# Equidistribution results vs transcendence degree

Consider $\alpha =(\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$ linearly independent over $\mathbb{Q}$, then the map $q \mapsto q \alpha:= ( q \alpha_1, \dots, q \alpha_n)$ gives a dense embedding of $\mathbb{Z}$ in the $n$ torus, i.e. $\mathbb{R}^n / \mathbb{Z}^n$, actually even more: it becomes equidistributed in the following sense $$\frac{1}{N}\sum\limits_{k \leq N} f(k\alpha) \rightarrow \int\limits_{\mathbb{R^N} / \mathbb{Z}^n} f( x) \mathrm{d} x,$$ where $\mathrm{d} x$ denotes the Haarmeasure. A similar result holds for the $\infty$ torus as well.

What can be said about the rate of convergence here? For $n=1$, can we distinguish wether $\alpha_1$ is transcendental or algebraic from the error term?

Motivation: I try to undestand effective version of universality theorems of L functions, see e.g. http://en.wikipedia.org/wiki/Zeta_function_universality.

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The rate of convergence is related to the discrepancy. Koksma's inequality says that if $f$ is of bounded variation $V(f)$, and $x_1,\dots,x_n$ are points in $[0,1)$ with discrepancy $D_n$, then $$\left|{1\over n}\sum_1^nf(x_i)-\int_0^1f(t)\,dt\right|\le V(f)D_n$$
The discrepancy, in turn, is a measure of departure from unifomity of distribution. It is defined by $$D_n=\sup_{0\lt a\le1}\left|{1\over n}\#\lbrace i\le n:x_i\lt a\rbrace-a\right|$$
The discrepancy of the sequence of fractional parts of $\alpha,2\alpha,3\alpha,\dots,n\alpha$ is closely related to the partial quotients in the continued fraction expansion of $\alpha$. For algebraic irrationals it is known that $nD_n=O(n^{\epsilon})$ for every $\epsilon\gt0$. But we don't know enough about continued fractions to distinguish transcendentals from algebraics (in general), so, in general, I don't think you'll be able to distinguuish the two classes by convergence rates.