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Recently there was a paper connecting the zeros of the GKW operator to the zeros of zeta, this is due to the fact that the Mellin transform of the Gauss map is a linear function of zeta . Transfer operator for the Gauss' continued fraction map. I. Structure of the eigenvalues and trace formulas and Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems

Why can the same relations not be used for the mapping associated with the Engel expansion formed by iterating the function

$\begin{array}{ll} g (x) & = x \left( \lfloor \frac{1}{x} \rfloor + 1 \right) - 1 \text{}\\ & = \sum_{n = 1}^{\infty} \left\{ \begin{array}{ll} x (n + 1) - 1 & \frac{1}{n + 1} < x \leqslant \frac{1}{n}\\ 0 & {otherwise} \end{array} \right. \end{array}$

whose Mellin transform is also a linear function of the Riemann $\zeta$ function ?

We have

$\begin{array}{ll} \int_0^{1} g (x) x^{s - 1} d x & = \sum_{n = 1}^{\infty} \int_{\frac{1}{n + 1}}^{\frac{1}{n}} (x (n + 1) - 1) x^{s - 1} x\\ & = \frac{\zeta (s + 1)}{s + 1} - \frac{1}{(s + 1) s}\\ & = \frac{\zeta (s + 1) s - 1}{(s + 1) s} \end{array}$

and the associated Transfer operator (I'll call it the Engel operator) analogous to the GKW operator is defined by \begin{equation} [\mathcal{L}_g f] (x) = \sum_{y : g (y) = x} \frac{f (y)}{\left| \frac{d}{d z} g (z) \right|_{z = y}} = \sum_{n = 1}^{\infty} \frac{f \left( \frac{x + 1}{n + 1} \right)}{n + 1} \end{equation} since $\frac{d}{d x} (x (n + 1) - 1) = n + 1$ and the inverse of the $n$-th component is $\frac{x + 1}{n + 1}$ which is found by solving $x (n + 1) - 1 = y$ for $x$.

The operator $[\mathcal{L}_g f] (x)$ applied to the function $x \rightarrow x^s$ results in

$[\mathcal{L}_g x \rightarrow x^s] (x^{}) = (x + 1)^s (\zeta (s + 1) - 1)$

The iteration function for the Gauss map is :

$h(x)=1/x-\lfloor 1/x \rfloor.$

where: $\lfloor 1/x \rfloor$ denotes the floor function

The Gauss–Kuzmin–Wirsing $G$ acts on functions $f$ as

$[Gf](x) = \sum_{n=1}^\infty \frac {1}{(x+n)^2} f \left(\frac {1}{x+n}\right).$

crow
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There is no polynomial basis for the transfer operator Lg of g(x). This is because the action of Lg on any nonzero constant diverges to infinity since sum(y/(n+1),n=1..inf) = sgn(y)*inf. There is however a polynomial basis for the solvable and related transfer operator of the map w(x)=floor(1/x)*g(x) which is mentioned in how to solve for eigenvectors given closed-form for infinite set of eigenvalues?

crow
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