Floating point multiplication in python

Gelonida N gelonida at gmail.com
Wed Sep 7 10:57:53 CEST 2011

On 09/07/2011 06:51 AM, Steven D'Aprano wrote:
> 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.

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