Since the dictionary defines the language, I think a better way of stating the issue is that the implementation incorrectly does something different?
Thanks, -- Raul On Thu, Jun 22, 2017 at 5:53 PM, Henry Rich <[email protected]> wrote: > What it actually is (when x and y are not infinite and x is not 0) is > > d =. |x > ab =. (+x) * y > q =. <.@(0.5&+)&.+. ab%d > y - x * q > > If my algebra is right, this is equivalent to > > (x|y)=y-x* <.@(0.5&+)&.+.y%x+0=x > > In other words, the Dictionary description incorrectly says that the complex > quotient will be rounded using complex floor, while in reality each > component of the quotient is (tolerantly) rounded independently. > > Henry Rich > > > > > > > > On 6/22/2017 4:53 PM, Louis de Forcrand wrote: >> >> Hi, >> >> There's been recent discussion on the GNU APL forums about how to define >> the residue function on complex arguments. I wondered how it was specified >> in J, but the dictionnary page seems kind of vague. >> >> Nuvoc says that >> (x|y)=y-x*<.y%x+0=x >> for all scalars x and y (including complex numbers). >> However, this is untrue for >> 'x y'=: 1j1 3j4 >> (tested on the old J for iOS) >> >> While it would seem like a logical extension, the dictionnary only says >> that this is true for rational numbers. >> >> So how is the residue of two complex numbers defined in J? >> >> Thanks, >> Louis >> ---------------------------------------------------------------------- >> For information about J forums see http://www.jsoftware.com/forums.htm > > > > --- > This email has been checked for viruses by AVG. > http://www.avg.com > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
