BR Maths Corner-1

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vina
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Re: BR Maths Corner-1

Postby vina » 13 Aug 2014 12:28

bhalluka wrote:Subhash Khot wins Nevanlinna prize

http://www.simonsfoundation.org/quanta/ ... mpossible/


Bade Mian! Take That and That! Madrassa B.Techs land a big one here ! Now kneel down and repent and ask for forgiveness for all the bad mouthing you did of those poor lil kids!

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Re: BR Maths Corner-1

Postby member_20292 » 13 Aug 2014 12:50

There was a good Indian mathematician that some people I know, knew personally, called Harish Chandra. He has an institute named after him in Allahabad University.

Good mathematician.

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Re: BR Maths Corner-1

Postby ramana » 14 Aug 2014 01:31

On Manjul Bhargav's achivement.


Hindu
The Midas Touch


The ‘Midas Touch’ mathematician
R. Ramachandran

Number theorist Manjul Bhargava wins Fields Medal
Manjul Bhargava, the Canadian-American number theorist from Princeton University, is one of the four who have been chosen for the highest award in mathematics, the Fields Medal, which is given once every four years by the International Mathematical Union (IMU) during the quadrennial International Congress of Mathematicians (ICM). The ICM2014 got underway on August 13 at Seoul, Republic of Korea.

Fields medal

Awarded in recognition of “outstanding mathematical achievement for existing work and for the promise of future achievement”, the Fields Medal is given to mathematicians of age less than 40 on January 1 of the year of the Congress. Born of Indian parents who migrated from Jaipur in the late 1950s, Bhargava, who turned 40 just last week, could not have hoped for a better birthday gift.

“Bhargava”, says the IMU citation, has been awarded the Fields Medal “for developing powerful new methods in the geometry of numbers, which he applied to count rings of small rank and to bound the average rank of elliptic curves.” (See Box for definitions of italicized terms)


• In ‘geometry of numbers’ one imagines a plane or a 3-dimensional space populated by a lattice whose grid points have integer co-ordinates.

• A ‘ring’ is an algebraic structure with two binary operations, commonly called addition and multiplication, which are generalizations of the familiar arithmetic operations with integers applied to algebraic objects. Examples of rings are polynomials of one variable with real coefficients, or square matrices of a given dimension. Algebraic number theory is the study of this and other algebraic structures.

• ‘Rank’ refers to the minimum number of objects required to generate the entire set of algebraic objects being studied; the dimension of a vector space, for example. The familiar 3-d vector space is of rank 3.

• ‘Elliptic curves’ are graphs generated by equations of the form y2 = a polynomial of degree 3, such as x3 + ax + b, where a and b are rational numbers.


A large body of work in number theory relates to the study of how numbers of interest, such as prime numbers, are distributed among the entire set of integers. Bhargava developed novel techniques to count objects in algebraic number theory that were previously considered completely inaccessible. His work has completely revolutionized the way in which fundamental arithmetic objects in algebraic number theory, such as number fields and elliptic curves, are now understood and studied, and this has given rise to wonderful applications.

About 200 years ago the German mathematician Carl Friedrich Gauss, one of the historical greats, had discovered a remarkable ‘composition law’ for binary quadratic forms, which are polynomials of the form ax2 + bxy + cy2, where a, b and c are integers. Using this law two binary quadratic forms could be combined to give a third one. Gauss’s law is a central tool in algebraic number theory. Bhargava discovered an ingenious and simpler geometrical technique to derive it and the technique allowed him to obtain composition laws for higher-degree polynomials as well.

Geometry of numbers

The technique reportedly dawned upon Bhargava one day while he was playing with Rubik’s cube. Implicit in Gauss’s method was the use of ‘geometry of numbers’ and it is this realization that enabled Bhargava to extend it to higher degrees. He then discovered 13 new composition laws for higher-degree polynomials. Until then, Gauss’s law was thought to be accidental and unique to binary quadratics. Nobody had even imagined that higher composition laws existed until Bhargava showed that Gauss’s law is part of a bigger theory applicable to polynomials of arbitrary degree. His approach has also broadened the canvas of applying geometry of numbers to address outstanding problems of algebraic number theory.

This work immediately led Bhargava to tackle a related problem, which was the counting of ‘number fields of fixed degree by discriminant’.

Discriminant

A number field is obtained by extending the rational numbers to include non-rational roots of a polynomial equation; if the polynomial equation is quadratic, such ax2+bx+c = 0, whose roots are given by the well-known formula [– b/2a ± √(b2 – 4ac)/2a], then one obtains a quadratic number field. The expression under the square root sign is called the ‘discriminant’ (defined appropriately for polynomials of different degrees). Higher degree number fields -- cubic, quartic, quintic etc. -- are correspondingly generated by higher degree polynomials.

The degree of the polynomial and its discriminant are two fundamental quantities associated with a polynomial. Despite number fields being one of the fundamental objects in algebraic number theory, answers to questions like how many number fields there are for a given degree n and a given determinant D were not known. If one has a quadratic polynomial, counting the number of lattice points in a certain region of 3-d space gives information about the associated quadratic number field. For example, using the geometry of numbers it can be shown that, for discriminant with absolute value less than D, there are approximately D quadratic number fields. The case of cubic number fields had been solved 40 years ago by Harold Davenport and Hans Heilbronn but since then the higher degree cases saw little progress until Bhargava came on the scene.

Quintic number fields

Armed with his new technique, Bhargava was able to solve the case of quartic and quintic number fields. The new composition laws and his new technique in using the geometry of numbers have together extended the reach and power of counting number fields. The cases of degrees greater than 5 still remain open as Bhargava’s composition laws alone seem inadequate to resolve these higher cases at present.

While the above work were al carried out between 2004 and 2008, more recently, Bhargava has employed his improved geometry of numbers technique to obtain striking results about ‘hyperellpitic curves’, which are graphs of equations of the form y2 = a polynomial with rational coefficients, the case where the degree of the polynomial is 3 being called the ‘elliptic curve’.

