- From: Stan Devitt <jsdevitt@radicalflow.com>
- Date: Wed, 12 Apr 2000 20:12:12 -0400
- To: "William F. Hammond" <hammond@csc.albany.edu>
- Cc: <www-math@w3.org>
The (D^2)(y) --> D(y)*D(y) interpretation is exactly the interpretation you get in, for example, Maple. Maple has the somewhat unique feature of supporting an algebra of functions so that (f+g)(x) -> f(x) + g(x), etc. and (1)(x) -> 1 . The operator precedence and binding is the same in both expressions $D^2 y$ and (D^2)(y). It is only the semantic interpretation that may be different. In the spirit of David's markup, it might have been written something more like \apply{D^2}{y} and then it is easier to see how the macro definitions for apply could take it either way, perhaps conditional on the presence of D or an operator. (see more below) Stan ----- Original Message ----- From: William F. Hammond <hammond@csc.albany.edu> To: <jsdevitt@radicalflow.com> Cc: <www-math@w3.org> Sent: Wednesday, April 12, 2000 6:04 PM Subject: Re: comments re draft version 2.0 > You wrote: > > > D^2 ( y ) could equally well mean D(y) * D(y) > > While both of these two expressions are consistent with the types, the > second is more naturally indicated with (D y)^2, which requires parentheses > in order to distinguish it from D (y^2), while parentheses in the > expression D^2 (y) do not serve to distinguish it from D^2 y . > > If a parser were to infer D(y)*D(y) from D^2 y , then wouldn't it be > via an earlier inference of (D y)^2 from D^2 y ? But doesn't True, (D y) * (D y ) could arise in this way, but it avoids the issue. The real question is "what meaning" do wish to associate with \apply{D^2}{y}. I claim that the answer may depend on the properties of D, and that even then, there is more than one reasonable meaning - at least one based on operator composition, and one based on product. So long as the author can say which definition is to be used (and we can) , we can over-ride whatever default meaning is chosen. The outcome can depend on the signature, for example, the presence of an operator versus a symbol. Stan.
Received on Wednesday, 12 April 2000 20:09:45 UTC