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Hello Everyone,
A warm welcome back to the Bharat Rakshak Forum.
Important Notice: Due to a corruption in the BR forum database we regret to announce that data records relating to some of our registered users have been lost. We estimate approx. 500 user details are deleted.
To ease the process of recreating the user IDs we request members that have previously posted on the BR forums to recognise and identify their posts, once the posts are identified please contact the BRF moderator team by emailing BRF Mod Team with your post details.
The mod team will be able to update your username, email etc. so that the user history can be maintained.
Unfortunately for members that have never posted or have had all their posts deleted i.e. users that have 0 posts, we will be unable to recreate your account hence we request that you reregister again.
We apologise for any inconvenience caused and thank you for your understanding.
Regards,
Seetal
BR Maths Corner1
Re: BR Maths Corner1
Yes, a=1, b=18 worked for me. So that means any multiples will also work  a=2, b=36; a=3, b=54; etc. (Of course not a=7, b=126, since that would break the condition that neither a nor b is divisible by 7).
What I did was to expand (a+b)^7 binomially, subtract out a^7+b^7, end up with:
7*a*b*(a^5+3*a^4*b+5*a^3*b^2+5*a^2*b^3+3*a*b^4+b^5)
I completed the fifth power, to end up with:
7*a*b*((a+b)^5  2*a^4*b  5*a^3*b^2  5*a^2*b^3  2*a*b^4)
which is:
7*a*b*(a+b)*((a+b)^4  a*b*(2*(a+b)^2a*b))
Simplifying and factoring gives:
7*a*b*(a+b)*(a^2+ab+b^2)^2
This should be a multiple of 7^7, which means (a^2+ab+b^2) has to be a multiple of 7^3. I just tried b=18 (because it's the largest square below 7^3), and the other number turned out to be 1.
But there could be other combinations of numbers which also satisfy (a^2+ab+b^2) being a multiple of 343 (7^3). How do you eliminate that possibility (if at all)?
What I did was to expand (a+b)^7 binomially, subtract out a^7+b^7, end up with:
7*a*b*(a^5+3*a^4*b+5*a^3*b^2+5*a^2*b^3+3*a*b^4+b^5)
I completed the fifth power, to end up with:
7*a*b*((a+b)^5  2*a^4*b  5*a^3*b^2  5*a^2*b^3  2*a*b^4)
which is:
7*a*b*(a+b)*((a+b)^4  a*b*(2*(a+b)^2a*b))
Simplifying and factoring gives:
7*a*b*(a+b)*(a^2+ab+b^2)^2
This should be a multiple of 7^7, which means (a^2+ab+b^2) has to be a multiple of 7^3. I just tried b=18 (because it's the largest square below 7^3), and the other number turned out to be 1.
But there could be other combinations of numbers which also satisfy (a^2+ab+b^2) being a multiple of 343 (7^3). How do you eliminate that possibility (if at all)?
Re: BR Maths Corner1
Good work. Few comments:
yes, (a+b)^7a^7b^7 = 7ab(a+b)(a^2+ab+b^2)^2
(Pattern, (a+b)^2a^2b^2=2ab; (a+b)^3a^3b^3=3ab(a+b);
(a+b)^5a^5b^5=5ab(a+b)(a^2+ab+b^2) ityadi..)
In fact for any prime p (>3) ; (a+b)^pa^pb^p has p, a,b,(a+b)(a^2+ab+b^2) as factors.
(This is easy to prove, as mentioned in the previous posts, if one substitute a=0 (or b=0), or (a+b)=0, or a=bw or a=bw^2 the expression comes out to be zero. Here w is cube root of 1 that is 1+w+w^2 = 0 and w^3=1.
(Thus for example (1+w)^71w^7 = (w^2)^71w^7 (remember 1+w=w^2 from 1+w+w^2=0)
or w^141w^7 = w^21w = 0 (remember w^3=1)) QED.
So factoring is actually very simple.
***
(18,1) works as a solution, as any multiple of those will too. This is same family of solutions.
There is another family (generate infinite for any one given above) which is again VERY easy to generate. Again that is trivial. (Hint: if (18,1) is solution, one get infinite more { Look it ONLY if you want to, It is rather trivial so you may be able to guess it right away } ===
(18, 243+1=244)
***
But there is another family of solution(s). Put in another way, there is/are other family of solution(s) even when b=1, Can one find it/them (without computer perhaps) in a systematic way.
yes, (a+b)^7a^7b^7 = 7ab(a+b)(a^2+ab+b^2)^2
(Pattern, (a+b)^2a^2b^2=2ab; (a+b)^3a^3b^3=3ab(a+b);
(a+b)^5a^5b^5=5ab(a+b)(a^2+ab+b^2) ityadi..)
In fact for any prime p (>3) ; (a+b)^pa^pb^p has p, a,b,(a+b)(a^2+ab+b^2) as factors.
(This is easy to prove, as mentioned in the previous posts, if one substitute a=0 (or b=0), or (a+b)=0, or a=bw or a=bw^2 the expression comes out to be zero. Here w is cube root of 1 that is 1+w+w^2 = 0 and w^3=1.
(Thus for example (1+w)^71w^7 = (w^2)^71w^7 (remember 1+w=w^2 from 1+w+w^2=0)
or w^141w^7 = w^21w = 0 (remember w^3=1)) QED.
So factoring is actually very simple.
***
(18,1) works as a solution, as any multiple of those will too. This is same family of solutions.
There is another family (generate infinite for any one given above) which is again VERY easy to generate. Again that is trivial. (Hint: if (18,1) is solution, one get infinite more { Look it ONLY if you want to, It is rather trivial so you may be able to guess it right away } ===
(18, 243+1=244)
***
But there is another family of solution(s). Put in another way, there is/are other family of solution(s) even when b=1, Can one find it/them (without computer perhaps) in a systematic way.
Re: BR Maths Corner1
sudarshan wrote:
But there could be other combinations of numbers which also satisfy (a^2+ab+b^2) being a multiple of 343 (7^3). How do you eliminate that possibility (if at all)?
As I said before, there indeed is another combination of numbers. If one thinks the "right way" the answer is easy to guess, but it may be not obvious otherwise.
I will give a hint, to see if people like to find the theory by themselves, but I will give one example:
10,180 works (as pointed in S's post  any multiple of (1,18) will work)
but so is (10,153) will (check it out yourself ..