Elliptic curves have important applications in pure as well as applied mathematics. Even though Fermat’s Last Theorem seems to be not even remotely connected with elliptic curves, it was key to its proof in 1995 by Andrew Wiles, who, incidentally, was also Bhargava’s thesis advisor. Operations using elliptic curves have become a core component of many of the cryptographic protocols that encode credit card numbers in online transactions. “Intellectual stimulation, beautiful structure, applications – elliptic curves have it all,” Bhargava has said.

An outstanding problem in algebraic number theory has been how to count the number of points on ‘hyperelliptic curves’ that have rational coordinates, which is the same as asking how many rational solutions does a hyperellptic equation have? The answer, it turns out, following Bhargava’s work, depends on the degree of the curve.

One can easily see that the number of rational solutions of a polynomial equation of degree 1, such as y = 9x + 4, is infinite: any rational value for x produces a rational value for y, and vice versa. Quadratics, such as, such as y2 = 2x2 + 5x – 3, have either no rational solutions or infinitely many. For curves of degree 1 and 2, there is an effective way of finding all the rational points. In 1983, Gerd Faltings, director of Max Planck Institute for Mathematics, Bonn, showed that for degree 5 and more there are only finitely many rational points. That left unresolved the cases of degree 3 – the elliptic curves – and of degree 4.

Finding rational points for elliptic curves is, however, not an easy matter. They can have zero, finitely many, or infinitely many rational solutions. When does a cubic equation have infinitely many solutions has been a central question in number theory since Pierre de Fermat in the 17th Century. In the recent past mathematicians have attempted to devise algorithms to decide whether a given elliptic curve has finitely many or infinite rational points but that route took them nowhere. They have only been able to guess how often these different possibilities arise.

But once you have found some rational points on an elliptic curve, it becomes possible to generate more by using the simple connecting-the-dots method. For example (see fig.), if you draw a line through two rational points, it usually intersects the elliptic at exactly one more point, which is again a rational point. But the opposite, namely given one rational point finding the two rational points that would generate it. This is what underlies the use of elliptic curves in cyber security.



Connecting-the dots method: Given two rational points of an elliptic curve y2 = x3 + 2x + 3, the point at which the line through those points intersects the curve at one more point is guaranteed to be a rational point. This connect-the-dots procedure is a means to generate all of an elliptic curve’s rational points starting from a small finite number. (Credit: Quanta, illustration by Manjul Bhargava)

Curve's rank

When the number of rational points of an elliptic curve is infinite, the smallest number of rational points that can generate essentially all the rational points is called the curve’s rank. When the infinite set of rational points can be generated essentially from just one point, the curve has rank 1, and so on. When the number of rational points is finite or none at all, the rank is 0.

In 1992 Armand Brumer showed that a 1965 conjecture made by Birch and Swinnerton-Dyer (BSD) implied that the average rank of the group of rational points of an elliptic curve defined over rational numbers is bounded. Later in 1979 Dorian Goldfeld conjectured that the bound is, in fact, is equal to ½. That is, in a statistical sense, half of all elliptic curves have rank 0 and half have rank 1. Previously, however, mathematicians did not even know that the average rank was finite (let alone ½).

The conjecture, of course, does not mean that curves of higher rank – 2, 3 and so on – do not exist, or even that there are only finitely many such. Indeed, computationally mathematicians have found such curves, the highest known rank till date is 28! But as the number of elliptic curves asymptotically becomes infinitely large, the curves with higher ranks approach a vanishingly small percentage of the whole.

Enter Bhargava and his collaborators, his doctoral student Arul Shankar (a 2007 Chennai Mathematical Institute graduate) in particular. Instead treading the beaten track of algorithms, they asked the question: what could be said about rational points on a typical curve? From this perspective they first showed that a sizeable fraction of elliptic curves has only one rational point (rank 0) and another sizeable proportion has infinitely many rational points (rank > 0). Using newly developed techniques, they were able to show that the average rank is, in fact, bounded. They have been further able to show that the bound is also less than 1, indicating that the conjecture is perhaps true.

“Bhargava introduced dramatically new ideas ​to study the average number of solutions and proved that the average rank of elliptic curves is bounded, and that the BSD Conjecture is true on the average, making it one of the most spectacular successes in number theory in recent years,” says Deependra Prasad, a number theorist from Tata Institute of Fundamental Research (TIFR).

Analogously, for the case of degree 4 too Bhargava and Shankar showed that a significant chunk of such curves has no rational points and another significant chunk positive proportion has infinitely many rational points. Using his expanded geometry of numbers technique Bhargava has also explored higher-degree curves in general.

While Faltings Theorem tells us that for curves of degree greater than 5, there are only finitely many rational points, it does not give a way to determine how many exactly there are. For the even degree case, Bhargava showed that the “typical” hyperelliptic curve had no rational points at all. The joint work of Bhargava and Benedict Gross, followed up by that of Bjorn Poonen and Michael Stoll, established the same result for the odd degree case as well. Bhargava’s work has thus clearly shown that the number of curves having rational points decreases rapidly as the degree increases. For example, for a typical 10 degree polynomial, there is a greater than 99 per cent chance that the curve has no rational points.

Bhargava’s work in number theory has had profound influence in the field. “A mathematician of extraordinary creativity, he has a taste for simple problems of timeless beauty, which he has solved by developing elegant and powerful new methods that offer deep insights,” said IMU’s information sheet on his work. “With his keen intuition, immense insight and great technical mastery, he seems to bring a ‘Midas touch’ to everything he works on,” it added.

Tabla player
Besides being one of the world’s leading mathematicians, Bhargava is also an accomplished Tabla player and plays at the concert level. He learnt the art initially from his mother and later came under the tutelage of the well-known tabla maestros Pandit Prem Prakash Sharma and Ustad Zakir Hussain. “Classical Indian music,” Bhargava told Princeton Weekly Bulletin when he was featured, “is very mathematical, but consciously thinking of the math would interfere with the improvisation and emotion of the playing. But somehow the connection is there. I often use music as a break, and many times I come back to the math later and things have cleared up." Indeed, Bhargava thinks of mathematics art. He is also keenly interested in linguistics in which he has published research work. It was his grandfather, a linguistics scholar, who taught him Sanskrit and developed his interest in linguistics.





