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 Tools annual

Global: Zendy Free Plan

Global: Zendy Tools monthly

Global: Zendy Plus monthly

We prove that if      { a n } n = 1 ∞  \{ {a_n}\} _{n = 1}^\infty       is such that      a n ↘  0 {a_n} \searrow 0       and  \[    lim n →  ∞  ( a n + 1 / a n ) = 1 , \lim \limits _{n \to \infty } ({a_n}_{ + 1}/{a_n}) = 1,     \]  then for the typical continuous function      f f       we have  \[    S n 0 := ∑  n = n 0 ∞  | f ( x n + 1 ) −  f ( x n ) | = + ∞  {S_{{n_0}}}: = \sum \limits _{n = {n_0}}^\infty | f({x_n}_{ + 1}) - f({x_n})| = + \infty     \]  whenever      x ∈  [ 0 , 1 −  a n 0 ] x \in [0,1 - {a_{{n_0}}}]       and      x n ∈  [ x + a n + 1 , x + a n ] {x_n} \in [x + {a_{n + 1}},x + {a_n}]      . Based on our result in a previous paper, we know that the above theorem fails to hold if      a n + 1 / a n = λ  &gt; 1 {a_{n + 1}}/{a_n} = \lambda &gt; 1      . We also prove that if      { a n } n = 1 ∞  \{ {a_n}\} _{n = 1}^\infty       is such that      a n ↘  0 {a_n} \searrow 0      , then for the typical continuous function      f f       we have      S n 0 = + ∞   if  x n = x + a n {S_{{n_0}}} = + \infty {\text { if }}{x_n} = x + {a_n}       and      x ∈  [ 0 , 1 −  a n 0 ] x \in [0,1 - {a_{{n_0}}}]      .

Continuous functions with everywhere infinite variation with respect to sequences