The case q=2 et nu=0 (Section 2)
The functional equation (we only count edges, so that we set w=1)
> | eqM2:=subs(nu=0,q=2,w=1,eqM);; |
> | Ker2:=-coeff(eqM2,M(x,y)); |
> | R2:=coeff(eqM2,M(x,y),0); |
> | subs(x=1+s*t,Ker2*(y-1)); |
> | factor(subs(t=0,%)); |
Expansion of Y_1 and Y_2
> | n:=6: res:=expand(convert(subs(M(1,y)=subs(x=1,Mser(n)),M(x,1)=subs(y=1,Mser(n)), x=1+t*s,q=2,nu=0,numer(Ker2)),polynom)): |
> | with(gfun): |
> | algeqtoseries(res,t,y,3,true); |
> |
> |
Construction of the invariants (performed for general q and nu further down in the session)
The two invariants
> | Idef; Jdef; |
The invariant equation (12).
> | eqinv:=Jdef-add(C[r]*II^r,r=0..4); |
Or, in terms of M(1,y)
> | eqinvM:=subs(II=Idef,eqinv): |
We now determine C_0, ..., C_4 by an expansion of the invariant equation around y=1.
> | factor(series(eqinvM,y=1,1)); |
> | factor(series(subs(C[4]=1,eqinvM),y=1,2)); |
> | factor(series(subs(C[4]=1,C[3]=-4,eqinvM),y=1,2)); |
> | factor(series(subs(C[4]=1,C[3]=-4,C[2]=4*t,eqinvM),y=1,3)); |
> | factor(series(subs(C[4]=1,C[3]=-4,C[2]=4*t,C[1]=8+8*t,eqinvM),y=1,4)); |
> | coeff(%,y-1,0); |
The equation on M(1,y)
> | collect(numer(factor(subs(C[4]=1,C[3]=-4,C[2]=4*t,C[1]=8+8*t,C[0]=-4-40*t-4*t^2+32*t^2*M(1,1),eqinvM)))/t/32,M,factor); |
Compare with (6):
> | eqref:=1+t*y^2*M(1,y)^2+t*y^2*(M(1,y)-M(1,1))/(y^2-1)-M(1,y); |
> | factor(%%/eqref); |
> |