While non secular folks write like this:

Fields Medal Winner


Finance More: Math Rubik's Cube A Winner Of Math's Most Prestigious Award Solved A 200-Year-Old Problem By Using A Rubik's Cube

Elena Holodny
Aug. 13, 2014, 10:40 AM

On Tuesday the winners of the most prestigious award in mathematics — the Fields Medal — were announced. The award is given to two to four mathematicians under age 40 every four years "to recognize outstanding mathematical achievement for existing work and for the promise of future achievement."

This award is typically given for mathematical research that solves or extends complicated problems that mathematicians have struggled with for decades or even centuries.

And, believe it or not, one of the winners, Manjul Bhargava, reformulated a 200-year-old number theory problem in a rather unconventional way: by using a Rubik's Cube and Sanskrit texts.

But before we get to his idea, let's go over some background.

Here's an interesting mathematical idea: If two numbers that are each the sum of two perfect squares are multiplied together, the resulting number will also be the sum of two perfect squares.

Let's take the numbers 25 (a perfect square of 5) and 36 (a perfect square of 6) and add them together as an example. The resulting equation would read:

(25 + 25)(36 + 36) = 3600, where 3600 equals ((36*36) + (48*48)), or (1296 + 2304).

You can keep trying this with all different number combinations — as long as the four numbers in the brackets are perfect squares — and the output will always also be a perfect square. (Yes, math is awesome.)

But let's get back to Bhargava. His grandfather, Purushottam Lal Bhargava, was the head of the Sanskrit department of the University of Rajasthan — and so "Bhargava grew up reading ancient mathematics and Sanskrit poetry texts," according to Quanta Magazine.

In one of these manuscripts, he discovered a "generalization [similar to the above math] developed in the year 628 by the great Indian mathematician Brahmagupta" that stated:

If two numbers that are each the sum of a perfect square and a given whole number times a perfect square are multiplied together, the product will again be the sum of a perfect square and that whole number times another perfect square.


Later on, when he was a graduate student, Bhargava learned that Carl Gauss — a leading mathematical figure of the 18th and 19th century "came up with a complete description of these kinds of relationships" — provided that the expressions had only two variables, and only quadratic forms (meaning x2, but not x3).

Essentially, Gauss came up with a "composition law" that would tell you which quadratic form you'd get if you multiplied two of these kinds of expressions together.

But here's the downside: Gauss' law took him approximately 20 pages to write out. Ouch. No one wants to read through all of that.

Bhargava was interested in uncovering an easier way to describe Gauss' law — and perhaps if he could tackle higher exponents, according to Quanta Magazine.

Then one day, Bhargava thought of a solution when he was sitting in his room, looking at a Rubik's Cube. He took a mini-cube (a 2x2 Rubik's Cube) and "realized that if he were to place numbers on each corner of the mini-cube and then cut the cube in half, the eight corner numbers could be combined in a natural way to produce a binary quadratic form," according to Quanta Magazine.

If you think about a Rubik's Cube, you'll note that there are three ways to cut the cube in half. You can split the top from the bottom; you can split the right from the left; you can split the front from the back.

In fancy math terms, that means that by cutting a Rubik's cube in three ways, you can generate "three quadratic forms."

The interesting thing about this, is that Bhargava discovered that these quadratic forms "add up to zero" — not when using normal addition, but "with respect to Gauss' method for composing quadratic forms."

And ta-da, by simply dividing a Rubik's Cube, he "gave a new and elegant reformulation of Gauss' 20-age law."


Bhargava said about this moment to the New Scientist:

Gauss's law says that you can compose two quadratic forms, which you can think of as a square of numbers, to get a third square. I was in California in the summer of 1998, and I had a 2 x 2 x 2 mini Rubik's cube. I was just visualising putting numbers on each of the corners, and I saw these binary quadratic forms coming out, three of them. I just sat down and wrote out the relations between them. It was a great day!

Afterward, Bhargava worked with a Rubik's Domino (a 2x3x3 shape) — and he realized that he could go beyond the limited quadratic forms of Gauss' law.

Eventually, he went on to discover 12 more compositional laws, which later became his Ph.D. thesis.

According to Quanta Magazine, Barghava said that he had always been interested in ideas like this — the “problems that are easy to state, and when you hear them, you think they’re somehow so fundamental that we have to know the answer.”


Read more: http://www.businessinsider.com/fields-m ... z3AJ46sKzO


Clearly his genius is in using simplicity to unravel complexity.

Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 14 Aug 2014 02:42

Congratulations to Manjul,

Yours truly predicted this 4 years ago.. when the last Fields Medal winner was to be announced...

See: my post from 2010
http://forums.bharat-rakshak.com/viewtopic.php?p=848505#p848505
Amber G. wrote:
joshvajohn wrote:India a world mathematics power, says professor Raghunathan
http://www.hindu.com/2010/04/01/stories ... 251200.htm

Rumors is that Kiran Kedlaya is in for Fields Medal. (Guys you heard it here first!). He is under 40, and is an invited speaker. People may recall I had talked about him in our math thread. He has won many IMO medals ... for those who don't know him, he is at present a prof at MIT.
(Of course, It is a guarded secret, and no one is suppose to know till the announcement is made)

Other I think may be Ngo or may be Lurie or Manjul Bhargava.. Lets us wait and see.



Ramana, also many aspects which you quoted in your post have been discussed in BRF math dhaga by me many times...

for example
In one of these manuscripts, he discovered a "generalization [similar to the above math] developed in the year 628 by the great Indian mathematician Brahmagupta" that stated:

If two numbers that are each the sum of a perfect square and a given whole number times a perfect square are multiplied together, the product will again be the sum of a perfect square and that whole number times another perfect square.


I have discussed it many times here, and actually presented simple proofs and problems..

