BR Maths Corner1
Re: BR Maths Corner1
errr....my first post here...
apologies if posted earlier.
interesting talk from manjul bhargava, explaining how hemachandra figure fibonocci centuries before
https://www.youtube.com/embed/2MCK3eVwTw4
apologies if posted earlier.
interesting talk from manjul bhargava, explaining how hemachandra figure fibonocci centuries before
https://www.youtube.com/embed/2MCK3eVwTw4
Re: BR Maths Corner1
Gus wrote:interesting talk from manjul bhargava, explaining how hemachandra figure fibonocci centuries before
https://www.youtube.com/embed/2MCK3eVwTw4
Thanks for posting this. I have posted about this many times before and if you have not seen those posts, you may enjoy.. for example
here in 2011 or this math problem
or Manjul and this
and the video and my comment..
As said before... it brings back good memories..I still remember the excitement when I learned how these numbers are related to golden ratio..(probably the third most popular (after pi and e) irrational/ transcendental number)
>> . When I was quite young.. I attended a public lecture titled "Beauty of Mathematics" which was given at the local university but I went even though I was much younger. Prof. Mehta, introduced these Fibonacci numbers in a very beautiful way and I still remember the excitement of learning new things...I was hooked.
Thanks again for posting this..
BTW Fibonacii, in his writings, himself admits that that work was not originally from him.. AFAIK he does not mention Hemchanda explicitly but he does say (and he certainly did) he had access to work/books of Indian Mathematician mathematician available to him.) .. (Personally I use the term Hemchand(r)a series when I teach this)
***
Somewhat related to this, Tao, recently "solved" a similar interesting type of sequence of numbers..Erodos first proposed the problem (with $500 prize) but it remained unsolved for last 70+ years..
The problem is extremely easy to state (and looks appealing!) .. but the "proof" is using computers and is too hard to check by humans!!
( For those who are interested, google "The Erdos discrepancy problem" in recent news or see this Nature's article:
Maths whizz solves a master's riddle  Terence Tao successfully attacks the Erdős discrepancy problem by building on an online collaboration.
Or click on the following image from Science News
83yearold math problem solved

 BRFite
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 Location: "Visa Officer", Indian Consulate #13,451, Khost Province, Afghanistan
Re: BR Maths Corner1
Hi,
Is the following problem wellknown and solved?
Given a set of nonzero integers S, and an integer n, let P(S,n) be the minimum size of S such that there exists at least one subset T of S, with T=n, such that the sum of the elements in T is 0 mod n.
Clearly, one can use the pigeonhole principle (php) to derive an upper bound for each n. By php, P(S,n) <= ceil((n1)*n+1) because if there are at least that many elements in S, then there will be at least n that are equal mod n, and the sum of those n will be 0 mod n. But by brute force, one can derive tighter bounds:
n  php bound  tight bound

2  3  3
3  7  5
4  13  7
5  21  9
which leads one to believe that the tight bounds are 2n1, but I can't prove it.
And how can this be extended to tuples? For instance, S has tuples of integers (a1,a2), and we want subsets of size n that sum to 0 mod n componentwise. php can be used to derive an upper bound (for instance, for n=4, there would be 4x4=16 combinations for the tuples mod 4, meaning that if there are at least 49 tuples, 4 will be same mod 4 which can then be added to give 0 mod 4. But is there a better bound than 49 since again we don't need to add same values mod 4 to get 0; can do 1+3 or 1+2+1 for each component, leading to a smaller bound.
Is the following problem wellknown and solved?
Given a set of nonzero integers S, and an integer n, let P(S,n) be the minimum size of S such that there exists at least one subset T of S, with T=n, such that the sum of the elements in T is 0 mod n.
Clearly, one can use the pigeonhole principle (php) to derive an upper bound for each n. By php, P(S,n) <= ceil((n1)*n+1) because if there are at least that many elements in S, then there will be at least n that are equal mod n, and the sum of those n will be 0 mod n. But by brute force, one can derive tighter bounds:
n  php bound  tight bound

2  3  3
3  7  5
4  13  7
5  21  9
which leads one to believe that the tight bounds are 2n1, but I can't prove it.
And how can this be extended to tuples? For instance, S has tuples of integers (a1,a2), and we want subsets of size n that sum to 0 mod n componentwise. php can be used to derive an upper bound (for instance, for n=4, there would be 4x4=16 combinations for the tuples mod 4, meaning that if there are at least 49 tuples, 4 will be same mod 4 which can then be added to give 0 mod 4. But is there a better bound than 49 since again we don't need to add same values mod 4 to get 0; can do 1+3 or 1+2+1 for each component, leading to a smaller bound.
Re: BR Maths Corner1
MurthyB garu: 108 stripes for not using TeX
Seriously, looks like an interesting problem. Please do not give away the solution till we
Seriously, looks like an interesting problem. Please do not give away the solution till we
Re: BR Maths Corner1
<dup> snipped
Last edited by Vayutuvan on 31 Oct 2015 01:01, edited 1 time in total.

 BRFite
 Posts: 699
 Joined: 18 Oct 2002 11:31
 Location: "Visa Officer", Indian Consulate #13,451, Khost Province, Afghanistan
Re: BR Maths Corner1
Ha ha, not a latex expert. Always used framemaker...
