Deriving $sqrt2 approx 1 + frac13 + frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$
3
$begingroup$
Here is a wierd expansion for $sqrt2$ found in the ancient Indian mathematical literature.
$$1 + frac13 + frac1{3 cdot 4} - frac1{3 cdot 4 cdot34} = frac {577}{408}$$
Today we know that the resulting fraction can be obtained using Pell numbers i.e. the recursion $frac{P_{n-1}}{P_n} - 1$ .
Can someone explain how can we come up with that particular expansion?
sequences-and-series approximation real-numbers
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
$endgroup$
add a comment |
3
$begingroup$
Here is a wierd expansion for $sqrt2$ found in the ancient Indian mathematical literature.
$$1 + frac13 + frac1{3 cdot 4} - frac1{3 cdot 4 cdot34} = frac {577}{408}$$
Today we know that the resulting fraction can be obtained using Pell numbers i.e. the recursion $frac{P_{n-1}}{P_n} - 1$ .
Can someone explain how can we come up with that particular expansion?
sequences-and-series approximation real-numbers
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
$endgroup$
2
$begingroup$
In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
$endgroup$
– Jack D'Aurizio
Oct 21 '18 at 6:18
4
$begingroup$
My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 8:22
$begingroup$
How could you come up with 12c.
$endgroup$
– Awe Kumar Jha
Oct 21 '18 at 8:26
$begingroup$
When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:46
add a comment |
3
3
3
$begingroup$
Here is a wierd expansion for $sqrt2$ found in the ancient Indian mathematical literature.
$$1 + frac13 + frac1{3 cdot 4} - frac1{3 cdot 4 cdot34} = frac {577}{408}$$
Today we know that the resulting fraction can be obtained using Pell numbers i.e. the recursion $frac{P_{n-1}}{P_n} - 1$ .
Can someone explain how can we come up with that particular expansion?
sequences-and-series approximation real-numbers
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
$endgroup$
Here is a wierd expansion for $sqrt2$ found in the ancient Indian mathematical literature.
$$1 + frac13 + frac1{3 cdot 4} - frac1{3 cdot 4 cdot34} = frac {577}{408}$$
Today we know that the resulting fraction can be obtained using Pell numbers i.e. the recursion $frac{P_{n-1}}{P_n} - 1$ .
Can someone explain how can we come up with that particular expansion?
sequences-and-series approximation real-numbers
sequences-and-series approximation real-numbers
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
edited Dec 24 '18 at 9:19
Lord_Farin
15.7k636110
15.7k636110
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
asked Oct 21 '18 at 5:33
Awe Kumar JhaAwe Kumar Jha
599113
599113
2
$begingroup$
In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
$endgroup$
– Jack D'Aurizio
Oct 21 '18 at 6:18
4
$begingroup$
My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 8:22
$begingroup$
How could you come up with 12c.
$endgroup$
– Awe Kumar Jha
Oct 21 '18 at 8:26
$begingroup$
When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:46
add a comment |
2
$begingroup$
In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
$endgroup$
– Jack D'Aurizio
Oct 21 '18 at 6:18
4
$begingroup$
My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 8:22
$begingroup$
How could you come up with 12c.
$endgroup$
– Awe Kumar Jha
Oct 21 '18 at 8:26
$begingroup$
When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:46
2
2
$begingroup$
In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
$endgroup$
– Jack D'Aurizio
Oct 21 '18 at 6:18
$begingroup$
In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
$endgroup$
– Jack D'Aurizio
Oct 21 '18 at 6:18
4
4
$begingroup$
My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 8:22
$begingroup$
My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 8:22
$begingroup$
How could you come up with 12c.
$endgroup$
– Awe Kumar Jha
Oct 21 '18 at 8:26
$begingroup$
How could you come up with 12c.
