BR Maths Corner-1

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Amber G.
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Re: BR Maths Corner-1

Post by Amber G. »

Wow! What a list for the Padma Awards

https://padmaawards.gov.in/Content/Padm ... es2023.pdf

I recognize quite a few and really feel happy for well deserved awards - and posted about Prof Dhar and Prof Mahalanabisin in other dhaga's.
I see Prof. S. R. S. Varadhan. Prof. Varadhan is an Abel laureate and a giant among Indian
(and American )mathematicians.

Congratulations!

Another notable recipient is Smt. Sudha Murthy (An rare - honorary DSc from IIT Kanpur). Her husband (Narayan Murthy - Infosys founder - also from IIT Kanpur) is quite famous too.
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Re: BR Maths Corner-1

Post by Amber G. »

Wow! Another mathematician...Sujatha Ramdorai with Pamdma award.
Sujatha Ramdorai is algebraic number theorist ( Iwasawa theory) - Proud moment for both Canada and India. Well deserved. She won ICTP Ramanujan Prize earlier.
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Re: BR Maths Corner-1

Post by ramana »

Nice article on Ramanujan's dreams:

https://kristinposehn.substack.com/p/ra ... dium=email
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Re: BR Maths Corner-1

Post by Amber G. »

Happy Valentine's Day.
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Vayutuvan
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Re: BR Maths Corner-1

Post by Vayutuvan »

Came across this beautiful excerpt from Best writings of Mathematics 2014 which is a collection of writings published in 2013.

https://www.degruyter.com/document/doi/ ... 7-018/html

A pdf can be downloaded from a link given on that web page. John H. Conway is one of my favorite mathematicians - ever.
Extreme Proofs I: The Irrationality of 2
John H. Conway and Joseph Shipman
Mathematicians often ask, “What is the best proof” of something, and
indeed Erdős used to speak of “Proofs from the Book,” meaning, of
course, God’s book. Aigner and Ziegler (1998) have attempted to re-
construct some of this Book.
Here we take a different and more tolerant approach. We shouldn’t
speak of “the best” proof because different people value proofs in differ-
ent ways. Indeed one person’s value might oppose another’s. For exam-
ple, a proof that quotes well- known results from Galois theory is valued
negatively by someone who knows nothing of that theory but positively
by the instructor in a course on Galois theory. Other “values” that have
been proposed include brevity, generality, constructiveness, visuality,
nonvisuality, “surprise,” elementarity, and so on. A single mathemati-
cian may hold more than one of the values dear. Clearly the ordering of
proofs cannot be a total order.
It is enjoyable and instructive to find proofs that are optimal with
respect to one or more such value functions not only because they tend
to be beautiful but also because they are more likely to point to possible
generalizations and applications.
In this respect, we can discard a proof C that uses all the ideas of
shorter proofs A and B because nobody should value it more highly
than both A and B. We model this by putting C on the line segment
AB, and it suggests that we think of proofs of a given result as lying in a
convex region in some kind of space, which in our pictures will be the
Euclidean plane.
...
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Re: BR Maths Corner-1

Post by Vayutuvan »

https://www.youtube.com/watch?v=UOuxo6SA8Uc

Math's pedagogical curse | Grant Sanderson JPBM Award Lecture, JMM 2023
By Grant Sanderson

This is an excellent lecture on visualizing higher math in addition to starting new topics through motivating examples rather than definitions -> lemma statement, proof -> theorem statement, proof -> counter-examples when the conditions are relaxed
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Re: BR Maths Corner-1

Post by Vayutuvan »

https://en.wikipedia.org/wiki/3Blue1Brown

Grant Sanderson runs this project.
Prem Kumar
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Re: BR Maths Corner-1

Post by Prem Kumar »

Love that YouTube channel!
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Re: BR Maths Corner-1

Post by Amber G. »

Congratulations to Professor C.R. Rao on being awarded the International Statistics Prize 2023.
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Re: BR Maths Corner-1

Post by Cyrano »

Fantastic to see such Indian origin women mathematiciennes.

