Richard Fateman wrote:
> I think this default would not be consistent with the general approach in
> Maxima. In particular, if sum(..) can be done exactly, and the result then
> converted to a bfloat, that is what I would expect the commands below to do.
>
I don't feel very strongly about this but since I think bfloat evaluates
its first argument
the expected behaviour would be automatic. Levin would be handed an
expression
that can not be simplified further (in the current situation that would
be almost all
indefinite sums but this could change if the closed_form package becomes
more and
more powerful!).
> Also you would have to consider what to do with (f(a)+sum (....)), bfloat.
>
>
Good question. I would imagine the behaviour to be the same as
f(a)+cos(1),bfloat;
f(a) + 5.403023058681397b-1
> How about this: use "nsum" instead of "sum" (Mathematica uses NSum vs
> Sum).
>
Yes this could be a good idea.
> Or do this:
>
> Simpsum:Levin, as a third option, other than true and false.
>
This I don't really like. Numerical evaluation is not the same as
simplification.
Michel
> RJF
>
>
>> -----Original Message-----
>> From: maxima-bounces at math.utexas.edu [mailto:maxima-
>> bounces at math.utexas.edu] On Behalf Of Michel Van den Bergh
>> Sent: Friday, January 05, 2007 6:04 AM
>> To: maxima at math.utexas.edu
>> Subject: Question: invoking levin by default
>>
>> I would like to experiment with invoking levin by default. The interface
>> would be something like
>>
>> load(levin);
>> fpprec:50;
>> sum(1/n-log(1+1/n),n,1,inf),bfloat;
>>
>> 5.7721566490153286060651209008240243104215933593992b-1
>>
>>
>> Question: is this achievable in the maxima language, or should one go down
>> to the lisp level?
>>
>> If the latter, can one do it in a localized way (i.e. purely within the
>> package levin)?
>>
>> Michel
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>>
>
>