Incorrect evaluation of logic statement

I'm new to MATLAB so I might be missing something obvious. I've read that MATLAB has floating point issues but this involves comparing two number with only a few digits.

Inside a while loop I have:

 disp(x)
 disp(B)
 disp(ind)
 x>=B^(ind+1)

after some loops I get:

 1.0000e-03
 10
 -4
 ans = 0

Entering the expression in the command window:

 >> 1.0000e-03>=10^(-4+1)
 ans =
      1

How do I get correct operation inside the while loop? Are there separate settings for precision inside loops? Thanks in advance.

[Sean de Wolski]

What are you doing after the displays in the loop? Please provide a whole minimal working example.

[Geoff Hayes]

But is x really 1.0000e-03 or is it just the rounded up number that is being displayed. Would be interesting to see what the following returns:

fprintf('x=%.24f\n',x);

for 24 decimal places...

[Sara]

If you type in the command window

format longg

you may discover that your variables are not what is shown with the disp command.

[Tommy]

Thanks for the answers. Initially, x=123.321 exactly but when I say x=x-100, I get x=23.32099... That seems like a pretty serious issue, and not one I've encountered in any other mathematics package. I know MATLAB is used in industry where precision is needed, so surely there's a way to fix this. Perhaps a different data type?

[Sara]

The error should be in the order of x*eps not that big. Can you attach your code?

Answer 1

[the cyclist]

I sincerely hope that you will not be offended by this, but the way you have phrased your question (and its title) indicates that you don't have a very sophisticated knowledge of floating point arithmetic in general. I can assure you that MATLAB is accurately evaluating the logical statement in both cases, but I am sympathetic that this can be hard to understand.

As Sean mentions, actual code that exhibits the difference would be helpful. But the gist is that x is probably not exactly 1.0000e-3, or B is not exactly 10, etc, inside your for loop (even if you expect them to be).

Your comparison is reliant on a condition that is finely tuned. For example,

x = 1.e-3
B = 10
ind = -4
x>=B^(ind+2*eps+1)

will give a result of 0, because I added just a tiny offset (approximately that of round-off error) to your equation.

If you display your variable with

format long

you may get a better handle on what is going on. Here is a good starting point for reading more about the trickiness of floating point arithmetic.