Thanks Stavros,
I want to group them just for better display, and your
> substpart(factor(piece),qq,[1,2])
> substpart(factor(piece),%,[2,3])
do it in the right way; ... I was just thinking Maxima could do it without my
help, finding by herself the terms with the same exponent. :)
Greetings from Italy,
Dionisio
Stavros Macrakis <macrakis <at> alum.mit.edu> writes:
>
> Why do you want to group them? ?For a nicer display? To perform some further
calculation?
> If it's just for a nicer display, you could do something like this:
>
>
> qq: 2*exp(-3*t/2)*
(18*sinh(sqrt(5)*t/2)/sqrt(5)-6*cosh((sqrt(5)*t/2)))+12,exponentialize,expand;
>
> ""(part(qq,[1,2])) + ""(part(qq,[3,4])) ? ? <<< apply the function with a null
name to each part (!)
>
>
> This is fine for display, but not very useful for further manipulation. (A
trickier way uses ' "("(...), but I don't want to confuse you too?much.)
>
> If you want to manipulate the parts separately, you might want to put them
into a list, like this:
>
>
> qql: [ part(qq,[1,2]), part(qq,[3,4]) ]
>
>
> Though another way to manipulate each part separately is with "substpart",
e.g.:
>
>
>
> substpart(factor(piece),qq,[1,2])
> substpart(factor(piece),%,[2,3]) ? ? ?<<< note that piece [1,2] became just
one term
>
> As I say, the correct answer to your question depends on why?you want to group
them.
>
> ? ? ? ? ? ? ?-s
>
> On Fri, Sep 7, 2012 at 10:02 AM, Dionisio De Zolt <ddzolt <at> libero.it>
wrote:
> Hi, just a simple question ... could you please tell me how to group first
andsecond term as well as third and fourth of the following expression(%i1)
2*exp(-3*t/2)*
(18*sinh(sqrt(5)*t/2)/sqrt(5)-6*cosh((sqrt(5)*t/2)))+12,exponentialize,expand;Th
anksRegardsDionisio
>
>
>
>
>
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