Re: [Maxima-discuss] solve in Maxima different from solve in SymPy

[Mike V. <mic...@gm...>]

Sorry to keep spamming you, but I think I know where you are going and I'm
going to give you some icing on the cake.

Consider:
res: (a + b/(1-x) + c/(1-x)^2) * (1-x)^2 - (1-2*x)^2;
(b/(1-x)+c/(1-x)^2+a)*(1-x)^2-(1-2*x)^2

Say we want to automagically make all coefficients vanish?
listCoeff: makelist( ratcoeff(res,x, (i-1)), i, 3);
[c+b+a-1,-b-2*a+4,a-4]

solve( listCoeff, [a,b,c]);
[[a=4,b=-4,c=1]]

I hope that helps, or at least demonstrates how to use bit of Maxima.



On Mon, May 18, 2015 at 8:18 PM, Mike Valenzuela <mic...@gm...>
wrote:

> There is the also the method: ratcoef(expr, variable, order), which also
> expands and simplifies first before attempting to extract its answer, so
> the results may not be identical coef's results.
>
> On Mon, May 18, 2015 at 8:14 PM, Mike Valenzuela <mic...@gm...>
> wrote:
>
>> You might want to look at coef(expr, variable, order)
>>
>> coef(expr, variable, 0), finds the subexpression free of the variable
>>
>> On Mon, May 18, 2015 at 8:00 PM, Marduk Bolaños <mar...@ma...> wrote:
>>
>>> Thank you for the detailed explanation. Now, in order to obtain the
>>> corresponding system of equations in an automated way I need to know
>>> what is Maxima's equivalent of SymPy's
>>>
>>> expr.coeff(a)
>>>
>>> which returns 2*b + m.
>>>
>>
>>
>