$endgroup$
– Awe Kumar Jha
Oct 21 '18 at 8:26
$begingroup$
When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:46
$begingroup$
When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:46
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion :
$$sqrt2 approx 1 + frac12 - frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$$
(an answer to the actual question is provided by Gerry Myerson in his first comment)
$$-$$
This (signed) Egyptian fraction may be obtained by starting with the 'exact' $sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder :
begin{array} {c|cc}
x&1/x&[1/x]\
sqrt{2}&0.707106781187&1\
sqrt{2}-1&2.41421356237&2\
sqrt{2}-1-frac 12&-11.6568542495&-12\
sqrt{2}-1-frac 12+frac 1{12}&-407.646752982&-408\
end{array}
This method may be generalized at wish...
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
$endgroup$
– Gerry Myerson
Oct 22 '18 at 21:42
1
$begingroup$
@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:50
1
$begingroup$
Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
$endgroup$
– Raymond Manzoni
Oct 23 '18 at 22:38
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion :
$$sqrt2 approx 1 + frac12 - frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$$
(an answer to the actual question is provided by Gerry Myerson in his first comment)
$$-$$
This (signed) Egyptian fraction may be obtained by starting with the 'exact' $sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder :
begin{array} {c|cc}
x&1/x&[1/x]\
sqrt{2}&0.707106781187&1\
sqrt{2}-1&2.41421356237&2\
sqrt{2}-1-frac 12&-11.6568542495&-12\
sqrt{2}-1-frac 12+frac 1{12}&-407.646752982&-408\
end{array}
This method may be generalized at wish...
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
$endgroup$
– Gerry Myerson
Oct 22 '18 at 21:42
1
$begingroup$
@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:50
1
$begingroup$
Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
$endgroup$
– Raymond Manzoni
Oct 23 '18 at 22:38
|
show 2 more comments
3
$begingroup$
EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion :
$$sqrt2 approx 1 + frac12 - frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$$
(an answer to the actual question is provided by Gerry Myerson in his first comment)
$$-$$
This (signed) Egyptian fraction may be obtained by starting with the 'exact' $sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder :
begin{array} {c|cc}
x&1/x&[1/x]\
sqrt{2}&0.707106781187&1\
sqrt{2}-1&2.41421356237&2\
sqrt{2}-1-frac 12&-11.6568542495&-12\
sqrt{2}-1-frac 12+frac 1{12}&-407.646752982&-408\
end{array}
This method may be generalized at wish...
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
$endgroup$
– Gerry Myerson
Oct 22 '18 at 21:42
1
$begingroup$
@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:50
1
$begingroup$
Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
$endgroup$
– Raymond Manzoni
Oct 23 '18 at 22:38
|
show 2 more comments
3
3
3
$begingroup$
EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion :
$$sqrt2 approx 1 + frac12 - frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$$
(an answer to the actual question is provided by Gerry Myerson in his first comment)
$$-$$
This (signed) Egyptian fraction may be obtained by starting with the 'exact' $sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder :
begin{array} {c|cc}
x&1/x&[1/x]\
sqrt{2}&0.707106781187&1\
sqrt{2}-1&2.41421356237&2\
sqrt{2}-1-frac 12&-11.6568542495&-12\
sqrt{2}-1-frac 12+frac 1{12}&-407.646752982&-408\
end{array}
This method may be generalized at wish...
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
EDIT: This is an answer for an algorithm to generate the unit fractions of the expansion :
$$sqrt2 approx 1 + frac12 - frac1{3 cdot 4} - frac1{3 cdot 4 cdot34}$$
(an answer to the actual question is provided by Gerry Myerson in his first comment)
$$-$$
This (signed) Egyptian fraction may be obtained by starting with the 'exact' $sqrt{2}$ and removing at each iteration the multiplicative inverse of the nearest integer of the remainder :
begin{array} {c|cc}
x&1/x&[1/x]\
sqrt{2}&0.707106781187&1\
sqrt{2}-1&2.41421356237&2\
sqrt{2}-1-frac 12&-11.6568542495&-12\
sqrt{2}-1-frac 12+frac 1{12}&-407.646752982&-408\
end{array}
This method may be generalized at wish...
