Thank you for the notes - I'll keep it in my bookmark as reference! I started out this morning with my pre-calculus book trying to practice J sentences with. I wanted the numeric answers for complex roots. Like:
// matlab version syms x eqn = x^6+1 == 0 solve(eqn, x) But, it seems J only gives the principal root (?), not all 6 of them; so, another opportunity for practise. But, I ended up writing like ... "matlab": ^ 0j1 * (1p1 + 2p1 * i.6) % 6 NB. 1st version That was why I was browsing NuVoc, looking for examples/ideas, hoping to see something to make my J sentence looks more ... "J-idiomatic" (while learning something out of the process). This is all I can I come up with today: ^ 0j1 * 6 %~ 1p1 + 2p1 * i.6 NB. 2nd version How would the same answer look like in the eyes of J Masters? thanks for your thoughts. On Sat, Oct 23, 2021 at 4:18 PM 'Pascal Jasmin' via Programming < [email protected]> wrote: > a more hollistic explanation, > > Most conjunctions, and including the & and @ famillies, produce verb > phrases when bound. A verb or verb phrase can/has to produce different > results/computations depending on monadic or dyadic cases. In u@v, u is > always monadic, and v is ambivalent. in u&v, v is always monadic, and u is > the valence of the verb phrase. > > A missing "composing conjunction" in J is ([ u v) where u is always > dyadic and v is ambivalent. But the fact that it is easy to write as a > fork suggests a dedicated conjunction is not needed. > > > On Saturday, October 23, 2021, 03:30:09 p.m. EDT, Raul Miller < > [email protected]> wrote: > > > > > > https://www.jsoftware.com/help/dictionary/d631.htm > > x u&.v y ↔ vi (v x) u (v y) > > Here: > u is + > v is *: > vi is %: (or *:inv) > x is 3 > y is 4 > > So these are equivalent > 3 +&.*: 4 > %: (*:3) + (*: 4) > *:inv (*:3) + (*: 4) > > I hope this makes sense. > > -- > Raul > > On Sat, Oct 23, 2021 at 3:03 PM More Rice <[email protected]> wrote: > > > > Hello, > > > > (Sorry for the previous empty email - web page problem) > > > > please excuse another newbie question ... > > > > Ref: https://code.jsoftware.com/wiki/Vocabulary/starco > > > > pythag =: +&.*: > > 3 pythag 4 > > 5 > > > > + operated dyadically and acted on both x and y - ok. > > > > but how does *: know to act on x as well? Isn't pythag using the monadic > > definition of *: to square y only? > > > > so magical ... > > > > thank you for the pointer and have a great weekend. > > > > > > Maurice > > ---------------------------------------------------------------------- > > 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 > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