Re: BR Maths Corner1
OT: never new mr tuvan was a latex expert... or he never revealed himself as such in the yell n yums.

 BRFite
 Posts: 699
 Joined: 18 Oct 2002 11:31
 Location: "Visa Officer", Indian Consulate #13,451, Khost Province, Afghanistan
Re: BR Maths Corner1
So noone knows this problem, or can prove it? I feel like I am missing something obvious for the singleton case. The tuples seems to be mode challenging. Below I should modify P(S,n) to just P(n) as it does not depend on S anyway.
MurthyB wrote:Hi,
Is the following problem wellknown and solved?
Given a set of nonzero integers S, and an integer n, let P(S,n) be the minimum size of S such that there exists at least one subset T of S, with T=n, such that the sum of the elements in T is 0 mod n.
Clearly, one can use the pigeonhole principle (php) to derive an upper bound for each n. By php, P(S,n) <= ceil((n1)*n+1) because if there are at least that many elements in S, then there will be at least n that are equal mod n, and the sum of those n will be 0 mod n. But by brute force, one can derive tighter bounds:
n  php bound  tight bound

2  3  3
3  7  5
4  13  7
5  21  9
which leads one to believe that the tight bounds are 2n1, but I can't prove it.
And how can this be extended to tuples? For instance, S has tuples of integers (a1,a2), and we want subsets of size n that sum to 0 mod n componentwise. php can be used to derive an upper bound (for instance, for n=4, there would be 4x4=16 combinations for the tuples mod 4, meaning that if there are at least 49 tuples, 4 will be same mod 4 which can then be added to give 0 mod 4. But is there a better bound than 49 since again we don't need to add same values mod 4 to get 0; can do 1+3 or 1+2+1 for each component, leading to a smaller bound.
Re: BR Maths Corner1
AmberG, VT and other math gurus please comment on this talk:
http://www.independentnews.com/labs_lin ... 269fe.html
http://www.independentnews.com/labs_lin ... 269fe.html
Re: BR Maths Corner1
http://swarajyamag.com/ideas/mathinin ... forgotten/
At some point we lost it all. When we today look at the classic ‘Mathematics can be fun’ by Yakov Perelman or ‘The Scientific American Book of Mathematical Puzzles & Diversions’ by Martin Gardner, one cannot but think of them as the updated versions of Kanakathikaram. And one cannot stop wondering with a bit of pain, why it does not happen anymore in India.
Re: BR Maths Corner1
Ramana garu: started watching but work interfered. The little bit I watched, the speaker seems to be very eloquent. Certainly provocative and thought provoking. But I am not stats guru.
Re: BR Maths Corner1
Meanwhile to get some idea of the utter stupidity, and lack of math skills of some , watch the following clip..
Chris Cuomo of CNN can not do elementary math. Because he read it somewhere, he repeats it here that 1200 divided by 300 is "few millions" ..
Chris Cuomo is a famous TV anchor, with a JD degree and comes from a family where both his brother and father were state governors. (Not to mention graduate of Yale etc)
For those who do not know the background, a few days ago some one started a rumor that the current mega lottery which is around 1.2 billion dollar, if divided equally among all US citizens (300 million), each one will get 4 million dollars (instead of 4) and Cuomo is one who fell for it.
Simply amazing!
http://theconcourse.deadspin.com/cnnanchortalkspowerballisextremelybadatmath1752932337
Or watch and enjoy this clip:
http://www.cc.com/videoclips/myupyp/thedailyshowwithtrevornoahcheckingthemathonthepowerball
Chris Cuomo of CNN can not do elementary math. Because he read it somewhere, he repeats it here that 1200 divided by 300 is "few millions" ..
Chris Cuomo is a famous TV anchor, with a JD degree and comes from a family where both his brother and father were state governors. (Not to mention graduate of Yale etc)
For those who do not know the background, a few days ago some one started a rumor that the current mega lottery which is around 1.2 billion dollar, if divided equally among all US citizens (300 million), each one will get 4 million dollars (instead of 4) and Cuomo is one who fell for it.
Simply amazing!
http://theconcourse.deadspin.com/cnnanchortalkspowerballisextremelybadatmath1752932337
Or watch and enjoy this clip:
http://www.cc.com/videoclips/myupyp/thedailyshowwithtrevornoahcheckingthemathonthepowerball
Re: BR Maths Corner1
ramana, Nandu (and anybody interested),
As promised here is the link to Chauvenet prize of MAA
In 2014 Ravi Vakil got it for his article titled The Mathematics of Doodling
Most prize winning writings are available as PDFs. Another interesting one for BRF forum readers is Granville's articl which he describes lucidly AKS polynomial time algorithm for primality testing which resolves a long standing question of whether primaility testing is in P affirmatively. AKS stands for Agarwala, Kayal, and Saxena of IITK.
As promised here is the link to Chauvenet prize of MAA
Chauvenet Prizes
The Chauvenet Prize, consisting of a prize of $1,000 and a certificate, is awarded at the Annual Meeting of the Association to the author of an outstanding expository article on a mathematical topic. First awarded in 1925, the Prize is named for William Chauvenet, a professor of mathematics at the United States Naval Academy. It was established through a gift in 1925 from J.L. Coolidge, then MAA President. Winners of the Chauvenet Prize are among the most distinguished of mathematical expositors.
