The Gauss map

The Gauss function (map) h is :

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

where ${\displaystyle \lfloor 1/x\rfloor }$ denotes the floor function.

It has an infinite number of jump discontinuities at x = 1/n, for positive integers n. It is hard to approximate it by a single smooth polynomial.^[1]

Operator on the maps

The Gauss–Kuzmin–Wirsing operator ${\displaystyle G}$ acts on functions ${\displaystyle f}$ as

${\displaystyle [Gf](x)=\int _{0}^{1}\delta (x-h(y))f(y)\,dy=\sum _{n=1}^{\infty }{\frac {1}{(x+n)^{2}}}f\left({\frac {1}{x+n}}\right).}$

it has the fixed point ${\displaystyle \rho (x)={\frac {1}{\ln 2(1+x)}}}$, unique up to scaling, which is the density of the measure invariant under the Gauss map.

Eigenvalues of the operator

The first eigenfunction of this operator is

${\displaystyle {\frac {1}{\ln 2}}\ {\frac {1}{1+x}}}$

which corresponds to an eigenvalue of λ₁ = 1. This eigenfunction gives the probability of the occurrence of a given integer in a continued fraction expansion, and is known as the Gauss–Kuzmin distribution. This follows in part because the Gauss map acts as a truncating shift operator for the continued fractions: if

${\displaystyle x=[0;a_{1},a_{2},a_{3},\dots ]}$

is the continued fraction representation of a number 0 < x < 1, then

${\displaystyle h(x)=[0;a_{2},a_{3},\dots ].}$

Because ${\displaystyle h}$ is conjugate to a Bernoulli shift, the eigenvalue ${\displaystyle \lambda _{1}=1}$ is simple, and since the operator leaves invariant the Gauss–Kuzmin measure, the operator is ergodic with respect to the measure. This fact allows a short proof of the existence of Khinchin's constant.

Additional eigenvalues can be computed numerically; the next eigenvalue is λ₂ = −0.3036630029... (sequence A038517 in the OEIS) and its absolute value is known as the Gauss–Kuzmin–Wirsing constant. Analytic forms for additional eigenfunctions are not known. It is not known if the eigenvalues are irrational.

Let us arrange the eigenvalues of the Gauss–Kuzmin–Wirsing operator according to an absolute value:

${\displaystyle 1=|\lambda _{1}|>|\lambda _{2}|\geq |\lambda _{3}|\geq \cdots .}$

It was conjectured in 1995 by Philippe Flajolet and Brigitte Vallée that

${\displaystyle \lim _{n\to \infty }{\frac {\lambda _{n}}{\lambda _{n+1}}}=-\varphi ^{2},{\text{ where }}\varphi ={\frac {1+{\sqrt {5}}}{2}}.}$

In 2018, Giedrius Alkauskas gave a convincing argument that this conjecture can be refined to a much stronger statement:^[2]

${\displaystyle {\begin{aligned}&(-1)^{n+1}\lambda _{n}=\varphi ^{-2n}+C\cdot {\frac {\varphi ^{-2n}}{\sqrt {n}}}+d(n)\cdot {\frac {\varphi ^{-2n}}{n}},\\[4pt]&{\text{where }}C={\frac {{\sqrt[{4}]{5}}\cdot \zeta (3/2)}{2{\sqrt {\pi }}}}=1.1019785625880999_{+};\end{aligned}}}$

here the function ${\displaystyle d(n)}$ is bounded, and ${\displaystyle \zeta (\star )}$ is the Riemann zeta function.

Continuous spectrum

The eigenvalues form a discrete spectrum, when the operator is limited to act on functions on the unit interval of the real number line. More broadly, since the Gauss map is the shift operator on Baire space ${\displaystyle \mathbb {N} ^{\omega }}$, the GKW operator can also be viewed as an operator on the function space ${\displaystyle \mathbb {N} ^{\omega }\to \mathbb {C} }$ (considered as a Banach space, with basis functions taken to be the indicator functions on the cylinders of the product topology). In the later case, it has a continuous spectrum, with eigenvalues in the unit disk ${\displaystyle |\lambda |<1}$ of the complex plane. That is, given the cylinder ${\displaystyle C_{n}[b]=\{(a_{1},a_{2},\cdots )\in \mathbb {N} ^{\omega }:a_{n}=b\}}$, the operator G shifts it to the left: ${\displaystyle GC_{n}[b]=C_{n-1}[b]}$. Taking ${\displaystyle r_{n,b}(x)}$ to be the indicator function which is 1 on the cylinder (when ${\displaystyle x\in C_{n}[b]}$), and zero otherwise, one has that ${\displaystyle Gr_{n,b}=r_{n-1,b}}$. The series

${\displaystyle f(x)=\sum _{n=1}^{\infty }\lambda ^{n-1}r_{n,b}(x)}$

then is an eigenfunction with eigenvalue ${\displaystyle \lambda }$. That is, one has ${\displaystyle [Gf](x)=\lambda f(x)}$ whenever the summation converges: that is, when ${\displaystyle |\lambda |<1}$.

