Aha, the singularity seems to occur .
[update] this hope didn't improve to knowledge... ;-( [/update]
By Eigenanalysis the eigenvalues of Bb are diag(1,u,u^2,...),
where u = log(t) and t is the fixpoint of b.
Now the h(b)-function gives a fixpoint for b=0.15.It gives [update]
t0=H(0.150,0) \\ = 0.436708722499
u0=log(t0) \\ = -0.828488845027
b = t0^(1/t0) =0.15
[/update]
Using Branch 1:
t1=H(0.150,1) \\ = -0.467424941871 - 2.38182350872*I
u1=log(t1) + 2*Pi*I \\ = 0.886761198657 + 4.51860497886*I
exp(u1/t1) \\ = 0.150
The eigenvalues of Mb should be diag(1/(1+1),1/(1+u),1/(1+u^2),... )
and I don't see a singularity occuring here. This, however does not use t but its logarithm u - but I'm possibly nearer to the core of the problem here.
Gottfried
[update] this hope didn't improve to knowledge... ;-( [/update]
By Eigenanalysis the eigenvalues of Bb are diag(1,u,u^2,...),
where u = log(t) and t is the fixpoint of b.
Now the h(b)-function gives a fixpoint for b=0.15.It gives [update]
t0=H(0.150,0) \\ = 0.436708722499
u0=log(t0) \\ = -0.828488845027
b = t0^(1/t0) =0.15
[/update]
Using Branch 1:
t1=H(0.150,1) \\ = -0.467424941871 - 2.38182350872*I
u1=log(t1) + 2*Pi*I \\ = 0.886761198657 + 4.51860497886*I
exp(u1/t1) \\ = 0.150
The eigenvalues of Mb should be diag(1/(1+1),1/(1+u),1/(1+u^2),... )
and I don't see a singularity occuring here. This, however does not use t but its logarithm u - but I'm possibly nearer to the core of the problem here.
Gottfried
Gottfried Helms, Kassel