4
$\begingroup$

Suppose that an irrational $x$ in $(0,1)$ has convergents $c(k,x)$, and let

$$d(x) = \sum_{k=0}^{\infty} \mid x - c(k,x)\mid.$$

What is the mean value of $d$?

$\endgroup$
  • $\begingroup$ The function $d(x)$ is continuous at any irrational $x$. So it is sufficient to know $d(x)$ for rational $x$. $\endgroup$ – Alexey Ustinov Jan 5 '16 at 6:04
  • $\begingroup$ Are there any reasons to study this constant? $\endgroup$ – Alexey Ustinov Jan 5 '16 at 11:41
  • $\begingroup$ Think of $d(x)$ as the deviance of x from its continued fraction, maximized by the golden ratio, with maximal value approximately $1.195955786017513596003474800021$. (One can also define upper and lower deviances, which are also maximized by the golden ratio.) I raised the question about the "mean deviance" because of the possibility that it doesn't exist; viz., is $d$ Lebesgue integrable? $\endgroup$ – Clark Kimberling Jan 12 '16 at 20:30
4
$\begingroup$

If $\frac{p_{2k}}{ q_{2k}}$ and $\frac{p_{2k+1}}{ q_{2k+1}}$ ($k\ge 0$) are consecutive convergents of the continued fraction expansion of $x$ then $$\left|x-\frac{p_{2k}}{ q_{2k}}\right|+\left|x-\frac{p_{2k+1}}{ q_{2k+1}}\right|=\frac{p_{2k+1}}{ q_{2k+1}}-\frac{p_{2k}}{ q_{2k}}=\frac{1}{q_{2k}q_{2k+1} }.$$

For example if $x=\frac{\sqrt{5}-1}{2 }$ then $$d(x)=\frac{1}{ F_1F_2}+\frac{1}{ F_3F_4}+\frac{1}{ F_5F_6}+\ldots=\frac{1}{ 1\times 1}+\frac{1}{2\times 3}+\frac{1}{5\times 8}+\ldots$$

Each interval $\left(\frac{a}{ b},\frac{c}{ d}\right)\subset(0,1)$ s.t. $ad-bc=-1$ may occur as $\left(\frac{p_{2k}}{ q_{2k}},\frac{p_{2k+1}}{ q_{2k+1}}\right)$. It means that $$D=\int_{0}^{1}d(x)dx=\sum_{{ad-bc=-1,b\le d\atop 0\le\frac{a}{ b}<\frac{c}{ d}\le 1}}\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)\frac{1}{bd },$$ where $\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)$ is a measure of the set $$\left\{x:\frac{a}{ b},\frac{c}{ d}\text{ are consecutive convergents of the continued fraction expansion of } x\right\}.$$ It is known that two fractions $\frac{P}{ Q}$ and $\frac{P'}{ Q'}$ are consecutive convergents of the continued fraction expansion of $x$ (and, moreover, the convergent $P/Q$ precedes the convergent $P'/Q'$ iff (see Lemma 1 here) $$0<\frac{Q'x-P'}{-Qx+P }<1.$$ For $P/Q<P'/Q'$ (and $Q\le Q'$) this condition defines the interval $\left(\frac{P+P'}{Q+Q' },\frac{P'}{Q' }\right)$ of the length $\frac{1}{ Q'(Q+Q')}$, so $\lambda\left(\frac{a}{ b},\frac{c}{ d}\right)=\frac{1}{ d(b+d)}$ and $$D=\sum_{{ad-bc=-1,b\le d\atop 0\le\frac{a}{ b}<\frac{c}{ d}\le 1}}\frac{1}{b(b+d)d^2}.$$ For each pair of denominatos $(b,d)$ s.t. $1\le b\le d$ and $(b,d)=1$ numerators $a$ and $c$ are uniquely defined (see for example section 3 here). Hence

$$D=\sum_{1\le b\le d,(b,d)=1}\frac{1}{b(b+d)d^2}=\frac{1}{ \zeta(4)}\sum_{1\le b\le d}\frac{1}{b(b+d)d^2}.$$

Calculation of this sum is another problem. It is better to ask people from MZV-community. I can only simplify it a little: $$\sum_{1\le b\le d}\frac{1}{b(b+d)d^2}=D_1-D_2,$$ where $$D_1=\sum_{1\le b\le d}\frac{1}{bd^3}=\frac{\pi^4}{ 72},$$ and $$D_2=\sum_{1\le b\le d}\frac{1}{(b+d)d^3}=\zeta(3)\log 2-\int_{0}^{1}\frac{\mathrm{Li}_3(t^2)}{1+t }dt=2\int_{0}^{1}\frac{\mathrm{Li}_2(t)}{t}\log(1+t)dt.$$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.