A safer alternative to list[index] is inpart(list, index). Another
possibility is
to use substitution:
(%i1) sol : linsolve([C1*3-2,C2*1+4],[C1,C2]);
(%o1) [C1=2/3,C2=-4]
(%i2) subst(sol, [C1, C2, C1 - C2]);
(%o2) [2/3,-4,14/3]
Why is inpart(list, index) safer than list[index]?
(%i5) ssol : linsolve([C1*3-2,C2*1+4],[C1,C2]);
(%o5) [C1=2/3,C2=-4]
Not what we want:
(%i6) ssol[1];
(%o6) ssol[1]
OK
(%i7) inpart(ssol,1);
(%o7) C1=2/3
Barton
-----maxima-bounces at math.utexas.edu wrote: -----
>To:?Michael?Anderl?<michael.anderl at gmail.com>
>From:?reyssat?<eric.reyssat at math.unicaen.fr>
>Sent?by:?maxima-bounces at math.utexas.edu
>Date:?12/15/2009?03:41PM
>cc:?maxima at math.utexas.edu
>Subject:?Re:?[Maxima]?access?results?of?linsolve
>
>Michael?Anderl?a??crit?:
>>?output:
>>?[C1=2/3,C2=-4]
>>
>>?how?to?access?C1?and?C2?
>>
>use?list[index]?to?extract?an?element?from?a?list,?and?the?function?rhs
>to?get?the?right?hand?side?(lhs?for?left)?of?an?equation?:
>rhs(%[1]);
>gives?2/3?since?it?is?the?right?hand?side?of?the?first?term?of?the?last
>computed?result?(%)
>This?is?C1.
>
>Eric
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