If I may say so this BR Math Corner has lot of gems... more than people realize :)

Also some may have mentioned it, but apart from a first rate mathematician, Manjul Bhargava is also known for his novel use of computers for mathematics exploration.. (Much like astronomers use telescope to help theoretical astrophysicists)

****

BTW this is the first time a woman won the fields medal..Maryam Mirzakhani from Stanford...(many guessed this correctly).

ramana
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Re: BR Maths Corner-1

Postby ramana » 14 Aug 2014 03:04

I know AmberG. I didn't want to burden you to explain Manjul's work. But if you want to please do so.
I found the Hindu's take dry.

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Re: BR Maths Corner-1

Postby Vayutuvan » 14 Aug 2014 06:30

There are several problems in VLSI that can be formulated as MILP (Mixed Linear Integer Programming) problems. LP is in P due ti the family of Ellipsoid algorithms which include Karmarkar's for LP and P.M. Vaidya's algorithms for LP and convex programs with the convex programs requiring an efficient oracle to calculate gradient or equivalently separating hyperplanes. ILP is NP complete which makes MILP NP-hard (NP complete?).

ILP is an optimization problem on an integer n dimensional lattice with constraints where n is the number of variables in the ILP. In an MILP, n is the number of integral variables.

Rubik's cube (of k-dimensions if one wishes) can be posed as an ILP.

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Re: BR Maths Corner-1

Postby Vayutuvan » 14 Aug 2014 08:13

AmberG ji: If I remember right, you asked me what "integral" means many moons ago. That missed my mind at that time but here is a short blurb from wikipedia.

Integral linear programs[edit]
A linear program in real variables is said to be integral if it has at least one optimal solution which is integral. Likewise, a polyhedron is said to be integral if for all bounded feasible objective functions c, the linear program has an optimum with integer coordinates. As observed by Edmonds and Giles in 1977, one can equivalently say that the polyhedron is integral if for every bounded feasible integral objective function c, the optimal value of the linear program is an integer.


That is generally accepted definition which is also used in min cut-max flow problems where it has been proved that if all lower and uppper bounds are integral and the cost function is integral, the optimal solution is integral. This problem is in P. OTH, it also has special structure, i.e. the constraints are of non-general form.

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Re: BR Maths Corner-1

Postby Eric Demopheles » 16 Aug 2014 10:35

Mirzakhani's and Avila's names were mentioned as possibilities prior to the last Fields Medal announcements too; but they did not win then.

Bhargava's Ph. D. was in the news for contributions to Gauss's composition laws, was somewhat silent for a few years, and then his subsequent newsbreaking major work started with Arul Shankar etc., and he was in the spotlight again, and (deservedly) got the Fields Medal now.

For a long time, almost all Fields Medalists were western. When I had this discussion once with an oriental(not Indian) mathematician a long time ago, he opined that this is because the judges were all from the West and that this will change in the future.

Now it has changed, and among the four medals this year, a majority is from sections/nations that were previous unrepresented, one for a Canadian Indian, one for an Iranian American and one for a Latin American. This is a most pleasing result.

Two other names that were around for years are Jacob Lurie's and Peter Scholze's. Among these, Lurie's Ph. D. thesis created a lot of interest from 2004 onwards and Scholze's undergraduate and graduate theses a few years ago were highly acclaimed.

Regarding Mirzakhani, she is the first woman and first Iranian to win. Iran had a good mathematical community for some time, comparable to India. Regarding the first woman to win the medal, Sophie Morel's name was also mentioned, but she did not win and Mirzakhani did.

There are several mathematicians of Indian origin, and some from India itself(but now abroad) of 20-40 age who are good and promising; but whether they will win a medal in future remains to be seen. We can say that future is not bad, although there is still much more to go.

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Re: BR Maths Corner-1

Postby Neshant » 18 Aug 2014 02:01

just try doing this in your head WITHOUT any paper :
you have 1000 coins facing up (heads)
you turn every other coin (so now 2, 4, 6..etc are tails)
then you turn every third coin down (so coin #6 is back to being heads)

how many heads are there in total?


This is supposed to be one of the supposedly "simple" questions math professors ask during an interview for students wishing to get into Cambridge Mathematics. Its to screen out candidates who can't think on their feet.

I tried doing this in my head and the answer i came up with is 333 heads. Is that correct?

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Re: BR Maths Corner-1

Postby Amber G. » 18 Aug 2014 06:41

you have 1000 coins facing up (heads)
you turn every other coin (so now 2, 4, 6..etc are tails)
then you turn every third coin down (so coin #6 is back to being heads)

how many heads are there in total?


A slightly more interesting question ... if you modify this to ...
Now turn every four coin down..
.. Now turn every fifth coin down...
and so on ..

Edited later: I removed the solution..so that people can have fun :)

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Re: BR Maths Corner-1

Postby Rahul Mehta » 18 Aug 2014 08:48

Amber G. wrote:
you have 1000 coins facing up (heads)
you turn every other coin (so now 2, 4, 6..etc are tails)
then you turn every third coin down (so coin #6 is back to being heads)

how many heads are there in total?


A slightly more interesting question ... if you modify this to ...
Now turn every four coin down..
.. Now turn every fifth coin down...
and so on ..

Edited later: I removed the solution..so that people can have fun :)


If I read correctly,

C1, C2, C3 ... C1000 are coins

In 1st round, flip coins numbered C(1i) i.e all coins
In 2nd round, flip coins C(2i) i.e. C2, C4, C6 ...
In 2rd round, flip coins numbered C(3i) i.e. C3, C6, C9 ...
...
in kth round , flip coin numbered C(ki) i.e. Ck, C(2k), ....

perfect squares will be all heads and rest will be all tails OR other ways. i.e. perfect squares will be become different from rest

Moral = when luck screws all in turns , the perfect ones become different from rest

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Re: BR Maths Corner-1

Postby ArmenT » 18 Aug 2014 11:42

Pretty easy to work out the solution.
Mega hint
Every number has even # of factors, except for square numbers. E.g. 6 has factors 1, 2, 3, 6; 7 has factors 1 and 7, whereas 25 has factors 1, 5, 25.

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Re: BR Maths Corner-1

Postby Neshant » 19 Aug 2014 03:10

yea so in your version of the question, the perfect square numbered coins that will be flipped an odd number of times will be in a new position from their original orientation.

while the rest will be back to their original position having been flipped an even number of times.

