[simonbyrne]

The key thing about floating point is that it maintains relative accuracy: in your case, if you have say f(x) and g(x) are both O(1e200), and are correct to some small relative tolerance, say 1e-10 (that is, the absolute error is 1e190). Then the relative for f(x)/g(x) stays nicely bounded to about 2e-10.

However if you do f(x) - g(x), the absolute error is on the order of 2e190: if f(x) - g(x) is small, then now the relative error can be huge (this is known as catastrophic cancellation).

[zoomablemind]

Both f(x) and g(x) could be calc'ed to proper machine precision (say, GSL double prec ~ 1E-15). Would this imply, that beyond the machine precision the values are effectively fixed to resp. zero or infinity, instead of carrying around the extreme orders of magnitude?

[simonbyrne]

I'm not exactly sure what you're asking here, but the point is that "to machine precision" is relative: if f(x) and g(x) are O(1e200), then the absolute error of each is still O(1e185). However f(x)/g(x) will still be very accurate (with absolute error O(1e-15)).