> 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