But none of the above relates to my original question where there are only 2 flips of C2, C4, C6.. and C3, C6, C9...

What's the numeric answer to my original question.

Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 19 Aug 2014 07:59

For the original question (Read only if interested)
you have 1000 coins facing up (heads)
you turn every other coin (so now 2, 4, 6..etc are tails)
then you turn every third coin down (so coin #6 is back to being heads)

how many heads are there in total?

Number of coins which were flipped (once or twice)
=1000/2+[1000/3]-[1000/6] (No double counting for 6th)
Number of coins which did not toggled at all
= 1000 - 1000/2-[1000/3]+[1000/6]
= 1000-500-333+166 = 333 (<==== This number of coins had NO flips)
We also have [1000/6] (=166) even number of flips...
So number remained in original position = 333+166 = 499
(The number is approx N/2)

HTH

Neshant
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Re: BR Maths Corner-1

Postby Neshant » 19 Aug 2014 18:27

499 does not sound correct to me.

It cannot possibly be roughly the same number as flipping every other coin which is 1000 / 2 = 500 heads.

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Re: BR Maths Corner-1

Postby Amber G. » 19 Aug 2014 18:56

Neshant wrote:499 does not sound correct to me.

It cannot possibly be roughly the same number as flipping every other coin which is 1000 / 2 = 500 heads.

Beautiful thing about math is any one can check the answer, and does not have to guess, and see if the answer is correct or not..:)
Try out 12 coins (or 6 coins) to see what happens..

(For example in 12 coins .. If I understand the problem correctly .. You have heads .., 1,5,6(flipped twice),7,11,12(flipped twice)... in all 6 coins (12/2=6). Correct?

(Actually a moments thought will convince you that, if the problem changed a little, and instead of "every third", every fifth (or every 37th for that matter) coin was flipped the second time.. the answer would still be N/2..:)

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Re: BR Maths Corner-1

Postby Eric Demopheles » 19 Aug 2014 23:32

Amber G. wrote:(For example in 12 coins .. If I understand the problem correctly .. You have heads .., 1,5,6(flipped twice),7,11,12(flipped twice)... in all 6 coins (12/2=6). Correct?



12 coins, flipping:

Flipped zero times: 1
Flipped Once: 2, 3, 5, 7, 11.
Flipped Twice: 4, 9.
Flipped Thrice: 6, 10.
Flipped Five Times: 12.

So flipped an even number of times : 1, 4, 9.

Indeed, these are the perfect squares, as per ArmenT's hint, which can be deduced very easily from the pairing that if a is a factor of an integer N, then N/a is also a factor.

Therefore the answer is the number of perfect squares less than or equal to x, where x is the total number of coins flipped. This is the greatest integer less than or equal to sqrt(x).

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Re: BR Maths Corner-1

Postby Amber G. » 20 Aug 2014 01:19

Eric - I think you are misunderstanding what Neshant asked.. (Or at least what I think he asked :).

There are two different problems.yy

For my version (modified from Neshant's.., and clearly stated). the correct answer was given by ArmentT (or RM)'s hint. Actually I had the solution and exactly the same hint as ArmentT posted right after the problem I posed, but as I said earlierr, I edited that part later. The problem, of course, is simple and quite well known..

(For details please re-read the problem(s))

Neshant's original problem, is different ... Only two rounds are taken here... every second and every third.

Eg: first flip .. coins effected ... 2,4,6,8, ...
Second flip ... coins effected ... 3,6,9,12...

(There is some unclarity of N's problem, as I see it..I assumed it as coins effected are every third..(in the original sequence)... but if he meant something different.. let him clear it .. In anycase the answer is simple once the problem is clearly stated...

HTH

Added later: BTW

Some may find it very interesting...

With the interprettion I described above.. and as I described before..
The answer does NOT change (for large N it is about N/2) if you change every third to any other odd number..(It is easy to understand that)

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Re: BR Maths Corner-1

Postby Eric Demopheles » 20 Aug 2014 06:46

Amber G. wrote:Eric - I think you are misunderstanding what Neshant asked.. (Or at least what I think he asked :).

There are two different problems.yy

For my version (modified from Neshant's.., and clearly stated). the correct answer was given by ArmentT (or RM)'s hint. Actually I had the solution and exactly the same hint as ArmentT posted right after the problem I posed, but as I said earlierr, I edited that part later. The problem, of course, is simple and quite well known..

(For details please re-read the problem(s))

Neshant's original problem, is different ... Only two rounds are taken here... every second and every third.

Eg: first flip .. coins effected ... 2,4,6,8, ...
Second flip ... coins effected ... 3,6,9,12...

(There is some unclarity of N's problem, as I see it..I assumed it as coins effected are every third..(in the original sequence)... but if he meant something different.. let him clear it .. In anycase the answer is simple once the problem is clearly stated...

HTH

Added later: BTW

Some may find it very interesting...

With the interprettion I described above.. and as I described before..
The answer does NOT change (for large N it is about N/2) if you change every third to any other odd number..(It is easy to understand that)


Ah, I see. Sorry for the misunderstanding then. I had thought that multiples of 2 are first flipped, then those of 3, then those of 4, 5, 6, .... and so on.

So, in Neshant's version, only multiples of 2 and 3 are ever flipped..

Then, you are right, because, just like in your computation for 1000, (1 - ( 1/2 + 1/3 - 1/6) ) + 1/6 = 3/6 = 1/2.

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Re: BR Maths Corner-1

Postby Neshant » 20 Aug 2014 08:32

yes its half.

the flipping of every 3rd coin essentially has no effect on the net orientation of the coins.

3 adds to the count while 6 subtracts from it
9 adds to the count while 12 subtracts from it.

the mistake I made was adding all the 3s and 6s but forgetting to subtract the 9s and 12s.

net effect is +1 and -1 essentially cancels itself out.

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Re: BR Maths Corner-1

Postby Saral » 20 Aug 2014 18:47

This is probably an "easy" problem, but one that I would appreciate an answer from the resident gurus.

