Re: Desired behaviour ?

["Karim Belabas" <Karim.Belabas@math.u-bordeaux1.fr>]

* Loic Grenie [2007-03-10 17:41]:
>     I'm just wondering if there is a strong reason why 1.*I*I has an
>   imaginary part (equal to 0., but it's there). For instance neither I*I*1.
>   nor 1.*(I*I) have any imaginary part.

1) Remember that floating point arithmetic is not associative, and need not
be commutative.

2) Using the standard formulas for complex arithmetic

  (a+bI)(c+dI) = (a*c - b*d) + I (a*d + b*c)

[ 4 multiplications ], we obtain:

  (0 + 1.*I) * (0 + I) = (0*0 - 1.) + I*(0*1. + 0*1) = -1.

and imaginary part is indeed an exact 0.

3) Unfortunately we use Karatsuba (aka 3M) instead of standard formulas,
and compute

  (a+bI)(c+dI) = (a*c - b*d) + I ((a+b)*(c+d) - a*c - b*d)

[ 3 multiplications ]. There the imaginary part becomes

  (0 + 1.)*(0 + 1) - (0*0) - (1.*1) = 0.,

NOT an exact 0 !

So, yes, this is expected.

Cheers,

    K.B.
-- 
Karim Belabas                  Tel: (+33) (0)5 40 00 26 17
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