Mixing
Continued fraction of a square root
$begingroup$
If I want to find the continued fraction of $sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.
If anyone has a good site that answers these questions either, please let me know. Thanks!
number-theory continued-fractions
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
$endgroup$
add a comment |
$begingroup$
If I want to find the continued fraction of $sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.
If anyone has a good site that answers these questions either, please let me know. Thanks!
number-theory continued-fractions
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
$endgroup$
add a comment |
$begingroup$
If I want to find the continued fraction of $sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.
If anyone has a good site that answers these questions either, please let me know. Thanks!
number-theory continued-fractions
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
$endgroup$
If I want to find the continued fraction of $sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.
If anyone has a good site that answers these questions either, please let me know. Thanks!
number-theory continued-fractions
number-theory continued-fractions
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
edited Dec 27 '12 at 4:39
MJD
47.3k29213396
47.3k29213396
asked Dec 27 '12 at 1:37
user51819user51819
5561314
asked Dec 27 '12 at 1:37
user51819user51819
5561314
asked Dec 27 '12 at 1:37
user51819user51819
5561314
5561314
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Let's just do an example. Let's find the continued fraction for $defsf{sqrt 5}sf$. $sfapprox 2.23$ or something, and $a_0$ is the integer part of this, which is 2.
Then we subtract $a_0$ from $sf$ and take the reciprocal. That is, we calculate ${1over sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So:
$$sf=2+cfrac{1}{4+cfrac1{vdots}}$$
Where we haven't figured out the $vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1over 4.23 - 4} approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = ldots = 4$:
$$sf=2+cfrac{1}{4+cfrac1{4+cfrac1{4+cfrac1vdots}}}$$
This procedure will work for any number whatever, but for $sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1over sf-2}$ stage, we apply algebra to convert this to ${1over sf-2}cdot{sf+2oversf+2} = sf+2$. So we could say that: $$begin{align}
sf & = 2 + cfrac 1{2+sf}\
2 + sf & = 4 + cfrac 1{2+sf}.
end{align}$$
If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ sf$, we get:
$$ begin{align}
2+ sf & = 4 + cfrac 1{4 + cfrac 1{2+sf}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}}} \
& = cdots
end{align}
$$
and it's evident that the fours will repeat forever.
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
$endgroup$
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
|
show 3 more comments
$begingroup$
Confirm the algebraic identity:
$$sqrt n=a+frac{n-a^2}{a+sqrt n}$$
Then chose whatever value of 'a' you want, and just keep on pluging in $sqrt n$
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
$endgroup$
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
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– Marc van Leeuwen
May 29 '18 at 10:50
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-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
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@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
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Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
add a comment |
$begingroup$
$a_0$ is simply the largest integer such that $a^2 le n$ . You can determine the continued fraction for a square root by performing the $frac1{sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.
I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
$endgroup$
add a comment |
$begingroup$
$a_0$ is the largest integer that is smaller than or equal to $sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.
If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 le n$ and $(g+1)^2 > n$, and then $a_0 = g$.
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
$endgroup$
add a comment |
$begingroup$
Here the easiest method to generate continued fraction for any square (or more) root. Lets take $sqrt{5}$:
$$sqrt{5} approx 2,2360679775...$$
$$sqrt{5} = 2 + 0,2360679775$$
So $2=left lfloor sqrt(5) right rfloor$
$$sqrt{5} =2 + x$$ with $x=0,2360679775$
$$sqrt{5}^{2} = (x + 2)^{2} Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$frac{1}{x}= x + 4 Rightarrow x = frac{1}{4+x}$$
So $x = frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$
Finally, we got: $sqrt{5} = 2 + frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$.
So for any x, $sqrt(x) = left lfloor sqrt(x) right rfloor + frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor + ddots}}}$
So to answer, $a_0 =left lfloor sqrt(x) right rfloor$ (because any periodic continued fraction is smaller than 1).
