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
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