Actually, then (u/k{.y) u (u/k}.y) produces the correct domain error
when y=:'!', doesn't it?On Mon, Apr 11, 2011 at 5:47 PM, Boyko Bantchev <[email protected]> wrote: > On 11 April 2011 23:43, Brian Schott <[email protected]> wrote: >> Boyko, >> >> The expression I cited applies to "... k e. 0,#y .", of which 1 >> is a possible value of #y. Isn't that good enough? > > If you mean that (u/k{.y) u (u/k}.y) is equivalent to u/y, and that > that equivalence could be used to define u/y for 1=#y, then note > that the said `equivalence' does not hold for many, many values > of u, y, and k. > > In your example, consider y=.'!' (keeping k=.1 and u=.-). > Then u/y appears to be '!' (so says the J interpreter), but > (u/k{.y) u (u/k}.y) is obviously erroneous (being equivalent > to (-/'!') - 0). ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
