Matlab Number Precision Affects the Results

Some numbers have 1 in their end after certain digits...

For example, -14.8 has following results if you format long. But some does not have this 1.

ans =

 -14.800000000000001

>> -14.7

ans =

 -14.699999999999999

>> -14.5

ans =

 -14.500000000000000

Right now, in my code, I have following problem: I am creating an array between -26 and 26 with step size 0.01. Part of the results are as follows:

   Columns 5182 through 5184

  25.809999999999999  25.820000000000000  25.829999999999998

  Columns 5185 through 5187

  25.840000000000000  25.850000000000001  25.859999999999999

  Columns 5188 through 5190

  25.870000000000001  25.879999999999999  25.890000000000001

  Columns 5191 through 5193

  25.899999999999999  25.910000000000000  25.920000000000002

So, I need to check some values whether if they are member of this array, however, due to accumulation of precision error, I end up with NOT MEMBER result. So, how can I solve my problem?

Answer 1

[dpb]

Use ismembertol instead. And read up on floating point precision at <What Need to Know About Floating Point>

[Berk Demir]

Thank you for your answer. Ismembertol works well for this case. However, is there a way to reduce this error? I do not need 15 digit precision, however, it accumulates

[Guillaume]

It's not clear what error you are talking about. As far as I can tell, there is no error, this is just the way floating point numbers work.

You could round the numbers to two decimals, or whatever you need. However, again, due to the way floating point numbers work this may not always give you the result you expect. ismembertol (or any other tolerance based method) as per dpb answer is the proper way to compare your numbers.

Possibly, a simpler explanation of floating point numbers.

[dpb]

"I am creating an array between -26 and 26 with step size 0.01. ... is there a way to reduce this error?"

'Pends on how you're doing it. If rounding is occurring and accumulating, you may be using loop or other inefficient technique that does have numerical issues. Show code, we can't diagnose what we can't see.

NB: that 0.01 cannot be held in floating point exactly no matter what precision; the lower the precision the more noticeable rounding will be if you accumulate.

NB2: the built-in linspace function is designed to minimize the discrepancies that occur in the alternative methods; for your case use

x1=-26;
x2=-x1;
dx=0.01;
x=linspace(x1,x2,(x2-x1)/dx);