Mixing
Algorithm for regular continued fraction of a square root
0
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Say I have a number $n$, and want to find the expression of $sqrt{n}$ as a regular continued fraction. How would I do such a thing systematically?
A naive computer algorithm wouldn't work, due to floating point errors.
algorithms roots radicals continued-fractions
asked Jan 24 at 18:21
J. DoeJ. Doe
1
$endgroup$
|
show 4 more comments
0
$begingroup$
Say I have a number $n$, and want to find the expression of $sqrt{n}$ as a regular continued fraction. How would I do such a thing systematically?
A naive computer algorithm wouldn't work, due to floating point errors.
algorithms roots radicals continued-fractions
asked Jan 24 at 18:21
J. DoeJ. Doe
1
$endgroup$
$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
1
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
1
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
2
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10
|
show 4 more comments
0
0
0
$begingroup$
Say I have a number $n$, and want to find the expression of $sqrt{n}$ as a regular continued fraction. How would I do such a thing systematically?
A naive computer algorithm wouldn't work, due to floating point errors.
algorithms roots radicals continued-fractions
asked Jan 24 at 18:21
J. DoeJ. Doe
1
$endgroup$
Say I have a number $n$, and want to find the expression of $sqrt{n}$ as a regular continued fraction. How would I do such a thing systematically?
A naive computer algorithm wouldn't work, due to floating point errors.
algorithms roots radicals continued-fractions
algorithms roots radicals continued-fractions
asked Jan 24 at 18:21
J. DoeJ. Doe
1
asked Jan 24 at 18:21
J. DoeJ. Doe
1
asked Jan 24 at 18:21
J. DoeJ. Doe
1
asked Jan 24 at 18:21
J. DoeJ. Doe
1
asked Jan 24 at 18:21
J. DoeJ. Doe
1
1
$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
1
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
1
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
2
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10
|
show 4 more comments
$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
1
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
1
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
2
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10
$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
1
1
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
1
1
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
2
2
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10
|
show 4 more comments
0
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$begingroup$
The algorithm is described here : en.wikipedia.org/wiki/…
$endgroup$
– Peter
Jan 24 at 18:33
$begingroup$
It doesn't always produce a regular continued fraction. for example when $S=7$
$endgroup$
– J. Doe
Jan 24 at 18:48
1
$begingroup$
it gives $a=sqrt4=2$, $r=3$ (because $S=7=lfloorsqrt7rfloor^2+3=4+3$), and that means $sqrt7=2+frac3{4+frac3{4+frac3{4+frac3{...}}}}$, while the desired result is one in which the first element is $lfloorsqrt7rfloor=2$, and all the numerators are 1. And yes, $n$ is a positive integer.
$endgroup$
– J. Doe
Jan 24 at 19:07
1
$begingroup$
Maybe, you have chosen the wrong algorithm. I have not further checked the site, but I still think that the algorithm for a regular continued fraction should be present.
$endgroup$
– Peter
Jan 24 at 19:09
2
$begingroup$
See the algorithm for which Wikipedia uses $sqrt{114}$ as an example.
$endgroup$
– Misha Lavrov
Jan 24 at 19:10