In 2014 Ravi Vakil got it for his article titled The Mathematics of Doodling
Most prize winning writings are available as PDFs. Another interesting one for BRF forum readers is Granville's articl which he describes lucidly AKS polynomial time algorithm for primality testing which resolves a long standing question of whether primaility testing is in P affirmatively. AKS stands for Agarwala, Kayal, and Saxena of IITK.
Re: BR Maths Corner1
VT:
Manindra Agrawal (Neeraj Kayal and Nitin Saxena, believe it or not, were his UG students at that time!) is very famous.
He has been mentioned in BRF *many* many times. I first posted about his "Primes in P" well before math dhaga's birth in Brf.(around 20022003). I remember taking my son to listen to his lecture about "Primes in P" after his famous paper and he was invited here in a US universities to present it. The fun part was, it is very rare that such papers can be understood (and appreciated) by a freshman math student like my son).
Along with many honors like PadmSri, Clay, Godel, he was first Infosys prize winner. He is one of the most distinguished IITK alumni and has been extremely active in alumni network and in inspiring others.
From what I know, he is also one of the top adviser/expert to NaMo's government for cyber security for terror intelligence etc. (Computational Complexity/number theory is quite useful here)
AKS stands for Agarwala, Kayal, and Saxena of IITK.
Manindra Agrawal (Neeraj Kayal and Nitin Saxena, believe it or not, were his UG students at that time!) is very famous.
He has been mentioned in BRF *many* many times. I first posted about his "Primes in P" well before math dhaga's birth in Brf.(around 20022003). I remember taking my son to listen to his lecture about "Primes in P" after his famous paper and he was invited here in a US universities to present it. The fun part was, it is very rare that such papers can be understood (and appreciated) by a freshman math student like my son).
Along with many honors like PadmSri, Clay, Godel, he was first Infosys prize winner. He is one of the most distinguished IITK alumni and has been extremely active in alumni network and in inspiring others.
From what I know, he is also one of the top adviser/expert to NaMo's government for cyber security for terror intelligence etc. (Computational Complexity/number theory is quite useful here)
Re: BR Maths Corner1
AmberG guruji: Yes I remember the announcement in 2003 time frame. Ron Graham at Lucent (nee Bell Labs) was the first one who checked the proof and after it checked out invited Prof. M. Agrawal to the labs. I went to my boss (exadviser) with the news and he was like "that is big news indeed. please download the paper for future reference". Unfortunately the constants in the asymptotics are so high that MillerRabin and Pollard's rho test heuristics are still the preferred methods. Both are probabilistic and have much faster running times. Prof. Agrawal is working on Cylindrical Decompositions (Mathematical logic related stuff) right now. I have seen a recent paper in a recent book. I think one of the two students is at MSR, Bangalore. Another has to come to the US for a PhD.
By the the way there is another fundamental result called AKS Sorting Network (AjtaiKomlosSzemeredi) which has a $O(\log n)$ depth. That was one of the fundamental results in size reduction. Again the bigO notation hides a large constant making the network impractical. The expander graphs used in the proof have connections to Ramanujan Graphs.
By the the way there is another fundamental result called AKS Sorting Network (AjtaiKomlosSzemeredi) which has a $O(\log n)$ depth. That was one of the fundamental results in size reduction. Again the bigO notation hides a large constant making the network impractical. The expander graphs used in the proof have connections to Ramanujan Graphs.
Re: BR Maths Corner1
^^^ Yes, he was also invited to, and spoke at a few other well known US schools at that time. The event also got in the news (in WSJ) because US state dept refused to grant Visa his undergrad students, who were coauthors, due to security concerns! (India and its scientists became less favorite post 1998 ) I am glad that much has change between India/US relationship!
To add: Manindra was quite well known even before his famous paper. The beauty about that paper I liked (got the preprint much before it became big news) was that I was able to fully explain to some bright highschoolers.
Of course, full details can be found in the original paper
(http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf, which is just 9 pages in length and almost entirely elementary in nature.
The MAIN point is AKS is the first provably deterministic algorithm to determine the primality of a given number with a run time which is guaranteed to be polynomial in the number of digits. ( BTW this order factor 12 in the original paper, can be as low as 3 if Agrawal conjecture is true  just a note if some one actually reads the original paper and is curious )
(Many previous primality testing algorithms, including the one you mentioned, existed, but they were either probabilistic in nature, had a running time slower than polynomial, or the correctness could not be guaranteed without additional hypotheses such as GRH  generalized Riemann hypothesis  which is still unproven.)
Miller–Rabin can test any given number for primality in polynomial time, but are known to produce only a probabilistic result. Miller test (Miller modified the above test) is fully deterministic and runs in polynomial time over all inputs, but its correctness depends on the truth of GRH. Other tests are similarly probabilistic or works with only certain type of numbers, like Mersenne Primes, which I have talked about in this dhaga.
