evaluation or simplification ?



Thanks for your answer. I was misleaded by the ev. 

Volker

Am 16 Dec 2008 um 14:56 hat Stavros Macrakis geschrieben:

> 
> It is simplification.
> 
> To re-simplify without evaluating, use:
> 
>   expand(xxx,expon,expop)
> 
> In this, case:
> 
> (%i1) listarith:false;
> (%o1)false
> (%i2) x: [1,2]+p*[3,4];
> (%o2)[3, 4] p + [1, 2]
> (%i3) expand(x);
> (%o3)[3, 4] p + [1, 2]
> (%i4) expand(x),listarith:true;
> (%o4)[3 p + 1, 4 p + 2]
> (%i5) block([listarith:true],expand(x));<<< ensure there are no weird ev effects
> (%o5)[3 p + 1, 4 p + 2]
> 
> On Tue, Dec 16, 2008 at 2:39 PM, van Nek <van.nek at arcor.de> wrote:
>     Hello,
>     
>     I have
>     
>     (%i1) listarith: false$
>     (%i2) x: [1,2]+p*[3,4];
>     (%o2)            [3, 4] p + [1, 2]
>     
>     When I do now
>     
>     (%i3) ev(x, listarith: true);
>     (%o3)            [3 p + 1, 4 p + 2]
>     
>     is this evaluation or simplification ? Can someone please explain the notation?
>     
>     Thanks
>     Volker van Nek
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> 
>