*****
Re: BR Maths Corner1
Some final comments on the above problem..
If we restrict ourselves to b=1, and a<243 we have two solutions, (18,1),(324,1).
(Of course, any multiple of both numbers is a solution, and also one can add 7^3=243 to any number and that will be a solution too.
Note that 324=18^2.
(Note that Hint I gave, (153,10) is same as (3240,10) because if you divide 3240 by 343 remainder is 153)
Basically the problem is to make students excited about modulo math. And just like there are three cube roots of 1 in complex numbers, there are three "cube roots' of 1 modulo 343.
For those who had no chance to study modular math in highschool or college, again any good book or this wiki article may be of interest.
https://en.wikipedia.org/wiki/Modular_arithmetic
One way to find the answer is to check that a^3==b^3 (mod 343) is a solution. You have to find that 18^3 == 1 (mod 343)
Theory here, if you have not read before, is quite interesting, and I will give a hint to read it in any standard higher math book, or this wiki article:
https://en.wikipedia.org/wiki/Euler%27s_theorem
So one way to do it to notice that phi(343)=6*7^2=98, so (anynumber prime to 7)^98 ==1 (mod 343), one can find, for example, 2^98
(which looks hard but really easy to do with hand as 2^10=1024==5 mod 343, 2^20==25, 2^40 ==625 == 61 etc)
Hope some find this of interest.
If we restrict ourselves to b=1, and a<243 we have two solutions, (18,1),(324,1).
(Of course, any multiple of both numbers is a solution, and also one can add 7^3=243 to any number and that will be a solution too.
Note that 324=18^2.
(Note that Hint I gave, (153,10) is same as (3240,10) because if you divide 3240 by 343 remainder is 153)
Basically the problem is to make students excited about modulo math. And just like there are three cube roots of 1 in complex numbers, there are three "cube roots' of 1 modulo 343.
For those who had no chance to study modular math in highschool or college, again any good book or this wiki article may be of interest.
https://en.wikipedia.org/wiki/Modular_arithmetic
One way to find the answer is to check that a^3==b^3 (mod 343) is a solution. You have to find that 18^3 == 1 (mod 343)
Theory here, if you have not read before, is quite interesting, and I will give a hint to read it in any standard higher math book, or this wiki article:
https://en.wikipedia.org/wiki/Euler%27s_theorem
So one way to do it to notice that phi(343)=6*7^2=98, so (anynumber prime to 7)^98 ==1 (mod 343), one can find, for example, 2^98
(which looks hard but really easy to do with hand as 2^10=1024==5 mod 343, 2^20==25, 2^40 ==625 == 61 etc)
Hope some find this of interest.
Re: BR Maths Corner1
dunno where our fizzic dhaaga disappeared.. my searches didn't result in any.
However, I found the GDF fizzic dhaaga. I am taking this oppty in math dhaaga to call all gurus to visit GDF and answer few questions.
TIA
However, I found the GDF fizzic dhaaga. I am taking this oppty in math dhaaga to call all gurus to visit GDF and answer few questions.
TIA
Re: BR Maths Corner1
xpost from Mangalyaan : ISRO's Mars Orbiter Mission thread. Interesting formula from Ramanujan regarding perimeter of an ellipse.
Amber G. wrote:Nowadays one just let a computer calculate the orbits but when we studied these there were not even calculators for us.. (as I tell my son, we used bamboo slide rules and stuffs like that ). The path where there is only one body to worry about (like when MoM is close to earth, or Mars, or in between but far enough from both planets)  aka elliptical orbits with "minor" perturbation are easier  that's what astronomers did using these methods.
One rather interesting tidbit. There is no easy formula to calculate the length (perimeter) of an ellipse. One can use series expansion which is okay for computers but not that convenient when one does by hand. One of my favorite and remarkable formula comes from Ramanujan. He just writes the formula, does not explain it, does not even give a clue how it came to him but is fairly accurate. ( Don't worry for Mangalyaan orbit my approximation is within a few percentage)
The formula is:
perimeter (path) = pi((a+b)+3(ab)^2/(10(a+b)+sqrt(a^2+b^2+14ab))) + small correction.
(See my previous post, a is AM, and b is GM of perigee and apogee)
Remarkably he gives the "correction" = 3ae^20/ 68719476736
As usual Ramanujan never explained his “empirical” method of obtaining this approximation, neither in his published work, nor in his Notebooks.. and kept many mathematician busy..(I still see papers trying to get insight in this approximation and numbers like 68719476736... How does one get that number ???.) (The formula is a final sentence of Ramanujan’s famous paper Modular Equations and Approximations to π, but he does not write anything else, and people who searched his notebooks did not find any clue what what his thinking!!
****
Re: BR Maths Corner1
(Could not see Physics Thread but Abrikosov was a mathematical Physicist too) Sad to see this news
NYTimes  Alexei Abrikosov, Nobel Laureate in Physics, Dies at 88
Happen to know him and learned a lot from him  (was a close friend of one of Math/physics guru) first time met in Delhi. (I Think he met his wife in Delhi around that time)
NYTimes  Alexei Abrikosov, Nobel Laureate in Physics, Dies at 88
Alexei Abrikosov, who shared the Nobel Prize in Physics in 2003 for important insights into how certain materials conduct electricity without resistance, died on Wednesday at his home in Sunnyvale, Calif. He was 88.
Happen to know him and learned a lot from him  (was a close friend of one of Math/physics guru) first time met in Delhi. (I Think he met his wife in Delhi around that time)
Re: BR Maths Corner1
One colleagues kids is preparing for math Olympiad. Can anyone suggest a few online resources for this? Thanks in advance.
Re: BR Maths Corner1
^^^ Are they in US or India? Are they qualified for USAMO or InMO?
Some like a few good online courses/resources (for example WOOT (Worldwide Online Olympiad Training).. Depending on if you are in US or India, best is if you have good/inspiring coach..
***
(Time flies like anything .. Just realized it is about 20 years since I got interested in working with these bright students  started when my eldest son participated and there very few people who were qualified and/or interested, I volunteered.)
If there is a good math person available to them, ask questions and work with old problems. There are books by T. Andreescu, Kiran. Kedlaya, P. Zeitz, Feng etc.. (Just google "IMO problems and solutions"  several books cover all the way from 70's to now) with problems and solutions with some tools/techniques/theory/methods not usually available in regular textbooks).