Image

Nalini Anantharaman (born 26 February 1976) is a French mathematician who has won major prizes including the Henri Poincaré Prize in 2012.

Nalini Florence Anantharaman was born in Paris in 1976 to two mathematicians. Her father and her mother are Professors at the University of Orléans. She entered Ecole Normale Supérieure in 1994. She completed her PhD in Paris under the supervision of François Ledrappier in 2000 at Université Pierre et Marie Curie (Paris 6).[2][3]
She became a full Professor, at the University of Paris-Sud, Orsay in 2009 following time out at the University of California in Berkeley in the year before as a Visiting Miller professor. From January to June 2013 she was in Princeton at the Institute for Advanced Study. She is now a Professor at Université de Strasbourg.[2]

In 2012 she won the Henri Poincaré Prize for mathematical physics that she shared with Freeman Dyson, Barry Simon and fellow Frenchwoman Sylvia Serfaty.[4] Anantharaman was included for her work in "quantum chaos, dynamical systems and Schrödinger equation, including a remarkable advance in the problem of quantum unique ergodicity".[5] In 2011 she won the Salem Prize which is awarded for work associated with the Fourier Series. She also took the Grand Prix Jacques Herbrand [fr] from the French Academy of Sciences in 2011.[3][6] In 2015, Nalini Anantharaman was elected to be a member of the Academia Europaea.[7] She was an invited plenary speaker at the 2018 International Congress of Mathematicians.[8]

In 2018, for her work related to “Quantum Chaos”, Anantharaman won the Infosys Prize (in Mathematical Sciences category), one of the highest monetary awards in India that recognize excellence in science and research.[9] In 2020 she received the Nemmers Prize in Mathematics.[10]

Selected writings[edit]
Anantharaman, Nalini (1 September 2008). "Entropy and the localization of eigenfunctions". Annals of Mathematics. Annals of Mathematics. 168 (2): 435–475. doi:10.4007/annals.2008.168.435. ISSN 0003-486X.
Anantharaman, Nalini; Koch, Herbert; Nonnenmacher, Stéphane (2009). "Entropy of Eigenfunctions". New Trends in Mathematical Physics. Dordrecht: Springer Netherlands. arXiv:0704.1564. doi:10.1007/978-90-481-2810-5_1. ISBN 978-90-481-2809-9. S2CID 39219368.
Anantharaman, Nalini; Nonnenmacher, Stéphane (2007). "Half-delocalization of eigenfunctions for the Laplacian on an Anosov manifold". Annales de l'Institut Fourier. Cellule MathDoc/CEDRAM. 57 (7): 2465–2523. doi:10.5802/aif.2340. ISSN 0373-0956. S2CID 41613433

https://en.wikipedia.org/wiki/Nalini_Anantharaman
sudarshan
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Re: BR Maths Corner-1

Post by sudarshan »

For the lurkers:



Amber G.
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Re: BR Maths Corner-1

Post by Amber G. »

I am very happy that Spirit of Ramanujan Fellow (2018) Derek Liu had a perfect score in International Math Olympiad 2023. He won a gold medal last year too.

Congratulations to USA and India for coming it top 10 in International Math Olympiad 2023.

USA IMO Team: (5 Golds, 1 Silver)
Derek Liu -Gold medal - Perfect Score -
Alexander Wang Gold medal
Jeff Lin Gold medal
Eric Shen Gold medal
Alex Zhao Gold medal
Maximus Lu Silver medal

India IMO Team (2 Gold 2 Silver 2 Bronze)

Atul Shatavart Nadig Gold medal
Arjun Gupta Gold medal
Ananda Bhaduri Silver medal
Siddharth Choppara Silver medal
Adhitya Mangudy Ganesh Bronze medal
Archit Manas Bronze medal


Here is a problem for you to try (Not too hard) and similar to some of the problems I posted here earlier:
Image
sudarshan
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Re: BR Maths Corner-1

Post by sudarshan »