A special case arises when one wishes to consider the Haar measure of the shift operator, that is, a function that is invariant under shifts. This is given by the Minkowski measure ${\displaystyle$ ?^{\prime }} . That is, one has that ${\displaystyle G?^{\prime }=?^{\prime }}$.^[3]

Entropy

The Gauss map, over the Gauss measure, has entropy ${\displaystyle {\frac {\pi ^{2}}{6\ln 2}}}$. This can be proved by the Rokhlin formula for entropy. Then using the Shannon–McMillan–Breiman theorem, with its equipartition property, we obtain Lochs' theorem.^[6]

Measure-theoretic preliminaries

A covering family ${\displaystyle {\mathcal {C}}}$ is a set of measurable sets, such that any open set is a disjoint union of sets in it. Compare this with base in topology, which is less restrictive as it allows non-disjoint unions.

Knopp's lemma. Let ${\displaystyle B\subset [0,1)}$ be measurable, let ${\displaystyle {\mathcal {C}}}$ be a covering family and suppose that ${\displaystyle \exists \gamma >0,\forall A\in {\mathcal {C}},\mu (A\cap B)\geq \gamma \mu (A)}$. Then ${\displaystyle \mu (B)=1}$.

Proof. Since any open set is a disjoint union of sets in ${\displaystyle {\mathcal {C}}}$, we have ${\displaystyle \mu (A\cap B)\geq \gamma \mu (A)}$ for any open set ${\displaystyle A}$, not just any set in ${\displaystyle {\mathcal {C}}}$.

Take the complement ${\displaystyle B^{c}}$. Since the Lebesgue measure is outer regular, we can take an open set ${\displaystyle B'}$ that is close to ${\displaystyle B^{c}}$, meaning the symmetric difference has arbitrarily small measure ${\displaystyle \mu (B'\Delta B^{c})<\epsilon }$.

At the limit, ${\displaystyle \mu (B'\cap B)\geq \gamma \mu (B')}$ becomes have ${\displaystyle 0\geq \gamma \mu (B^{c})}$.

The Gauss map is ergodic

Fix a sequence ${\displaystyle a_{1},\dots ,a_{n}}$ of positive integers. Let ${\displaystyle {\frac {q_{n}}{p_{n}}}=[0;a_{1},\dots ,a_{n}]}$. Let the interval ${\displaystyle \Delta _{n}}$ be the open interval with end-points ${\displaystyle [0;a_{1},\dots ,a_{n}],[0;a_{1},\dots ,a_{n}+1]}$.

Lemma. For any open interval ${\displaystyle (a,b)\subset (0,1)}$, we have

$${\displaystyle \mu (T^{-n}(a,b)\cap \Delta _{n})=\mu ((a,b))\mu (\Delta _{n})\underbrace {\left({\frac {q_{n}(q_{n}+q_{n-1})}{(q_{n}+q_{n-1}b)(q_{n}+q_{n-1}a)}}\right)} _{\geq 1/2}}$$

Proof. For any ${\displaystyle t\in (0,1)}$ we have ${\displaystyle [0;a_{1},\dots ,a_{n}+t]={\frac {q_{n}+q_{n-1}t}{p_{n}+p_{n-1}t}}}$ by standard continued fraction theory. By expanding the definition, ${\displaystyle T^{-n}(a,b)\cap \Delta _{n}}$ is an interval with end points ${\displaystyle [0;a_{1},\dots ,a_{n}+a],[0;a_{1},\dots ,a_{n}+b]}$. Now compute directly. To show the fraction is ${\displaystyle \geq 1/2}$, use the fact that ${\displaystyle q_{n}\geq q_{n-1}}$.

Theorem. The Gauss map is ergodic.

Proof. Consider the set of all open intervals in the form ${\displaystyle ([0;a_{1},\dots ,a_{n}],[0;a_{1},\dots ,a_{n}+1])}$. Collect them into a single family ${\displaystyle {\mathcal {C}}}$. This ${\displaystyle {\mathcal {C}}}$ is a covering family, because any open interval ${\displaystyle (a,b)\setminus \mathbb {Q} }$ where ${\displaystyle a,b}$ are rational, is a disjoint union of finitely many sets in ${\displaystyle {\mathcal {C}}}$.