For those, who are curious and interested in this aspect of prime numbers, and math and computerscience, but do not have time or resources to read technical papers, I highly recommend the polymath series.
http://michaelnielsen.org/polymath1/index.php?title=Finding_primes
To add: Manindra was quite well known even before his famous paper. The beauty about that paper I liked (got the preprint much before it became big news) was that I was able to fully explain to some bright highschoolers.
Of course, full details can be found in the original paper
(http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf, which is just 9 pages in length and almost entirely elementary in nature.
The MAIN point is AKS is the first provably deterministic algorithm to determine the primality of a given number with a run time which is guaranteed to be polynomial in the number of digits. ( BTW this order factor 12 in the original paper, can be as low as 3 if Agrawal conjecture is true  just a note if some one actually reads the original paper and is curious )
(Many previous primality testing algorithms, including the one you mentioned, existed, but they were either probabilistic in nature, had a running time slower than polynomial, or the correctness could not be guaranteed without additional hypotheses such as GRH  generalized Riemann hypothesis  which is still unproven.)
Miller–Rabin can test any given number for primality in polynomial time, but are known to produce only a probabilistic result. Miller test (Miller modified the above test) is fully deterministic and runs in polynomial time over all inputs, but its correctness depends on the truth of GRH. Other tests are similarly probabilistic or works with only certain type of numbers, like Mersenne Primes, which I have talked about in this dhaga.
For those, who are curious and interested in this aspect of prime numbers, and math and computerscience, but do not have time or resources to read technical papers, I highly recommend the polymath series.
http://michaelnielsen.org/polymath1/index.php?title=Finding_primes
Re: BR Maths Corner1
posted in burkha.. gurus, pliss enlighten me. just wanted to know what he is up to?

 BRF Oldie
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Re: BR Maths Corner1
Guru loge, please to see a problem posted in the GDF MAth thread and help there onlee..
Re: BR Maths Corner1
Many people have heard about the story of Hardy and Ramanujan and the taxi cab numbered 1729 (see Taxicab number and 1729 if you don't remember the story). Most people think that it was just some random fact that Ramanujan knew about from memory. Turns out that there might be more to that story  from a study of Ramanujan's notes, it appears he was actually up to something deeper before Hardy showed up in that cab. It turns out he had written 1^3 + 12^3 = 9^3 + 10^3 in his notes a while back and also a whole other series of numbers where x^3 + y^3 = z^3 + 1 or x^3 + y^3 = z^3  1 (i.e.) numbers that are very near misses to fermat's last theorem and it appears that he was investigating the theory of elliptic curves (which were used by Andrew Wiles 90 years later to prove Fermat's last theorem). Turns out Ramanujan had proved something else in the theory of mathematics about 40 years before anyone else had thought about it: K3 surfaces. And because of that research he was doing, he knew about 1729.
More details are here: https://plus.maths.org/content/ramanujan
More details are here: https://plus.maths.org/content/ramanujan
Re: BR Maths Corner1
Problem from 538.com:
http://fivethirtyeight.com/features/can ... neighbors/
http://fivethirtyeight.com/features/can ... neighbors/
The misanthropes are coming. Suppose there is a row of some number, N, of houses in a new, initially empty development. Misanthropes are moving into the development one at a time and selecting a house at random from those that have nobody in them and nobody living next door. They keep on coming until no acceptable houses remain. At most, one out of two houses will be occupied; at least one out of three houses will be. But what’s the expected fraction of occupied houses as the development gets larger, that is, as N goes to infinity?
Re: BR Maths Corner1
http://www.thehindu.com/scitech/science/ramanujanencyclopedialaunched/article8426583.ece
The Ramanujan Encyclopedia is a comprehensive reference book that will contain information.
An encyclopedia of Srinivasa Ramanujan and his mathematics is being launched by Springer. This was announced by Marc Strauss, Editorial Director, Mathematics, of Springer, North America, at an international conference of mathematics held at the University of Florida. “The comprehensive encyclopedia of about 1000 pages, in two volumes, will contain everything important about Ramanujan’s life and mathematics,” said Mr. Strauss during his announcement at the recently held International Conference on Number Theory at the University of Florida. “We have assembled a team of leading researchers as EditorsinChief, who are experts on Ramanujan, and who have considerable editorial experience, to ensure the success of this massive project,” he added. The EditorsinChief of the Ramanujan encyclopedia are Professors Krishnaswami Alladi (University of Florida), George Andrews (The Pennsylvania State University), Bruce Berndt (University of Illinois), and Ken Ono (Emory University). “
The Ramanujan Encyclopedia is a comprehensive reference book that will contain information on all the mathematical contributions of Ramanujan and their impact on scientific fields, as well as on important aspects Ramanujan’s life including the individuals who have played significant roles in his life and with regard to his work. The themes that the encyclopedia will include are: (1) Ramanujan’s life, (2) persons closely connected with Ramanujan’s life or mathematics, (3) Ramanujan’s notebooks and work in India, (4) Ramanujan’s letters to Hardy, (5) Ramanujan in England, (6) Ramanujan’s published papers, (7) Ramanujan’s lost notebook, (8) Ramanujan’s work and its influence, (9) Books/expositions on Ramanujan’s life and work, (10) Ramanujan in the media, (11) honouring and preserving Ramanujan’s legacy, (12) modern developments in research and (13) Ramanujan’s health.
