On Sun, 15 Oct 2006 20:03:18 +1000, Ben Finney wrote: > "Paddy" <[EMAIL PROTECTED]> writes: > >> Ben Finney wrote: >> > Your calculator is probably doing rounding without you asking for >> > it. >> > >> > Python refuses to guess what you want, and gives you the >> > information available. >> >> I don't think Python should take too much credit here. > > I don't understand what you could mean by this. Credit for "giv[ing] > you the information available"? That's exactly what it's doing. > >> Floating point calcuations are subject to rounding. Sometimes it >> shows. > > And Python is showing it, rather than hiding it. It certainly isn't > doing any rounding unless asked to do so.
Python simply exposes whatever the C maths library does. The C maths library is almost certainly doing some rounding: floats have only a finite precision, which generally means the designer of the math library has two choices: just truncate (chop) the calculation, or carry extra guard digits (or bits) and round down. Most systems these days use guard digits, as that is more accurate than truncating to a fixed precision. If you mean that Python isn't doing *extra* rounding, above and beyond what the C library is doing, you're correct. But rounding is happening. This is a useful resource: "What every computer scientist should know about floating-point arithmetic" http://docs.sun.com/source/806-3568/ncg_goldberg.html To go back to the Original Poster's problem, he pointed out that his "16 bit calculator" gave a more accurate (but less precise) answer for sin(pi), namely 0, instead of the more precise (but less accurate) 1.2246063538223773e-016 that Python reported. That just goes to show that, sometimes, extra precision in floating point maths is a bad thing -- a less precise library would actually have given a more correct answer. This isn't strictly a Python question, but if there is anybody out there who knows what the C library is doing, I'd appreciate an answer: since sine is periodic, doesn't it make sense to reduce the argument modulo pi before calculating the sine? Something like this: def _sin(x): x = x % math.pi return math.sin(x) Yes, you lose precision for large values of x because of the modulo, and because math.pi isn't precisely pi, but that's got to be better than losing precision for moderate values of x, surely? Have I missed something? -- Steve. -- http://mail.python.org/mailman/listinfo/python-list
