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## Re: [Axiom-developer] Re: "Factoring" Expression Integer/ simplifying po

 From: Martin Rubey Subject: Re: [Axiom-developer] Re: "Factoring" Expression Integer/ simplifying powers (an example) Date: 06 Mar 2006 12:42:25 +0100 User-agent: Gnus/5.09 (Gnus v5.9.0) Emacs/21.4

```Dear Francois,

> 1/ with mupad I can define an operator *%* (with any priority),
>    so I type x *%* y and not etoile (x, y), it's nicer to read

Yes, the infix operator precedences seem to be hardcoded in
interp/newaux.lisp.pamphlet.

>    solve (x*y=0, [x,y]) doesn't work, I must type solve ([x*y=0],[x,y])
>
> 5/ With axiom I can't compute x^(n) for the indeterminate n
> With other CAS I can transform :
> (1+x)^n*(1+x) + x + 1 into (1+x)^n*(1+x)

I don't really understand. (1+x)^n*(1+x) + x + 1 is not equal to (1+x)^n*(1+x),
is it?

> a/ Either with a solve (rec (u(n+1)=u(n)*(1+x)+x+1, u(0)=1, u(n)))

> b/ or factor ((1+x)^n*(1+x) + x + 1) computes (1+x)^n*(1+x)

I don't believe this.

> a/ It seems that solve (rec ...)) doesn't exist in axiom.

not yet. As you might know I did already propose a hierarchy for these things
and Antoine Hersen said that he'd be interested in implementing. Unfortunately,
he did not do it yet.

I can help, but I'm unable to do it all by myself, I don't have enough time. If
you come to the axiom workshop, I guess we could get it started in very little
time.

> // Is it an axiom bug in axiom ?
>   I don't understand why (x+y+1)^2 ::DMP([x],Integer) gives me an
>   Expression Integer, not an error.  //

Well, even (x+y+1)^2 ::DMP([x,y],Integer) gives an EXPR INT. The reason is,
that :: has a very high precedence. So you are converting 2 to a DMP([x],INT)
-- or a DMP([x,y], INT) -- and then you compute (1+x+y) to the power of that
polynomial...

> But when I use Expression I only have ONE type : Expression Integer.
> On this example the (1+x)^n term force me to use Expression Integer.
>
> So I lose (almost) all the power of axiom.

very true. It is one of my main goals to introduce a good hierarchy of common
functions. I hope that this won't take too long. Help (with coding) is very
much appreciated.

Unfortunately, I don't have the time to go through your worksheet right now,
sorry.

Martin

```