3 ms·
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 som
by simonbyrne 5y ago
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 5y agoBoth 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 5y agoI'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)).