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Topic: Maxima - simplification/evaluation
Replies: 5   Last Post: Mar 1, 2013 2:24 PM

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 Mate Posts: 389 Registered: 8/15/05
Re: Maxima - simplification/evaluation
Posted: Feb 28, 2013 3:03 PM

On Feb 28, 6:56 am, Richard Fateman <fate...@cs.berkeley.edu> wrote:
> On 2/27/2013 3:32 AM, Mate wrote:
>

> > In Maxima, the single quote operator ' does not
> > prevent evaluation for '(2*3); this _is_ evaluated to 6.

>
> No, 2*3 is simplified to 6.
>

> > We can set
>
> > simpl:false;
> > n : 2*3;

>
> that would be  simp:false.
>
>
>

> > to keep n unevaluated
>
> no, n is unsimplified. It is evaluated though the evaluation
> leaves it unchanged.
> try  x:2
> n:x*3
> to see that n is evaluated.
>
>   but the function factor(6)> which also returns an unevaluated 2*3,
>
> no, it is marked as already simplified by the factor command.
> try
> factor(6);
> ?print(%);
>
>   uses some other method
>

> > because simp is not involved.
>
> Wrong again.  The simplifier notices that there is a "simp"
> flag on the expression, and so it is not simplified to 6.
>

> > (I would be interested in a "pure Maxima" solution, without Lisp.)

>
> If you said what your objective was, perhaps a solution is possible.
> You haven't said what you that is.
>
> If you want to see 2*3 displayed, you could do print("2*3");
>
> While it is possible to turn the simplifier off by simp:false,
> it is almost always a terrible idea.
>
> RJF

Thank you Richard, for clarifying all these aspects.
I'd like to know how factor() displays the unsimplified results.
So, what would be the definition of a function f(m,n)
such that:

f(2,3)
2*3
%+1;
2*3+1
ratsimp(%);
7

Thank you.
M.

Date Subject Author
2/27/13 Mate
2/27/13 Richard Fateman
2/28/13 Mate
3/1/13 Richard Fateman
3/1/13 Mate
3/1/13 Richard Fateman