Le cas nu=0 (triangulations bien 3-coloriées, donc eulériennes)
When nu=0
> | subs(nu=0,eqcatQ); |
Let us first express Q1 and Q3 in terms of Q2 and Q4, using the following expressions of the second and fourth derivative:
> | D2Qsol; |
> | D4Qsol; |
> | isolate(D4Qsol-`@@`(D,4)(Q)(0), `@@`(D,3)(Q)(0)); |
> | eqcatmixte:=collect(factor(subs(%,D(Q)(0)= nu*t*`@@`(D,2)(Q)(0)/2,eqcatQ/nu)),Q,factor); |
> | eqcatnu0:=collect(op(3,factor(subs(nu=0,eqcatmixte))),Q,factor); |
We apply the quadratic method : the discriminant of the above equation (in Q(y)) has a double root (in y)
> | e1:=op(4,factor(discrim(subs(Q(y)=Qy,eqcatnu0),Qy))); |
We now take the discriminant wrt y
> | e2:=factor(discrim(subs(`@@`(D,2)(Q)(0)=2*Q2,e1),y)); |
> | alg1:=collect(op(1,op(3,e2)),Q2,factor); alg2:=collect(op(1,op(4,e2)),Q2,factor); |
> | n:=6: factor(series(subs(Q2=coeff(subs(q=3,w=1,z=1,x=0,nu=0,Qser(n)),y,2),alg1),t,n+1)); |
> | n:=3: factor(series(subs(Q2=coeff(subs(q=3,w=1,z=1,x=0,nu=0,Qser(n)),y,2),alg2),t,n+1)); |
C'est donc alg1
> | factor([solve(alg1,Q2)]); |
> | series(op(1,%)/2,t,32); |
C'est bien les cartes bicubiques
> |