Motivation for Ramanujan's mysterious $pi$ formula
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The following formula for $pi$ was discovered by Ramanujan:
$$frac1{pi} = frac{2sqrt{2}}{9801} sum_{k=0}^infty frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}!$$
Does anyone know how it works, or what the motivation for it is?
calculus sequences-and-series approximation pi
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show 1 more comment
$begingroup$
The following formula for $pi$ was discovered by Ramanujan:
$$frac1{pi} = frac{2sqrt{2}}{9801} sum_{k=0}^infty frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}!$$
Does anyone know how it works, or what the motivation for it is?
calculus sequences-and-series approximation pi
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1
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This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
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– Adrián Barquero
Dec 13 '10 at 1:46
17
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That crazy Ramanujan guy never can tell us anything useful... :D
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– J. M. is not a mathematician
Dec 13 '10 at 3:08
33
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One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
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– Mariano Suárez-Álvarez
May 22 '11 at 15:15
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I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
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– Masacroso
Sep 17 '14 at 7:24
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I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
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– timtfj
Dec 24 '18 at 0:20
|
show 1 more comment
$begingroup$
The following formula for $pi$ was discovered by Ramanujan:
$$frac1{pi} = frac{2sqrt{2}}{9801} sum_{k=0}^infty frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}!$$
Does anyone know how it works, or what the motivation for it is?
calculus sequences-and-series approximation pi
$endgroup$
The following formula for $pi$ was discovered by Ramanujan:
$$frac1{pi} = frac{2sqrt{2}}{9801} sum_{k=0}^infty frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}!$$
Does anyone know how it works, or what the motivation for it is?
calculus sequences-and-series approximation pi
calculus sequences-and-series approximation pi
edited Nov 25 '11 at 14:15
Srivatsan
21k371126
21k371126
asked Dec 13 '10 at 1:28
Nick AlgerNick Alger
9,91363267
9,91363267
1
$begingroup$
This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
$endgroup$
– Adrián Barquero
Dec 13 '10 at 1:46
17
$begingroup$
That crazy Ramanujan guy never can tell us anything useful... :D
$endgroup$
– J. M. is not a mathematician
Dec 13 '10 at 3:08
33
$begingroup$
One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:15
$begingroup$
I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
$endgroup$
– Masacroso
Sep 17 '14 at 7:24
$begingroup$
I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
$endgroup$
– timtfj
Dec 24 '18 at 0:20
|
show 1 more comment
1
$begingroup$
This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
$endgroup$
– Adrián Barquero
Dec 13 '10 at 1:46
17
$begingroup$
That crazy Ramanujan guy never can tell us anything useful... :D
$endgroup$
– J. M. is not a mathematician
Dec 13 '10 at 3:08
33
$begingroup$
One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:15
$begingroup$
I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
$endgroup$
– Masacroso
Sep 17 '14 at 7:24
$begingroup$
I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
$endgroup$
– timtfj
Dec 24 '18 at 0:20
1
1
$begingroup$
This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
$endgroup$
– Adrián Barquero
Dec 13 '10 at 1:46
$begingroup$
This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
$endgroup$
– Adrián Barquero
Dec 13 '10 at 1:46
17
17
$begingroup$
That crazy Ramanujan guy never can tell us anything useful... :D
$endgroup$
– J. M. is not a mathematician
Dec 13 '10 at 3:08
$begingroup$
That crazy Ramanujan guy never can tell us anything useful... :D
$endgroup$
– J. M. is not a mathematician
Dec 13 '10 at 3:08
33
33
$begingroup$
One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:15
$begingroup$
One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:15
$begingroup$
I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
$endgroup$
– Masacroso
Sep 17 '14 at 7:24
$begingroup$
I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
$endgroup$
– Masacroso
Sep 17 '14 at 7:24
$begingroup$
I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
$endgroup$
– timtfj
Dec 24 '18 at 0:20
$begingroup$
I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
$endgroup$
– timtfj
Dec 24 '18 at 0:20
|
show 1 more comment
4 Answers
4
active
oldest
votes
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Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = frac{5+sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$big(U_{29}big)^3=70+13sqrt{29},quad text{thus};;color{blue}{70}^2-29cdotcolor{blue}{13}^2=-1$$
$$big(U_{29}big)^6=9801+1820sqrt{29},quad text{thus};;color{blue}{9801}^2-29cdot1820^2=1$$
$$2^6left(big(U_{29}big)^6+big(U_{29}big)^{-6}right)^2 =color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$frac{1}{pi} =frac{2 sqrt 2}{color{blue}{9801}} sum_{k=0}^infty frac{(4k)!}{k!^4} frac{29cdotcolor{blue}{70cdot13},k+1103}{color{blue}{(396^4)}^k}$$
See also this MO post.