Actually this dhaga (Math thread), if I may say so, is good resource. Look from very beginning. It has *many* good problems (many did appear in olympiad exams) with fun solutions.
***
Some like a few good online courses/resources (for example WOOT (Worldwide Online Olympiad Training).. Depending on if you are in US or India, best is if you have good/inspiring coach..
***
(Time flies like anything .. Just realized it is about 20 years since I got interested in working with these bright students  started when my eldest son participated and there very few people who were qualified and/or interested, I volunteered.)
If there is a good math person available to them, ask questions and work with old problems. There are books by T. Andreescu, Kiran. Kedlaya, P. Zeitz, Feng etc.. (Just google "IMO problems and solutions"  several books cover all the way from 70's to now) with problems and solutions with some tools/techniques/theory/methods not usually available in regular textbooks).
Actually this dhaga (Math thread), if I may say so, is good resource. Look from very beginning. It has *many* good problems (many did appear in olympiad exams) with fun solutions.
***
Re: BR Maths Corner1
Amber G. wrote:^^^ Are they in US or India? Are they qualified for USAMO or InMO?
Some like a few good online courses/resources (for example WOOT (Worldwide Online Olympiad Training).. Depending on if you are in US or India, best is if you have good/inspiring coach..
***
(Time flies like anything .. Just realized it is about 20 years since I got interested in working with these bright students  started when my eldest son participated and there very few people who were qualified and/or interested, I volunteered.)
If there is a good math person available to them, ask questions and work with old problems. There are books by T. Andreescu, Kiran. Kedlaya, P. Zeitz, Feng etc.. (Just google "IMO problems and solutions"  several books cover all the way from 70's to now) with problems and solutions with some tools/techniques/theory/methods not usually available in regular textbooks).
Actually this dhaga (Math thread), if I may say so, is good resource. Look from very beginning. It has *many* good problems (many did appear in olympiad exams) with fun solutions.
***
Thank you very much Amberji.
They are in the US. I'll pass on your information to my colleague.
Re: BR Maths Corner1
I need some help from you high rolling mathematicians as I'm stumped.
I have 8 points in blue from a set of points that forms the above graph.
4 blue points correspond to the point where the graph from a level position starts dipping.
4 blue points correspond to where the graph starts rising back up and leveling off.
I've kind of drawn the graph with perfectly straight lines.
But the red & blue points are actually more scattered than what's shown. i.e. its a signal
Nevertheless the points when connected with straight lines forms the graph shape shown reasonably well.
Lets call a set of 4 blue points :
X1, Y1
X2, Y2
X3, Y3
X4, Y4
Now here's the question :
How do i figure out the exact span of the dip shown in green. i.e. from edge to edge?
My algorithm can figure out the span to the nearest (X,Y) point on either side of the dip. But that's not accurate enough.
How do i interpolate between those 4 blue points on either side of the dip to figure out the exact span of the dip.
Any better way to determine the span of the dip or any other ideas welcome.
Re: BR Maths Corner1
I am no mathematician by any stretch but it seems to me you can solve this problem by finding the equation of 3 lines and then solving for 2 intersections of these lines and finally finding the distance between these intersection points.
1. find equation of line (L1) the two level points on either side of the the trough ( X1 Y4)
2. Find equation of the dipping line (L2) on left (X2,X3 etc.)
3. Find equation of the dipping line (L3) on right (y2,y3 etc.)
4. find intersection I1 of L1 and L2.
5. find intersection I2 of L1 and L3.
6. Find distance between I1 and I2
Note:
Since points are supposed to be scattered
1. In step 2 you can have 3 lines (X2,X3), (X2,X4), (X3,X4), so you will get 3 intersections with L1
2. Similarly in Step 3 you can have 3 different lines. (Y3,Y2), (Y3,Y1), (Y2,Y1)
3. This approach will give 3X3 solutions, you can pick the smallest/largest which you think is appropriate.
4. In step 2 and 3 you can also solve for parabolas (quadratic) passing through the 3 points if you think
that will be more appropriate. Then find the intersections of the parabolas with the level line. You will get one solution in this case.
O.k. I know this is two simplistic, but I told you I was not a mathematician!
1. find equation of line (L1) the two level points on either side of the the trough ( X1 Y4)
2. Find equation of the dipping line (L2) on left (X2,X3 etc.)
3. Find equation of the dipping line (L3) on right (y2,y3 etc.)
4. find intersection I1 of L1 and L2.
5. find intersection I2 of L1 and L3.
6. Find distance between I1 and I2
Note:
Since points are supposed to be scattered
1. In step 2 you can have 3 lines (X2,X3), (X2,X4), (X3,X4), so you will get 3 intersections with L1
2. Similarly in Step 3 you can have 3 different lines. (Y3,Y2), (Y3,Y1), (Y2,Y1)
3. This approach will give 3X3 solutions, you can pick the smallest/largest which you think is appropriate.
4. In step 2 and 3 you can also solve for parabolas (quadratic) passing through the 3 points if you think
that will be more appropriate. Then find the intersections of the parabolas with the level line. You will get one solution in this case.
O.k. I know this is two simplistic, but I told you I was not a mathematician!
Re: BR Maths Corner1
One can perhaps rephrase the problem as follows: Given ANY two sets E and F, (in your case E is say the set of 4 blue points to the left, F is the set of 4 blue points to the right), what is the HAUSDORFF distance between the two sets E, F? The Hausdorff distance is a measure of the largest separation between two sets. A way/algorithm of computing this distance and examples with pictures can be seen here:
https://en.wikipedia.org/wiki/Hausdorff_distance
This works in any metric space and is the preferred tool in many problems like in Computer vision problems. In effect better estimates of the separation can be made by choosing E, F wisely in your case. For example if you allow some red dots in E say, the Hausdorff distance will go up obviously, so you must exclude them.
https://en.wikipedia.org/wiki/Hausdorff_distance
This works in any metric space and is the preferred tool in many problems like in Computer vision problems. In effect better estimates of the separation can be made by choosing E, F wisely in your case. For example if you allow some red dots in E say, the Hausdorff distance will go up obviously, so you must exclude them.
Re: BR Maths Corner1
Dipanker wrote:I am no mathematician by any stretch but it seems to me you can solve this problem by finding the equation of 3 lines and then solving for 2 intersections of these lines and finally finding the distance between these intersection points.
