As a computer experiment I did a few Riemannian sums to see if I could quantify the density statement $\overline{\mathbb{Q}(\sqrt{2}, \sqrt{3})} = \mathbb{R}$ :

$$ \Big| \frac{1}{N^2} \sum_{0 \leq m,n \leq N} \{ \sqrt{2} m + \sqrt{3} n \}^5 - \frac{1}{6} \Big| \stackrel{?}{<} \frac{1}{N^2} $$

A log-plot shows the correct exponent is a bit less than $2$. Is it a Hausdorff dimension of some kind?

enter image description here

The general quantitative statement looks like some error term to an average:

$$ \Big| \frac{1}{N^2} \sum_{0 \leq m,n \leq N} f \big( \{ \sqrt{2} m + \sqrt{3} n \} \big) - \int_0^1 f(x) \, dx \Big| \stackrel{?}{\ll} \frac{1}{N^2} $$

This is certainly false... what might the a good exponent be? The statement could be more general - I have some kind of totally real number field - but then we get a worse exponent.

  • 5
    $\begingroup$ I think you can get an upper bound by standard bounds for the relevant exponential sums combined with the Erdős-Turán-Koksma inequality. $\endgroup$ – GH from MO Mar 19 '18 at 12:54
  • $\begingroup$ @GHfromMO this is vanilla discrepancy theory full. the wikipedia article has a statement that's somewhat hard to read. I also notice their use of measure theory. In a way, I'm just asking to work out the Discrepancy upper bound in that case of $\mathbb{Z}[\sqrt{2}, \sqrt{3}]/\mathbb{Z}$. $\endgroup$ – john mangual Mar 19 '18 at 13:04
  • 4
    $\begingroup$ Yes, I understand, and I gave a hint how to work out an upper bound. $\endgroup$ – GH from MO Mar 19 '18 at 13:12
  • $\begingroup$ "A log-plot" ... of what, exactly? $\endgroup$ – Greg Martin Mar 19 '18 at 16:44
  • $\begingroup$ @GregMartin the horizontal axis is $N$. The vertical axis is the log of the (absolute value) of the error term. $\endgroup$ – john mangual Mar 19 '18 at 17:54

From Koksma-Hlawka inequality (see here), the following holds for functions with bounded variation in Hardy-Krause sense.

If $f:[0,1]\rightarrow\mathbb{R}$ and $g:[0,1]\times [0,1]\rightarrow\mathbb{R}$ defined by $g(x,y)=f(\{x+y\})$ is bounded variation in Hardy-Krause sense, then $$ \left|\frac1{N^2} \sum_{0\leq m,n \leq N} f(\{\sqrt 2 m + \sqrt 3 n\})-\int_0^1 f(x)dx\right|\leq C V(g) D_{N^2} $$ where $C>0$ is absolute, $V(g)$ is the Hardy-Krause variation of $g$, and $D_{N^2}$ is the discrepancy of the double sequence $\{(\{\sqrt 2 m\},\{\sqrt 3 n\} ) \ | \ 0\leq m,n\leq N\}$.

As @GH from MO suggested, Erdos-Turan-Koksma inequality gives an upper bound of $D_{N^2}$. The crucial estimate here is the bound of $||x||$ the distance between $x$ and its nearest integer, for certain numbers $x$.

Let $1\leq H\leq N^2$, we have $$\begin{align} D_{N^2}&\ll \frac1H + \sum_{h_1<H, \ h_2<H} \frac1{h_1h_2} \frac1{N^2}\left|\sum_{m\leq N, \ n\leq N} e^{2\pi i (h_1 m\sqrt 2 +h_2 n \sqrt 3)}\right|\\ &\ll \frac1H+ \frac1{N^2}\sum_{h_1<H} \frac1{h_1 \| h_1 \sqrt 2 \|} \sum_{h_2<H} \frac1{h_2 \|h_2\sqrt 3\|} \end{align} $$

If $\alpha$ is an irrational number with bounded partial quotients in its continued fraction, then $$ \sum_{h<H} \frac1{h\| h\alpha\|} \ll \log^2 H. $$ Since $\sqrt 2$ and $\sqrt 3$ both have bounded partial quotients in their continued fractions, we have $$ D_{N^2} \ll \frac1H+ \frac1{N^2} \log^4 H. $$ Taking $H=N^2$, we obtain $$ D_{N^2} \ll \frac{\log^4 N}{N^2}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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