Consider the n*(n-1)/2 entries in a lower triangular matrix (entries below the main diagonal).
Assume the cells are indexed sequentially in a one dimensional 2-tuple array as follows (example for n=6)

[ (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6) ]

So there are 15 indexed positions, going from 1 to 15. The cells are represented using (x,y) where x < y and 1 <= x < y <=6 in this case.

Question 1: Is there an arithmetic function (no loops please) that converts the ordered 2-tuples to the index.
examples: (5,6) --> 15 and (2,4) --> 7 and so on.

Question 2: Is there a simple inverse arithmetic function (no loops) that would return a 2-tuple, given the index.
examples: 15 --> (5,6) and 7 --> (2,4) and so on.

The functions should work for positive n > 1. Requirement for no loops is that it is a constant time operation.

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Re: BR Maths Corner-1

Postby Eric Demopheles » 20 Aug 2014 19:07

Saral wrote:This is probably an "easy" problem, but one that I would appreciate an answer from the resident gurus.

Consider the n*(n-1)/2 entries in a lower triangular matrix (entries below the main diagonal).
Assume the cells are indexed sequentially in a one dimensional 2-tuple array as follows (example for n=6)

[ (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6) ]

So there are 15 indexed positions, going from 1 to 15. The cells are represented using (x,y) where x < y and 1 <= x < y <=6 in this case.

Question 1: Is there an arithmetic function (no loops please) that converts the ordered 2-tuples to the index.
examples: (5,6) --> 15 and (2,4) --> 7 and so on.

Question 2: Is there a simple inverse arithmetic function (no loops) that would return a 2-tuple, given the index.
examples: 15 --> (5,6) and 7 --> (2,4) and so on.

The functions should work for positive n > 1. Requirement for no loops is that it is a constant time operation.


Here you are using a bit of non-standard terminology; but it does not hurt the solvability issues.

It is not fully clear what you mean by "arithmetic function" and "inverse arithmetic function" in these cases. Is a polynomial function going to be good enough?

If so, you can check out Newton Polynomials. If your background is engineering, you might have studied them during your engineering mathematics classes and Wikipedia will refresh you.

This will give a polynomial function from ordered pairs to single indexes. The other way is a bit more problematic in the general case.

Perhaps matrimc and other more practically experienced guys can chime in.

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Re: BR Maths Corner-1

Postby Saral » 20 Aug 2014 19:27

Eric Demopheles wrote:It is not fully clear what you mean by "arithmetic function" and "inverse arithmetic function" in these cases. Is a polynomial function going to be good enough?

If so, you can check out Newton Polynomials. If your background is engineering, you might have studied them during your engineering mathematics classes and Wikipedia will refresh you.

This will give a polynomial function from ordered pairs to single indexes. The other way is a bit more problematic in the general case.

Perhaps matrimc and other more practically experienced guys can chime in.


Right, Arithmetic was a loose qualifier. I am cool with polynomials so long as they are constant time. Will check Wiki Newton Polynomials. Thanks for the Pointer. I think Rahul Mehta-Ji might also have something useful to say here. Maybe I should post on his SMS thread (ducks for cover).

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Re: BR Maths Corner-1

Postby Amber G. » 20 Aug 2014 21:20

Saral - First, thanks for a very well defined problem (with example so one can understand what is meant).

Second - You have the answer in the strict sense, because, if you can define how to get an answer, the function is well defined.

Third - If you want "explicit" (or in familiar terms) value, again for one moments thought will give you the answer... For example in case of 6, for (x,y) your "function" is something like:

f(x,y) = (11x-x^2)/2 +y - 6

(You can check for x=5,y=6 this becomes => (55-25)/2+6-6 = 15 and
(for (2,4) ==> we have (22-4)/2+4-6 = 7)

(Of course, for a general case instead of 6, the answer is equally easy
It becomes ((2n-1)x-x^2)/2+y-n))

(In case you are wondering how one gets the polynomial - in practical sense - One of the standard way to do this is to recognize that this is quadratic function in x and linear function in y..and plug in the values...)

just try ax^2+bx+c+y for a few (in this case 3) cases and solve for a,b,c :)
Last edited by Amber G. on 21 Aug 2014 08:07, edited 2 times in total.

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Re: BR Maths Corner-1

Postby Saral » 20 Aug 2014 21:29

Amber,

Thanks Saar or is it Maam? What about the inverse function that gives the tuple given n (say 6) and an index (say 7) ? Is this function (which returns a 2-tuple) easily obtained?

Yeah, I got the notion that it has to quadratic in x and linear in y (good observation, Amber-ji). Thanks for the insight. You need to edit the /2 factor in the general formula.
Last edited by Saral on 21 Aug 2014 02:05, edited 1 time in total.

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Re: BR Maths Corner-1

Postby Vayutuvan » 20 Aug 2014 21:43

Saral: are you doing matrix vector multiply (where the matrix is symmetric and diagonal stored separately) or solve with a lower triangular matrix? If so think in terms of vector ops like saxpy or inner product and shifting a pointer. Or do you need to access the entries randomly or need to get the upper part of the symmetric matrix where only the lower part is stored in column major (fortran) like format? Are you testing for symmetry?

In any case unless your need is to truly random access into say a lower matrix (I suppose the diagonals are either in a separate vector or they are some known value usually 1.0) stored in a long vector column major order fortran fashion, most cases where you need Blas 2 or 3 operations can be handled without an explicit formula.

That said it is simple enough to figure out the formula though. I haven't checked but AmberG's formula but probably he is right as usual.

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Re: BR Maths Corner-1

Postby Saral » 20 Aug 2014 21:49

vina wrote:
bhalluka wrote:Subhash Khot wins Nevanlinna prize

http://www.simonsfoundation.org/quanta/ ... mpossible/


Bade Mian! Take That and That! Madrassa B.Techs land a big one here ! Now kneel down and repent and ask for forgiveness for all the bad mouthing you did of those poor lil kids!