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add a comment |
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Here is a nice article Entitled: "Ramanujan's Series for $displaystylefrac{1}{pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf
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@Nick:Hopefully this will help.
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– user9413
May 22 '11 at 11:57
2
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You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
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– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
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@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
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– user9413
May 22 '11 at 15:24
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
$endgroup$
– Mariano Suárez-Álvarez
May 23 '11 at 1:51
$begingroup$
@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
|
show 2 more comments
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This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.
Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.
By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$frac{1}{pi} = sum_{n = 0}^{infty}(a + bn)d_{n}c^{n}tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.
There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.
In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.
@Derek Jennings
The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.
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Would the downvoter care to comment?
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– Paramanand Singh
Aug 2 '16 at 20:20
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A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
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– Mark Viola
Feb 25 '17 at 18:52
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Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
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– Paramanand Singh
Feb 25 '17 at 20:47
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Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
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– Mark Viola
Feb 25 '17 at 21:37
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@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
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– Paramanand Singh
May 10 '17 at 10:29
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show 3 more comments
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The explanation for the existence of this series is given here. Search for the phrase
"The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.
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Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
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– Nick Alger
Dec 14 '10 at 2:03
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = frac{5+sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$big(U_{29}big)^3=70+13sqrt{29},quad text{thus};;color{blue}{70}^2-29cdotcolor{blue}{13}^2=-1$$
$$big(U_{29}big)^6=9801+1820sqrt{29},quad text{thus};;color{blue}{9801}^2-29cdot1820^2=1$$
$$2^6left(big(U_{29}big)^6+big(U_{29}big)^{-6}right)^2 =color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$frac{1}{pi} =frac{2 sqrt 2}{color{blue}{9801}} sum_{k=0}^infty frac{(4k)!}{k!^4} frac{29cdotcolor{blue}{70cdot13},k+1103}{color{blue}{(396^4)}^k}$$
See also this MO post.
$endgroup$
add a comment |
$begingroup$
Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = frac{5+sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$big(U_{29}big)^3=70+13sqrt{29},quad text{thus};;color{blue}{70}^2-29cdotcolor{blue}{13}^2=-1$$
$$big(U_{29}big)^6=9801+1820sqrt{29},quad text{thus};;color{blue}{9801}^2-29cdot1820^2=1$$
$$2^6left(big(U_{29}big)^6+big(U_{29}big)^{-6}right)^2 =color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$frac{1}{pi} =frac{2 sqrt 2}{color{blue}{9801}} sum_{k=0}^infty frac{(4k)!}{k!^4} frac{29cdotcolor{blue}{70cdot13},k+1103}{color{blue}{(396^4)}^k}$$
See also this MO post.
$endgroup$
add a comment |
$begingroup$
Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = frac{5+sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$big(U_{29}big)^3=70+13sqrt{29},quad text{thus};;color{blue}{70}^2-29cdotcolor{blue}{13}^2=-1$$
$$big(U_{29}big)^6=9801+1820sqrt{29},quad text{thus};;color{blue}{9801}^2-29cdot1820^2=1$$
$$2^6left(big(U_{29}big)^6+big(U_{29}big)^{-6}right)^2 =color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$frac{1}{pi} =frac{2 sqrt 2}{color{blue}{9801}} sum_{k=0}^infty frac{(4k)!}{k!^4} frac{29cdotcolor{blue}{70cdot13},k+1103}{color{blue}{(396^4)}^k}$$
See also this MO post.
