Just found a msg in sci.math.reseach of last year where I looked at the trace of the carleman-matrix. This may be another element of a proof.
if \( B_b = W*D*W^{-1} \) where D is diagonal, containing [1,u,u^2,u^3,...]
It must hold (from finite matrices): trace(Bb) = trace(D)
We have: \( trace(D) = \frac{1}{1-u} \)
The structure of Bb is simple; the sum of its diagonal elements are (using \( b=t^{1/t}, u=log(t) \)
\( trace(B_b) = \sum_{r=0}^{\infty} ( \log(b)^r*\frac{r^r}{r!} ) \\[12pt]
\)
\( \hspace{64}\\[12pt]
= \sum_{r=0}^{\infty} ( (\frac{u*r}{t} )^r /r! ) \\[12pt]
\)
some numerical checks:
if \( B_b = W*D*W^{-1} \) where D is diagonal, containing [1,u,u^2,u^3,...]
It must hold (from finite matrices): trace(Bb) = trace(D)
We have: \( trace(D) = \frac{1}{1-u} \)
The structure of Bb is simple; the sum of its diagonal elements are (using \( b=t^{1/t}, u=log(t) \)
\( trace(B_b) = \sum_{r=0}^{\infty} ( \log(b)^r*\frac{r^r}{r!} ) \\[12pt]
\)
\( \hspace{64}\\[12pt]
= \sum_{r=0}^{\infty} ( (\frac{u*r}{t} )^r /r! ) \\[12pt]
\)
some numerical checks:
Code:
[ u , t , b , trace(D) , trace(Bb) , trace(D)-trace(Bb]
[0.500000000000, 1.64872127070, 1.35427374603, 2.00000000000, 2.00000000000, 4.31794217767 E-60]
[0.600000000000, 1.82211880039, 1.38997669617, 2.50000000000, 2.50000000000, 2.412917620 E-86]
[0.700000000000, 2.01375270747, 1.41567962006, 3.33333333333, 3.33333333333, 3.217223494 E-86]
[0.800000000000, 2.22554092849, 1.43256016868, 5.00000000000, 5.00000000000, 6.43444698 E-86]
[0.900000000000, 2.45960311116, 1.44182935647, 10.0000000000, 10.0000000000, 1.608611747 E-85]
Gottfried Helms, Kassel