pell's equation converge
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Explain Why pell's equation $x_n+ny_n=1$, $(x_n/y_n)^2$ is coverge to n as n increase
For example, n=11 the answer $(x_n/y_n)^2$ is very close to 11 when n increase.
algebraic-topology pell-type-equations
asked Dec 9 '18 at 1:54
WendyWendy
1
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add a comment |
$begingroup$
Explain Why pell's equation $x_n+ny_n=1$, $(x_n/y_n)^2$ is coverge to n as n increase
For example, n=11 the answer $(x_n/y_n)^2$ is very close to 11 when n increase.
algebraic-topology pell-type-equations
asked Dec 9 '18 at 1:54
WendyWendy
1
$endgroup$
$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
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– reuns
Dec 9 '18 at 2:03
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if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
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– Will Jagy
Dec 9 '18 at 2:17
add a comment |
$begingroup$
Explain Why pell's equation $x_n+ny_n=1$, $(x_n/y_n)^2$ is coverge to n as n increase
For example, n=11 the answer $(x_n/y_n)^2$ is very close to 11 when n increase.
algebraic-topology pell-type-equations
asked Dec 9 '18 at 1:54
WendyWendy
1
$endgroup$
Explain Why pell's equation $x_n+ny_n=1$, $(x_n/y_n)^2$ is coverge to n as n increase
For example, n=11 the answer $(x_n/y_n)^2$ is very close to 11 when n increase.
algebraic-topology pell-type-equations
algebraic-topology pell-type-equations
asked Dec 9 '18 at 1:54
WendyWendy
1
asked Dec 9 '18 at 1:54
WendyWendy
1
asked Dec 9 '18 at 1:54
WendyWendy
1
asked Dec 9 '18 at 1:54
WendyWendy
1
asked Dec 9 '18 at 1:54
WendyWendy
1
1
$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– reuns
Dec 9 '18 at 2:03
$begingroup$
if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
$endgroup$
– Will Jagy
Dec 9 '18 at 2:17
add a comment |
$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– reuns
Dec 9 '18 at 2:03
$begingroup$
if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
$endgroup$
– Will Jagy
Dec 9 '18 at 2:17
$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– reuns
Dec 9 '18 at 2:03
$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– reuns
Dec 9 '18 at 2:03
$begingroup$
if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
$endgroup$
– Will Jagy
Dec 9 '18 at 2:17
$begingroup$
if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
$endgroup$
– Will Jagy
Dec 9 '18 at 2:17
add a comment |
1 Answer
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Let $mathcal{P}_d(mathbb{Z})^+$ be the set of the positive integer solutions to the Pell equation $x^2-dy^2=1$, we have that, if $(a,b)inmathcal{P}_d(mathbb{Z})^+$, then, $a/b$ is a convergent of $sqrt d$, when you square $a/b$ you are getting ridiculously good rational approximations of $d$, remember convergents give you the "best" rational approximations
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
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1 Answer
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1 Answer
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$begingroup$
Let $mathcal{P}_d(mathbb{Z})^+$ be the set of the positive integer solutions to the Pell equation $x^2-dy^2=1$, we have that, if $(a,b)inmathcal{P}_d(mathbb{Z})^+$, then, $a/b$ is a convergent of $sqrt d$, when you square $a/b$ you are getting ridiculously good rational approximations of $d$, remember convergents give you the "best" rational approximations
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
$endgroup$
add a comment |
$begingroup$
Let $mathcal{P}_d(mathbb{Z})^+$ be the set of the positive integer solutions to the Pell equation $x^2-dy^2=1$, we have that, if $(a,b)inmathcal{P}_d(mathbb{Z})^+$, then, $a/b$ is a convergent of $sqrt d$, when you square $a/b$ you are getting ridiculously good rational approximations of $d$, remember convergents give you the "best" rational approximations
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
$endgroup$
add a comment |
$begingroup$
Let $mathcal{P}_d(mathbb{Z})^+$ be the set of the positive integer solutions to the Pell equation $x^2-dy^2=1$, we have that, if $(a,b)inmathcal{P}_d(mathbb{Z})^+$, then, $a/b$ is a convergent of $sqrt d$, when you square $a/b$ you are getting ridiculously good rational approximations of $d$, remember convergents give you the "best" rational approximations
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
$endgroup$
Let $mathcal{P}_d(mathbb{Z})^+$ be the set of the positive integer solutions to the Pell equation $x^2-dy^2=1$, we have that, if $(a,b)inmathcal{P}_d(mathbb{Z})^+$, then, $a/b$ is a convergent of $sqrt d$, when you square $a/b$ you are getting ridiculously good rational approximations of $d$, remember convergents give you the "best" rational approximations
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
answered Dec 17 '18 at 2:24
Bruno AndradesBruno Andrades
1309
1309
add a comment |
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$begingroup$
take a look at en.wikipedia.org/wiki/Pell%27s_equation
$endgroup$
– reuns
Dec 9 '18 at 2:03
$begingroup$
if you are going to call the fixed coefficient $n,$ you should change the index. Other errors as well. Correct: $$ x_j^2 - n y_j^2 = 1 $$
$endgroup$
– Will Jagy
Dec 9 '18 at 2:17