Eigenvalues of the Carleman matrix of b^x

[Gottfried]

Lets transform this a bit:
\( \sum_{n=0}^\infty \left(\frac{u*n}{e^u}\right)^n \frac{1}{n!}= \sum_{n=0}^\infty \frac{n^n}{n!} (ue^{-u})^n = \sum_{n=0}^\infty \frac{(-n)^n}{n!} ((-u)e^{-u})^n \)

This looks already damn like the power series of the LambertW function which is
\( W(x)=\sum_{n=0}^\infty \frac{(-n)^{n-1}}{n!} x^n \).

index-error: we start at n=1

\( W(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!} x^n \).

Hence \( W'(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{(n-1)!} x^{n-1}=\sum_{n=0}^\infty \frac{(-n)^{n}}{n!} x^{n} \)

Then:

\( W'(x)=\sum_{n=0}^\infty \frac{(-(1+n))^{n}}{n!} x^{n} \)


rewrite derivative of Lambert-W
\( \hspace{24}
W' (x) = \sum_{n=1}^{\infty}\frac{(-n)^{n-1}n}{n!}x^{n-1} \)

\( \hspace{36}
=-\frac1x \sum_{n=1}^{\infty}\frac{(-n)^n}{n!}x^n \)

rewrite trace
let c= u/e^u
\( \hspace{48}
\sum_{n=0}^{\infty}\frac{n^n}{n!}c^n \hspace{12} =\sum_{n=0}^{\infty}\frac{(-n)^n}{n!}(-c)^n \\
\\[12pt]

\hspace{36}
=1 + \sum_{n=1}^{\infty}\frac{(-n)^n}{n!}(-c)^n \hspace{12} = 1 - (-c)*W'(-c)
\)



I cannot follow the formal derivative at the moment,

And we want to show that \( W'((-u)e^{-u})=\frac{1}{1-u} \).
Ok, we have \( W(x)e^{W(x)}=x \). Take the derivative:
\( W'(x)e^{W(x)}+W(x)e^{W(x)}W'(x)=1 \)
\( W'(x)=\frac{e^{-W(x)}}{1+W(x)} \)
now is \( (-u)e^{-u}=W^{-1}(-u) \) so
\( W'((-u)e^{-u})=W'(W^{-1}(-u))=\frac{e^u}{1-u} \)

Hm this is not quite \( \frac{1}{1-u} \) so where is the error?

but the last identity helps...
\( \\[12pt]

\hspace{24}
W'(-c)= \frac{e^u}{1-u} \\
\\[12pt]

\hspace{12}
1 + cW'(-c) = 1 + \frac{u}{e^u}\frac{e^u}{1-u} \\
\\[12pt]

\hspace{100}
= 1 + \frac{u}{1-u} \\
\\[12pt]

\hspace{100}
= \frac{(1-u)+ u}{1-u} \\
\\[12pt]

\hspace{100}
= \frac{1}{1-u}
\)

Then it would be interesting, how this looks:
a) for higher powers of the carleman-matrix
b) for different fixpoints/different branches of W