“The encyclopedia will contain several hundred entries in the form of articles by experts that will provide a detailed treatment of these themes. But all entries will be listed alphabetically to facilitate easy reference,” said Professor Alladi. “The encyclopedia will be updated periodically to keep abreast of current developments,” he added.
The Ramanujan encyclopedia is planned not only as a historical document, but also as a valuable reference for those pursuing research on, or related to, Ramanujan’s work. It will be of interest to experts and the lay person alike.
“Springer is already playing a big role in the world of Ramanujan with the publication of the edited versions of Ramanujan’s Notebooks authored by Bruce Berndt, and Ramanujan’s Lost Notebook authored by George Andrews and Bruce Berndt, as well as The Ramanujan Journal, for which Alladi is the EditorinChief. The Ramanujan encyclopedia is the next big step in Springer’s commitment to fostering the legacy of Ramanujan,” said Mr. Strauss.
Re: BR Maths Corner1
ENCOUNTER WITH THE INFINITE
HOW DID THE MINIMALLY TRAINED, ISOLATED SRINIVASA RAMANUJAN, WITH LITTLE MORE THAN AN OUTOFDATE ELEMENTARY TEXTBOOK, ANTICIPATE SOME OF THE DEEPEST THEORETICAL PROBLEMS OF MATHEMATICS—INCLUDING CONCEPTS DISCOVERED ONLY AFTER HIS DEATH?
Re: BR Maths Corner1
My Search for Ramanujan: How I Learned to Count; Ken Ono & Amir D. Aczel
http://www.thehindu.com/scitech/ramanujansgiftsolutionstoellipticcurves/article7927190.ece
Mathematician Ken Ono on how a chance letter from Ramanujan’s wife changed the way he looked not only at the world of numbers but at life itself.
In the pursuit of sharplydefined goals, we can at times forget that the journey itself holds riches that, by far, outweigh the goals. Life’s journeys hold all kinds of signs and messages that reveal new destinations that are far more intimately connected with the self than with preset goals. My Search for Ramanujan is the story of one such journey.
The first author of the book, mathematician Ken Ono, is a professor at Emory University. Born to Sachiko and Takashi Ono, who was himself a leading mathematician, Ono’s future in mathematics appeared set. However (and that’s what this book is about), his life was far from being smooth or dictated by logic. Ono had to undertake his own version of a hero’s journey before coming in contact with a mathematics he could call his own; before, in fact, being able to embrace with love his “tiger parents”.
Ono was, even as a child, groomed to become a firstclass mathematician, and the demands played havoc with his selfesteem. Being a prodigiously talented teenager did not help either. Violin lessons and mathematics were the two activities he was encouraged to focus upon. And focus he did, at great expense to his selfworth and motivation. As it happened, when the pressure rose to an intolerable degree, he decided to quit music lessons and school.
The day he decided to quit and break the news to his father, a letter arrived from Janaki Ammal, widow of renowned mathematician Srinivasa Ramanujan. In the letter, Janaki thanked Ono’s father for making a contribution towards building a small statue of Ramanujan.
Ono harboured no real hope of his father permitting him to leave school. Strangely, perhaps triggered by Janaki’s letter, his father — with his thoughts now dwelling on how Ramanujan’s life was tragically cut short by neglect, malnutrition and tuberculosis —halfheartedly accepted Ono’s idea of taking a break and going bike tour.
Ono undertook his journey around the world, where he ultimately found mentors, mathematics and freedom from the voices of disapproval he had internalised. His journey took him full circle, and he finally returned home to make peace with his parents. Despite being brought up to deny anything irrational, he even went on a ‘pilgrimage’ to Kumbakonam, Ramanujan’s birthplace, where more signs and treasures awaited him.
The journey is, at one level, literally that, as Ono moved from place to place, institute to institute, learning, teaching, failing, recovering and, most importantly, discovering the mathematics of Ramanujan.
There are many other threads to Ono’s complex story — an AsianAmerican teenager growing up at the interface of incompatible cultures and breaking free; an arc of recovery of an individual suffering from low selfesteem and lack of motivation; a young, bright mathematician moving from bookish knowledge to an appreciation of what is divine about math.
The long sections devoted to describing his life story apart, Ramanujan makes his presence felt in other ways too. Beckoning from his world of numbers, Ramanujan slowly reveals the secrets of his mathematics to Ono. Subtly, the reader is drawn to understand the difference between problemsolving and theorybuilding in mathematics, and how Ramanujan fits into neither category, but is an anticipator of mathematics.
In his short lifetime and brief stint at Trinity College, Ramanujan spewed out ideas and theorems on number theory at a furious rate. Ramanujan was a supernova among mathematicians, throwing out multiple ideas, which no one recognised then and few do now.
Guided by this presence, and nurtured by mathematicians such as Basil Gordon and Paul Sally, to whom this book is dedicated, Ono chooses to work on the underlying theories in the apparently disconnected statements made by Ramanujan, and not just work on his ideas. Ramanujan used to jot down his ideas methodically in notebooks, in green ink. One of these, called the ‘lost notebook’, was discovered in the Trinity College library by mathematician George Andrews in 1976, and later published as a book. Ono made a discovery about elliptic curves recently that was prompted by Ramanujan’s comments in his notebooks.
