Largest representable floating point / Tiny mistake in "What Every Computer Scientist Should Know About Floating-Point Arithmetic".

Question

I believe there is a tiny mistake in the paper "What Every Computer Scientist Should Know About Floating-Point Arithmetic".

It claims that

A less common situation is that a real number is out of range, that is, its absolute value is larger than
http://img219.imageshack.us/img219/7396/screenshot2011052714105.png http://img219.imageshack.us/img219/7396/screenshot2011052714105.png.

This is almost exact, the maximum representable floating point number is slightly less than that, and the real number is out of range when it is larger than
http://img707.imageshack.us/img707/9236/screenshot2011052714045.png http://img707.imageshack.us/img707/9236/screenshot2011052714045.png

Right?

• ShreevatsaR: Perhaps "larger than" should be "larger than or equal to".
• jadn: Yes that would be correct as well, but not as precise.
• Nemo: Looks right to me. Also, you can ditch the 1 + from your expression if you just make the sum start at i=0.
• Nemo: Wait, that's not right. Beta is the base, right? So for example, if beta=10, you want a number like .9999 * 10^n. I think you need a (beta-1) factor in your sum.

Answer 1

I cannot be bothered with images, so I will write b for "beta" and m for "e_max".

So say b is the base, p is the precision, and m is the max exponent.

Then I think the expression you want is:

(1 - b^(-p)) * b^m

For example, for base-10 with 4 digits of precision and a max exponent of 12, this gives:

.9999 * 10^12

...which is correct.

Note that this is not exactly right for IEEE floating point, because there the leading "1" bit is implicit. And I vaguely recall some oddities when the exponent is all 1's.