machine precision and max and min value of a double-precision type

Question

(1) I have met several cases where epsilon is added to a non-negative variable to guarantee nonzero value. So I wonder why not add the minimum value that the data type can r

Answer 1

You need to understand how floating point numbers are represented in the CPU. In the data type, 1 bit is reserved for the sign, i.e. whether it is a positive or negative number, (yes you can have positive and negative 0 in floating point numbers,) then a number of bits is reserved for the significand (or mantissa,) these are the significant digits in the floating point number and finally a number of bits is reserved for the exponent. The value of the floating point number now is:

-1^sign * significand * 2^exponent

1. This means the smallest number is a very small value, namely the smalles significand with the lowest exponent. The rounding error however is much larger and depends on the magnitude of the number, namely the smallest number with a given exponent. The epsilon is the difference between 1.0 and the next representable larger value. That's why epsilon is used in code that is robust for rounding errors, and really you should scale the epsilon with the magnitude of the numbers you work with if you do it right. The smallest representable value is not really of any significant use normally.

2. You're seeing the difference between the normalized and denormalized minimum. The problem is that due to the way the significand is used it is possible to make a larger negative exponent than a positive one, say the bit pattern of the significand is all zeros except the last bit, which is one, then the exponent is effectively lowered by the number of bits in the significand. For the maximum you cannot do this, even if you set the significand to all ones, the effective exponent will still only be the exponent that is given. i.e. think of the difference between 0.000001e-10 and 9.999999e+10, the first is much smaller than the second is big. The first is actually 1e-16 while the second is approx 1e+11.

3. It depends on the precision of the floating point number of course. In the case of double precision, the difference between the maximum and the next smaller value is already huge, (along the lines of 10^292,) so your rounding errors will be very big. If the value is too small you will simply get inf instead, as you already saw. Really, there is no strict answer, it depends entirely on the precision of numbers you need. Given that the rounding error is approx epsilon*magnitude, the reciprocal of (1/epsilon) already has a rounding error of around 1.0 if you need numbers to be accurate to 1e-3 then even epsilon would be too big to divide by.

See these wikipedia pages on IEEE754 and Machine epsilon for some background info.