Raul, I can finally get to the problem I'm having with t.
f=: 1 2 1&p.
g=: 1 3 3 1&p.
x=: 10%~i=: i.8
pd=: 13 :'1 2 1&p. y'
pd
1 2 1&p.
5!:4 <'pd'
┌─ 1 2 1
── & ─┴─ p.
pd x
1 1.21 1.44 1.69 1.96 2.25 2.56 2.89
NB.c p. x ↔ +/c*x^i.#c from vocabulary
X=:10%~I=:i.8
PD=: 13 :'+/"1 x*"1 y^/i.#x'
PD
[: +/"1 [ *"1 ] ^/ [: i. [: # [
5!:4 <'PD'
┌─ [:
│ ┌─ / ─── +
├─ " ─┴─ 1
│
──┤ ┌─ [
│ │ ┌─ *
│ ├─ " ─┴─ 1
│ │
└─────┤ ┌─ ]
│ ├─ / ─── ^
│ │
└─────┤ ┌─ [:
│ ├─ i.
└─────┤ ┌─ [:
└────┼─ #
└─ [
1 2 1 PD X
1 1.21 1.44 1.69 1.96 2.25 2.56 2.89
j=: 13 :'(f*g) y'
j
f * g
5!:4 <'j'
┌─ f
──┼─ *
└─ g
j x
1 1.61051 2.48832 3.71293 5.37824 7.59375 10.4858 14.1986
J=: 13 :'(1 2 1 PD y)*1 3 3 1 PD y'
J
(1 2 1 PD ]) * 1 3 3 1 PD ]
5!:4 <'J'
┌─ 1 2 1
┌───┼─ PD
│ └─ ]
──┼─ *
│ ┌─ 1 3 3 1
└───┼─ PD
└─ ]
J X
1 1.61051 2.48832 3.71293 5.37824 7.59375 10.4858 14.1986
]c=:j t. i
1 5 10 10 5 1 0 0
c
1 5 10 10 5 1 0 0
5!:4 <'c'
── 1 5 10 10 5 1 0 0x
c
1 5 10 10 5 1 0 0
]C=:J t. I
ran with error:
|rank error: scriptd
| ]C=: J t.I
#$j x
1
#$J X
1
Note the little x that shows up in the tree version of c
I don’t know how to insert that x in C
Linda
from] On Behalf Of Linda Alvord
Sent: Sunday, November 04, 2012 1:31 AM
To: 'Programming forum'
Subject: Re: [Jprogramming] Taylor coefficients dyad
You explain what ^t. does, but what does t. do?
t. i.6
|syntax error
| t.i.
1 t. i.6
|length error
| 1 t.i.6
^t.
%@!
t.
t.
Linda
-----Original Message-----
From: programming-boun...@jsoftware.com
[mailto:programming-boun...@jsoftware.com] On Behalf Of Raul Miller
Sent: Monday, April 04, 2011 11:31 AM
To: Programming forum
Cc: Zsbán Ambrus
Subject: Re: [Jprogramming] Taylor coefficients dyad
On Mon, Apr 4, 2011 at 10:33 AM, Zsbán Ambrus <amb...@math.bme.hu> wrote:
> Let me ask some question on the builtins for Taylor coefficients.
>
> Firstly, I don't understand how diadic (u t.) works. Of the following
> two phrases, the first one gives the correct result, and the second
> one should give the same result according to the dictionary if I read
> it right, but I get something else. Why?
>
> ^t.i.6
> 1 1 0.5 0.166667 0.0416667 0.00833333
> 1 ^t.i.6
> _ 1 0.5 0.333333 0.25 0.2
^t.
%@!
This is all that t. does -- it's up to you how you use it.
%@! i.6
1 1 0.5 0.166667 0.0416667 0.00833333
1 %@! i.6
_ 1 0.5 0.333333 0.25 0.2
The first sentence is an example how this mechanism was intended to be used.
The second sentence is just coincidence -- the result of t. was not designed to
deal with a left argument.
--
Raul
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