Ono and his coauthor Amir D. Aczel have woven reallife incidents into a story that is engaging while prodding the reader to explore the world. The book will make a great turning point not only for young and aspiring mathematicians but also for others who have a voice in their heads telling them they are on the wrong road.
http://www.thehindu.com/scitech/ramanujansgiftsolutionstoellipticcurves/article7927190.ece
The anecdote goes that once when Hardy visited Ramanujan who was sick, Hardy remarked: “I had ridden in taxicab number 1729, and it seems to me a rather dull number. I hope it was not an unfavourable omen.” To this Ramanujan had replied, “No, it is a very interesting number. It is the smallest number expressible as the sum of two cubes in two different ways.”
Yes, 1729 = 93 + 103 and also 1729 = 123 +13
This story is often narrated to explain Ramanujan’s familiarity with numbers but not more than that. Recent discoveries have brought to light that it was far from coincidence that Ramanujan knew the properties of 1729. There are now indications that he had, in fact, been looking at more general structures of which this number was but an example.
Mathematicians Ken Ono and Andrew Granville were leafing through Ramanujan’s manuscripts at the Wren Library in Cambridge University, two years ago, when they came across the equation 93 + 103 = 123 +13, scribbled in a corner. Recognising the representations of the number 1729, they were amused at first; then they looked again and found that there was another equation on the same page that indicated Ramanujan had been working, even then, on a famous seventeenth century problem known as Fermat’s Last Theorem (proved by Andrew Wiles in 1994).
“I thought I knew all of the papers there, but to my surprise, we found one page with near misses to the Fermat equation,” writes Dr Ono, who is also a Ramanujan scholar, in an email to this correspondent. Having a sneaking suspicion that Ramanujan had a secret method that gave him his amazing formulas, Dr Ono returned to Emory University and started working on these leads with his PhD student Sarah Trebat Leder.
“Together we worked backwards through Ramanujan’s notes, and we figured out his secret…[Ramanujan] arrived at the formulae on this page by producing a much more general identity. One which I recognised as a K3 surface (a concept that mathematician Andrew Wiles used for solving Fermat’s last theorem), an object that mathematicians did not discover until the 1960s,” notes Dr Ono.
Ramanujan died in 1920, long before mathematicians discovered the K3 surfaces, but from research done by Ono and Trebat Leder, it transpires that he knew of these functions long before. Dr Ono continues, “Ramanujan produced so many mysterious formulas, which can be misunderstood at first glance. We have come to learn that Ramanujan was perhaps the greatest anticipator of mathematics. His bizarre methods and formulas have repeatedly offered hints of the future in mathematics. In this case, we have added to Ramanujan’s legend.”
Commenting on their own work on this, he says, “He [Ramanujan] anticipated the theory of K3 surfaces before anyone had the merest glimpse. These surfaces are now at the forefront of research in mathematics and physics. In addition to adding to Ramanujan’s legacy, Sarah and I were able to apply his formulas to a problem in number theory (finding large rank elliptic curves), and his formulas immediately set the record on the problem. We hardly had any work to do. Ramanujan’s formula was a gift to us.”
Re: BR Maths Corner1
^^^ Saw the movie (Men Who Knew Infinity) with family a few months ago. Liked it. Cambridge part (Hardy, Littlewood etc) is quite good IMO (kids who spent some time in Cambridge agreed). India part (specifically his life/family etc) could have been better IMO (wish they had an actor who is familiar with India or had good coach), Math was good (and not silly as most movies tend to be)  not surprising they had Manjul Bhargava type help. Recommend the movie.
So in his honor let me ask a "simple" problem... Hope you enjoy. As usual using a computer program or internet search may spoil the fun  the problem is expected to be solved without the help of calculator..
If there are many solutions can you find the all such numbers..
So in his honor let me ask a "simple" problem... Hope you enjoy. As usual using a computer program or internet search may spoil the fun  the problem is expected to be solved without the help of calculator..
Problem: find two whole natural numbers (positive integers) a and b such that 
(1) a is NOT divisible by 7
(2) b is NOT divisible by 7
(3) (a+b) is NOT divisible by 7
but
(4) (a+b)^7  a^7  b^7 is divisible by 7^7.
If there are many solutions can you find the all such numbers..
Re: BR Maths Corner1
xpost from another brf thread ..
Problem: Factor the number (5^19851) into product of three integers, each of which is greater than 10^60.
(Hint: No calculators or computers are expected to solve this)
Problem: Factor the number (5^19851) into product of three integers, each of which is greater than 10^60.
(Hint: No calculators or computers are expected to solve this)
Re: BR Maths Corner1
Answers please? And kindly explain how to solve it?
Amber G. wrote:^^^ Saw the movie (Men Who Knew Infinity) with family a few months ago. Liked it. Cambridge part (Hardy, Littlewood etc) is quite good IMO (kids who spent some time in Cambridge agreed). India part (specifically his life/family etc) could have been better IMO (wish they had an actor who is familiar with India or had good coach), Math was good (and not silly as most movies tend to be)  not surprising they had Manjul Bhargava type help. Recommend the movie.
