rule for nn^3*K3(xx,yy) --> K3(nn*xx, nn*yy)?

• Subject: rule for nn^3*K3(xx,yy) --> K3(nn*xx, nn*yy)?
• From: Robert Dodier
• Date: Sat, 27 Nov 2010 11:27:45 -0700

On 11/26/10, RLP <rhyslpratt at yahoo.com.cn> wrote:

> Why does not the pattern variable aa match the whole expression, i.e.
> the whole (n/a)^3*K3((b*log(n))/n^2+a/n,(b*log(n-1))/(n-1)^2+a/(n-1)). I
> have learned from "? matchdeclare" that the matching algorithm is
> greedy.

I'm pretty sure that's because of the other operators in the match
pattern ("^" and K3). If aa takes all, there's nothing left to match them.

> (I am not native to English, so, sometimes, I really feel
> English language is disturbing. :-))

Hmm, Rhys Pratt is a very Welsh name, isn't it? Are you from Wales?

best,

Robert Dodier

• Prev by Date: New package for drawing direction fields with gnuplot
• Next by Date: Is a parallel build of Maxima (using make -j N), OK?
• Previous by thread: rule for nn^3*K3(xx,yy) --> K3(nn*xx, nn*yy)?
• Next by thread: rules for non-commutative-multiplication
• Index(es):
  • Date
  • Thread