying to perform the vector sum below. Why am I getting 0.0000 in the 3rd entry when I should be getting 0

[-1/14 -1/7 -3/14 0] - 1/5*[9/14 9/7 -15/14 0]

= -0.2000 -0.4000 0.0000 0

Thank you.

Answer 1

[Walter Roberson]

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

The fractions 1/5, 1/7 and 1/14 are not exactly representable in finite binary floating point, just the same way that 1/7 and 1/14 cannot be exactly represented in finite decimal expansion.

It is much the same problem as representing 1/3 as 0.33 . Add that together three times and you get 0.99 instead of 1, as the truncation losses build up.

[Eddie]

I see. I am trying to write up the Gramm Schmidt orthogonalization process, but I am running into this issue. I remember studying tolerance in my beginning CS course. Thanks again.

[Walter Roberson]

If I recall correctly, round-off can lead to significant problems in Gramm Schmidt orthogonalization

You might want to consider switching to symbolic computations.

[Eddie]

I looked into symbolic computations. Thank you for the help, this solved the problem. However, I read that the computations are the slowest, so maybe I should try and be efficient about which variables I make symbolic:

%%variables
n = size(X,1);
mu = sym(zeros(n));
Xstar = sym(zeros(n));
gstar = sym(zeros(1,n));

for i = 1:n
    mu(i,i) = 1;
    Xstar(i,:) = X(i,:);
        for j = 1:i - 1
            mu(i,j) = (X(i,:)*Xstar(j,:)') / gstar(j);
            Xstar(i,:) = Xstar(i,:) - mu(i,j) * Xstar(j,:);
        end
        gstar(i) = Xstar(i,:)*Xstar(i,:)'; 
end

end

Thanks again! Also comments on my write up would be gladly accepted. I'm a math grad student trying to catch up on good programming practices.

[Walter Roberson]

In MATLAB we do not recommend that you use i or j as variable names, as they are used as the imaginary units, equal to sqrt(-1) .

It can get confusing to readers as to whether you mean the imaginary unit or not.

Also if you accidentally forget to initialize i or j, or if there is a code path that does not assign to them, then you do not get a nice message about them being unassigned: eventually when you try to use them to subscript you get an error about positive integers or logicals being required. It can be pretty difficult to see the source of the problem, especially if you have logic like

if A > t
  i = 7
elseif A <= t
  i = 8
end

it seems obvious that this code must assign to i, but it does not necessarily do so. It is possible for both A > t and A <= t to be false in MATLAB, if either A or t are not scalars, because in MATLAB an "if" is not considered true unless all of the conditions are true. "if [-1 1] > 0" is false because only one of the two results of the comparison is true; "if [-1 1] <= 0" is false because only one of the two comparisons is true.

[Eddie]

Wow thank you so much for the input. Nope I did not initialize i and j, and only because I didn't see a need to, and now I do. I'll make the changes you recommended, take care.

[Walter Roberson]

Your "for" loops are initializing i and j okay; I was discussing in general how it can be hard in code to notice that you did not initialize variables. Failure to initialize becomes more obvious if you are using a variable name that is not i or j or pi (or inf or NaN or nan !), but if you happen to use one of those as a variable name then you can end up staring at the code for a long time trying to figure out what is going wrong. So good coding practice is to avoid using any of those as variable names.