Raul - I get this too. With a little experimentation to try to simplify the expression, here's what I came up with - it appears to be a tacitizing problem:
e=: '(x+t) + (1+x) + t=. y' 2 (13 :e, 4 :e) 3 10 11 e=: '+/x,t,1,x,t=. y' 2 (13 :e, 4 :e) 3 11 11 NB. Parens matter: e=: '(t+x) + 1+x + t=. y' 2 (13 :e, 4 :e) 3 11 11 e=: 't+x + (1+x) + t=. y' 2 (13 :e, 4 :e) 3 11 11 NB. So does assignment e=: '(x+y) + (1+x) + y' 2 (13 :e, 4 :e) 3 11 11 e=: '+/(x,t),(1,x),t=. y' 2 (13 :e, 4 :e) 3 10 11 On Mon, Sep 8, 2008 at 12:23 AM, Raul Miller <[EMAIL PROTECTED]> wrote: > e=: '(x^t) * (1-x) ^ >:t=. y' > 2 (13 :e, 4 :e) 3 > 128 8 > > FYI, > > -- > Raul > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > -- Devon McCormick, CFA ^me^ at acm. org is my preferred e-mail ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
