And then the next line says:
The definition y-x*<. y % x+0=x extends the residue to a zero left
argument, and to negative and fractional finite arguments. For
example:
That said, I believe I provided a case where it is inconsistent with
positive arguments. That said, it was in my follow-up message.
I will re-state it, here, for clarity:
one=: {. 1 0.j1
one
1
2 | 1
1
2 | one
_1
Note also that this result is inconsistent with the expression from
the dictionary which appeared in the sentence that I quoted above:
Y=: one
X=: 2
Y-X*<. Y % X+0=X
1
--
Raul
On Wed, Dec 28, 2011 at 7:06 PM, Linda Alvord <lindaalv...@verizon.net> wrote:
> The definition of residue in the Vocabulary doesn't promise to deal with
> negative or imaginary numbers:
>
> "The familiar use of residue is in determining the remainder on dividing a
> non-negative integer by a positive:
>
> Linda
>
> -----Original Message-----
> From: programming-boun...@jsoftware.com
> [mailto:programming-boun...@jsoftware.com] On Behalf Of Raul Miller
> Sent: Wednesday, December 28, 2011 4:10 PM
> To: Programming forum
> Subject: [Jprogramming] inconsistent result from |
>
> {. 2 | _1
> 1
> {. 2 | _1 1j1
> _1
>
> --
> Raul
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