So in his honor let me ask a "simple" problem... Hope you enjoy. As usual using a computer program or internet search may spoil the fun  the problem is expected to be solved without the help of calculator..Problem: find two whole natural numbers (positive integers) a and b such that 
(1) a is NOT divisible by 7
(2) b is NOT divisible by 7
(3) (a+b) is NOT divisible by 7
but
(4) (a+b)^7  a^7  b^7 is divisible by 7^7.
If there are many solutions can you find the all such numbers..
Re: BR Maths Corner1
Bhaskar_T wrote:Answers please? And kindly explain how to solve it?
Please don't give the answer yet (sorry, Bhaskar ji). Please give a hint instead. I tried solving this, but my thinking is too conventional and straitjacketed, and I'm unable to think in the innovative ways required to tackle a problem like this. I'm not a mathematician either, but this kind of problem really intrigues me.
I got tantalizingly close (I think) by doing a binomial expansion, and figuring out that the final number ( (a+b)^7a^7b^7) would have to be:
1. Even
2. A multiple of (a+b)*a*b*(7^7)
Actually, condition 2. automatically implies condition 1. Also, the numbers a and b will have to be different (i.e., a=/=b).
Re: BR Maths Corner1
Amber G. wrote:xpost from another brf thread ..
Problem: Factor the number (5^19851) into product of three integers, each of which is greater than 10^60.
(Hint: No calculators or computers are expected to solve this)
is one of the factor a^4 + a^3 + a^2 +a + 1 where a = 5^397 ? though even then I don't have the full answer as in all factors > 10^60
Edited later to correct an error.
Last edited by chilarai on 16 Sep 2016 12:26, edited 2 times in total.
Re: BR Maths Corner1
sudarshan wrote:
I got tantalizingly close (I think) by doing a binomial expansion, and figuring out that the final number ( (a+b)^7a^7b^7) would have to be:
1. Even
2. A multiple of (a+b)*a*b*(7^7)
Actually, condition 2. automatically implies condition 1. Also, the numbers a and b will have to be different (i.e., a=/=b).
Yes, you are very close, if you proceed, as my son used to say about "math fairy" will make rest of the expression turn out to be lucky.. (most of such problems which I ask have neat solutions if one looks at it the right way and continue.. complex expressions have a tendency to become simpler )
Re: BR Maths Corner1
Bhaskar_T wrote:Answers please? And kindly explain how to solve it?Expression is (a+b)^7a^7b^7
we want to find (a,b) such that above is divisible by 7^7 while (a,b,(a_b)) are not divisible by 7
Some hints:
Hint#0 (nonmathematical hint)  such questions are such that a Ramajujan would have solved in a second or so. (It is not a bruteforce problem and has a simple solution).
Hint#1  The expression is divisible by a  (As S has pointed out)
Why? (put a=0 in the expression and whole expression becomes 0, hence a is a factor)
Hint#2  The expression is divisible by b  same reason
Hint#3  The expression is divisible by (a+b)  (again put a=b, the expression becomes zero)
So you might as well divide the above expression by ab(a+b), and since this is not divisible by 7, the quotient will be.
****
Less known, but very important learning is (a+b)^7a^7b^7 is divisible by 7.
In fact for any prime p : (a+b)^p  a^p b^p is divisible by p.
This is a very important result in math, and such results are the basis of Prof Manindra Agrawal (and his under graduate students) the famous AKS Primality Test.
This is easy to see, even using elementary school algebra, and though many bright student learn about it in highschool, most Indian students (and US students) hear about it only in graduate level math courses.
In fact if one puts a=b=1, the above becomes (2^p2)  the famous Fermat's Little theorem. (For any prime p, 2^p2 is divisible by p).
***
Thia ought to help.
Re: BR Maths Corner1
I didnt know it was a prominent result. Just expanded (a+b)7  a^7  b^7 and 7 is a factor in the result due to binomial expansion.
Thanks for the hints. Will try.
Thanks for the hints. Will try.
Re: BR Maths Corner1
I got those from the binomial theorem (that (a+b)^7a^7b^7 is divisible by a, b, (a+b), and by 7). The quotient you mention is now a fourth order polynomial, which I can't factor any further . But this polynomial would now have to be divisible by 7^6.
Re: BR Maths Corner1
chilarai wrote:Amber G. wrote:xpost from another brf thread ..
Problem: Factor the number (5^19851) into product of three integers, each of which is greater than 10^60.
(Hint: No calculators or computers are expected to solve this)
is one of the factor a^4 + a^3 + a^2 +1 where a = 5^397 ? though even then I don't have the full answer as in all factors > 10^60
Another factor is (5^397  1) i guess. If we factorize your factor it must be giving one more required factor.
Re: BR Maths Corner1
JayS wrote:
Another factor is (5^397  1) i guess. If we factorize your factor it must be giving one more required factor.
5^397 1 should be divisible by 4 ...but that is definitely smaller than 10^60 ..hence i was not sure
Re: BR Maths Corner1
Yayavar wrote:I didnt know it was a prominent result. Just expanded (a+b)7  a^7  b^7 and 7 is a factor in the result due to binomial expansion.
Thanks for the hints. Will try.
Yes it is easy to see from binomial expression, and also easy to see why this will be true for not only 7 but any prime number. This is why I said that the part of the proof is quite elementary in that sense. But the result has deeper meaning..and application(s).
