The case q=2 (m=4,k=1) (Section 12.1)
The Chebyshev polynomials
> | 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); |
> | I2:=normal(subs(C=CC,II=I1,q=2,yinv=1/y,M(1,y)=M(y),normal(sqrt(C)^m*subs(x=xsubs,T(m))))): |
> | indets(I2); |
> | factor(series(normal(I2),y=1,1)); |
> | eqinv:=subs(M(1,y)=M(y),q=2,I2-add(c[i]*I1^i,i=0..4)): |
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> | indets(eqinv); |
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Determination of the series c_i. An equation with a single catalytic variable and two additional unknown functions , M(1,1) et M'(1,y=1)
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> | eqinvM; |
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We will now solve this equation in one catalytic variable, following the approach used by Tutte for colored planar triangulations
> | RHS:=subs(c[4]=c4sol,c[3]=c3sol,c[2]=c2sol,c[1]=c1sol,c[0]=c0sol,add(c[i]*X^i,i=0..4)); |
> | factor(T(4)-1);factor(T(4)+1); |
The following two polynomials in X have a double root named P_+ and P_- in the paper)
> | pol1:=factor(subs(q=2,II=X, RHS-CC^2)); pol2:=factor(subs(q=2,II=X,RHS+CC^2)); |
The first discriminant
> | eq1:=factor(discrim(pol1,X)); |
> | nops(eq1); |
> | seq(op(i,eq1),i=1..5); |
> | nops(op(6,eq1)); |
Here is the factor of eq1 that vanishes
> | eq1:=op(6,eq1): |
The second discriminant
> | eq2:=factor(discrim(pol2,X)): |
> | nops(eq2); |
> | op(6,eq2); |
> | nops(op(3,eq2)); |
> | eq2:=op(3,eq2): |
Let us check that eq1 and eq2 are indeed 0, using the expansion of M
> | N:=7:normal(series(subs(M(1)=subs(x=1,y=1,q=2,z=1,Mser(N)), D(M)(1)=subs(y=1,diff(subs(x=1,q=2,z=1,Mser(N)),y)), eq1),t,N)); |
> | N:=7:normal(series(subs(M(1)=subs(x=1,y=1,q=2,z=1,Mser(N)), D(M)(1)=subs(y=1,diff(subs(x=1,q=2,z=1,Mser(N)),y)), eq2),t,N)); |
We eliminate M'(1). The series M(1) is now denoted M0 (it is the zero-th derivatiove of M....)
> | res:=factor(resultant(subs(D(M)(1)=M1, M(1)=M0,eq1),subs(D(M)(1)=M1,M(1)=M0,eq2),M1)): |
> | nops(res); |
> | seq( nops(op(i,res)),i=1..7); |
> | seq(op(i,res),i=1..5); |
There is one big factor (the 6th one). The 7th one is the square of a polynomial that might vanish
> | op(6,res): op(7,res); |
Which of these two factors vanishes? Here is the big guy:
> | alg1:=collect(op(6,res),M0,factor): nops(alg1); |
and the smaller one:
> | alg2:=collect(op(1,op(7,res)),M0,factor); |
> | degree(alg1,M0);degree(alg2,M0); |
To decide, we expand alg1 and alg2 in t. Alas... the correct equation for M0 turns out to be the big guy, alg1
> | n:=6:normal(series(subs(M0=subs(x=1,y=1,q=2,z=1,Mser(n)),alg1),t,n+1)); |
> | n:=4:factor(series(subs(M0=subs(x=1,y=1,q=2,z=1,Mser(n)),alg2),t,n+1)); |
> | algM0:=collect(alg1,M0,factor): degree(algM0,M0); |
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Let us not count vertices
> | res:=factor(subs(w=1,algM0)):nops(res); |
> | op(1,res);op(2,res);op(4,res); |
here is an equation of degree 6 for M0
> | algMw1:=collect(op(3,factor(subs(w=1,algM0))),M0,factor); |
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> | degree(algMw1,M0); |
> | n:=6:normal(series(subs(M0=subs(x=1,y=1,q=2,w=1,z=1,Mser(n)),algMw1),t,n+1)); |
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Parametrization for w=1 (degree 6)
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Asymptotics and phase transition (w=1)
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