$endgroup$
Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".
http://sites.google.com/site/tpiezas/0013
Update: Just to make this more intriguing, define the fundamental unit $U_{29} = frac{5+sqrt{29}}{2}$ and fundamental solutions to Pell equations,
$$big(U_{29}big)^3=70+13sqrt{29},quad text{thus};;color{blue}{70}^2-29cdotcolor{blue}{13}^2=-1$$
$$big(U_{29}big)^6=9801+1820sqrt{29},quad text{thus};;color{blue}{9801}^2-29cdot1820^2=1$$
$$2^6left(big(U_{29}big)^6+big(U_{29}big)^{-6}right)^2 =color{blue}{396^4}$$
then we can see those integers all over the formula as,
$$frac{1}{pi} =frac{2 sqrt 2}{color{blue}{9801}} sum_{k=0}^infty frac{(4k)!}{k!^4} frac{29cdotcolor{blue}{70cdot13},k+1103}{color{blue}{(396^4)}^k}$$
See also this MO post.
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Dec 15 '10 at 16:22
Tito Piezas IIITito Piezas III
27.1k365170
27.1k365170
add a comment |
add a comment |
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Here is a nice article Entitled: "Ramanujan's Series for $displaystylefrac{1}{pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf
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@Nick:Hopefully this will help.
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– user9413
May 22 '11 at 11:57
2
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You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
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– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
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@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
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– user9413
May 22 '11 at 15:24
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
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– Mariano Suárez-Álvarez
May 23 '11 at 1:51
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@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
|
show 2 more comments
$begingroup$
Here is a nice article Entitled: "Ramanujan's Series for $displaystylefrac{1}{pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf
$endgroup$
$begingroup$
@Nick:Hopefully this will help.
$endgroup$
– user9413
May 22 '11 at 11:57
2
$begingroup$
You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
$begingroup$
@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
$endgroup$
– user9413
May 22 '11 at 15:24
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
$endgroup$
– Mariano Suárez-Álvarez
May 23 '11 at 1:51
$begingroup$
@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
|
show 2 more comments
$begingroup$
Here is a nice article Entitled: "Ramanujan's Series for $displaystylefrac{1}{pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf
$endgroup$
Here is a nice article Entitled: "Ramanujan's Series for $displaystylefrac{1}{pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.
- http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.158.2533&rep=rep1&type=pdf
edited May 26 '11 at 14:10
answered May 22 '11 at 10:46
user9413
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@Nick:Hopefully this will help.
$endgroup$
– user9413
May 22 '11 at 11:57
2
$begingroup$
You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
$begingroup$
@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
$endgroup$
– user9413
May 22 '11 at 15:24
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
$endgroup$
– Mariano Suárez-Álvarez
May 23 '11 at 1:51
$begingroup$
@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
|
show 2 more comments
$begingroup$
@Nick:Hopefully this will help.
$endgroup$
– user9413
May 22 '11 at 11:57
2
$begingroup$
You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
$begingroup$
@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
$endgroup$
– user9413
May 22 '11 at 15:24
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
$endgroup$
– Mariano Suárez-Álvarez
May 23 '11 at 1:51
$begingroup$
@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
$begingroup$
@Nick:Hopefully this will help.
$endgroup$
– user9413
May 22 '11 at 11:57
$begingroup$
@Nick:Hopefully this will help.
$endgroup$
– user9413
May 22 '11 at 11:57
2
2
$begingroup$
You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:18
$begingroup$
You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :)
$endgroup$
– Mariano Suárez-Álvarez
May 22 '11 at 15:18
1
1
$begingroup$
@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
$endgroup$
– user9413
May 22 '11 at 15:24
$begingroup$
@Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him.
