[GiNaC-list] collecting on similar series expansion

[Sheplyakov Alexei]

Hello,

On Sun, Jan 28, 2007 at 09:31:18AM -0300, Charlls Quarra wrote:
> I'm just trying out ginac for the first time. I have
> an expression with sumands of this type (in latex
> format):
[snipped the expression]
 
> so i want to collect terms that have the same
> coefficients of the form D[0..i times] (F)(q) * D[ 0..
> j times] (F)(q) .... but the polynomial collect won't
> work.

Could you please post the complete example which illustrates this?

>  Initially i thought about substituting each D[ ]
> (F)(q) with a different symbol, then collect on the
> multivariate polynomials of these symbols and then
> finally substitute back the D[] expressions, however,
> I don't know how the collect call is supposed to
> receive the symbols that it must collect,

Reading the manual _does_ help:

"Another useful representation of multivariate polynomials is as a
univariate polynomial in one of the variables with the coefficients
being polynomials in the remaining variables.  The method `collect()'
accomplishes this task:

     ex ex::collect(const ex & s, bool distributed = false);

The first argument to `collect()' can also be a list of objects in which
case the result is either a recursively collected polynomial, or a
polynomial in a distributed form with terms like c*x1^e1*...*xn^en, as
specified by the `distributed' flag.

Note that the original polynomial needs to be in expanded form (for the
variables concerned) in order for `collect()' to be able to find the
coefficients properly."


> the code
> looks like this:
> 
>        ex jacobian = diff( q + dq_new*dt , q )*diff( p
> + dp_new*dt , p ) - diff( q + dq_new*dt , p )*diff( p
> + dp_new*dt , q );
>         cout << " the jacobian of the integrator is "
> << expand(jacobian) << endl;
>         exmap diff_map , inv_diff_map;
>         symbol foo[10];
>         diff_map[ F(q) ] = foo[0]; inv_diff_map[
> foo[0] ] = F(q);
>         diff_map[ F(q).diff(q) ] = foo[1];
> inv_diff_map[ foo[1] ] = F(q).diff(q);
>         diff_map[ F(q).diff(q,2) ] = foo[2];
> inv_diff_map[ foo[2] ] = F(q).diff(q,2);
>         diff_map[ F(q).diff(q,3) ] = foo[3];
> inv_diff_map[ foo[3] ] = F(q).diff(q,3);
>         diff_map[ F(q).diff(q,4) ] = foo[4];
> inv_diff_map[ foo[4] ] = F(q).diff(q,4);
>         ex rord = collect( jacobian.subs( diff_map ) ,
> { foo[0] , foo[1] , foo[2] , foo[3] , foo[4] } ); // i
> know this is an error in C++, but this is the idea
>         rord = rord.subs( inv_diff_map );
>         cout << "reordered we obtain " << rord <<
> endl;
> 
> any thoughts on this?

I tnink this should do the job:

ex rord = jacobian.expand().collect(lst(F(q), F(q).diff(q),
          F(q).diff(q, 2), F(q).diff(q, 3), F(q).diff(q, 4)),
					true /* treat the expression as Z[x,y,...] 
					        (as opposed to Z[x][y]...) */ );

Best regards,
 Alexei

-- 
All science is either physics or stamp collecting.

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