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
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I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
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– Marc van Leeuwen
May 29 '18 at 10:48
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@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's just do an example. Let's find the continued fraction for $defsf{sqrt 5}sf$. $sfapprox 2.23$ or something, and $a_0$ is the integer part of this, which is 2.
Then we subtract $a_0$ from $sf$ and take the reciprocal. That is, we calculate ${1over sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So:
$$sf=2+cfrac{1}{4+cfrac1{vdots}}$$
Where we haven't figured out the $vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1over 4.23 - 4} approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = ldots = 4$:
$$sf=2+cfrac{1}{4+cfrac1{4+cfrac1{4+cfrac1vdots}}}$$
This procedure will work for any number whatever, but for $sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1over sf-2}$ stage, we apply algebra to convert this to ${1over sf-2}cdot{sf+2oversf+2} = sf+2$. So we could say that: $$begin{align}
sf & = 2 + cfrac 1{2+sf}\
2 + sf & = 4 + cfrac 1{2+sf}.
end{align}$$
If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ sf$, we get:
$$ begin{align}
2+ sf & = 4 + cfrac 1{4 + cfrac 1{2+sf}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}}} \
& = cdots
end{align}
$$
and it's evident that the fours will repeat forever.
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
$endgroup$
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
|
show 3 more comments
$begingroup$
Let's just do an example. Let's find the continued fraction for $defsf{sqrt 5}sf$. $sfapprox 2.23$ or something, and $a_0$ is the integer part of this, which is 2.
Then we subtract $a_0$ from $sf$ and take the reciprocal. That is, we calculate ${1over sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So:
$$sf=2+cfrac{1}{4+cfrac1{vdots}}$$
Where we haven't figured out the $vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1over 4.23 - 4} approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = ldots = 4$:
$$sf=2+cfrac{1}{4+cfrac1{4+cfrac1{4+cfrac1vdots}}}$$
This procedure will work for any number whatever, but for $sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1over sf-2}$ stage, we apply algebra to convert this to ${1over sf-2}cdot{sf+2oversf+2} = sf+2$. So we could say that: $$begin{align}
sf & = 2 + cfrac 1{2+sf}\
2 + sf & = 4 + cfrac 1{2+sf}.
end{align}$$
If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ sf$, we get:
$$ begin{align}
2+ sf & = 4 + cfrac 1{4 + cfrac 1{2+sf}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}}} \
& = cdots
end{align}
$$
and it's evident that the fours will repeat forever.
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
$endgroup$
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
|
show 3 more comments
$begingroup$
Let's just do an example. Let's find the continued fraction for $defsf{sqrt 5}sf$. $sfapprox 2.23$ or something, and $a_0$ is the integer part of this, which is 2.
Then we subtract $a_0$ from $sf$ and take the reciprocal. That is, we calculate ${1over sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So:
$$sf=2+cfrac{1}{4+cfrac1{vdots}}$$
Where we haven't figured out the $vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1over 4.23 - 4} approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = ldots = 4$:
$$sf=2+cfrac{1}{4+cfrac1{4+cfrac1{4+cfrac1vdots}}}$$
This procedure will work for any number whatever, but for $sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1over sf-2}$ stage, we apply algebra to convert this to ${1over sf-2}cdot{sf+2oversf+2} = sf+2$. So we could say that: $$begin{align}
sf & = 2 + cfrac 1{2+sf}\
2 + sf & = 4 + cfrac 1{2+sf}.
end{align}$$
If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ sf$, we get:
$$ begin{align}
2+ sf & = 4 + cfrac 1{4 + cfrac 1{2+sf}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}}} \
& = cdots
end{align}
$$
and it's evident that the fours will repeat forever.
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
$endgroup$
Let's just do an example. Let's find the continued fraction for $defsf{sqrt 5}sf$. $sfapprox 2.23$ or something, and $a_0$ is the integer part of this, which is 2.