Sounds like a good idea.
Note:
Since points are supposed to be scattered
1. In step 2 you can have 3 lines (X2,X3), (X2,X4), (X3,X4), so you will get 3 intersections with L1
2. Similarly in Step 3 you can have 3 different lines. (Y3,Y2), (Y3,Y1), (Y2,Y1)
3. This approach will give 3X3 solutions, you can pick the smallest/largest which you think is appropriate.
I was thinking of doing a least squares algorithm to get a "line of best fit" from those dipping points. Ditto for the "level" region of the graph  which is jagged rather than perfectly level as drawn.
4. In step 2 and 3 you can also solve for parabolas (quadratic) passing through the 3 points if you think
that will be more appropriate. Then find the intersections of the parabolas with the level line. You will get one solution in this case.
Yes i had the vague notion of having an arc roughly fit through those points on the level & dip.
And then do some Calculus black magic to get the tangent to the arc.
But I have no clue how to derive the equation of an arc or parabola from x,y cartesian points.
Anyway this method would probably be slower than the method up top.
O.k. I know this is two simplistic, but I told you I was not a mathematician!
That is about the maximum level of math I can digest!
Thanks for your help.
Re: BR Maths Corner1
vsunder wrote:One can perhaps rephrase the problem as follows: Given ANY two sets E and F, (in your case E is say the set of 4 blue points to the left, F is the set of 4 blue points to the right), what is the HAUSDORFF distance between the two sets E, F? The Hausdorff distance is a measure of the largest separation between two sets. A way/algorithm of computing this distance and examples with pictures can be seen here:
https://en.wikipedia.org/wiki/Hausdorff_distance
Thanks, I will look up Hausdorff separation distance algorithm.
It seems mathematicians have solve all kinds of problems and have ready solutions for them.
But 99.9% of us have no idea these solutions even exists.
Re: BR Maths Corner1
Neshant wrote:I was thinking of doing a least squares algorithm to get a "line of best fit" from those dipping points. Ditto for the "level" region of the graph  which is jagged rather than perfectly level as drawn.
I had that in mind too but thought there were not enough points. But theoretically least square fit sounds like better option than my somewhat arbitrary approach.
Yes i had the vague notion of having an arc roughly fit through those points on the level & dip.
And then do some Calculus black magic to get the tangent to the arc.
But I have no clue how to derive the equation of an arc or parabola from x,y cartesian points.
Anyway this method would probably be slower than the method up top.
A second degree polynomial of the form a * x**2 + b * x + c ( parabola ) passing through 3 point can be analytically solved since you have 3 unknown (a,b,c) and you have 3 points giving you 3 equations to solve for a,b,c. In matrix notation this is equivalent to solving for X in AX = B i.e X = Inverse(A) *B.
Matrix A can be set up by having x**2 in first column, x in second column, and 1's in third column. B's are just y coordinates.
For a more general least square curve fitting the equation in matrix form is:
X = Inverse (transpose(A) * A) * transpose(A) * B
Finding equation of a tangent to a curve at any given point is easy as slope at any point on the curve is first derivative of the curve evaluated at that point and then using the point/slope formula for a line.
But in your case how will be you decide the point to evaluate the tangent at ?
Yes, least square curve fitting as opposed to least square line will be computationally somewhat more expensive as it would involve 2 extra matrix multiplication and 1 transpose operation. You can try both, the least square lines and the least square curve to see which gives you better solution.
Of course If you think your problem is akin to finding the Hausdorff_distance then that should give the best answer.
Re: BR Maths Corner1
vnms wrote:Thank you very much Amberji.
They are in the US. I'll pass on your information to my colleague.
You are welcome.
So I assume they are participating in USAMO. Good luck. I believe the test starts tomorrow (4/19 and 4/20). Just curious, how did they do in their AIME.
Keep in mind, if they have qualified for USAMO, it is quite an achievement. (Schools like MIT give quite a bit of weight for scores in AIME  in fact everyone I know who has been qualified for USAMO had no trouble getting into schools like MIT).
For people in US and India, if you know anyone who is bright in Math and Science, let them appear in AMC (in US open to *all* and good high schools regularly participate  but many schools and parents do not know about it) or its equivalent in India. If you qualified you are invited for AIME (RMO in India) and USAMO (InMO in India) which selects the team for IMO.
FWIW: Here are some practice problems
***
Many problems posted by me here in Math dhaga are from (or inspired or based on) such competitions.
Just for fun, let me (re)post two problems from GDF Math dhaga as that problem has appeared in such a contest. (The solutions and/or hinrs are in GDF dhaga so don;t look there). This is a relatively "easy" problem as IMO problems go. (No calculators, computer searches are allowed)
 There are three numbers. All are whole numbers (greater than 0, positive integers). If you multiply any two of them and subtract the third, you get a perfect power of 2. Find all such numbers.
 Factor the number (5^19851) into product of three integers, each of which is greater than 10^60.
Re: BR Maths Corner1
One of the professor who inspired me and many others to learn computers was a math/physics professor (In those days there were no "computer scientists) who was a pioneer to start computer science education in India.
Professor Harry D Huskey, an Honorary Fellow of the Computer Society of India, founder of IIT Kanpur Computer Center, passed away on April 9, 2017, at the age of 101 at Santa Cruz, California. (See wiki entry for him for back ground)
Like many students then my first interaction with computers was 60's in IIT Kanpur. I learned to program in FORTRAN and enjoyed it.
Kanpur Indo American Programme (KIAP) in 60's  the task of assisting India to set up an Institute of Technology at Kanpur. Professor Huskey lead a team to establish a computer centre at IIT/Kanpur. IBM 1620 arrived at IIT Kanpur in 1963. Y. (Prof. Huskey is at bottom right).
Like many good things, KIAP died in 71 or soon after when USA foolishly supported murderers in Bangaladesh. US/India relationship took such a nose drive that even good programs which benefitted all came to an end.
A nice writeup about him is from the computer science museum in Mountain View, California near Google campus. (Link: http://www.computerhistory.org/atchm/harryhuskey2013chmfellow/)
I am glad that many of IITK people were able to meet him recently as he remembered his India days very fondly. Here is a tribute by Dr Rajaraman.
link
Re: BR Maths Corner1
Okay, This was day one of USAMO (and USAJMO) (US Math olympiad and US Junior Math Olympiad for high school and Junior High students.
