On a Conjecture of Cusick on a sum of Cantor sets
In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as a sum of two continued fractions with no partial quotients equal to $1$. In other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{ for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also...
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creator | Shulga, Nikita |
description | In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as
a sum of two continued fractions with no partial quotients equal to $1$. In
other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{
for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also
conjectured that this result is unique in the sense that if you exclude partial
quotients from $1$ to $k-1$ with $k\geq3$, then the Lebesgue measure $\lambda$
of the set of numbers which can be expressed as a sum of two continued
fractions with no partial quotients from $\{1,\ldots,k-1\}$ is equal to $0$,
that is $$\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3.$$ In this
paper, we disprove the conjecture of Cusick by showing that $$ S(k)+S(k)
\supseteq \left[0,\frac{1}{k-1}\right]. $$ The proof is constructive and does
not rely on ideas from previous works on the topic. We also show the existence
of countably many 'gaps' in $S(k)+S(k)$, that is intervals, for which the
endpoints lie in $S(k)+S(k)$, while none of the elements in the interior do so.
Finally, we prove several results on the sums $$ S(m)+S(n) $$ for $m\neq n$. |
doi_str_mv | 10.48550/arxiv.2411.17379 |
format | Article |
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a sum of two continued fractions with no partial quotients equal to $1$. In
other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{
for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also
conjectured that this result is unique in the sense that if you exclude partial
quotients from $1$ to $k-1$ with $k\geq3$, then the Lebesgue measure $\lambda$
of the set of numbers which can be expressed as a sum of two continued
fractions with no partial quotients from $\{1,\ldots,k-1\}$ is equal to $0$,
that is $$\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3.$$ In this
paper, we disprove the conjecture of Cusick by showing that $$ S(k)+S(k)
\supseteq \left[0,\frac{1}{k-1}\right]. $$ The proof is constructive and does
not rely on ideas from previous works on the topic. We also show the existence
of countably many 'gaps' in $S(k)+S(k)$, that is intervals, for which the
endpoints lie in $S(k)+S(k)$, while none of the elements in the interior do so.
Finally, we prove several results on the sums $$ S(m)+S(n) $$ for $m\neq n$.</description><identifier>DOI: 10.48550/arxiv.2411.17379</identifier><language>eng</language><subject>Mathematics - Number Theory</subject><creationdate>2024-11</creationdate><rights>http://creativecommons.org/licenses/by/4.0</rights><oa>free_for_read</oa><woscitedreferencessubscribed>false</woscitedreferencessubscribed></display><links><openurl>$$Topenurl_article</openurl><openurlfulltext>$$Topenurlfull_article</openurlfulltext><thumbnail>$$Tsyndetics_thumb_exl</thumbnail><link.rule.ids>228,230,780,885</link.rule.ids><linktorsrc>$$Uhttps://arxiv.org/abs/2411.17379$$EView_record_in_Cornell_University$$FView_record_in_$$GCornell_University$$Hfree_for_read</linktorsrc><backlink>$$Uhttps://doi.org/10.48550/arXiv.2411.17379$$DView paper in arXiv$$Hfree_for_read</backlink></links><search><creatorcontrib>Shulga, Nikita</creatorcontrib><title>On a Conjecture of Cusick on a sum of Cantor sets</title><description>In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as
a sum of two continued fractions with no partial quotients equal to $1$. In
other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{
for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also
conjectured that this result is unique in the sense that if you exclude partial
quotients from $1$ to $k-1$ with $k\geq3$, then the Lebesgue measure $\lambda$
of the set of numbers which can be expressed as a sum of two continued
fractions with no partial quotients from $\{1,\ldots,k-1\}$ is equal to $0$,
that is $$\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3.$$ In this
paper, we disprove the conjecture of Cusick by showing that $$ S(k)+S(k)
\supseteq \left[0,\frac{1}{k-1}\right]. $$ The proof is constructive and does
not rely on ideas from previous works on the topic. We also show the existence
of countably many 'gaps' in $S(k)+S(k)$, that is intervals, for which the
endpoints lie in $S(k)+S(k)$, while none of the elements in the interior do so.
Finally, we prove several results on the sums $$ S(m)+S(n) $$ for $m\neq n$.</description><subject>Mathematics - Number Theory</subject><fulltext>true</fulltext><rsrctype>article</rsrctype><creationdate>2024</creationdate><recordtype>article</recordtype><sourceid>GOX</sourceid><recordid>eNpjYJA0NNAzsTA1NdBPLKrILNMzMjE01DM0Nza35GQw9M9TSFRwzs_LSk0uKS1KVchPU3AuLc5MzlbIB8kUl-aChRLzSvKLFIpTS4p5GFjTEnOKU3mhNDeDvJtriLOHLtjw-IKizNzEosp4kCXxYEuMCasAAPwzMAM</recordid><startdate>20241126</startdate><enddate>20241126</enddate><creator>Shulga, Nikita</creator><scope>AKZ</scope><scope>GOX</scope></search><sort><creationdate>20241126</creationdate><title>On a Conjecture of Cusick on a sum of Cantor sets</title><author>Shulga, Nikita</author></sort><facets><frbrtype>5</frbrtype><frbrgroupid>cdi_FETCH-arxiv_primary_2411_173793</frbrgroupid><rsrctype>articles</rsrctype><prefilter>articles</prefilter><language>eng</language><creationdate>2024</creationdate><topic>Mathematics - Number Theory</topic><toplevel>online_resources</toplevel><creatorcontrib>Shulga, Nikita</creatorcontrib><collection>arXiv Mathematics</collection><collection>arXiv.org</collection></facets><delivery><delcategory>Remote Search Resource</delcategory><fulltext>fulltext_linktorsrc</fulltext></delivery><addata><au>Shulga, Nikita</au><format>journal</format><genre>article</genre><ristype>JOUR</ristype><atitle>On a Conjecture of Cusick on a sum of Cantor sets</atitle><date>2024-11-26</date><risdate>2024</risdate><abstract>In 1971 Cusick proved that every real number $x\in[0,1]$ can be expressed as
a sum of two continued fractions with no partial quotients equal to $1$. In
other words, if we define a set $$ S(k):= \{ x\in[0,1] : a_n(x) \geq k \text{
for all } n\in\mathbb{N} \}, $$ then $$ S(2)+S(2) = [0,1]. $$ He also
conjectured that this result is unique in the sense that if you exclude partial
quotients from $1$ to $k-1$ with $k\geq3$, then the Lebesgue measure $\lambda$
of the set of numbers which can be expressed as a sum of two continued
fractions with no partial quotients from $\{1,\ldots,k-1\}$ is equal to $0$,
that is $$\lambda\Bigl( S(k)+S(k) \Bigl)= 0 \text{ for }k\geq 3.$$ In this
paper, we disprove the conjecture of Cusick by showing that $$ S(k)+S(k)
\supseteq \left[0,\frac{1}{k-1}\right]. $$ The proof is constructive and does
not rely on ideas from previous works on the topic. We also show the existence
of countably many 'gaps' in $S(k)+S(k)$, that is intervals, for which the
endpoints lie in $S(k)+S(k)$, while none of the elements in the interior do so.
Finally, we prove several results on the sums $$ S(m)+S(n) $$ for $m\neq n$.</abstract><doi>10.48550/arxiv.2411.17379</doi><oa>free_for_read</oa></addata></record> |
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title | On a Conjecture of Cusick on a sum of Cantor sets |
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