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Vayutuvan
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Re: BR Maths Corner-1

Post by Vayutuvan »

Wow. How much time did you spend on this?
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Re: BR Maths Corner-1

Post by sudarshan »

Vayutuvan wrote: 20 Jul 2023 22:03 Wow. How much time did you spend on this?
I played with it in Excel for ten to 15 mins and managed to figure out the basic principle of x_n+2 being the same as x_n+1, if one wanted the a's to be consecutive integers. Then I slept on it, and in the morning I opened up MS Word, and the whole thing wrote itself. So overall - 15 mins of Excel, 15 to 20 mins in Word, and the few hours of sleeping on it in between.

Does the solution look right?
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Re: BR Maths Corner-1

Post by Cyrano »

https://www.quantamagazine.org/two-stud ... -20230810/

NUMBER THEORY
Two Students Unravel a Widely Believed Math Conjecture
By
MAX G. LEVY
August 10, 2023

Mathematicians thought they were on the cusp of proving a conjecture about the ancient structures known as Apollonian circles. But a summer project would lead to its downfall.
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Re: BR Maths Corner-1

Post by Amber G. »

sudarshan wrote: 21 Jul 2023 05:06
Vayutuvan wrote: 20 Jul 2023 22:03 Wow. How much time did you spend on this?
I played with it in Excel for ten to 15 mins and managed to figure out the basic principle of x_n+2 being the same as x_n+1, if one wanted the a's to be consecutive integers. Then I slept on it, and in the morning I opened up MS Word, and the whole thing wrote itself. So overall - 15 mins of Excel, 15 to 20 mins in Word, and the few hours of sleeping on it in between.

Does the solution look right?
Nice. Solution is right. (I think all Indians and US students got it right too. Easy to see a_1 = 1 (= (x_1)(1/x-1) :) ) and AM-GM inequality gives right away that a_n > n ... finding relationship between x_(n+2) and x_n could be little tricky but the fact all x's are different, inequality is not too hard to find.

----

Just for fun, here is one rather simple problem .. (I posed it to my grandchildren and six year old was able to solve it). May be fun to ask - to encourage out of box thinking.

Four grandchildren - two boys, two girls . While looking for jugnu (fireflies) each caught some. (Obviously a positive integer :)).. the product of Fireflies caught by boys is equal to product of fireflies caught by girls and this is also equal to sum of *all* fireflies.

How many fireflies each grandchild caught?


Intesting (to me) is how one starts to attack such problems which are a little different than what is being taught in the school.
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Re: BR Maths Corner-1

Post by sudarshan »

Let boys be <a,b>, girls be <c,d>.

a*b=c*d=a+b+c+d

General case: a, b, c, d are all different.

Or: a=c, b=d

If a=c and b=d, one has: a*b=2*(a+b)

a*b can be interpreted as the area of the rectangle of sides a and b. Which means the perimeter of the rectangle, being 2*(a+b), has to be equal to the area. This means the number of 1x1 squares in the rectangle is a*b.

The number of 1x1 squares along the perimeter of a rectangle of sides a and b, is 2*(a+b)-4 (since the corner squares are on two sides each). This means, the number of squares along the perimeter of this rectangle, is four less than the total number of squares in the rectangle, i.e., there have to be 4 squares in the interior of the rectangle. These 4 squares have to be arranged in a rectangle, so either 1x4 or 2x2.

If the interior is 1x4, that means the sides are 3 and 6. If the interior is 2x2, the sides are 4 and 4.

So boys=<3,6> and girls=<3,6>

(or)

Boys=<4,4> and girls=<4,4>.

I don't yet have a solution for the general case (a, b, c, d all being different).
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Re: BR Maths Corner-1

Post by sudarshan »

Oh, I see.

If a, b, c, d are all different, then we can think of two different rectangles.

One of size a x b, other of size c x d.

Then we have: a*b=c*d=(a+b+c+d), or (a*b)+(c*d)=2*(a+b+c+d), which means the sum of the areas of both rectangles has to be the sum of perimeters of both the rectangles.