Since the first day of the contest is over and it is past 7PM local time, we are allowed to discuss problems now.
(For those who are not familiar, there are three problems given each day, time 4.5 hours  no calculators, computers etc)
Also din't I mention that that brf math dhaga is a good resource !... As I said..
If one solved the problem at given at the top of this page one would have an easy time solving prob #2 (which curiously appeared for USAJMO  easy enough for Junior Highschoolers (In my opinion, it is easier than the one discussed on the top of this page having similar ideas )
Here is the problem: Try it for fun 
Since the first day of the contest is over and it is past 7PM local time, we are allowed to discuss problems now.
(For those who are not familiar, there are three problems given each day, time 4.5 hours  no calculators, computers etc)
Also din't I mention that that brf math dhaga is a good resource !... As I said..
Amber G. wrote:
You are welcome.
So I assume they are participating in USAMO. Good luck. I believe the test starts tomorrow (4/19 and 4/20).
***
*** Many problems posted by me here in Math dhaga are from (or inspired or based on) such competitions.
.
If one solved the problem at given at the top of this page one would have an easy time solving prob #2 (which curiously appeared for USAJMO  easy enough for Junior Highschoolers (In my opinion, it is easier than the one discussed on the top of this page having similar ideas )
Here is the problem: Try it for fun 
Consider the equation
[(3x^3+xy^2)(x^2y+3y^3)=(xy)^7]
(a) Prove that there are infinitely many pairs (x,y) of positive integers satisfying the equation.
(b) Describe all pairs (x,y) of positive integers satisfying the equation
Re: BR Maths Corner1
Let me also post the prob #1 which is ought to be even easier
Given a and b are two relatively prime integers (both > 1 ) prove that there are infinite pairs (a,b) such that (a^b+b^a) is divisible by (a+b).
Given a and b are two relatively prime integers (both > 1 ) prove that there are infinite pairs (a,b) such that (a^b+b^a) is divisible by (a+b).
Re: BR Maths Corner1
I was watching a documentary on the computer built by MIT for NASA for the first manned landing on the moon.
Computers of the day were huge and they had the task of shrinking it down into something the size of a briefcase to fit it on board the space craft.
Check it out :
Computers of the day were huge and they had the task of shrinking it down into something the size of a briefcase to fit it on board the space craft.
Check it out :
Re: BR Maths Corner1
I posted two problems from this years USA(J)MO  just for perspective, here is one problem from India's National Math olympiad, this year 
Problem is relatively easy (2nd problem) as InMO problems go,
Given n >= 0 is an integer and all the roots of
x^3 + a x + 4  ( 2 *2016^n = 0)
are integers. Find all possible values of a
Problem is relatively easy (2nd problem) as InMO problems go,
Given n >= 0 is an integer and all the roots of
x^3 + a x + 4  ( 2 *2016^n = 0)
are integers. Find all possible values of a
Re: BR Maths Corner1
vsunder wrote:One can perhaps rephrase the problem as follows: Given ANY two sets E and F, (in your case E is say the set of 4 blue points to the left, F is the set of 4 blue points to the right), what is the HAUSDORFF distance between the two sets E, F? The Hausdorff distance is a measure of the largest separation between two sets. A way/algorithm of computing this distance and examples with pictures can be seen here:
While I think this is the best bet, first Neshant has to come up with a separating plane (in this case a line) that separates the points into two sets. May be he can by eyeballing and drawing a line between those two sets. Any line will do. So a vertical line right in the middle of the trough, for example.
Re: BR Maths Corner1
Vayutuvan wrote:
While I think this is the best bet, first Neshant has to come up with a separating plane (in this case a line) that separates the points into two sets. May be he can by eyeballing and drawing a line between those two sets. Any line will do. So a vertical line right in the middle of the trough, for example.
Given the nature of the problem i.e. presence of a trough in the middle, the partitioning the points into two sets can be done by topologically sorting the points from left to right on x coordinate and then picking two points with min. y coordinates the left one (and all the points preceding it) belonging to the left set and the right one ( and all the points following it) belonging to the right set.
Of course that will be the ideal case assuming a convex bottom! Heuristical treatment will be needed if the bottom turns out to be concave i.e. there are points with higher y coordinates in the middle of two min. y coordinate points. But may be the problem domain is restricted to only convex bottom for the trough?
Re: BR Maths Corner1
Yes these calculations to determine width have to be done at high speeds. It puts a limit on how many floating point operations (i.e. divisions or multiplication of decimal numbers) I can do for each measurement.
Finding the separating plane gets back to the question of where does the right & left set terminate.
What I drew was an idealized picture of the signal. The actual signal is somewhat jagged and the "trough" may look like an inverted (jagged) plateau/mountain of varying depths with overshoots on edges. This is the pitfall of passive rather than active detection. But that's the sensor i have to work with. I can run a smoothing filter on the signal, but all that's going to do is degrade the accuracy further.
On top of that, particles of dust and other goodies flying by might momentarily distort the signal in random ways. My algorithm has to watch for that and reject that signal set.
Some other thoughts i had : Since the separating plane may not be easy to consistently define, I was thinking of collecting a set of 8 signals very rapidly, calculating widths of 8 sets as best i can and then doing some kind of Bell Curve statistical magic on it. The position of the material being detected will not have changed during the time taken to acquire those 8 signal sets.
Due to variation in separating plane identification from one set to another and other stuff, some widths might be slightly longer and others slightly shorter. But hopefully some good old statistics will yield the actual width answer to some degree of accuracy.
Finding the separating plane gets back to the question of where does the right & left set terminate.
What I drew was an idealized picture of the signal. The actual signal is somewhat jagged and the "trough" may look like an inverted (jagged) plateau/mountain of varying depths with overshoots on edges. This is the pitfall of passive rather than active detection. But that's the sensor i have to work with. I can run a smoothing filter on the signal, but all that's going to do is degrade the accuracy further.
On top of that, particles of dust and other goodies flying by might momentarily distort the signal in random ways. My algorithm has to watch for that and reject that signal set.
Some other thoughts i had : Since the separating plane may not be easy to consistently define, I was thinking of collecting a set of 8 signals very rapidly, calculating widths of 8 sets as best i can and then doing some kind of Bell Curve statistical magic on it. The position of the material being detected will not have changed during the time taken to acquire those 8 signal sets.