Then we subtract $a_0$ from $sf$ and take the reciprocal. That is, we calculate ${1over sf-2}$. If you're using a calculator, this comes out to 4.23 or so. Then $a_1$ is the integer part of this, which is 4. So:
$$sf=2+cfrac{1}{4+cfrac1{vdots}}$$
Where we haven't figured out the $vdots$ part yet. To get that, we take our $4.23$, subtract $a_1$, and take the reciprocal; that is, we calculate ${1over 4.23 - 4} approx 4.23$. This is just the same as we had before, so $a_2$ is 4 again, and continuing in the same way, $a_3 = a_4 = ldots = 4$:
$$sf=2+cfrac{1}{4+cfrac1{4+cfrac1{4+cfrac1vdots}}}$$
This procedure will work for any number whatever, but for $sf$ we can use a little algebraic cleverness to see that the fours really do repeat. When we get to the ${1over sf-2}$ stage, we apply algebra to convert this to ${1over sf-2}cdot{sf+2oversf+2} = sf+2$. So we could say that: $$begin{align}
sf & = 2 + cfrac 1{2+sf}\
2 + sf & = 4 + cfrac 1{2+sf}.
end{align}$$
If we substitute the right-hand side of the last equation expression into itself in place of $ 2+ sf$, we get:
$$ begin{align}
2+ sf & = 4 + cfrac 1{4 + cfrac 1{2+sf}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}} \
& = 4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{4 + cfrac 1{2+sf}}}} \
& = cdots
end{align}
$$
and it's evident that the fours will repeat forever.
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
edited Dec 27 '12 at 4:39
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
answered Dec 27 '12 at 1:59
MJDMJD
47.3k29213396
47.3k29213396
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
|
show 3 more comments
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
I always thought they were in the form a0 + 1/(a1 + 1/ (a2 + 1/( ...
$endgroup$
– user51819
Dec 27 '12 at 2:01
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
@user51819 Quite so (although $a_0$ could be 0.) I had the typesetting wrong, and have corrected it.
$endgroup$
– MJD
Dec 27 '12 at 2:03
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
$begingroup$
Sorry, one last question; how do we know, in the general sense, when we've hit the point where it's periodic? When you encounter an $a_k$ you've seen before? Or when a reciprocal equals a reciprocal you've seen before? I am guessing the latter?
$endgroup$
– user51819
Dec 27 '12 at 2:09
1
1
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
When the reciprocal is one you have seen before. It couldn't be when you reach an $a_k$ you've seen before or else you couldn't have a repeating sequence like $1,1,2,1,1,2,1,1,2,ldots$.
$endgroup$
– MJD
Dec 27 '12 at 2:15
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
$begingroup$
Incredible, really.
$endgroup$
– Meitar
Jun 13 '15 at 12:19
|
show 3 more comments
$begingroup$
Confirm the algebraic identity:
$$sqrt n=a+frac{n-a^2}{a+sqrt n}$$
Then chose whatever value of 'a' you want, and just keep on pluging in $sqrt n$
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
$endgroup$
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
add a comment |
$begingroup$
Confirm the algebraic identity:
$$sqrt n=a+frac{n-a^2}{a+sqrt n}$$
Then chose whatever value of 'a' you want, and just keep on pluging in $sqrt n$
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
$endgroup$
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
add a comment |
$begingroup$
Confirm the algebraic identity:
$$sqrt n=a+frac{n-a^2}{a+sqrt n}$$
Then chose whatever value of 'a' you want, and just keep on pluging in $sqrt n$
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
$endgroup$
Confirm the algebraic identity:
$$sqrt n=a+frac{n-a^2}{a+sqrt n}$$
Then chose whatever value of 'a' you want, and just keep on pluging in $sqrt n$
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
edited Dec 27 '12 at 2:06
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
answered Dec 27 '12 at 1:51
EthanEthan
6,87512164
6,87512164
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
add a comment |
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
2
2
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
$begingroup$
With this you end up with generalized continuous fraction. I mean $n-a^2$ not necessarily $1$.