This means that the total number of 1 x 1 squares in the interior of both rectangles, has to be 8. On top of that, a*b has to be equal to c*d, which means both rectangles have to have the same area.

Divide these interior 8 squares between the two rectangles as: <0,8>, <1,7>, <2,6>, <3,5>, or <4,4>.

<0,8> means one rectangle has sides <2 x 2>, the other has <3 x 10> or <4 x 6>. The areas don't match in either case.

<1,7> means one rectangle has sides <3 x 3>, the other has <3 x 9>, and the areas don't match.

<2,6> means one rectangle has sides <3 x 4>, the other has either <3 x 8> or <4 x 5>. The areas don't match in either case.

<3,5> means one rectangle has sides <3 x 5>, the other has <3 x 7>. The areas don't match.

<4,4> is the case that was already covered before: a=c, b=d. So this case leads to the only two solutions:

<3,6> for boys, and <3,6> for girls

(or)

<4,4> each for boys and girls.
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Re: BR Maths Corner-1

Post by Amber G. »

sudarshan wrote: 13 Aug 2023 20:07 Oh, I see.

If a, b, c, d are all different, then we can think of two different rectangles.

<snip>

<4,4> is the case that was already covered before: a=c, b=d. So this case leads to the only two solutions:

<3,6> for boys, and <3,6> for girls

(or)

<4,4> each for boys and girls.
Very Nice.. but how about boys (2,15), girls (3,10) ? :) )
(There may be more :)
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Re: BR Maths Corner-1

Post by sudarshan »

Amber G. wrote: 13 Aug 2023 20:58 Very Nice.. but how about boys (2,15), girls (3,10) ? :) )
(There may be more :)
Yes, I just came here to admit that there was an error in my previous reasoning. This specific jump is not justified:
Then we have: a*b=c*d=(a+b+c+d), or (a*b)+(c*d)=2*(a+b+c+d), which means the sum of the areas of both rectangles has to be the sum of perimeters of both the rectangles.

This means that the total number of 1 x 1 squares in the interior of both rectangles, has to be 8.
The first statement says that the sum of areas of both rectangles has to be the sum of the perimeters of both rectangles.

The second statement says that the area of each rectangle has to be the same as the perimeter of each rectangle. This does not necessarily follow from the first statement (i.e. - there is a loss of generality right here).

Will think about it some more.
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Re: BR Maths Corner-1

Post by sudarshan »

Actually, my <0,8> case above leads to two more solutions:
<0,8> means one rectangle has sides <2 x 2>, the other has <3 x 10> or <4 x 6>. The areas don't match in either case.
This isn't quite right. If one rectangle has 0 interior 1 x 1 squares, then its sides can be <1 x N> or <2 x N>, where N can be any positive integer.

The other rectangle is either <3 x 10> or <4 x 6> which means the first rectangle can be <2 x 15> or <2 x 12>.

So the solutions I have right now are:

Boys........Girls

<2,15>.....<3,10>
<2,12>.....<4,6>
<3,6>......<3,6>
<4,4>.......<4,4>
(or any permutations of the above 4)
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Re: BR Maths Corner-1

Post by Cyrano »

Factorisation problem !
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Re: BR Maths Corner-1

Post by Amber G. »

sudarshan wrote: 13 Aug 2023 22:15 Actually, my <0,8> case above leads to two more solutions:<
So the solutions I have right now are:
<snip>
Boys........Girls

<2,15>.....<3,10>
<2,12>.....<4,6>
<3,6>......<3,6>
<4,4>.......<4,4>
(or any permutations of the above 4)
Yes, all the solutions.
FWIW: Solution for my grandchildren -- It takes more time to write it down, then to get answers logically .. (While chasing jugnus it is difficult to use pen/paper for young boys/girls :)...But using Sudarshn's variables here are the key points.

