Lochs's theorem explained

In number theory, Lochs's theorem concerns the rate of convergence of the continued fraction expansion of a typical real number. A proof of the theorem was published in 1964 by Gustav Lochs.

The theorem states that for almost all real numbers in the interval (0,1), the number of terms m of the number's continued fraction expansion that are required to determine the first n places of the number's decimal expansion behaves asymptotically as follows:

\lim _{n\toinfty} (m)/(= n) (6ln ( 2 ) ln ( 10 ))/(\pi ^{2}) ≈ 0 . 97027014 .

As this limit is only slightly smaller than 1, this can be interpreted as saying that each additional term in the continued fraction representation of a "typical" real number increases the accuracy of the representation by approximately one decimal place. The decimal system is the last positional system for which each digit carries less information than one continued fraction quotient; going to base-11 (changing

ln ( 10 ) to ln ( 11 ) in the equation) makes the above value exceed 1.

The reciprocal of this limit,

(\pi ^{2})/(6ln ( 2 ) ln ( 10 )) ≈ 1 . 03064083 ,

is twice the base-10 logarithm of Lévy's constant.

A prominent example of a number not exhibiting this behavior is the golden ratio—sometimes known as the "most irrational" number—whose continued fraction terms are all ones, the smallest possible in canonical form. On average it requires approximately 2.39 continued fraction terms per decimal digit.^[1]

Proof

The proof assumes basic properties of continued fractions.Let

T : x \mapsto 1 /x \mod 1 be the Gauss map.

Let

\rho(t) = (1)/(( 1+ t) ln2) be the probability density function for the Gauss distribution, which is preserved under the Gauss map.

Since the probability density function is bounded above and below, a set is negligible with respect to the Lebesgue measure iff to the Gauss distribution.

Lemma

Lemma. $\backslash frac\; 1n\; \backslash ln\; T^n\; x\; \backslash to\; 0$.

Proof. Since $T^nx\; \backslash leq\; 1$, we have $\backslash frac\; 1n\; \backslash ln\; T^n\; x\; \backslash to\; 0$ iff$$\backslash liminf\backslash frac\; 1n\; \backslash ln\; T^n\; x\; =\; 0$$Let us consider the set of all $x$ that have . That is,$$\backslash$$ $$=\; \backslash cup\_\; \backslash cap\_\; \backslash cup\_\; [0;\; \backslash N,\; \backslash dots,\; \backslash N,\; a\_n\; >\; e^\{cn\},\; \backslash N,\; \backslash dots]$$where

[ 0 ; \N, ... , \N, a _{n} > e ^{cn} , \N, ... ] denotes the set of numbers whose continued fraction expansion has a _{n} > e ^{cn} , but no other constraints. Now, since the Gauss map preserves the Gauss measure,$$[0;\; \backslash N,\; \backslash dots,\; \backslash N,\; a\_n\; >\; e^\{cn\},\; \backslash N,\; \backslash dots]$$has the same Gauss measure as $[0;\; a\_n\; >\; e^\{cn\},\; \backslash N,\; \backslash dots]$, which is the same as

$$\backslash int\_0^\; \backslash rho(t)dt\; =\; \backslash log\_2(1+e^)\backslash sim\; \backslash frac$$The union over $\backslash cup\_$ sums to $\backslash sim\; \backslash frac$, which at the $N\backslash to\backslash infty$ limit is zero.

Thus the set of such $x$ has Gauss measure zero.

Finish the estimate

Now, expand the term using basic continued fraction properties:$$\backslash ln\backslash left|\; x\; -\; \backslash frac\; \backslash right|\; =\; \backslash ln\; \backslash frac\; =\; -\; 2\backslash ln\; q\_n\; +\; \backslash ln\; T^n\; x\; -\; \backslash ln\backslash left(1\; +\; \backslash fracT^n\; x\; \backslash right)$$The second is $o(n)$. The third term is $\backslash in\; [\backslash ln\; 1,\; \backslash ln\; 2]$. Both disappear after dividing by

n .
Thus$$\backslash lim\_\; \backslash frac\; 1n\; \backslash ln\backslash left|\; x\; -\; \backslash frac\; \backslash right|\; =\; -2\; \backslash lim\_n\; \backslash frac\; 1n\; \backslash ln\; q\_n\; =\; -\backslash frac$$where we used the result from Lévy's constant.

References

1. News: Cooper. Harold. Continued Fraction Streams. Exist . 17 August 2016 . 30 August 2016.