For example, little Fermat's theorem, was proved by Euler much later and he understood it's significance, he also generalized it. Everyone knows, Fermat thought F5= 2^(2^5)+1 was prime. Euler proved that Fermat was wrong. In fact if Fermat understood the significance of the theorem named after him, he could have easily proved that F5 was not prime. (Yes, all he has to do is to check (3^F53),which is not divisible by F5 hence F5 is not prime  something which he could have done with pencil and paper).
The reason I mentioned AKS is that it is one of the paper which one can follow even if you are not an expert in number theory.
Re: BR Maths Corner1
JayS wrote:chilarai wrote:
is one of the factor a^4 + a^3 + a^2 +1 where a = 5^397 ? though even then I don't have the full answer as in all factors > 10^60
Another factor is (5^397  1) i guess. If we factorize your factor it must be giving one more required factor.
I think chilarai left out an "a"  the factor is a^4+a^3+a^2+a+1 where a=5^397. This can't be factorized polynomially, I don't think. The other factor is 5^3971, as JayS pointed out. Can this be factored further? 4 is obviously a factor, but then it becomes messy after that.
But the above two factors are definitely > 10^60.
Re: BR Maths Corner1
Thanks.Indeed I left out an 'a' like I always did in my exams. Will correct when I check brf on bigger screen.
Re: BR Maths Corner1
sudarshan wrote:I got those from the binomial theorem (that (a+b)^7a^7b^7 is divisible by a, b, (a+b), and by 7). The quotient you mention is now a fourth order polynomial, which I can't factor any further . But this polynomial would now have to be divisible by 7^6.
I am sure you have factored it by now. Maha HINT: It is much much simpler than you think!
Take a break, think about cube root of 1 (complex kind) and you will have the answer in front of you.
For others Hints are given below: stop reading (or peek and stop when you think you got it).
Hint1: Just assume b=1, now you have a polynomial in a only. This is actually same thing as the original problem (as you will easily see why) except that perhaps it is a little less messy.
Hint 2: (This technique must be taught in every school with basic equation solving  (finding roots and factoring is one and the same thing, BTW).. If in a fourthorder polynomial coefficient of (a^4) is same as the constant (or a^0 term), and coefficient of (a^3) is same as (a) then essentially this is a quadratic equation.
(which is easily solvable)
(In general any 2n degree polynomial becomes n degree polynomial if one has this kind of symmetry) :
To this, divide the whole thing by (a^2) and you will get:
something(a^2+1/a^2)+ something (a+1/a)+ something
Now if you put y=a+1/a, so that y^2 = a^2+1/a^2+2
Substituting above you essentially have a simple quadratic equation in y. Once you know "y" you can solve for "a" (another quadratic equation).
And surprise! if you do the above! (Yes, the your polynomial (which you need to be divisible by 7^6) comes out to be a perfect square!!
Hint3: BTW it is even easier if you think about cube root of 1, where w^3=1 (but w is not 1 and is = ((1+sqrt(3))/2).. you will see that x=w is always a solution. (which is true for any expression (a+b)^p  a^p b^p where p is prime>3... This is why I said that Ramanujan (or even someone who worked out his problems) would have gotten all the factors in a second )
Re: BR Maths Corner1
I factored that fourth order polynomial (it did turn out to be very simple, once I noticed the symmetry WRT a & b). I can now give you a long string of numbers which will work, but I still haven't proved that this is the only such set of numbers which will work. Should I post what I have so far, or are there people still trying, who don't want spoilers? This is for the 7^7 problem.
Re: BR Maths Corner1
sudarshan wrote:I factored that fourth order polynomial (it did turn out to be very simple, once I noticed the symmetry WRT a & b). I can now give you a long string of numbers which will work, but I still haven't proved that this is the only such set of numbers which will work. Should I post what I have so far, or are there people still trying, who don't want spoilers? This is for the 7^7 problem.
Nice! I am interested in seeing those numbers, specially what pattern you have discovered and how.
Some interesting background on this problem:
This is an old problem from a contest where problem asked for only one set of numbers  didn't want to make the problem too hard in an exam type environment where people had only short time to tackle the problem. Interestingly most contestants (they were top math students in a nation) solved it by educated guess (see note#1 below) and the poser of the problem said later on that it would have been much better problem if it had asked the family of solutions. (I have to confess, when I saw the problem for the first time, I solved it using a general way (more complex) getting a general solution using the same theme the poser intended the problem to be. Mine (and intended) method was even simpler  it was really to find cube root of 1, using modulo math (just knowing things like 2^3=8==1 mod 7, hence 2 is a cube root of 1 mod 7 ). (Will post the solution in separate post unless someone already posts it).
Just looking at the problem I just realize another method ... very easy if some one studied Bhramgupta's identity.
(Wiki link: https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity)
and noticing that 2^2+3*1^2 = 7.
(I will post a separate post after few days to give people chance to try)
This is what makes math fun for me.
Note 1# Most contested realized (see previous posts here) that (a^2+ab+b^2) must be divisible by 7^3=343. They also realized that (a+b)^2 > (a^2+ab+b^2)>=343 ==> (a+b)>18, there first guess, a=18, b=1 worked.
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