Due to variation in separating plane identification from one set to another and other stuff, some widths might be slightly longer and others slightly shorter. But hopefully some good old statistics will yield the actual width answer to some degree of accuracy.
Re: BR Maths Corner1
Here is another hack to try! IMO the problem basically boils down to identifying the beginning and the end of the left and right slope of the trough which can be done in linear time and thus will be really fast to execute.
We can scan the points left to right and compute slope S of a pair of points. Now if abs(S) is smaller than a certain THRESHOLD_VALUE that would be mean the points are on the level line and we move over by one point and repeat the procedure. If abs(S) is greater than the THRESHOLD_VALUE then we will assume that the last point does not belong to the level line but to the left slope of the trough. This identifies the point X1 where the left slope starts and point X2 which is on the left slope of the trough.
Similarly for identifying the right slope of the trough, we can repeat the above procedure but scan the points right to left and come up with with points Y1 and Y2 (right to left) similar to X1 and X2 above.
At this stage we need some heuristic to compute the distance, I can think of two approaches. One approach can to simply find the distance between X1 and Y1. Another approach can be to find the distance between the mid point of X1,X2 and midpoint of Y1,Y2. My hunch is this approach should produce acceptable results on filtered data.
We can pick THRESHOLD_VALUE of 1/10 or less which translates to a slope of about 5 degree or less, basically the idea is to pick a value which is suitable for problem at hand.
We can scan the points left to right and compute slope S of a pair of points. Now if abs(S) is smaller than a certain THRESHOLD_VALUE that would be mean the points are on the level line and we move over by one point and repeat the procedure. If abs(S) is greater than the THRESHOLD_VALUE then we will assume that the last point does not belong to the level line but to the left slope of the trough. This identifies the point X1 where the left slope starts and point X2 which is on the left slope of the trough.
Similarly for identifying the right slope of the trough, we can repeat the above procedure but scan the points right to left and come up with with points Y1 and Y2 (right to left) similar to X1 and X2 above.
At this stage we need some heuristic to compute the distance, I can think of two approaches. One approach can to simply find the distance between X1 and Y1. Another approach can be to find the distance between the mid point of X1,X2 and midpoint of Y1,Y2. My hunch is this approach should produce acceptable results on filtered data.
We can pick THRESHOLD_VALUE of 1/10 or less which translates to a slope of about 5 degree or less, basically the idea is to pick a value which is suitable for problem at hand.
Re: BR Maths Corner1
Essentially the above is what I currently have.
In the first pass, I sweep from left to right looking for two consecutive slopes surpassing a certain threshold followed by a flatish region.
In the second pass, I sweep from right to left doing the same.
I compute distance between X1 and Y1 and take an average of a set of 8 of them (without introducing decimal computation).
Tests on my desktop with my sensor prototype and a target I move with my finger some distance away indicates it is able to track slight variations in width down to the millimeter level. I realized I have to also do a tangent (tan) calculation on the width to correct for when the target is off the center axis from the midpoint of the sensor.
In the first pass, I sweep from left to right looking for two consecutive slopes surpassing a certain threshold followed by a flatish region.
In the second pass, I sweep from right to left doing the same.
I compute distance between X1 and Y1 and take an average of a set of 8 of them (without introducing decimal computation).
Tests on my desktop with my sensor prototype and a target I move with my finger some distance away indicates it is able to track slight variations in width down to the millimeter level. I realized I have to also do a tangent (tan) calculation on the width to correct for when the target is off the center axis from the midpoint of the sensor.
Re: BR Maths Corner1
Sounds like you already have a working solution more or less! You could incorporate filtering/smoothing the data before the slope calculations, but it seems you are not in favor of doing that?
My hunch says that if you smooth the data just the right amount by tweaking the filtering parameters ( under smoothing preferred to over smoothing ), you will get better results.
My hunch says that if you smooth the data just the right amount by tweaking the filtering parameters ( under smoothing preferred to over smoothing ), you will get better results.
Re: BR Maths Corner1
Looks like Neshant's problem requires dynamic update of the point set. What kind of compuational resources are available processing? Do you have to do the calcs on an embedded system or you have some reasonable amount of RAM? How many points? If your time frame is like a decisecond (stock market tick happens every 0.1 sec) then you have a really challenging problem on hand. Also you may have to run a clustering algorithm first to clump points together, compute each cluster's geometric center and possibly construct a convex hull of each cluster. This will help in classifying new points that come in. Most new points might fall into already existing clusters. It is easy (IIRC) to detect whether a point is inside or outside a convex polygon. We are talking about a 2D point set, i.e. each point is defined by a pair of real numbers, right? right? If it is a 2D, convex hull computations are fast. I am not a computational geometer so I do not have all the computational complexity results in my head. May be you should look at LEDA library by Kurt Melhorn which is free for research purposes, IIRC. There is another library also from INRIA that is open source. While they have all these algorithms builtin, they are quite complex to use.
Re: BR Maths Corner1
Neshant, look at Computational Geometry Algorithms LIbrary  CGAL
Re: BR Maths Corner1
One more thing I forgot to say. Unless one has lots and lots of data points, one would not be able to smooth away the discrete nature of the data without losing some information. Losing some key information might be the difference between predictions being correct or totally wrong.
One thumb rule of designing asymptotically fast algorithms is to choose those kinds of models which gives one the ability to employ efficient algorithms to solve subproblems. The domain expert who models a physical/abstract problem to be solved has several options while modeling where each option has certain good and bad points and bounds on the accuracy with which the variable values the modeler wants to know.
So it behooves the domain expert and the algorithmicist to work very closely and consult a mathematician when in doubt about the mathematics.
One thumb rule of designing asymptotically fast algorithms is to choose those kinds of models which gives one the ability to employ efficient algorithms to solve subproblems. The domain expert who models a physical/abstract problem to be solved has several options while modeling where each option has certain good and bad points and bounds on the accuracy with which the variable values the modeler wants to know.
So it behooves the domain expert and the algorithmicist to work very closely and consult a mathematician when in doubt about the mathematics.

 BRF Oldie
 Posts: 7630
 Joined: 11 Aug 2016 06:14
Re: BR Maths Corner1
Sorry I could not find the Physics thread (pbuh). This should establish my credentials in both physics and mathematics though. I was copied on this, no doubt a testament to said credentials and fame emanating therefrom.
