First, let me summarize continued fractions and Liouville numbers.

Continued fractions.

We can represent each irrational number as a (simple) continued fraction by $$[a_0;a_1,a_2,\cdots\ ]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$$ where for natural numbers $i$ we have $a_i\in\mathbb{N}$, and we also have $a_0\in\mathbb{Z}$. Each irrational number has a unique continued fraction and each continued fraction represents a unique irrational number.

Liouville numbers.

An irrational number $\alpha$ is a Liouville number such that, for each positive integer $n$, there exist integers $p,q$ (where $q$ is nonzero) with $$\left|\alpha-\frac pq\right|<\frac1{q^n}$$ The important thing here is that you can approximate Liouville numbers well, and the side effect is that these numbers are transcendental.

Now if we look at the Liouville's constant, that is, $L=0.1100010\ldots$ (where the $i!$-th digit is a $1$ and the others are $0$), then we can write $$L=[0;9,1,99,1,10,9,999999999999,1,\cdots\ ]$$ The large numbers in the continued fraction make the convergents very close to the actual value, so that the number it represents is in that sense "well approximatable".
My question now is, can we bound the numbers in the continued fraction below to be sure that the number it represents is a Liouville number?


Bounding the error.

The error between a continued fraction $[a_0;a_1,a_2,\ldots]$ and its truncation to the rational number $[a_0;a_1,a_2,\ldots,a_n]$ is given by $$ |[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]|=\left|\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots]}\right) - \left(a_0 + \frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\right|=\left|\frac{[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]}{[a_1;a_2,a_3,\ldots]\cdot[a_1;a_2,a_3,\ldots,a_n]}\right| \le \frac{\left|[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]\right|}{a_1^2}, $$ terminating with $\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;]\right|\le 1/a_1$; by iterating this recursive bound we conclude that $$ \left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]\right| \le \frac{1}{a_1^2 a_2^2 \cdots a_n^2}\cdot \frac{1}{a_{n+1}}. $$ Let $D([a_0;a_1,a_2,\ldots,a_n])$ be the denominator of the truncation $[a_0;a_1,a_2,\ldots,a_n]$ (in lowest terms). Then we have a Liouville number if for any $\mu > 0$, the inequality $$ a_{n+1} \ge \frac{D([a_0;a_1,a_2,\ldots,a_n])^\mu}{a_1^2 a_2^2 \cdots a_n^2} $$ holds for some $n$. To give a more explicit expression, we need to bound the growth of $D$.

Bounding the denominator.

Let $D(x)$ and $N(x)$ denote the denominator and numerator of a rational number $x$ in lowest terms. Then $$ D([a_0;a_1,a_2,\ldots, a_n])=D\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =D\left(\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)=N([a_1;a_2,a_3,\ldots,a_n]), $$ and $$ N([a_0;a_1,a_2,\ldots, a_n])=N\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =N\left(a_0+\frac{D([a_1;a_2,a_3,\ldots,a_n])}{N([a_1;a_2,a_3,\ldots,a_n])}\right) = a_0 N([a_1;a_2,a_3,\ldots,a_n]) + D([a_1;a_2,a_3,\ldots,a_n]) \\ = a_0 D([a_0;a_1,a_2,\ldots, a_n]) + D([a_1;a_2,a_3,\ldots,a_n]). $$ So $$ D([a_0;a_1,a_2,\ldots,a_n]) = a_1 D([a_1;a_2,a_3,\ldots,a_n]) + D([a_2;a_3,a_4,\ldots,a_n]), $$ and the recursion terminates with $D([a_0;])=1$ and $D([a_0;a_1])=D(a_0+1/a_1)=a_1$. Since we have $D([a_0;a_1,a_2,\ldots,a_n]) \ge a_1 D([a_1;a_2,a_3,\ldots,a_n])$, we can say that $D([a_1;a_2,a_3\ldots,a_n]) \le \frac{1}{a_1}D([a_0;a_1,a_2,\ldots,a_n])$, and so $$ D([a_0;a_1,a_2,\ldots,a_n]) \le \left(a_1 +\frac{1}{a_2}\right) D([a_1;a_2,a_3,\ldots,a_n]) \le (a_1 + 1)D([a_1;a_2,a_3,\ldots,a_n]). $$ An explicit bound on the size of the denominator is therefore $$ D([a_0;a_1,a_2,\ldots,a_n]) \le (a_1+1)(a_2+1)\cdots(a_n+1). $$


We conclude the following theorem:

The continued fraction $[a_0;a_1,a_2,\ldots]$ is a Liouville number if, for any $\mu > 0$, there is some index $n$ such that $$a_{n+1} \ge \prod_{i=1}^{n}\frac{(a_i + 1)^{\mu}}{a_i^2}.$$

  • $\begingroup$ This is a really nice answer, exactly what I was looking for. I have one question however: should the last equation in "Bounding the error" not have a "greater than" ($>$) sign instead of a "greater or equal" ($\geq$) sign? Or am I missing something here? $\endgroup$ – vrugtehagel Mar 8 '16 at 21:18
  • $\begingroup$ It doesn't matter... the definition of a Liouville number is equivalent if $\le$ is used, since $\varepsilon \le q^{-(n+1)} \implies \varepsilon < q^{-n}$. $\endgroup$ – mjqxxxx Mar 8 '16 at 21:38
  • $\begingroup$ Right, of course. Silly of me. Thanks a lot! $\endgroup$ – vrugtehagel Mar 8 '16 at 21:40

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