[Prev][Next][Index][Thread]
Re: comments re draft version 2.0
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.
References: