English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

In this work we proof the following theorem which is, in addition tosomeother lemmas, our main result:\noindent \textbf{theorem}. Let$\ X=\{ (x\_{1}\text{, }%t\_{1}) \text{, }( x\_{2}\text{, }t\_{2}) \text{, ...,}(x\_{n}\text{, }t\_{n})\} $ be a finite part of $\mathbb{R}\times\mathbb{R}^{\ast +}$, then there exist a finite part $R$ of $\mathbb{R}%^{\ast+}$ such that for all $\varepsilon > 0$ there exists $r\in R$ such that if $0 <\varepsilon \leq r$ then there exist rational numbers $( \dfrac{p\_{i}}{q})\_{i=1,2,...,n}$ such that:\{c}| x\_{i}-\dfrac{p\_{i}}{q}| \leq \varepsilont\_{i} \varepsilon q\leq t\_{i}|\text{, }i=1,2,...,n\text{.} \tag{*}\noindentIt is clear that the condition $\varepsilon q\leq t\_{i}$ for $%i=1,2,...,n$ isequivalent to $\varepsilon q\leq t=\underset{i=1,2,...,n}{Min%}$ $( t\_{i}) $.\Also, we have (*) for all $\varepsilon $verifying $0 < \varepsilon \leq\varepsilon \_{0}=\min R$.The previous theorem is the classical equivalent ofthe following one whichis formulated in the context of the nonstandard analysis($[ 2] $%, $[ 5] $, $[ 6] $, $[ 8] $).\noindent \textbf{theorem. }For everypositive infinitesimal real $\varepsilon$, there exists an unlimited integer$q$ depending only of $\varepsilon $, such that $\forall ^{st}x \in \mathbb{R}\\exists p_{x} \in \mathbb{Z}$ $:\{ \{ccc}x & = \& \dfrac{p_{x}}{q}+\varepsilon\phi \varepsilon q \& \cong \& 0.$ For this reason, to prove the nonstandardversion of the main result and to get its classical version we place ourselvesin the context of the nonstandard analysis.

Diophantine approximation with improvement of the simultaneous control of the error and of the denominator