$endgroup$
– Yola
Oct 24 '16 at 17:59
1
1
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
This is a sad case of changing the question to get a simpler answer (and that is assuming the question was how to find the continued fraction of $sqrt n$; in fact it was rather uninterestingly just about its first term, the integer part of $sqrt n$).
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:50
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
-1 As @Yola points out, this doesn’t (necessarily) give you a continued fraction
$endgroup$
– HelloGoodbye
Aug 30 '18 at 22:55
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
@HelloGoodbye OP changed his question, I answered what he originally asked.
$endgroup$
– Ethan
Sep 1 '18 at 1:06
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
$begingroup$
Okay. If so, shouldn't it be possible to see that here?
$endgroup$
– HelloGoodbye
Sep 1 '18 at 23:39
add a comment |
$begingroup$
$a_0$ is simply the largest integer such that $a^2 le n$ . You can determine the continued fraction for a square root by performing the $frac1{sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.
I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
$endgroup$
add a comment |
$begingroup$
$a_0$ is simply the largest integer such that $a^2 le n$ . You can determine the continued fraction for a square root by performing the $frac1{sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.
I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
$endgroup$
add a comment |
$begingroup$
$a_0$ is simply the largest integer such that $a^2 le n$ . You can determine the continued fraction for a square root by performing the $frac1{sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.
I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
$endgroup$
$a_0$ is simply the largest integer such that $a^2 le n$ . You can determine the continued fraction for a square root by performing the $frac1{sqrt n - a_0}$ step and then using the conjugate to remove the square root from the denominator, and repeating.
I recommend Ron Knott's site: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html . Good Luck.
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
answered Dec 27 '12 at 1:49
half-integer fanhalf-integer fan
840413
840413
add a comment |
add a comment |
$begingroup$
$a_0$ is the largest integer that is smaller than or equal to $sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.
If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 le n$ and $(g+1)^2 > n$, and then $a_0 = g$.
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
$endgroup$
add a comment |
$begingroup$
$a_0$ is the largest integer that is smaller than or equal to $sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.
If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 le n$ and $(g+1)^2 > n$, and then $a_0 = g$.
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
$endgroup$
add a comment |
$begingroup$
$a_0$ is the largest integer that is smaller than or equal to $sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.
If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 le n$ and $(g+1)^2 > n$, and then $a_0 = g$.
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
$endgroup$
$a_0$ is the largest integer that is smaller than or equal to $sqrt n$. Or put another way, you want $a_0^2$ to be smaller than or equal to $n$, and $(a_0+1)^2$ to be bigger than $n$.
If you really have no idea what integer to use, then you find it by guessing an integer $g$. Then you calculate $g^2$. If $g^2$ is bigger than $n$, your guess $g$ was too big, and you try a smaller guess. If $g^2$ is much smaller than $n$, your guess $g$ was too small, and you try a bigger guess. You keep doing this until you find a guess $g$ where $g^2 le n$ and $(g+1)^2 > n$, and then $a_0 = g$.
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
answered Dec 27 '12 at 1:51
MJDMJD
47.3k29213396
47.3k29213396
add a comment |
add a comment |
$begingroup$
Here the easiest method to generate continued fraction for any square (or more) root. Lets take $sqrt{5}$:
$$sqrt{5} approx 2,2360679775...$$
$$sqrt{5} = 2 + 0,2360679775$$
So $2=left lfloor sqrt(5) right rfloor$
$$sqrt{5} =2 + x$$ with $x=0,2360679775$
$$sqrt{5}^{2} = (x + 2)^{2} Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$frac{1}{x}= x + 4 Rightarrow x = frac{1}{4+x}$$
So $x = frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$
Finally, we got: $sqrt{5} = 2 + frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$.
