what is a bit confusing though is:

     $ 3 (i.@{.@$) i. 4 5
0 1 2 3 4
    # 3 (i.@{.@$) i. 4 5             ---returns nothing
0

Assuming you meant to apply $ monadically, and just got lucky with the sample 
data :)

Lucky?, how about:


x (f g) y means x f (g y)

or in this case:

x f~ (g y)

--
Met vriendelijke groet,
@@i = Arie Groeneveld

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