They shouldn't agree, and they are both correct.

u@:v y  is u v y

([: u v) y   is  u v y

u@v y   is  u@:v"v  y

which differs in rank: |:@:*:"*: is not the same as |:@:":

Henry Rich

On 2/5/2012 11:00 PM, Linda Alvord wrote:
> Shouldn't these agree?
>
>      (|: @ *:)i. 2 2
> 0 1
> 4 9
>     ([: |: *:)i. 2 2
> 0 4
> 1 9
>
> And isn't the second one correct?
>
> Linda
>
>
> -----Original Message-----
> From: programming-boun...@jsoftware.com
> [mailto:programming-boun...@jsoftware.com] On Behalf Of Brian Schott
> Sent: Sunday, February 05, 2012 4:25 PM
> To: Programming forum
> Subject: Re: [Jprogramming] Another early morning exercise
>
> 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?
>
>
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