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If $f$ is a continuous periodic fonction on $[0,1]$ and $a\not\in\mathbb{Q}$, the Weyl's equidistribution theorem states that $$\frac{1}{n}\sum_{k=0}^{n-1}f(ak)\rightarrow \int_0^1 f(x)dx.$$ Can we say something about the speed of convergence ? Thank you for your help !

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3 Answers 3

up vote 4 down vote accepted

As a complement to Gerry Myerson answer, you can bound the discrepancy $D_N$ using the Erdos-Turan inequality $$ D_{N} \leq \frac{\log 2}{\pi (H + 1)} + \frac{1}{\pi N} \sum_{h = 1}^{H} \frac{1}{h} \bigg | \sum_{n = 1}^{N} \exp(2\pi i h x_n) \bigg | $$ with $H$ arbitrary. The Erdos-Turan inequality is a reasonably easy consequence of Poisson-summation. To get nice constants as above one uses Beurling-Selberg majorants.

See: http://www.math.northwestern.edu/~mlerma/papers/charla/appl_of_ef.pdf and the references there-in (in particular the survey of Valeer in the Bulletin). The result cited above is Theorem 3.2 in the link.

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Koksma's inequality says $$\left|N^{-1}\sum_1^Nf(x_n)-\int_0^1f(t)\,dt\right|\le V(f)D^*_N$$ where $V(f)$ is the variation of $f$ (we're assuming $f$ is of bounded variation) and $D^*_N$ is the "discrepancy" of $\{x_1,\dots,x_N\}$. See Kuipers and Niederreiter, Uniform Distribution of Sequences, for this and more.

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As you see from the above two answers the rate of convergence will depend on the diophantine nature of $\alpha$. Indeed $$ \sum_{n = 1}^{N} \exp(2\pi i h \alpha n)\ll \min(N,1/||h \alpha||) $$ where $||\xi||$ is the distance of $\xi$ to the nearest integer. Therefore the speed of convergence will be bounded by $$ V(f) \cdot \bigg ( \frac{1}{H} + \frac{1}{N} \sum_{h = 1}^{H} \frac{\min(N,1/||h \alpha||)}{h} \bigg ) $$ and thus depend highly on the algebraic nature of $\alpha$.

Case: Algebraic numbers For example if $\alpha$ is algebraic then by Roth's theorem there are only finitely many solutions to $||q \alpha|| < q^{-1 - \varepsilon}$, therefore for any given $\varepsilon$ we will have $|| h \alpha || > c(\varepsilon) h^{-1 - \varepsilon}$ therefore $1 / ||h \alpha|| \leq c'(\varepsilon) h^{1 + \varepsilon}$. Therefore we get the following bound for the discrepancy $$ \ll_{\varepsilon} V(f) \cdot \bigg ( \frac{1}{H} + \frac{H^{1 + \varepsilon}}{N} \bigg ) $$ provided that $H \ll_{\varepsilon} N^{1 - \varepsilon}$. Choosing $H = \sqrt{N}$ we conclude that the discrepancy for algebraic numbers is bounded by $$ \ll_{\varepsilon} V(f) N^{-1/2 + \varepsilon} $$

Case: Transcendental numbers In general you can pick a transcendental number that will make convergence as slow as one wishes. If you are interested in specific transcendental numbers such as $\pi$ and $e$ then you might use results on the measure of transcendence to obtain a rate of convergence, much in the same manner as done above for algebraic numbers.

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The exponent is $-(2+\epsilon)$ in Roth's theorem, so I suppose one would pick $H = N^{1/3}$ to minimise the right-hand size. –  user58955 Sep 12 at 23:00

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