On 09/07/2011 06:51 AM, Steven D'Aprano wrote: 11258999068426240 > > Error in float 1.1*1.1: > >>>> b = F(11, 10)**2 >>>> y = F.from_float(1.1**2) >>>> f = y - b >>>> print f > 21/112589990684262400 > > which is slightly more than double e above, and slightly less than our > estimate of 2*a*e = 11/56294995342131200 > > So we can conclude that, at least for 1.1**2, Python floats are more > accurate than we would expect from a simple application of the binomial > theorem. (For implementations using IEEE doubles.)
The reason why the error is different from the 2*a*e is, that we encounter two problems. first problem is, that x = a + e e exists because a float does have a limited number (let's call it N) of digits and a has an infinite amount of non zero digits in the binary format. second problem is, that the result of the multiplication is not (a+e) * (a+e) but a 'rounded' version of it, because the floating point representation of the result would require about 2*N digits, whereas only N digits will be stored in the result. depending on the rounding which happened (up or down) the error will be bigger or smaller than the estimated one. -- http://mail.python.org/mailman/listinfo/python-list
