That x [EMAIL PROTECTED]&0 y is equivalent to f^:x y follows from the defn x (f&n) y <=> (f&n)^:x y , not the other way around.
----- Original Message ----- From: Roger Hui <[EMAIL PROTECTED]> Date: Friday, August 31, 2007 11:10 Subject: Re: [Jprogramming] bug or feature? To: Programming forum <[email protected]> > > x (f&n) y <=> (f&n)^:x y > > > > but I can't quite see how that follows from what > > the Vocabulary says: > > > > The phrase x [EMAIL PROTECTED]&0 y is equivalent to f^:x y , apply the > > monad > f > > x times to y . > > Since x [EMAIL PROTECTED]&0 y is [EMAIL PROTECTED]&0^:x y according to the > defn, > it comes down to the meaning of [EMAIL PROTECTED]&0 z . But: > > [EMAIL PROTECTED]&0 z > z [EMAIL PROTECTED] 0 defn of u&n > f z[0 defn of @ > f z > > That is, the monad [EMAIL PROTECTED]&0 is the same as the monad f, > whence [EMAIL PROTECTED]&0^:x y is the same as f^:x y. > > > > ----- Original Message ----- > From: Nollaig MacKenzie <[EMAIL PROTECTED]> > Date: Friday, August 31, 2007 10:56 > Subject: Re: [Jprogramming] bug or feature? > To: Programming forum <[email protected]> > > > > > On 2007.08.31 16:47:16, you, > > the extraordinary Yuri Burger, emitted: > > > > > > > > how it works ??? > > > > > > p=:*&0.1 > > > p 10 > > > 1 > > > 1 p 10 > > > 1 > > > 2 p 10 > > > 0.1 > > > 3 p 10 > > > 0.01 > > > 4 p 10 > > > 0.001 > > > 5 p 10 > > > 0.0001 > > > > > > > > > > > > > > > (,&'-') '|' > > > |- > > > 2 (,&'-') '|' > > > |-- > > > 10 (,&'-') '|' > > > |---------- > > > > x (f&n) y <=> (f&n)^:x y > > > > but I can't quite see how that follows from what > > the Vocabulary says: > > > > The phrase x [EMAIL PROTECTED]&0 y is equivalent to f^:x y , apply the > > monad > f > > x times to y . ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