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
edited Oct 23 '18 at 12:29
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
answered Oct 21 '18 at 9:02
Raymond ManzoniRaymond Manzoni
37.3k563117
37.3k563117
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
$endgroup$
– Gerry Myerson
Oct 22 '18 at 21:42
1
$begingroup$
@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:50
1
$begingroup$
Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
$endgroup$
– Raymond Manzoni
Oct 23 '18 at 22:38
|
show 2 more comments
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
$endgroup$
– Gerry Myerson
Oct 22 '18 at 21:42
1
$begingroup$
@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:50
1
$begingroup$
Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
$endgroup$
– Raymond Manzoni
Oct 23 '18 at 22:38
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
$begingroup$
Does this amount to the same thing as in my "greedy algorithm" comment?
$endgroup$
– Gerry Myerson
Oct 21 '18 at 23:48
1
1
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
$begingroup$
@Gerry Myerson: Well it appears to be a (signed) egyptian fraction algorithm and thus a greedy algorithm even if wikipedia sources Lambert $(1770)$ for the signed version instead of an ancient Indian. I had noticed your comment but also your $+3^{-1}$ second term : the best unit fraction to get the nearest to $sqrt{2}$ while I searched the nearest integer to the reciprocal of the remainder and got $2$ for $+2^{-1}$. Near but somewhat different... Cheers,
$endgroup$
– Raymond Manzoni
Oct 22 '18 at 21:04
1
1
$begingroup$
In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
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– Gerry Myerson
Oct 22 '18 at 21:42
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In the original post, the second term was $3^{-1}$. An editor changed it to $2^{-1}$ for reasons unclear to me. I believe that $3^{-1}$ is correct, if you want the sum to be $577/408$.
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– Gerry Myerson
Oct 22 '18 at 21:42
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1
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@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
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– Raymond Manzoni
Oct 22 '18 at 21:50
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@GerryMyerson: I saw only the $2^{-1}$ version and as I indicated this requires the sign to be changed in front of $dfrac 1{3cdot 4}$ : we have indeed $;dfrac 12-dfrac 1{3cdot 4}=dfrac 13+dfrac 1{3cdot 4}$. Your answer seems to be the correct one to the initial question (and explains better the factorization of the denominators) : you should have answered! :-)
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– Raymond Manzoni
Oct 22 '18 at 21:50
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1
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Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
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– Raymond Manzoni
Oct 23 '18 at 22:38
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Glad you liked it @Awe Kumar Jha : the number of significant figures is doubled at each step following the number of digits at the denominators (this applies too to the usual Egyptian fractions). The drawback is that you have to know the digits of $pi$ at the start! Ramanujan's formula is still faster for $,pi,$ and $sqrt{S}$ may be computed as quickly using Newton's method. Excellent continuation anyway,
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– Raymond Manzoni
Oct 23 '18 at 22:38
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In terms a continued fractions, $sqrt{2}=[1;2,2,2,2,ldots]$. $frac{577}{408}$ is one of the convergents. For the continued fraction of quadratic surds, see Lagrange's theorem.
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– Jack D'Aurizio
Oct 21 '18 at 6:18
4
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My guess is a greedly algorithm. Choose $a=3$ to make $1+a^{-1}$ as close as possible to $sqrt2$; then choose $b=4$ to make $1+3^{-1}+(3b)^{-1}$ as close as possible to $sqrt2$; then choose $c=-34$ to make $1+3^{-1}+(3b)^{-1}+(12c)^{-1}$ as close as possible to $sqrt2$.
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– Gerry Myerson
Oct 21 '18 at 8:22
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How could you come up with 12c.
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– Awe Kumar Jha
Oct 21 '18 at 8:26
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When I chose $b=4$, that made the third term have denominator 12, so I wanted the fourth denominator to be a multiple of that 12.
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– Gerry Myerson
Oct 21 '18 at 23:46