One gender <a,b>, other be <c,d>.
ab = cd = a+b+c+d
Let us assume (WLOG) a ≤ b ; c ≤ d ; and further a ≤ c
(we can always call the smaller of numbers a , if a=b it does not matter which we call a )
Obviously, if a=c then b=d, and if a<c then b>d in either case we can say a ≤ c ≤ d ≤ b

(IOW NONE is smaller than a and NONE is larger than b (If

So ab = a+b+c+d ≤ b + b +b +b = 4b
so ab ≤ 4b (Since b is positive, not equal to zero we can cancel it out and get
a≤ 4

Rest is super easy!

Equality holds only when a=b=c=d s os the case when a = 4 gives solution a=b=c=d = 4

So, only cases remaining are a=1, a=2, a=3
a=1 gives 1*b = a+b+c+d this gives a+c+d = 0 which is not possible. (a,c,d all are positive integers)
a=2 gives 2*b = a+b+c+d gives b = 2+c+d , so cd = a+b+c+d gives cd = 4 + 2c + 2d which is easy to solve (see note 1)
a = 3 gives a=c=3, b=d=6

Note 1:
Standard way to solve cd = 4 + 2c+2d (Brahmgupta's way) is to write it as:
(c-2)(d-2) = cd -2c -2c + 4 = 8
Factors of 8 are 1*8 and 2*4
c-2 = 1 and d-2 = 8 which gives c=3, d=10
c-2 = 2 dnd d -2 = 4 which gives c=4, d = 6

(This could easily be solved by trial and error or other simple methods too as we have to consider *very* few possibilities, but Brhahmgupta's method is standard one it can solve equation of the type xy = mx +ny in integers for bigger values of m,n )

:)
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Re: BR Maths Corner-1

Post by Amber G. »

Decided to reply to my own post from other dhaga -- as there may be more interest here -
Amber G. wrote: 27 Aug 2023 00:37 While I was listening to PM Modi's speech (Youtube link is here in many of the posts), I was happy that he asked newer generation to study scientific texts like Surya Sidhanta - *and* modern scientific data and how these can be brought into modern scientific language. So just for fun, here is a challenge:

In Surya Sidhant we are given a few different figures of length of Lunar month : (I am giving approximate values in modern (commonly know) units (All in days); (There figures are given, how many cycles it will make in a whole yuga etc.)

1 - 29.53059 days
2 - 27.32158 days
3 - 27.32166 days
4 - 27.55455 days
5 - 27.21222 days

These are the *average( length of different cycles* (all in days) for 'full rotation' to calculate the exact position (naxtra) of the Moon, if you knew a position at one time... or to help you make a panchng of a particular year.

Challenge Q: What these different cycles represent? Why they are (slightly but noticeably) different from each other?
(If you know the answer, wait for a day or so before explaining -- if you did not know the answer before , and figured it out, please do comment)

(These values are pretty close to modern values, BTW)

Don't peek if you want to think about it, but if you wish go ahead..
:
----
Here they are:
Image






The sidereal month (27.3217 d)
- the most intuitive definition of the lunar orbit period (a lunar month) Sidereal is used to reference an inertial frame of reference, “relative to the stars”. Time it takes for the moon to complete one orbit relative to a fixed stars.
Image

Synodic Month (29.5306 days) - Average Value
From the standpoint of an earth viewer, a synodic month is the most intuitive definition. The phase, rather than precise position among stats. As such, the synodic month is the *average* time between new moons .However, any particular phase cycle may vary from the mean by up to seven hours
Image
(There is a typo in the picture, there should be a "-" sign instead of "+")

Tropical Month (27.3216 days)
A tropical month is very minor variation of a sidereal month ..It uses the March equinox as the reference frame rather than the fixed stars. Because the equinoxes slowly precess (-26000 years), the moon arrives at this reference slightly faster .
Image

Anomalistic Month (27.5545 days)
An anomalistic month is easily understood as a variation. Time taken from perigee to perigee. I The line of apsides of the moon’s orbit precesses forward with a period of about 8.85 years, so it takes longer for the moon to reach the same point.
Image
(Typo: Sign is reversed )

Draconitic Month (27.2122 days)
An draconitic month uses nodes (processes in 18.6 years) - intersection of the plane of the moon’s orbit around the earth and the plane of the earth’s orbit around the sun, shorter than a sidereal month. Helpful in predicting eclipses.