He is onlee the second Madrassa B.Tech to get the prize. See http://en.wikipedia.org/wiki/Madhu_Sudan

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Re: BR Maths Corner-1

Postby Vayutuvan » 20 Aug 2014 22:26

Saral: Check AmberG's formula using induction on (i,j) where i is the column and j is the row.
Check for (1,1).
Assume the formula true for (i,j), i<= j;
Check for (i,j+1) (obviously i<= j+1)
Check for (i+1,j) (in case i+1 > j it is in the upper part so you have to check for (j,i+1))

Are you forming a correlation matrix?

(or is it assembly of a finite difference matrix coming out of mesh-less methods? :wink: . If so, I will tell you a better method to do it)
Last edited by Vayutuvan on 20 Aug 2014 23:37, edited 1 time in total.

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Re: BR Maths Corner-1

Postby Amber G. » 20 Aug 2014 22:41

Saral wrote:Amber,

.. What about the inverse function that gives the tuple given n (say 6) and an index (say 7) ? Is this function (which returns a 2-tuple) easily obtained?

Yeah, I got the notion that it has to quadratic in x and linear in y (good observation, Amber-ji). Thanks for the insight.


Yes of course..and I think you have answered your own question, (and am pretty sure, you will not need what I write below as you have figured it out already). "Easy" depends on your definition of "easily"..If one knows how to solve a quadratic equation and take an "integer function" (very much like finding quotient and remainder of an positive integer). Just solve the equation (ignore y) to find "x", take the integer value of "x" and solve for y. Fairly simple.

(Change of coordinates - a standard technique in physics - makes the problem even simpler)

***

Allow me to make a comment -
The "quadratic form" (as in above relationship) (to find integer solution of a general quadratic form and studying some interesting patterns -) and its relationship with permutation group (aka 2x2 Rubiks cube) is what got Manus Bhargava .. this years Fields Medal...:) (Yours truly in 1980's used Rubiks cube to teach some of the finer points of higher Math so to me it was very interesting)

If there is interest I may comment here further, but some may find this news item interesting
Rubik’s Quadratic Forms --
How a toy inspired an award-winning mathematician’s Eureka moment.

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Re: BR Maths Corner-1

Postby Amber G. » 20 Aug 2014 23:57

Neshant wrote:yes its half.

the flipping of every 3rd coin essentially has no effect on the net orientation of the coins.

3 adds to the count while 6 subtracts from it
9 adds to the count while 12 subtracts from it.

the mistake I made was adding all the 3s and 6s but forgetting to subtract the 9s and 12s.

net effect is +1 and -1 essentially cancels itself out.

Yes, one way to think is that multiples of 3 are evenly distributed between odd and even numbers.
(So after initial "every second" flip which leaves half coins in "head" position.. the "flips by third" does not change much.. equal number of heads are turned into tails and vice-ver-sa so net effect is nil)

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Re: BR Maths Corner-1

Postby Saral » 21 Aug 2014 01:33

matrimc, amberg: My application is very humble onlee, just finding the easy and obvious way to index a lower half correlation matrix (using 1d array). Re meshless my birather is supposedly in this area and you may know of him (herr professor at uc davis) and I have no use for these things yet.

amberg: While I will try to code the inverse function, could you take me through the case of index=7, n=6 to get (x, y) to be (2, 4) ?

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Re: BR Maths Corner-1

Postby Amber G. » 21 Aug 2014 05:11

Saral wrote:matrimc, amberg: My application is very humble onlee, just finding the easy and obvious way to index a lower half correlation matrix (using 1d array). Re meshless my birather is supposedly in this area and you may know of him (herr professor at uc davis) and I have no use for these things yet.

amberg: While I will try to code the inverse function, could you take me through the case of index=7, n=6 to get (x, y) to be (2, 4) ?


Saral - I am also trying to be "humble only" :) ...(Matrims's ..."vectors matrix multiplications" and other impressive sounding names confuses me too :)... In any case the answer to yours is simple only...

In general case of n (your example was for n=6)
We have f(x,y) = -x^2/2+ (n-1/2)x+y-n gives you the answer (check out, f(5,6)=15 and f(2,4)=7 when n=6)
Inverse function is simple onlee... just solve for x (in terms of "f")
***
How to get this is a STANDARD tool of math..
1. Logic - It is linear function in "y" (obvious .. if you add one to y you have to add 1 to f as long as you keep x the same)
2. It is quadratic in x
3. So assume it is ax^2+bx+c+y

Now you know that f(1,2)=1,
f(2,3) = n
f(3,4)=2n-2
Rest is simple solve by putting these values eg
a+b+c+2=1
4a+2b+c+3=n
9a+3b+c+4=2n-2
and you get the above...

****
What's make if fun (for me.. and those who become interested in Math) is simple changing the "coordinate system" (or change of variables --- or "rotate it 90 degrees.. so to speak..") makes it extremely SIMPLE (That's how I got the answer.. in about 10 seconds without paper or pencil)

Basically you count in REVERSE...
so let g= n(n-1)/2 - f
u = n -x
v = n-y

So instead of saying when n=6, f(5,6)=15 you say ... g(1,0) = 0 (Note that 1=6-5, 0=6-6 and 0=15-15)
And instead of saying f(2,4)=7 you say g(4,2)=8

In this case the formula is MUCH simple...

g = u(u-1)/2 +v
(Beauty of this is, there is NO "n" in the formula!)
While I will try to code the inverse function, could you take me through the case of index=7, n=6 to get (x, y) to be (2, 4) ?

In this case the reverse function is MUCH simpler .. and easy...

u = Integer( sqrt(2*g+1/4)+1/2) or pretty close... (You can prove that (u+1/2) >= sqrt(2g+1/4)>=u-1/2 etc)
So when you want to find the value when f=15, you have g = 15-15=0, so u=(sqrt(1/4)+1/2)=1 ==> x=5 (rest is easy)
Similarly when f=7, we have g=15-7 = 8 so u = Int(sqrt(16.25)+1/2) = 4 so x=2... rest is easy..