Suppose a set ${\displaystyle B}$ is ${\displaystyle T}$-invariant and has positive measure. Pick any ${\displaystyle \Delta _{n}\in {\mathcal {C}}}$. Since Lebesgue measure is outer regular, there exists an open set ${\displaystyle B_{0}}$ which differs from ${\displaystyle B}$ by only ${\displaystyle \mu (B_{0}\Delta B)<\epsilon }$. Since ${\displaystyle B}$ is ${\displaystyle T}$-invariant, we also have ${\displaystyle \mu (T^{-n}B_{0}\Delta B)=\mu (B_{0}\Delta B)<\epsilon }$. Therefore,

$${\displaystyle \mu (T^{-n}B_{0}\cap \Delta _{n})\in \mu (B\cap \Delta _{n})\pm \epsilon }$$

By the previous lemma, we have

$${\displaystyle \mu (T^{-n}B_{0}\cap \Delta _{n})\geq {\frac {1}{2}}\mu (B_{0})\mu (\Delta _{n})\in {\frac {1}{2}}(\mu (B)\pm \epsilon )\mu (\Delta _{n})}$$

Take the ${\displaystyle \epsilon \to 0}$ limit, we have ${\displaystyle \mu (B\cap \Delta _{n})\geq {\frac {1}{2}}\mu (B)\mu (\Delta _{n})}$. By Knopp's lemma, it has full measure.

Matrix elements

Consider the Taylor series expansions at x = 1 for a function f(x) and ${\displaystyle g(x)=[Gf](x)}$. That is, let

${\displaystyle f(1-x)=\sum _{n=0}^{\infty }(-x)^{n}{\frac {f^{(n)}(1)}{n!}}}$

and write likewise for g(x). The expansion is made about x = 1 because the GKW operator is poorly behaved at x = 0. The expansion is made about 1 − x so that we can keep x a positive number, 0 ≤ x ≤ 1. Then the GKW operator acts on the Taylor coefficients as

${\displaystyle (-1)^{m}{\frac {g^{(m)}(1)}{m!}}=\sum _{n=0}^{\infty }G_{mn}(-1)^{n}{\frac {f^{(n)}(1)}{n!}},}$

where the matrix elements of the GKW operator are given by

${\displaystyle G_{mn}=\sum _{k=0}^{n}(-1)^{k}{n \choose k}{k+m+1 \choose m}\left[\zeta (k+m+2)-1\right].}$

This operator is extremely well formed, and thus very numerically tractable. The Gauss–Kuzmin constant is easily computed to high precision by numerically diagonalizing the upper-left n by n portion. There is no known closed-form expression that diagonalizes this operator; that is, there are no closed-form expressions known for the eigenvectors.

Riemann zeta

The Riemann zeta can be written as

${\displaystyle \zeta (s)={\frac {s}{s-1}}-s\sum _{n=0}^{\infty }(-1)^{n}{s-1 \choose n}t_{n}}$

where the ${\displaystyle t_{n}}$ are given by the matrix elements above:

${\displaystyle t_{n}=\sum _{m=0}^{\infty }{\frac {G_{mn}}{(m+1)(m+2)}}.}$

Performing the summations, one gets:

${\displaystyle t_{n}=1-\gamma +\sum _{k=1}^{n}(-1)^{k}{n \choose k}\left[{\frac {1}{k}}-{\frac {\zeta (k+1)}{k+1}}\right]}$

where ${\displaystyle \gamma }$ is the Euler–Mascheroni constant. These ${\displaystyle t_{n}}$ play the analog of the Stieltjes constants, but for the falling factorial expansion. By writing

${\displaystyle a_{n}=t_{n}-{\frac {1}{2(n+1)}}}$

one gets: a₀ = −0.0772156... and a₁ = −0.00474863... and so on. The values get small quickly but are oscillatory. Some explicit sums on these values can be performed. They can be explicitly related to the Stieltjes constants by re-expressing the falling factorial as a polynomial with Stirling number coefficients, and then solving. More generally, the Riemann zeta can be re-expressed as an expansion in terms of Sheffer sequences of polynomials.

This expansion of the Riemann zeta is investigated in the following references.^{[7][8][9][10][11]} The coefficients are decreasing as

${\displaystyle \left({\frac {2n}{\pi }}\right)^{1/4}e^{-{\sqrt {4\pi n}}}\cos \left({\sqrt {4\pi n}}-{\frac {5\pi }{8}}\right)+{\mathcal {O}}\left({\frac {e^{-{\sqrt {4\pi n}}}}{n^{1/4}}}\right).}$