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Revision difference : Weyl's criterion
Version current Version 3
Let $\{u_n\}$ be a sequence of real numbers. Then $\{u_n\}$ is Let $\{u_n\}$ be a sequence of real numbers. Then $\{u_n\}$ is
uniformly distributed modulo $1$ if and only if uniformly distributed modulo $1$ if and only if
\begin{equation*} \begin{equation*}
\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N e(k u_n)=0 \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N e(k u_n)=0
\end{equation*} \end{equation*}
for every nonzero integer $k$, where $e(x)=\exp(2\pi i x)$. for every nonzero integer $k$, where $e(x)=\exp(2\pi i x)$.
Weyl's criterion reduces the problem of uniform distribution of Weyl's criterion reduces the problem of uniform distribution of
sequences to the problem of estimating certain exponential sums. sequences to the problem of estimating certain exponential sums.
Whereas the problem of estimating a family of exponential sums Whereas the problem of estimating a family of exponential sums
might seem harder at first, the exponential map has the might seem harder at first, the exponential map has the
multiplicative property which often makes the problem easier. multiplicative property which often makes the problem easier.
\emph{Example:} If $x$ is irrational, then the sequence $\{nx\}$ \emph{Example:} If $x$ is irrational, then the sequence $\{nx\}$
is uniformly distributed modulo $1$. Proof: is uniformly distributed modulo $1$. Proof:
\begin{equation*} \begin{equation*}
\abs{\sum_{n=1}^{N} e(k n \abs{\sum_{n=1}^{N} e(k n
x)}=\abs{\frac{e(k(N+1)x)-e(kx)}{e(kx)-1}}\leq x)}=\abs{\frac{e(k(N+1)x)-e(kx)}{e(kx)-1}}\leq
\frac{2}{\abs{\,e(kx)-1}}=O_k(1) \frac{2}{\abs{\,e(kx)-1}}=O_k(1)
\end{equation*} \end{equation*}
because the irrationality of $x$ implies $e(kx)\neq 1$. because the irrationality of $x$ implies $e(kx)\neq 1$.
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