(Once x is known, y is trivial to find)

Hope this helps.
(for larger g, u is pretty close to sqrt(2g). )



***

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Re: BR Maths Corner-1

Postby Saral » 21 Aug 2014 07:39

Amber G. Yes, I too had the intuition about reversing the array because N is taken out of the picture. You can visualize it easily enough as the matrix gets bigger the coordinates do not change and N becomes irrelevant as you show. Thanks for working out and typing all of this painstakingly. I remain humbled and awed by your responsiveness and insights.

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Re: BR Maths Corner-1

Postby Saral » 22 Aug 2014 18:34

For some closure here is some Scala psuedo-code that illustrates the two functions. What AmberG provided will make its way into a system.

Code: Select all

function to get linear index position from (x, y) indices,  linearindex(x,y,n)
val u = n - x
val v = n - y
val g = u(u-1)/2 +v
val linearindex = n(n-1)/2 - g

The inverse function is to get the x and y indices from the linear index value and n
Given: m is the linear index and n is the number of variables.
val p = n(n-1)/2 - m
val u = (sqrt(2p+1/4) + 1/2).floor // floor of the double so 4.7 becomes 4 after truncation
val x = n - u
val y = n - p + u(u-1)/2
So the required coordinates for linearindex m with n variables is (x, y).

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Re: BR Maths Corner-1

Postby Amber G. » 26 Aug 2014 19:24

Saral - Nice work.
Also
X-post - Thanks Prem Kumar for this link:
http://www.simonsfoundation.org/quanta/20140812-the-musical-magical-number-theorist/

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Re: BR Maths Corner-1

Postby Eric Demopheles » 28 Aug 2014 09:54

Quanta Magazine usually has good articles on mathematics, suitable for general audience, yet without too much dumbing down. Erica Klarreich is an especially good journalist in this regard.

Unfortunately, for cutting edge stuff, the situation is always like this:

http://simonsingh.net/wp-content/upload ... arris3.jpg

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Re: BR Maths Corner-1

Postby Vayutuvan » 28 Aug 2014 11:05

Timothy Gowers (a past Fields Medalist) on Manjul Bhargava's work. From there one can click on links to his posts on other three Fields medalists as well. He mentioned in one of these blogs that he would blog about Subhash Khot's work which should be interesting as Gowers is working in Algorithms and Complexity as well recently.

ICM2014 — Bhargava laudatio

Do not miss Gowers' blog on the first day of ICM. There is some heated discussion about Church, Religion, etc. with some perceptive comments by a non-mathematician in the comments section.

Gowers also edited (and wrote a large number of chapters) of the encyclopedic "The Princeton Companion to Mathematics" as well as a short book on Mathematics. I did not find the short book too interesting - probably that money will be better spent in aquiring "What is Mathematics" by Courant and Robbins which is a truly great book. Every high school student getting into Mathematics, Science or Engineering should read the book from cover to cover.

HTH

(edited to improve language and spelling etc.)
Last edited by Vayutuvan on 29 Aug 2014 02:26, edited 1 time in total.

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Re: BR Maths Corner-1

Postby Eric Demopheles » 28 Aug 2014 11:48

Terrence Tao's posts are also interesting:

http://terrytao.wordpress.com/2014/08/1 ... irzakhani/

http://terrytao.wordpress.com/2014/08/1 ... griffiths/

Re: Gowers, there are criticisms about him, for example:

http://owl-sowa.blogspot.in/2013/05/abo ... owers.html

Manjul Bhargava getting a Fields Medal is a notable achievement; but it is also not good to get carried away in the enthusiasm. Fields Medals are not always going to the best mathematicians and everyone can tell instances of worthy people who did not get it.

For that matter, Fields Medals always rewards fads. It is difficult to recall the importance of the work of some of the awardees. For example, not many people know about the work of the first Fields Medalist Jesse Douglas. The co-awardee Lars Ahlfors is more famous for his Complex Analysis textbook.

Most people who got the Medal are good mathematicians, though.

Courant and Robbins is a good book although a bit dated. Being partial to the platonic philosophy, ,my own personal philosophy of mathematics differs from the application-oriented approach of Courant. Hardy's "Apology" is more interesting to me.

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Re: BR Maths Corner-1

Postby Amber G. » 29 Aug 2014 01:46

Interesting discussion here...

Also interesting is that I just talked about M. Mehta (a few post above) and Random Matrix theory.. Eric Demopheles link about Tao's post references this..

From: http://terrytao.wordpress.com/2014/08/12/avila-bhargava-hairer-mirzakhani/
Manjul Bhargava produces amazingly beautiful mathematics, though most of it is outside of my own area of expertise. One part of his work that touches on an area of my own interest (namely, random matrix theory) is his ongoing work with many co-authors on modeling (both conjecturally and rigorously) the statistics of various key number-theoretic features of elliptic curves (such as their rank, their Selmer group, or their Tate-Shafarevich groups)

M. Mehta's pioneering work is getting exposure. Tao also proved ((about a year or so ago ) fairly well known Mehta's conjucture. (also known as Mehta-wigner-dyson conjucture - google for this, if interested). M. Mehta (referenced a few posts ago) conjuctured it alsmot 50 years ago and recently lot of progress is being made here. Tao, BTW, is right now one of the expert in Random Matrix theory and has done lot of work... So the techneques devloped by M. Mehta is now finding applications in many fields (Not only nuclear physics but also string theory, number theory (pure math), neuroscience, random matrices are used to model the network of synaptic connections between neurons in the brain !)... Coincidentally I was visiting my old school and had lunch with some old professors and talk about M. mehta came up...these guys are/were great admirers of M. Mehta)...

Also, I find it VERY interesting that some part of what has been discussed in above few blogs have been previously discussed (some times in many posts) in BRF. For example relationship between Fibonachi numbers and Sanskrit shlokas (Manjul Bhargava's interest), Brhamgupta work.. and some aspects of Gauss's work has been discussed (or explained in simpler terms) here.

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Re: BR Maths Corner-1

Postby Vayutuvan » 29 Aug 2014 02:29

EricD: I have not read Hardy's apology but what I read about it is that it is more of an explanation piece rather than the book length exposition which is What is Mathematics?. I will have to read it one of these days. Also I have downloaded the short article "the unreasonable effectiveness of mathematics" and syet to read it :(


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