FWIW, this alternate phrasing of the rt verb is also possible.
rt=: (%: -)~ * ^@(%~ 0j1p1+0j2p1*i.)@]
Elijah Stone <[email protected]> wrote:
> - is always a verb, which negates its right argument. -1 0 1 is - (1 0
> 1), or _1 0 _1; so p. solves -x^2 - 1 = 0. I expect you want _1 0 1.
>
> -E
>
> On Sun, 24 Oct 2021, More Rice wrote:
>
> > Thank you Elijah. There is a lot for me to unpack in your approach. I
> > need a cup of coffee and chew on it deeply in the morning.
> >
> > Thanks again Raul. I looked at the p. verb before - this verb feels kind
> > of strange.
> >
> > For this specific case (x^6+1=0) and the example in NuVoc, they work
> > nicely! I tried something even simpler when I first saw it: x^2 - 1 = 0.
> > The answer looks very strange.
> >
> > p. -1 0 1
> > +--+--------+
> > |_1|0j1 0j_1|
> > +--+--------+
> >
> > So, I have always thought the p. verb is designed for a special kind of
> > polynomial which I don't currently understand, and opted to use the Euler
> > formula based approach instead.
> >
> > Or am I using the p. wrong?
> >
> >
> > Maurice
> >
> > On Sat, Oct 23, 2021 at 9:00 PM Elijah Stone <[email protected]> wrote:
> >
> >> Here is a fun party trick:
> >>
> >> rt=. (] %: -@[) * [: ^ [: j. ] %~ 1p1 + 2p1 * i.@]
> >> pw=. ^ :. rt
> >> f=. 1 + ] pw 6:
> >> (f^:_1) 0
> >> 0.866025j0.5 6.12323e_17j1 _0.866025j0.5 _0.866025j_0.5 _1.83697e_16j_1
> >> 0.866025j_0.5
> >> f (f^:_1) 0
> >> _2.22045e_16j6.10623e_16 0j3.67394e_16 _2.22045e_16j_6.10623e_16
> >> _2.22045e_16j2.05391e_15 0j1.10218e_15 0j3.10862e_15
> >>
> >> (Sadly, the inverter is not smart enough to invert e.g. 1 + pw&3 + pw&6,
> >> so p. is probably the more practical solution.)
> >>
> >> -E
> >>
> >> On Sat, 23 Oct 2021, More Rice wrote:
> >>
> >> > 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
> >> >>
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