Mixing
Representation of Transcendental number via continued fractions
My question is quite simple.
As far as I know, rational numbers has finite continued fraction representation. And if for any number we have infinite continued fraction, then it necessarily is irrational.
This help us to make distinction between rationals and irrationals.
My question is, are their any criteria (of form of continued fraction) which will give distinction between algebraic irrationals and transcendental irrationals.(Maybe some pattern is prohibited for transcendental numbers, or some pattern is required.)
Sorry if this question sound stupid.
Thank you.
continued-fractions transcendental-numbers
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
add a comment |
My question is quite simple.
As far as I know, rational numbers has finite continued fraction representation. And if for any number we have infinite continued fraction, then it necessarily is irrational.
This help us to make distinction between rationals and irrationals.
My question is, are their any criteria (of form of continued fraction) which will give distinction between algebraic irrationals and transcendental irrationals.(Maybe some pattern is prohibited for transcendental numbers, or some pattern is required.)
Sorry if this question sound stupid.
Thank you.
continued-fractions transcendental-numbers
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
add a comment |
My question is quite simple.
As far as I know, rational numbers has finite continued fraction representation. And if for any number we have infinite continued fraction, then it necessarily is irrational.
This help us to make distinction between rationals and irrationals.
My question is, are their any criteria (of form of continued fraction) which will give distinction between algebraic irrationals and transcendental irrationals.(Maybe some pattern is prohibited for transcendental numbers, or some pattern is required.)
Sorry if this question sound stupid.
Thank you.
continued-fractions transcendental-numbers
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
My question is quite simple.
As far as I know, rational numbers has finite continued fraction representation. And if for any number we have infinite continued fraction, then it necessarily is irrational.
This help us to make distinction between rationals and irrationals.
My question is, are their any criteria (of form of continued fraction) which will give distinction between algebraic irrationals and transcendental irrationals.(Maybe some pattern is prohibited for transcendental numbers, or some pattern is required.)
Sorry if this question sound stupid.
Thank you.
continued-fractions transcendental-numbers
continued-fractions transcendental-numbers
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
edited Nov 21 '18 at 15:35
rtybase
10.4k21433
10.4k21433
asked Nov 21 '18 at 9:27
kolobokish
40438
asked Nov 21 '18 at 9:27
kolobokish
40438
asked Nov 21 '18 at 9:27
kolobokish
40438
40438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $frac{a+bsqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
From here
It is conjectured that all infinite continued fractions with bounded
terms that are not eventually periodic are transcendental (eventually
periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $frac{a+bsqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
From here
It is conjectured that all infinite continued fractions with bounded
terms that are not eventually periodic are transcendental (eventually
periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
add a comment |
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $frac{a+bsqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
From here
It is conjectured that all infinite continued fractions with bounded
terms that are not eventually periodic are transcendental (eventually
periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
add a comment |
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $frac{a+bsqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
From here
It is conjectured that all infinite continued fractions with bounded
terms that are not eventually periodic are transcendental (eventually
periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
The question is still being researched, some results are known, some are conjectured.
For examples if continued fraction is periodic then it represents a quadratic irrational number
Theorem 8.11 (Lagrange) Every periodic regular continued fraction is a quadratic irrationality (i.e., $frac{a+bsqrt{c}}{d}$ for some integers $a,b,c,$ and $d$, where $b ne0, c>1, d>0$ anf $c$ is square-free). The converse is also true: every quadratic irrationality has a periodic regular continued fraction.
from this book, page 94. Same result in Khinchin's famous book, page 48. To some extent - spread here.
From here
It is conjectured that all infinite continued fractions with bounded
terms that are not eventually periodic are transcendental (eventually
periodic continued fractions correspond to quadratic irrationals).
and here is a good paper highlighting some of the latest developments.
And of course Liouville's theorem, quoting Khinchin's book, page 46
Liouville's theorem shows that algebraic numbers do not admit rational-fraction approximations of greater than a certain order of accuracy.
this includes the best rational approximations (generated by simple continued fractions) as well.
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
answered Nov 21 '18 at 12:38
rtybase
10.4k21433
10.4k21433
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
add a comment |
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
1
1
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
Thank you very much. Everything except last part, were unknown to me. Thank you.
– kolobokish
Nov 21 '18 at 13:04
add a comment |
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