So for any x, $sqrt(x) = left lfloor sqrt(x) right rfloor + frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor + ddots}}}$
So to answer, $a_0 =left lfloor sqrt(x) right rfloor$ (because any periodic continued fraction is smaller than 1).
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
$endgroup$
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
add a comment |
$begingroup$
Here the easiest method to generate continued fraction for any square (or more) root. Lets take $sqrt{5}$:
$$sqrt{5} approx 2,2360679775...$$
$$sqrt{5} = 2 + 0,2360679775$$
So $2=left lfloor sqrt(5) right rfloor$
$$sqrt{5} =2 + x$$ with $x=0,2360679775$
$$sqrt{5}^{2} = (x + 2)^{2} Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$frac{1}{x}= x + 4 Rightarrow x = frac{1}{4+x}$$
So $x = frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$
Finally, we got: $sqrt{5} = 2 + frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$.
So for any x, $sqrt(x) = left lfloor sqrt(x) right rfloor + frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor + ddots}}}$
So to answer, $a_0 =left lfloor sqrt(x) right rfloor$ (because any periodic continued fraction is smaller than 1).
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
$endgroup$
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
add a comment |
$begingroup$
Here the easiest method to generate continued fraction for any square (or more) root. Lets take $sqrt{5}$:
$$sqrt{5} approx 2,2360679775...$$
$$sqrt{5} = 2 + 0,2360679775$$
So $2=left lfloor sqrt(5) right rfloor$
$$sqrt{5} =2 + x$$ with $x=0,2360679775$
$$sqrt{5}^{2} = (x + 2)^{2} Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$frac{1}{x}= x + 4 Rightarrow x = frac{1}{4+x}$$
So $x = frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$
Finally, we got: $sqrt{5} = 2 + frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$.
So for any x, $sqrt(x) = left lfloor sqrt(x) right rfloor + frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor + ddots}}}$
So to answer, $a_0 =left lfloor sqrt(x) right rfloor$ (because any periodic continued fraction is smaller than 1).
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
$endgroup$
Here the easiest method to generate continued fraction for any square (or more) root. Lets take $sqrt{5}$:
$$sqrt{5} approx 2,2360679775...$$
$$sqrt{5} = 2 + 0,2360679775$$
So $2=left lfloor sqrt(5) right rfloor$
$$sqrt{5} =2 + x$$ with $x=0,2360679775$
$$sqrt{5}^{2} = (x + 2)^{2} Rightarrow 5= x^2 + 4x + 4$$
$$1= x^2 + 4x$$
$$frac{1}{x}= x + 4 Rightarrow x = frac{1}{4+x}$$
So $x = frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$
Finally, we got: $sqrt{5} = 2 + frac{1}{4+frac{1}{4+frac{1}{4+frac{1}{4+ddots}}}}$.
So for any x, $sqrt(x) = left lfloor sqrt(x) right rfloor + frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor+frac{x-left lfloor sqrt(x) right rfloor^{2}}{2left lfloor sqrt(x) right rfloor + ddots}}}$
So to answer, $a_0 =left lfloor sqrt(x) right rfloor$ (because any periodic continued fraction is smaller than 1).
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
answered Nov 21 '17 at 17:48
Julien BlanchonJulien Blanchon
412
412
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
add a comment |
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
$begingroup$
I am getting better at mentally doing square roots... generally to slide rule accuracy. But your answer mentions other roots, Could you amplify this a bit for cube roots?
$endgroup$
– richard1941
Apr 29 '18 at 0:57
2
2
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
How does this even work for say $sqrt 7$? In a continued fraction the numerators have to be $1$, which you don't get by this method.
$endgroup$
– Marc van Leeuwen
May 29 '18 at 10:48
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
$begingroup$
@MarcvanLeeuwen see math.stackexchange.com/q/3075250/432081
$endgroup$
– CopyPasteIt
Jan 16 at 2:49
add a comment |
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