Image
(Typo sign is reversed)
Last edited by Amber G. on 01 Sep 2023 10:47, edited 1 time in total.
sudarshan
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Re: BR Maths Corner-1

Post by sudarshan »

Great explanation, thanks. But mathematically, I think three of the signs got reversed, the numbers don't work out:
Amber G. wrote: 31 Aug 2023 23:52 Synodic Month (29.5306 days) - Average Value
Image

Anomalistic Month (27.5545 days)
Image

Draconitic Month (27.2122 days)
Image
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Re: BR Maths Corner-1

Post by Amber G. »

sudarshan wrote: 01 Sep 2023 07:51 Great explanation, thanks. But mathematically, I think three of the signs got reversed, the numbers don't work out:

YES! I Noticed it just after posting -- too lazy to edit the pictures now.. wish BRF allowed LaTex.. (Copied/pasted from an old article, nice pictures but signs are reversed -)

But the errors are obvious (+ sign should become - etc)
I was able to edit the text. Thanks.
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Re: BR Maths Corner-1

Post by Amber G. »

Similar lengths for the year:

Tropical : 365.24212
Sidereal : 365.25636
Anomalistic: 365.25964
Eclipse: 346.62003
(Values are obvious - google the terms)
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Re: BR Maths Corner-1

Post by sudarshan »

Question for knowledgeable folks:

Can one do a FFT (or equivalent transform) on a signal sampled on a logarithmic grid (if that's the right term?)

I mean - instead of the grid representing points at (for example): 1, 2, 3, 4..., it instead represents points at (for example) 1, 2, 4, 8....
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Re: BR Maths Corner-1

Post by Amber G. »

Sharing this nice interview with Prof. Apoorva Khare, a winner of this year's Bhatnagar Prize for Mathematics.

Congrats to Prof Khare. Recommend to read the article.
Today, mathematics is not only necessary in daily life but pervasive
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Re: BR Maths Corner-1

Post by Amber G. »

Some tidbits:
-The 2024 SASTRA Ramanujan Conference was during Dec 20-22 in Kumbakonam, Ramanujan’s hometown. For this conference Prof. (Ms). Yunqing Tang (2022 SASTRA Ramanujan Prize Winner), and Prof. Ruixiang Zhang (2023 SASTRA Ramanujan Prize Winner), both from University of California, Berkeley, were present -- along with may other speakers.

- Some IITK classmates (Graduating in late 60's - Physics & Mathematics - MS and PhD) got together after nearly 55 years gap (from US and India) and spend some quality time and also decided to make pilgrimage to Kumbakonam .
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Re: BR Maths Corner-1

Post by Amber G. »

Sharing: Glad to see, some of the people ( and friends) I have discussed in this dhaga are elected an Honorary Fellow of the Indian Academy of Sciences.

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Re: BR Maths Corner-1

Post by Vayutuvan »

Very nice. Ken Ono is a Ramanujan expert among other things.
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Re: BR Maths Corner-1

Post by Prem Kumar »

https://techcrunch.com/2024/01/17/deepm ... -problems/

👆This is Huge! DeepMind's AI program can solve IMO Geometry problems at a Gold medalist level!!
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Re: BR Maths Corner-1

Post by Amber G. »

Comments on <this> from other thread here:
___

^^^ For those who are really interested in the subject -- please read. Ignore otherwise - zero interest in pointless debate.

As they say: Without logic, mere quoting some 'sastras' is silly..
यस्य नास्ति स्वयं प्रज्ञा, शास्त्रं तस्य करोति किं।
लोचनाभ्याम विहीनस्य, दर्पण:किं करिष्यति
Rough translation: A person who does not have his own prudence/logic - ethology is of no use, just as mirror is useless for a blind person.