Your Lecture at the University of Queensland
To
Your Lecture at the University of Queensland
To
denis.mourard@oca.eu;fathi.namouni@oca.eu;nicolas.nardetto@oca.eu;roula.nassif@oca.eu;gilles.niccolini@unice.fr;alain.noullez@oca.eu;pascal.oberti@oca.eu;omasahiro@oca.eu;christophe.ordenovic@oca.eu;sebastien.ottogalli@oca.eu;jose.pacheco@oca.eu;pahlevan@oca.eu;thierry.passot@oca.eu;cpere@oca.eu;romain.petrov@unice.fr;bernard.pichon@oca.eu;yannick.ponty@oca.eu;olivier.preis@oca.eu;janine.provost@oca.eu;ulan.batori@chenghishkanu.ac.mg; yves.rabbia@oca.eu;dbewsher@uclan.ac.uk;bkgibson@uclan.ac.uk;dwkurtz@uclan.ac.uk;dward.thompson@uclan.ac.uk;rwwalsh@uclan.ac.uk;dbewsher@uclan.ac.uk;jbibby2@uclan.ac.uk;kbowman@uclan.ac.uk;dsbrown@uclan.ac.uk;ibutchart@uclan.ac.uk;tvcawthorne@uclan.ac.uk;rgclowes@uclan.ac.uk;sdalla@uclan.ac.uk;vpdebattista@uclan.ac.uk;spseyres@uclan.ac.uk;bjmhassall@uclan.ac.uk;ckestner@uclan.ac.uk;andrew.iskauskas@durham.ac.uk;s.m.fearn@durham.ac.uk;peter.bowcock@durham.ac.uk;t.s.winyard@durham.ac.uk
Professor David Reitze,
Director of LIGO,
Dear Sir,
You will recall that I made a number of points directly to you, from the audience, after your public lecture on LIGO's alleged detection of black holes and gravitational waves, at the University of Queensland on the 6th December 2016. It was very clear to me that you did not understand what I said to you. I must therefore put the arguments in writing.
1. The LIGOVirgo Collaboration asserted in its 'discovery' paper that the speed of propagation of Einstein's gravitational waves is the speed of light, and attributed the theoretical discovery of black holes to Karl Schwarzschild. Both claims are demonstrably false. The speed of propagation of Einstein's gravitational waves is coordinate dependent. This is a mathematical issue pertaining to the derivation of a wave equation, from the linearised form you displayed in one of your slides. The mathematical proof that the speed of propagation is arbitrary, subject to choice of coordinates, is in the Appendix of this paper:
Crothers, S.J., A Critical Analysis of LIGO's Recent Detection of Gravitational Waves Caused by Merging Black Holes, Hadronic Journal, n.3, Vol. 39, 2016, pp.271302, http://vixra.org/pdf/1603.0127v4.pdf
Karl Schwarzschild did not breathe a word about black holes, and his solution precludes them. As I said to you, this is verified by reading Schwarzschild's paper. By your remarks at your lecture, you are obviously ignorant of Schwarzschild's paper. Here is Schwarzschild's paper: {I love the respectful discourse between pisskists!}
http://www.sjcrothers.plasmaresources.c ... schild.pdf
2. General Relativity cannot localise its gravitational energy. Consequently all talk of Einstein's gravitational waves is meaningless (see the paper above).
3. General Relativity violates the usual conservation of energy and momentum for a closed system and is thereby in conflict with a vast array of experiments (see the above paper). Einstein's attempt to satisfy the usual conservation laws fails because his pseudotensor produces, by contraction, a firstorder intrinsic differential invariant. The pure mathematicians proved in 1900 that firstorder intrinsic differential invariants do not exist. And how many experiments are sufficient to invalidate a theory? One will do.
4. The mathematical theory of black holes contains a latent violation of the rules of pure mathematics. It requires, for example, that the square of a real number must take on negative values. Equivalently, it requires that the length of a hypotenuse take on negative lengths, in violation of the Theorem of Pythagoras. Consequently the mathematical theory of black holes is false. The proof is in the paper above.
5. The two 4 K loads of the Low Frequency Instrument (LFI) of the Planck satellite were attached to the shield of the High Frequency Instrument (HFI) by means of metal washers and screws. The shield was cooled to 4 K. Although this attachment ensured that the 4 K loads were at 4 K, the metal connexions produced conduction paths from the loads to the shield. Consequently the 4 K loads did not operate as blackbody emission sources at 4 K. There is no blackbody when conduction is present. Since heat was shunted from the loads into the shield by conduction, the 4 K loads emitted negligible or no photons. The signal from each of the two sky horns of the LFI were subtracted from the two reference horns for the 4 K loads respectively. Since the loads did not operate as blackbody sources, owing to conduction, they effectively presented at ~0 K to the reference horns. The Planck Team reported better than expected response from the LFI. The only means by which this could have been achieved is that the sky is also at ~0 K. This means that there is no monopole signal at L2, and hence, no anisotropies. This also proves that the socalled 'CMB' is not of cosmic origin and therefore that Big bang cosmology is false.
