I think that was a good decision.
A couple of points regarding d. :
0. The 0 in Dictionary's entry for d. [0] means, as far as I know, that the
product of u d. n is (monadic) verb with rank 0. However,
(1 + %) d.0 NB. OK
(1 + %)"0
(1 + %) d.1 NB. OK
-@%@*:
(1 + %) d._1 NB. Not OK
1&* + ^.
Is my understanding of "Derivative u d. n 0" correct? If it is, is the d.
entry inaccurate or is the implementation of d. (and other related entries)
incorrect?
1. The entry for d. points to the entry for D. and its description does not
make clear if derivative is meant in the sense of the usual (real)
derivative or the complex derivative. Does anybody know? I guess I could
try to find out by testing it more but I thought I should ask first.
[0] Derivative u d. n 0
http://www.jsoftware.com/help/dictionary/dddot.htm
On Tue, Oct 3, 2017 at 3:32 PM, Henry Rich <henryhr...@gmail.com> wrote:
> I checked in this fix for the next release.
>
> I added more commentary to the original function, and after doing so I
> realized that a couple of phrases could be removed.
>
> I resisted the urge to define (0&^ d. _1) as (% >:&0) .
>
> Henry Rich
>
>
>
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