To put "दीर्घचतुरस्रस्याक्ष्णया रज्जुः पार्श्वमानी तिर्यग् मानी च यत् पृथग् भूते कुरूतस्तदुभयं करोति ॥" in context from someone who can understand Sanskrit, and has taught Math..

Here is the translation:
"The diagonal of an oblong produces by itself both the areas which the two sides of the oblong produce separately.
Apart from using more precise word like "आयत" (Rectangle) one uses a sloppy word meaning 'oblong' ... yet as most interpretation like it is 'equivalent' to ' Pythagorus theorem' on the basis:

-- The diagonal and sides referred to are those of a rectangle (oblong), and the areas are those of the squares having these line segments as their sides. Since the diagonal of a rectangle is the hypotenuse of the right triangle formed by two adjacent sides, the statement is seen to be equivalent to the Pythagorean theorem"

-- There are more such imprecise shlokas-- (eg "The cord which is stretched across a square produces an area double the size of the original square"

There is no 'proof' given..
Even worse some of the sholkas from the *same* place (Baudhāyana ) gives: (easy to check - easy to check the originals - see for example: https://mathshistory.st-andrews.ac.uk/ (for archives)
- Absurd value of pi
- Absurd value of square root 2: (= 577/408 - exactly)

Note that there are good/respected Indian math and sources -- but to quote such - and especially pick these is IMO silly.
The emphasis on rectangles and squares; this arises from the need to specify yajña bhūmikās—i.e. the altar on which rituals were conducted, including fire offerings were of practical values to present them as proof of xyz theorem is silly.
Note that Pythagorean Theorem-- rules was known *much* before Pythagorean - to a number of ancient civilizations, Not only in India but also the Greek and the Chinese, and was recorded as far back as 2000 BCE.
Amber G.
BRF Oldie
Posts: 9265
Joined: 17 Dec 2002 12:31
Location: Ohio, USA

Re: BR Maths Corner-1

Post by Amber G. »

Prem Kumar wrote: 18 Jan 2024 11:40 https://techcrunch.com/2024/01/17/deepm ... -problems/

👆This is Huge! DeepMind's AI program can solve IMO Geometry problems at a Gold medalist level!!
Thanks. Will be interesting to see.
FWIW: I have played a little with chatGPT (and google's, MS version's version etc)..despite lot of claims, at present most of (the one easily available to us) them do very poorly for math related questions - even those types which I have posted in BRF. Things are going to change but IMO I think it will still take some time - (Of course my opinion only :))
Amber G.
BRF Oldie
Posts: 9265
Joined: 17 Dec 2002 12:31
Location: Ohio, USA

Re: BR Maths Corner-1

Post by Amber G. »

From Nature:Mathematician who tamed randomness wins Abel Prize
exceprts:
A mathematician who developed formulas to make random processes more predictable and helped to solve an iconic model of complex phenomena has won the 2024 Abel Prize, one of the field’s most coveted awards. Michel Talagrand received the prize for his “contributions to probability theory and functional analysis, with outstanding applications in mathematical physics and statistics”, the Norwegian Academy of Science and Letters in Oslo announced on 20 March.
At CNRS, there was/were (a few) Indian Theoretical Physicists/mathematicians (for example working in Random Matrix theory) worked with him. Also:
Talagrand’s journey to becoming a top researcher was unconventional. Born in Béziers, France, in 1952, he lost vision in his right eye at age five because of a genetic predisposition to detachment of the retina. Although while growing up in Lyon he was a voracious reader of popular science magazines, he struggled at school, particularly with the complex rules of French spelling. “I never really made peace with orthography,” he told an interviewer in 2019.

His turning point came at age 15, when he received emergency treatment for another retinal detachment, this time in his left eye. He had to miss almost an entire year of school. The terrifying experience of nearly losing his sight — and his father’s efforts to keep his mind busy while his eyes were bandaged — gave Talagrand a renewed focus. He became a highly motivated student after his recovery, and began to excel in national maths competitions.
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