6. The COBE satellite reported a very strong monopole signal from an altitude of ~900km. The COBE shield was incapable of protecting the satellite's detectors from microwave diffraction over the shield, owing to its inadequate design. Water is a good absorber of microwaves, as microwave ovens in the home and submarines at sea prove. A good absorber is also a good emitter, and at the same frequencies. Approximately 70% of the surface of Earth is covered by water. This water is not microwave silent. Microwaves are emitted by water via the hydrogen bond. This emission from the oceans is scattered by the atmosphere to produce isotropy. The scattered microwave emissions diffracted over the COBE shield right into its detectors, no matter which direction COBE pointed its sky horn. COBE reported a monopole signal at 2.725 K. COBE detected the microwave emission from the oceans. The temperature of the oceans is not 3 K. There are two bonds in water: (a) the hydroxyl bond, (b) the hydrogen bond. Energy bound within the hydroxyl bond is not available to microwave emission. Only the much lower energy in the hydrogen bonds is available to microwave emission. The temperature extracted from the scattered hydrogen bond emission spectrum in the atmosphere reports 3 K. This is an example of the fact that Kirchhoff's Law of Thermal Emission is false and that Planck's equation for thermal spectra is not universal. Any temperature extracted from a thermal spectrum that is not from a true blackbody, such as soot, is only an apparent temperature, not the true temperature of the emitter. The walls of an isolated arbitrary cavity at thermal equilibrium always contain energy that is not available to thermal emission, unless the cavity is made from a black material such as carbon. Kirchhoff and Planck however, in their theorising, and contrary to experimental facts known even in their time, incorrectly permitted all the energy of the walls of an arbitrary cavity at thermal equilibrium to be available to thermal emission. In doing so they made all cavities black. Cavities are not always black. The nature of the walls cannot be ignored, contrary to Kirchhoff and Planck. Physical proof of this fact is at hand. NMR and MRI are thermal processes, facilitated by spinlattice relaxation. This means that there is energy in the walls of an arbitrary cavity that is not available to thermal emission. If Kirchhoff and Planck were right, then NMR and MRI could not exist. The clinical existence of MRI proves that Kirchhoff and Planck are wrong. LIGO did not detect black holes or gravitational waves; and it never will. I refer you to the following papers:
Robitaille P.M., WMAP: A Radiological Analysis, Progress in Physics, v.1, pp.318, (2007), http://vixra.org/pdf/1310.0121v1.pdf
Robitaille P.M., COBE: A Radiological Analysis, Progress in Physics, v.4, pp.1742, (2009), http://vixra.org/pdf/1310.0125v1.pdf
Robitaille P.M., Crothers S. J. "The Theory of Heat Radiation" Revisited: A Commentary on the Validity of Kirchhoff's Law of Thermal Emission and Max Planck's Claim of Universality, Progress in Physics, v. 11, p.120132, (2015),
http://vixra.org/pdf/1502.0007v2.pdf
I look forward to receiving your considerations on these issues.
Yours faithfully,
Stephen J. Crothers.
11th December 2016
Re: BR Maths Corner1
Mirzakhani, math superstar from Iran (a Prof at Stanford at present) died.
I mentioned her here in Brf when she won the Fields Medal along with India's Manjul Bhargava.
(She won IMO gold medals  first Iranian women to do so  She won Fields Medal most recently (2014)  First Women to get this prize, also the first from Iran)
Iranian math genius Mirzakhani passes away
Interestingly Mirzakhani is perhaps famous because when Hassan Rouhani tweeted congratulations to her (being the first Iranian to win the prize) he also used a photo of her without a hijab!
Some may like : A Tenacious Explorer of Abstract Surfaces
I mentioned her here in Brf when she won the Fields Medal along with India's Manjul Bhargava.
(She won IMO gold medals  first Iranian women to do so  She won Fields Medal most recently (2014)  First Women to get this prize, also the first from Iran)
Iranian math genius Mirzakhani passes away
Interestingly Mirzakhani is perhaps famous because when Hassan Rouhani tweeted congratulations to her (being the first Iranian to win the prize) he also used a photo of her without a hijab!
Some may like : A Tenacious Explorer of Abstract Surfaces
Re: BR Maths Corner1
International Math Olympiad 2017 is about to start
Good luck to Indian Team (Sutanay Bhattacharya,Tarush Goyal, Anant Mudgal, Aditya Prakash, Shubham Saha Yash Sanjeev)
(IIRC Anant and Sutanay won Bronze in IMO 2016)
The US Team (Junyao Peng, Zachary Chroman, James Lin, Andrew Gu, Ankan Bhattacharya, and
Vincent Huang)
Ankan has won gold in IMO 2016.
Nepal will be participating for the first time.
Good luck to Indian Team (Sutanay Bhattacharya,Tarush Goyal, Anant Mudgal, Aditya Prakash, Shubham Saha Yash Sanjeev)
(IIRC Anant and Sutanay won Bronze in IMO 2016)
The US Team (Junyao Peng, Zachary Chroman, James Lin, Andrew Gu, Ankan Bhattacharya, and
Vincent Huang)
Ankan has won gold in IMO 2016.
Nepal will be participating for the first time.
Re: BR Maths Corner1
Past Results of the International Math Olympiad
https://www.imoofficial.org/results.aspx
https://www.imoofficial.org/results.aspx
Re: BR Maths Corner1
IMO 2017  Contest is over, result will be out soon.
Two day contest, 3 problems each day. (6 problems, each one can get you 7 points)
Generally prob 1 (and 4) easy, Prob 2 (and 5) mediumhard, Prob 3 (and 6) are hardest..
This year Prob. 1 was much easier than usual (relatively speaking) perhaps since there are many new countries this time, they wanted everyone to solve at least one problem.
and 3 was harder much harder IMO (even for US students  a strong team)..
Day 2  Prob 4 was easier, and 6 was very hard..
I think the cutoff for medals will be very low this year as it seems very few people would have solved prob 3 and 6..we will see when results come out..personally I will be impressed by anyone who have solved more than 2 problems.. and is getting more than 14 points...
Two day contest, 3 problems each day. (6 problems, each one can get you 7 points)
Generally prob 1 (and 4) easy, Prob 2 (and 5) mediumhard, Prob 3 (and 6) are hardest..
This year Prob. 1 was much easier than usual (relatively speaking) perhaps since there are many new countries this time, they wanted everyone to solve at least one problem.
and 3 was harder much harder IMO (even for US students  a strong team)..
Day 2  Prob 4 was easier, and 6 was very hard..
I think the cutoff for medals will be very low this year as it seems very few people would have solved prob 3 and 6..we will see when results come out..personally I will be impressed by anyone who have solved more than 2 problems.. and is getting more than 14 points...
Re: BR Maths Corner1
Have any of you guys looked into Srinivasa Ramanujan's work? I would like to read a technical book on his discoveries some day.
Re: BR Maths Corner1
IMO 2017 problems are there on their official site  (I don't see pdf version yet.. may be there shortly or my link is not working).. here is english pdf version.
2017 IMO problems
2017 IMO problems
Re: BR Maths Corner1
KJo wrote:Have any of you guys looked into Srinivasa Ramanujan's work? I would like to read a technical book on his discoveries some day.
There is lot of discussion about Ramanujan's work (and life) in this thread.. I have given some links for his notebooks and other works. In fact there are many problems from his in dhaga  simplified in layman's terms, I have posted here with solutions and insights.. just do a search and that may be of your interest.
There are many very good references, such as:
An overview of his notebooks
Hardy's book "A Mathematician's Apology" is classic.
Another classic is by Hardy,Aiyangar, Seshu Aiyar, Wilson  "Collected Papers of Srinivasa Ramanujan"
There are, of course, many more!
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