Yes, you want x=._ (infinity). Thus, f^:(<_) y or you can use the convenient shorthand form, f^:a: y .
-Dan On Nov 20, 2011, at 5:36 AM, David Vaughan <[email protected]> wrote: > If I'm doing f^:(<x) y, can I write an expression so that f is repeatedly > applied to y until the result has come up before? > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
