Mireille Bousquet-Mélou's message dated Feb. 13, 2015

Dear Doron,

thank you for quoting me!

Still, if Pol(x0,x1,t,x) is a 4-variate polynomial, and you want to solve
Pol(F(x),F(1),t,x)=0, and obtain an algebraic equation for F(1),
I would recommend the following (one-line!) Maple session:

discrim(discrim(Pol(x0,F1,t,x),x0),x)).

This iterated discriminant is an algebraic equation for F1, as explained,
and generalized, in my paper with Arnaud Jehanne:
http://fr.arxiv.org/abs/math/0504018

Actually I have typed this Maple line a few times this week, with colleagues
who had written such equations for map problems. I hope you like it!

Bien amicalement,

mireille

Mireille Bousquet-Mélou's message dated Feb. 14, 2015

From mireille.bousquet@labri.fr  Sat Feb 14 05:26:06 2015


>
>  Your one-liner works instantaneously for Alpha and Beta trees, but
> when I tried it for the 2-stack-sortable functional equation
>
>                     1      x y (f(x, 1) - y f(x, y)) (f(x, 1) - f(x, y))
>        f(x, y) = ------- + ---------------------------------------------
>                  1 - x y                            2
>                                              (1 - y)
>
>
> I got zero.

Hem, that's what it is when one wants to have a single Maple line! My  
fault. In the first
discriminant (in F(x,y)) one has to take the relevant factor. In this  
case this
first discriminant has a factor (y-1)^2, and, sure, its discriminant  
w.r.t. y vanishes...
I typed things at a slower pace in the attached Maple session.


>
> I also got zero, for the Functional Equation
>                              3  2          2  2  2
> f(x, t) = (-f(x, 1) f(x, t) t  x  + f(x, 1)  t  x  + f(x, 1) f(x, t) t x
>
>                 2              2
>      - f(x, 1) t  x + f(x, t) t  x - f(x, 1) f(x, t) x - t x f(x, 1)
>
>      - f(x, t) t x + f(x, 1) x + f(x, t) x + t - 1)/((-1 + x) (-1 + t)
>
>     (t x f(x, 1) - 1))
>
> that was the subject of my email (the one that came up in the  
> Bloom-Burstein article).

This one is linear in F(x,y). No discriminant! In this case the  
technique of our paper specializes
to the so-called kernel method: il the equation reads a*F(x,y)+b=0,  
and equation
for F(x,1) is given by the resultant of a and b with respect to y. See  
also the
Maple session.

I hope this is neater.

All the best,

mireille