$endgroup$
– user9413
May 22 '11 at 15:24
4
4
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
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– Mariano Suárez-Álvarez
May 23 '11 at 1:51
$begingroup$
Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though!
$endgroup$
– Mariano Suárez-Álvarez
May 23 '11 at 1:51
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@Mariano: Sorry Mariano. If I managed to do like that.
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– user9413
May 23 '11 at 4:51
$begingroup$
@Mariano: Sorry Mariano. If I managed to do like that.
$endgroup$
– user9413
May 23 '11 at 4:51
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show 2 more comments
$begingroup$
This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.
Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.
By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$frac{1}{pi} = sum_{n = 0}^{infty}(a + bn)d_{n}c^{n}tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.
There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.
In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.
@Derek Jennings
The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.
$endgroup$
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
1
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
|
show 3 more comments
$begingroup$
This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.
Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.
By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$frac{1}{pi} = sum_{n = 0}^{infty}(a + bn)d_{n}c^{n}tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.
There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.
In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.
@Derek Jennings
The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.
$endgroup$
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
1
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
|
show 3 more comments
$begingroup$
This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.
Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.
By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$frac{1}{pi} = sum_{n = 0}^{infty}(a + bn)d_{n}c^{n}tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.
There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.
In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.
@Derek Jennings
The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.
$endgroup$
This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.
Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.
By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$frac{1}{pi} = sum_{n = 0}^{infty}(a + bn)d_{n}c^{n}tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.
There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.
In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.
@Derek Jennings
The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.
edited Oct 24 '17 at 20:16
answered Jul 2 '13 at 5:16
Paramanand SinghParamanand Singh
50k556163
50k556163
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
1
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
|
show 3 more comments
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
1
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
Would the downvoter care to comment?
$endgroup$
– Paramanand Singh
Aug 2 '16 at 20:20
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
A big (+1)! A very thoughtful synopsis on a very interesting and robust topic.
$endgroup$
– Mark Viola
Feb 25 '17 at 18:52
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics.
$endgroup$
– Paramanand Singh
Feb 25 '17 at 20:47
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
$begingroup$
Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) ....
$endgroup$
– Mark Viola
Feb 25 '17 at 21:37
1
1
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
$begingroup$
@JaumeOliverLafont: the approximation for $pi^{4}$ is indeed based on numerical values. Borwein brothers in their book Pi and the AGM hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd...
$endgroup$
– Paramanand Singh
May 10 '17 at 10:29
|
show 3 more comments
$begingroup$
The explanation for the existence of this series is given here. Search for the phrase
"The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.
$endgroup$
1
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
add a comment |
$begingroup$
The explanation for the existence of this series is given here. Search for the phrase
"The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.
$endgroup$
1
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
add a comment |
$begingroup$
The explanation for the existence of this series is given here. Search for the phrase
"The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.
$endgroup$
The explanation for the existence of this series is given here. Search for the phrase
"The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.
answered Dec 13 '10 at 10:30
Derek JenningsDerek Jennings
12k3054
12k3054
1
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
add a comment |
1
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
1
1
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
$begingroup$
Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58").
$endgroup$
– Nick Alger
Dec 14 '10 at 2:03
add a comment |
protected by t.b. Jul 31 '11 at 14:12
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
1
$begingroup$
This article by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ pi$.
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– Adrián Barquero
Dec 13 '10 at 1:46
17
$begingroup$
That crazy Ramanujan guy never can tell us anything useful... :D
$endgroup$
– J. M. is not a mathematician
Dec 13 '10 at 3:08
33
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One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities…
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– Mariano Suárez-Álvarez
May 22 '11 at 15:15
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I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians.
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– Masacroso
Sep 17 '14 at 7:24
$begingroup$
I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4cdot 99$ do jump out)
$endgroup$
– timtfj
Dec 24 '18 at 0:20