The case q=2 (m=4,k=1) (Section 12.2)
> | T:=proc(m) subs(cos(x)=x,expand(cos(m*x))): end: |
> | T(4); |
> | m:=4: k:=1: 2+2*cos(2*k*Pi/m); |
The second invariant
> | I2:=normal(subs(C=CC,II=I1,q=2,sqrt(C)^m*subs(x=xsubs,T(m)))): |
> | indets(I2); |
> | factor(series(normal(I2),y,1)); |
The invariant equation
> | eqinv:=subs(q=2,Q(0,y)=Q(y),beta=nu-1,I2-add(c[i]*(t*I1)^i,i=0..4)): |
> |
Determination of the series c_i. An equation with one catalytic variable and 2 additional unknown series Q'(0) and Q'''(0)
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Solution using Tutte's trick
> | RHS:=map(factor,collect(subs(c[4]=c4sol,c[3]=c3sol,c[2]=c2sol,c[1]=c1sol,c[0]=c0sol,add(c[i]*(t*X)^i,i=0..4)),Q)); |
> | CC; |
Each of the following polynomials in X has a double root.
We denote by Q_i the series counting quasi-triangulations with outer degree i.
> | pol1:=collect(op(3,factor(subs(`@@`(D,3)(Q)(0)=6*Q3,D(Q)(0)=Q1,beta=nu-1,factor(subs(q=2,II=X,RHS-CC^2))))),[Q1,Q3,X],factor); pol2:=collect(op(3,factor(subs(`@@`(D,3)(Q)(0)=6*Q3,D(Q)(0)=Q1,beta=nu-1,factor(subs(q=2,II=X,RHS+CC^2))))),[Q1,Q3,X],factor); |
> | eq1:=factor(discrim(pol1,X)): |
> | nops(eq1); |
> | seq(nops(op(i,eq1)),i=1..5); |
> | seq(op(i,eq1),i=1..4);; |
A first equation between Q1 and Q3
> | eq1:=collect(op(5,eq1),[Q1,Q3],factor); |
> | eq2:=factor(discrim(pol2,X)): |
> | nops(eq2); |
> | seq(nops(op(i,eq2)),i=1..4); |
> | seq(op(i,eq2),i=1..3);; |
Second equation between Q1 and Q3
> | eq2:=collect(op(4,eq2),[Q1,Q3],factor); |
We finally eliminate Q3
> | res:=factor(resultant(eq1,eq2,Q3)): |
> | nops(res); |
> | seq(nops(op(i,res)),i=1..5); |
> | op(1,res); op(2,res);op(3,res); |
> | op(4,res); |
> | op(5,res); |
Two candidates for the minimal polynomial of Q1
> | alg1:=collect(op(4,res),Q1,factor); |
> | degree(alg1,Q1); |
The second candidate
> | alg2:=collect(op(1,op(5,res)),Q1,factor); |
> | degree(alg2,Q1); |
Let us decide
> | n:=2:normal(series(subs( Q1=normal(collect(coeff(subs(x=0,w=1,z=1,q=2,Qser(3*n)),y,1),t)),alg2),t,n+1)); |
> | n:=4:normal(series(subs( Q1=normal(collect(coeff(subs(x=0,w=1,z=1,q=2,Qser(3*n)),y,1),t)),alg1),t,n+2)); |
Hence the minimal polynomial of Q1 has degree 5. Let us check
> | with(gfun): |
> | factor(algeqtoseries(alg1,t,Q1,10,true)); |
> | n:=4:map(factor,collect(coeff(subs(x=0,q=2,w=1,z=1,Qser(3*n)),y,1),t)); |
In the series t*Q1, denoted Q1t below, all exponents of t are multiples of 3. We denote t^3=w.
> | alg1t:=collect(subs(t=w^(1/3),factor(numer(subs(Q1=Q1t/t,alg1)))),Q1t,factor); |
> |
> |
> | vnu:=v=(nu+1)/(nu-1); |
> | nuv:=isolate(vnu,nu); |
The parametrization of the paper (with v=(nu+1)/(nu-1))
> | wS := 1/64*(-2*v-2*S+4*S^3-S^2+v^2)*(-S+v)*(S-2+v)/S^2/(1+v)^3; |
let us check that it indeed factors the minimal polynomial of Q1
> | res:=numer(factor(subs(w=wS,nuv,alg1t))): |
> | nops(res); |
> | op(1,res); |
> | op(2,res); |
> | degree(op(3,res),Q1t); |
The right factor is not op(3,res), but the linear one
> | factor(subs(S=v,Q1t=0,op(3,res))); |
> | Q1tsol:=factor(solve(op(2,res),Q1t)); |
Expression of Q3? We get back to one of the equation we had between Q1 and Q3, and introduce Q3t:= t^3 Q3
> | eq2; |
> | subs(t=w^(1/3),numer(factor(subs(Q1=Q1t/t,Q3=Q3t/t^3,eq2)))); |
> | res:=numer(factor(subs(nuv,Q1t=Q1tsol,w=wS,%))); |
> | nops(res); |
> | degree(op(3,res),Q3t);degree(op(2,res),Q3t); |
The right factor is op(3,res):
> | factor(subs(S=v,Q3t=0, op(3,res)));factor(subs(S=v,Q3t=0, op(2,res))); |
> | Q3tsol:=factor(solve(op(3,res),Q3t)); |
> | factor(Q3tsol*wS); |
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Asymptotics
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