Kip may be onto something that at least for me shows a little of the
problem of going from an expression with @ to one without. For example
in the first 3 execution below, the first two produce the same result,
but the third, does not.
(|: @ *:) i. 2 2
0 1
4 9
13 : '|: @ *: y' i. 2 2
0 1
4 9
13 : '|: *: y' i. 2 2
0 4
1 9
To me, the way to make the errant third one above conform is to
include the rank operator in the defintion as follows.
13 : '|:"*: *: y' i. 2 2
0 1
4 9
On the other hand the third of these next 3 *does* produce the same
result because @: automatically implies the rank of the previous
example.
(|: @: *:) i. 2 2
0 4
1 9
13 : '|: @: *: y' i. 2 2
0 4
1 9
On Sun, Feb 5, 2012 at 2:41 PM, km <k...@math.uh.edu> wrote:
> (|: @ *:) i. 2 2
> 0 1
> 4 9
>
> Composition?
>
> Sent from my iPad
>
>
> On Feb 5, 2012, at 8:16 AM, Boyko Bantchev <boyk...@gmail.com> wrote:
>
>> On 5 February 2012 14:11, Linda Alvord <lindaalv...@verizon.net> wrote:
>>> My goal has been to translate from expressions with @ to ones without it.
>>
>> You also mentioned eliminating @ in another thread.
>> Why do you consider it important?
>> @ is the composition of functions – and is composition not the most
>> natural operation on functions that one could think of?
--
(B=)
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