# 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>
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: