BR Maths Corner-1

[sudarshan]

Does such a number exist?

In the general case, with:

P * n^2 + 1 is a perfect square

Obviously, for P being a perfect square, there are no integer solutions.

For any other P, we want:

P * n^2 = k^2 - 1 = (k-1) * (k+1)

Since P is not a perfect square, k-1 and k+1 have to share the factors of P

If P is additionally prime (as is the number 61), then either k-1 or k+1 have to be divisible by P

If k-1 = a * P, then (k+1)/a has to be a perfect square (and vice versa)

Or a itself has to be a perfect square, with k+1 (or k-1) also being a perfect square

I set this up in Excel and tried with P being any number from 2 to 61 (leaving out perfect square values of P). For most of the other numbers, the value of n was easy to obtain within a few tries. The number 52 was an exception, it took many tries. The number 61 seems to be special in some way, even after a humongous amount of tries (not by hand, automated in Excel), I still couldn't find a solution.

The number 61 seems to be specially selected to inflict maximal pain on the aspirant . Even assaulting it with many intelligent fruit boars (oops - intelligent brute force) doesn't seem to yield results. I can see why Fermat had trouble with it, and my respect for Brahmagupta went up a couple of notches.

Hint?

[sudarshan]

After thinking about this problem some more, I'm getting a lot more gyan on it.

In the general case, when we want:

P * C^2 + 1 should be a perfect square, say k^2

If P is a perfect square, there are no integer solutions for C or k

Otherwise:

P * C^2 = (k-1) * (k+1)

For the special case where P is prime (which will work for our problem right now, since 61 is a prime), and we designate m = k-1, n = k+1

This automatically implies: m = n-2 (or) n = m+2

It also implies: k = m + 1, C = sqrt(m * n / P)

(C is the number we want, when we finally substitute P=61)

Either:

Case 1:

m = a * P, and:

Case 1.1: n/a is a perfect square, or

Case 1.2: a is a perfect square and n is also a perfect square

Or:

Case 2:

n = a * P, and:

Case 2.1: m/a is a perfect square, or

Case 2.2: a is a perfect square and m is also a perfect square

Cases 1.1 and 2.1 - a cannot be greater than 2 (since if m = a * P, n = m + 2 implies that for a > 2, n will not be divisible by a, and similarly for n = a * P, m = n - 2 implies that for a > 2, m will not be divisible by a)

So for both cases 1.1 and 2.1, a can only take the values of 1 and 2.

The case where a=1 can equally go to Cases 1.2 and 2.2, since 1 is a perfect square, so it can be handled in those cases.

So focus on a=2, with cases 1.1 and 2.1.

If a=2, what values of P will that work for?

m = 2 * P, with n/a = (m + 2)/a = 2 * (P+1) / a being a perfect square

(Or)

n = 2 * P, with m = n-2 = 2 * (P-1) / a being a perfect square

So a=2 will work for any P, where either P-1 is a perfect square, or P+1 is a perfect square

So for P=0, 2, 3, 5, 8, 10, ... we can derive m/n as 0/2, 2/4, 6/8, 8/10, 16/18, 18/20, ... we can derive k as 1, 3, 7, 9, 17, 19, ... with C being 0, 2, 4, 4, 6, 6, ...

[sudarshan]

Now for prime P, where neither P-1, nor P+1 is a perfect square (the number 61 falls in this category):

Here, a has to be a perfect square.

So either:

m = a * P, with n = a * P + 2 being a perfect square,

Or

n = a * P, with m = a * P - 2 being a perfect square

So for P = 7, a = 1 will work, we get m = 7, n = 9, k = 8, C = 3
For P = 13, a = 50 works, even though 50 is not a perfect square - there is some error in my reasoning somewhere, however, if there is a perfect square value of a such that a * P + 2 is a perfect square, or a * P - 2 is a perfect square, that value of a will work, I'm just not able to find one.

So for P = 61, if I can find a perfect square a such that a * P + 2 is a perfect square, or a * P - 2 is a perfect square, then the problem is solved. But it seems other values of a, which are not perfect squares, might also work, will have to think some more about it.

[sudarshan]

AH HA!!! I found the error in my reasoning, and I also found a number which works with 61:

61 * C^2 + 1 is a perfect square (k^2) when:

C = 226153980
k = 1766319049

Still brute force, but I was able to narrow down the search values for "a" by a lot through the above reasoning.

The error in my reasoning was in cases 1.2 and 2.2 above, a can also be 2 times a perfect square, with either n/2 being a perfect square, or m/2 being a perfect square.

As a consequence of case 2.2, modified as above, if one can find a value of "a" which yields a solution C, then 2*C^2 will serve as a new "a" to get one more value of C. And so on. So if one can find one solution, then infinitely many solutions can be found (unless C ends up being sqrt(a/2)).

[Amber G.]

^^^ Nice.
Meanwhile I was talking about this problem with a former student (I coached him in his high-school years - some 20 years ago, now he has a PhD and is fairly well known in his field)..he used brute-force while we were carrying out our conversation, It took 10 minutes of CPU time. In our chat - I told him if we were still doing the math as we did in the old days .. would be much faster. In our chat I told him "My method - VERY intuitive, takes about 1 minute with just pen and paper - seconds only if calculator which can add/multiply .. and obvious enough that you would have understood it even 20 years ago...
Meanwhile he was also looking at his usual sources and obviously found the right reference but he found the historic aspect very interesting. Here is some cut and paste from that talk ..

The Brahmgupta's method (written 1500 years ago):
(And "61" is just an example .. it works for *any* other number)



(This work was translated in Arabic but from Arabic to English was don in 1700's so western word did not know any easy solution )

Here is my method: (Based on above, but done in modern math):


This is: (1766319049 + 226153980 sqrt(61)) (1766319049 - 226153980 sqrt(61)) = 1
Which means 1766319049^2 - 61* 226153980 ^2 = 1
(These are the same numbers as given in Sudershan's post. (These are the lowest numbers - except 1^2 - 0*61 = 1))


Of course, just take any power of the above, (n=1, 2, 3 ... etc) .. and you get infinite set of solutions.
(You have to calculate (1766319049 + 226153980 sqrt(61))^n )

[sudarshan]

I couldn't figure out the Sanskrit text. Why did you pick 8 in the first step, and 7 in the second? Does the text say to do that? I'm guessing because 8 is the smallest number whose square is larger than 61, and 7 is the largest whose square is smaller than 61?

How do you decide when to square/ cube etc.?

I'm asking because I tried with P=2 and P=3 instead of P=61. I couldn't follow the steps like you had above, instead of cubing, I had to square and do some different operations to get 1 on the RHS. Also, the solutions I got for P=2 and P=3 worked, but they were not the lowest possible solutions for either case. For example, for P=2, I got 2*12^2+1=17^2, and for P=3 I got 3*15^2+1=26^2.

But for P=2, the lowest possible solution seems to be 2*2^2+1=3^2, and for P=3, it seems to be 3*4^2+1=7^2.

It's ingenious though. However, one question that I have is - what happens when P is a perfect square? Looking at the equation: P * C^2 + 1 = k^2, one can immediately see that when P is a perfect square, there are no solutions where both C and k are integers (C=0/k=1 doesn't count). How does that work with the above procedure?

[sudarshan]

Also, I'm curious as to what Euler's solution would have looked like. The above solution seems to be procedural, based on observing numbers, selecting them and working them to the required terms on the LHS and RHS (manipulation based on observing the terms and actively massaging them to the desired end, if you know what I mean). An intuitive, right-brain kind of approach.

Whereas, I'm guessing Euler's solution would have been more along the lines of European mathematics, with formal algebraic steps and proofs and "rigorous" analysis? More left-brained.

Two different approaches, pretty fascinating. Right-brain vs. left brain (mine was the hare-brained approach I think, relying on brute force mostly ).

P.S.: Nothing personal against hares, I've heard they're pretty adept at multiplication.

[Vayutuvan]

I found an interesting link for sudarshan ji or anybody who wants to do semi-bruteforce NT. You don't need to write your own programs but use this prebuilt functions. Only downside is it is in lisp/scheme like syntax and functional programming language called racket.

https://docs.racket-lang.org/math/number-theory.html

[Vayutuvan]

Also, I'm curious as to what Euler's solution would have looked like. The above solution seems to be procedural, based on observing numbers, selecting them and working them to the required terms on the LHS and RHS (manipulation based on observing the terms and actively massaging them to the desired end, if you know what I mean). An intuitive, right-brain kind of approach. ... Euler ...

A lot, I mean really a lot, of math nowadays is done as mix of these two approaches. Most examples come from Linear Algebra and diffeq. You guess a solution and then massage till it fits through a calculation to verify. In fact, iterative methods for solving a system of linear equations is one such.

In Sciences, it is called an ansatz. The same concept can be applied in Mathematics too.

One famous example comes to mind.

1. Positive resolution of Kadison-Singer Conjecture/Problem by Daniel Speilman, Adam Marcus, and Nikhil Srivatsava.

The previous one is Strassen's matrix multiplication algorithm. Four Color Theorem was also proved using computers but with a lot of simplifications coming from Euler/Hilbert/Jacobi type formality.

2. Fields Medalists Stephen Smale and David Mumford have done work in this areas, especially Mumford. Stephen Smale, Narendra Karmarkar, and several others started a program todo experimental mathematics and also started a journal on Experimental Mathematics. One of the problems Karmarkar was working on was Hilbert-Smith Conjecture which is the final resolution of Hilbert's fifth problem.

3. Daniel Spielman gave an expository article (lecture?) on how mathematical experiments led them to give a positive proof to Kaddison-Singer Problem. It is quite a deep result which touches on "hidden variable problem" of Quantum Mechanics.

@Amber G ji might be able to throw some light on the "Hidden Variable Theory" of Physics for us laypeople.

[Vayutuvan]

https://en.wikipedia.org/wiki/Experimental_mathematics

As expressed by Paul Halmos: "Mathematics is not a deductive science—that's a cliché. When you try to prove a theorem, you don't just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork. You want to find out what the facts are, and what you do is in that respect similar to what a laboratory technician does."

Most of the time the guesswork fails, unless one is able to go from one branch of mathematics to another and make connections.

[Amber G.]

Few comments will appear in separate posts ..

The number 61 seems to be specially selected to inflict maximal pain on the aspirant . Even assaulting it with many intelligent fruit boars (oops - intelligent brute force) doesn't seem to yield results. I can see why Fermat had trouble with it, and my respect for Brahmagupta went up a couple of notches.

Hint?

Well if I was really mean (to inflict maximal pain) I would have put 109 -- where the smallest solution has about 15 digits. . I have seen some of Fermat's letters about this problem ( with 61) to his fellow mathematicians and they are somewhat amusing as he (Fermat's was getting really frustrated and was really challenging and daring his math friends "Isn't there a true mathematician here".. One of his friends (IIRC Wallis - the Famous English Mathematician) supplied with an answer: (1523/2 + (195/2) and Fermat pointed out, 1523/2 is not an integer ..Wallis responded with something akin to "that damned Frenchman").. ( BTW Fermat was not a professional Mathematician so he was free to make fun of "professional" mathematicians.)..

I remember a story about prime number 313 -- Frenicle challenged Wallis to solve x^2 -313y^2 =1 The smallest solution is (32188120829134849, 1819380158564160).. (Both these guys are famous mathematicians ... one of them read the translation of Brhamgupta's work)..

Jaydev, an Indian Mathematician, who lived about 500 years after Brahmgupta but formalized his work and explained his methods beautifully writes about this problem " If someone solves this problem within one year, (without reading his method) is a "true mathematician"..He (Jaidev) not only gives a general solution but proves that those are the smallest solution, and the method will always work within a few short steps.

Some current history books suggests that some English Mathematicians (eg Wallis etc) had access to some of this work - translated and carried to West by south-India missionaries.
(I am really glad that in school and from my father - I learnt Sanskrit - not an expert, but good enough to read Mathematics).

[Amber G.]

Sudarsha et all - (Hope this is useful and inspires some young readers to be interested in math..

I couldn't figure out the Sanskrit text. Why did you pick 8 in the first step, and 7 in the second? Does the text say to do that? I'm guessing because 8 is the smallest number whose square is larger than 61, and 7 is the largest whose square is smaller than 61?

How do you decide when to square/ cube etc.?

I'm asking because I tried with P=2 and P=3 instead of P=61. I couldn't follow the steps like you had above, instead of cubing, I had to square and do some different operations to get 1 on the RHS. Also, the solutions I got for P=2 and P=3 worked, but they were not the lowest possible solutions for either case. For example, for P=2, I got 2*12^2+1=17^2, and for P=3 I got 3*15^2+1=26^2.

But for P=2, the lowest possible solution seems to be 2*2^2+1=3^2, and for P=3, it seems to be 3*4^2+1=7^2.

It's ingenious though. However, one question that I have is - what happens when P is a perfect square? Looking at the equation: P * C^2 + 1 = k^2, one can immediately see that when P is a perfect square, there are no solutions where both C and k are integers (C=0/k=1 doesn't count). How does that work with the above procedure?

The Sanskrit text is very clear and it answers why one picks 8 etc..( when and why to do square/ cube etc might not be clear there - those were my short-cuts)
If Sanskrit is rusty - One can google or see standard math text - the methods are called Brahmgupta's bhavana method ... you can see wiki or for example (https://bhavana.org.in/the-bhavana-in-mathematics/ (If someone does enough googling - they may find my decades old class notes - or others work - used for coaching - using modern methods and in simpler terms )

Bhaskara wrote about this extensively - good text books or wiki - do google on " cakravāla or chakravala (= cyclic) method of Bhaskara" -- This is basically recursive formula - each step (starting with 8^2 ) takes you closer.

Bhaskara gives fairly rigorous treatment - proving how this will work for every number (except perfect square) to give minimal solution in very few steps.

(For those interested - there is *plenty* of literature to see how Bhavana or Chakravala works but basically you start with a number - whose square is as close to 61 - with certain property. .. you work with x^2 - 61 y^2 = k , and keep getting better values of k until you find k=1.

*****
Euler and many others in western world, used continued fraction method - *again* developed by Indian Mathematicians but that work has been translated into English. See my comments about Ramanujan and continued fraction method.

Basically here, one finds continued fraction of sqrt(61) (which is [7+ : 1, 4,3,1,2,2,1,3,4,1,14)].. you do the same computation as Ramanujan did for sqrt(2) ... you get rations like ...7/1, 8/1, 39/5, 125/16, 164/21, 453/58, 1070/137, 1523/195, 5639/722, 24079/3083, 29718/3805, 440131/56353.. etc.. at 22nd term, you will hit the correct answer, and then again 44nd term etc... )
The method is equivalent to Brahmgupta's method but a little slower..

I like my method - posted here in BRF (from a screen of my chat) .. which essentially used only 3-4 steps. ..(First 2 steps are essentially chakravala but "cubing" / squaring etc are my short-cuts not that well known - they save may be dozen or half a dozen steps from pure charavala method)

****
There are *many* fine-points which are very interesting in modern math - basically algebraic fields. Manjul Bhargava (India's first Field Medalist) often talks about Brahmgupta's contribution in his work and credits him.
For us physicists group theory is used in QM and modern physics so this kind of math is quite useful - practically speaking ..

Hope this is all fun and even educational.....

(For your information, the minimal solution to x^2–9739y^2=1is

2004678915287129865051784235972681465598817784993286048703987347587076305931512617412027372229926151890^2 -
9739*20313634766038988908316803031882483814532877863673358783668788287930891909883492238723036864710413171^2 = 1 )

[sudarshan]

More comments as I get to them, but what I found interesting in that Sanskrit text was that there seemed to be a lot of abbreviations/ short forms. "Ka" "Kshe" "Jye" etc. There was also a mention of "kuttarth" which I think means "simplification" as in Aryabhatta's "kuttaka?" The rest I could read, but not comprehend, my Sanskrit works at the level of slokas and such, this was a different beast.

Also, is that an "=" sign which I spy in the text, or is it something else?

[sudarshan]

A lot, I mean really a lot, of math nowadays is done as mix of these two approaches. Most examples come from Linear Algebra and diffeq. You guess a solution and then massage till it fits through a calculation to verify. In fact, iterative methods for solving a system of linear equations is one such.

In Sciences, it is called an ansatz. The same concept can be applied in Mathematics too.
...
@Amber G ji might be able to throw some light on the "Hidden Variable Theory" of Physics for us laypeople.

Did not know that, I always thought of pure math as being deductive, more so than applied science. Nowadays in science one uses techniques like neural networks or optimization algorithms like GA, PGSL, or Particle Swarm approaches, which are simply ways of aping natural processes, rather than any techniques with mathematical backing behind them. Basically, we have computers, so just throw those at the problem.

I thought pure math would be much more rigorous, and that all proofs would be totally (or mostly) traceable to the axioms. Guess not.

[SriKumar]

A lot, I mean really a lot, of math nowadays is done as mix of these two approaches. Most examples come from Linear Algebra and diffeq. You guess a solution and then massage till it fits through a calculation to verify. In fact, iterative methods for solving a system of linear equations is one such.

In Sciences, it is called an ansatz. The same concept can be applied in Mathematics too.
...
@Amber G ji might be able to throw some light on the "Hidden Variable Theory" of Physics for us laypeople.

Did not know that, I always thought of pure math as being deductive, more so than applied science. Nowadays in science one uses techniques like neural networks or optimization algorithms like GA, PGSL, or Particle Swarm approaches, which are simply ways of aping natural processes, rather than any techniques with mathematical backing behind them. Basically, we have computers, so just throw those at the problem.

I thought pure math would be much more rigorous, and that all proofs would be totally (or mostly) traceable to the axioms. Guess not.

NOt speaking for Vayutuvan here but I suppose he is talking about mathematical solutions of real-world physics problems (like for example weather modeling, rocket dynamics, and maybe stock market predictions). As I understand it, most of these (certainly the first 2 items in my list) are modeled using differential equations to which analytical solutions do not exist for many real-world problems, and therefore iterative solutions are applied with an initial guess, followed by a corrected guess and so on till you get a solution with an acceptably low error value (or set it up as an energy minimization problem that gives you a set of simultaneous linear equations to be solved iteratively).

On an unrelated note, the numbers in Amber.G's post above has the largest numbers I have seen anywhere used in meaningful manner.
Typically, for large numbers, I'll use scientific notation and truncate after 5 (or 10) places after decimal point (of course, the calculations use double precision, but I dont need to see all the digits).

[Amber G.]

^^^^Isaac Asimov in his preface to Carl Boyer and Uta Merzbach's "History of Mathematics"

Now we can see what makes mathematics unique. Only in mathematics is there no significant correction - only extension...there, and only there, do we touch the human mind at its peak.

****
As I have mentioned, the encryption in your iPhone, depends on those cubic curves, I mentioned in brf .. point is "one way" computation .. pretty easy to check that the answer is correct but not that easy to get to the answer - so you have basis of public key/private key encryption.

[Amber G.]

Speaking of "largest numbers" a *very* famous problem from antiquity (mentioned by Archimedes - Do google on Archimedes's cattle problem) wants to find the solution of Brahmgupta's equation ..
x^2 - 410 286423 278424 y^2 = 1
. The solution has 206545 *digits* and required about 47 pages to just print the answer. (The solution appeared in an article in 1980's where CRAY-1 super computer was used)

****
I just read the link I posted here (https://bhavana.org.in/the-bhavana-in-mathematics/) for Bhavana carefully - (Initially I just did a google and found the paper good enough to post here) .. It is a very nice article and recommend people to read it.. It does a fairy good historical review and fairly accurate in math.

In Europe, Brahmagupta’s identity would reappear in the works of L. Euler during the eighteenth century. Euler called the identity theorema eximium (a theorem of capital importance) and theorema elegantissimum (a most elegant theorem). {Euler, I think saw translated work and was not aware of the original authors who did the work and thus some work eg "Pell's" name was attached here although Pell's work was just translation and publish the collected work}

While this specific identity on the important binary quadratic form y^2-N x^2=1
is now a basic result in number theory, the real greatness of Brahmagupta’s bhāvanā lies in its manifesting the very principle of composition in mathematics, a principle that pervades the whole of modern algebra and number theory. But, as is the case with many outstanding mathematical discoveries in ancient India, this master stroke of Brahmagupta and its significance have not always been adequately highlighted in accounts on history of mathematics.

Reading Brhamgupta (and some modern mathematicians account) gives a new respect for his work. The quadratic fields and his concept of "composition" is nearly 1500 years ahead of his time.

[SriKumar]

AmberG (or anyone else who cares to take a whack at it), what do you think of the 'new' techniques that fall under the AL/Machine learning categoies. A lot of it seems to be old (Well-known techniques like regression, linear algebra and probability). Are there any actual new mathematics that came up in the last 10-15 years that fall in the realm of machine learning?

Also is: AI/ML seems to be mainly about detecting and recognizing patterns where an algorithm is 'trained' by subjecting it to a real LOT of examples and 'told' right/wrong till it comes up with its own set of 'rules'. It does not need or rely on physics.

How (i)useful, and (ii) applicable is it in trying to solve real-world problems, where laws of physics are the fundamental driver. All physical phenomena are driven (ultimately) by forces (electromagnetic, graviational, strong/weak etc .)

Engineering (aerospace/electronics) is physics too albeit at a more gross/coarse level, and here the physics laws are further simplified/made coarse (e.g. Newtom's laws, Consv. of Energy, Cons. of Momentum, say in rocket dynamics) and applied to solve real world problems. Is there a role for AI-type solutions in this realm. THAnks for any insights.

[sudarshan]

^ I don't know all the latest developments, but the way this "big data" is typically handled in many places, is pretty farcical. A lot of people are just looking to bandwagon on the concept. Likewise with AI/ ML, what I observe (which may be pretty far from the true cutting edge in the field) is that these little companies just download Python or other readily available routines, and just try "potluck" with those. No real understanding or application of the leetle grey cells, but lots of fancy jargon thrown around. As you say, many of the techniques are decades old, extremely simple at heart, and/ or just plain improvised for specific problems way back then, but treated as "cutting edge" now - like KNN, BBNs, or even NNs and Fuzzy Logic. Likewise with a lot of these "advanced" optimization routines, which simply have no mathematical backing, but are instead fancified search strategies. But in conferences and journals, everybody plays along with treating those as the leading edge of science, I suspect because they themselves are deep into all that and don't want to admit that what they're doing ain't all that great.

If you're talking of "laws of physics being the fundamental driver" then there's still model-based reasoning, which uses hard-core physical, engineering, or mathematical concepts. But it's possible to hybridize that with AI/ ML as well.

Somebody else might have a less cynical view of all this, willing to learn.

[Prem Kumar]

There's certainly bandwagoning, no doubt. But after decades in the doldrums, thanks to new algorithmic breakthroughs and computing tipping points, we have a situation where some real AI applications are possible (example: look how good Google Translate is). Sooner or later we will hit the limit and maybe get stuck there for God knows how long.

Where I think there might be *new insights* from AI/ML and in general the ability to crunch massive amounts of data to look for patterns, will be areas like Feigenbaum's constants in chaos theory. There might be hidden constants that such number crunching might uncover. Lets see....

[Amber G.]

Those who want to check out if their favorite AI program - see if it can prove this ...
(The AI program can even look at all my posts in brf about continued fractions ..

[Vayutuvan]

@sudarshan and @SriKumar

http://nemocas.github.io/Nemo.jl/latest/index.html

Please check this out for number theoretic experimentation. Would it be useful to solve the above problem given by @Amber G.?

AI is not going to solve that problem or even simpler problems in boolean algebra

[SriKumar]

^ I don't know all the latest developments, but the way this "big data" is typically handled in many places, is pretty farcical. A lot of people are just looking to bandwagon on the concept. Likewise with AI/ ML, what I observe (which may be pretty far from the true cutting edge in the field) is that these little companies just download Python or other readily available routines, and just try "potluck" with those. No real understanding or application of the leetle grey cells, but lots of fancy jargon thrown around. As you say, many of the techniques are decades old, extremely simple at heart, and/ or just plain improvised for specific problems way back then, but treated as "cutting edge" now - like KNN, BBNs, or even NNs and Fuzzy Logic. Likewise with a lot of these "advanced" optimization routines, which simply have no mathematical backing, but are instead fancified search strategies. But in conferences and journals, everybody plays along with treating those as the leading edge of science, I suspect because they themselves are deep into all that and don't want to admit that what they're doing ain't all that great.

If you're talking of "laws of physics being the fundamental driver" then there's still model-based reasoning, which uses hard-core physical, engineering, or mathematical concepts. But it's possible to hybridize that with AI/ ML as well.

Somebody else might have a less cynical view of all this, willing to learn.

Saw your response and sorry for the delayed response to your response. Agree with the band-wagon thing and learning jargon and bandying it about. I see this in some of the IT-ivty talking AI but no inclination for math. On the flip side, as you say all these softwares Matlab/Python etc. have the algorithms already inside of them (Matlab was built on linear algebra suboutines (BLAS, LINPACK etc.- traditional linear algebra subroutines)) so one does not need to know the details, only the ability to hit some buttons, get some plots and put out an interpretation to appear knowledgeable about it.

What I am trying to figure out is what changed in the last 10 years that it has become the rage now, and is there anything real about it. From what I can see, tons of data bas become available (people's behaviors, images, people's responses etc) and this has created more data to 'train' the algorithms. Of course, we (the public) see a biased version of the efficacy of the techniques- the successes are bragged about on TV and the failures are not broadcast or published. Only the researchers and developers would know.

My point of interest is about modeling of physical phenomena, and this relies entirely on having mathematical equations to describe behavior of nature. A lot of equations do a pretty good job of representing physical phenomena and have been for decades now (whether it is orbital mechananics, material response, fluid mechanics, wave propagation- useful in fluid mechanics, earthquake engineering, electromagnetic telecommunications, astronomy etc.). The use of A.I.-type methods applied to physics may provide limited benefits here IMHO. Since by definition, A.I. does not rely on a mathematical description of nature, rather it is a 'regression' (To over-simplify it), physicists might be wary of using A.I. (other than perhaps to tell them 'hey, this looks different- take a second look at this').

vayutuvan- thanks for the link but right now I am still trying to wrap my mind around how the square root (!) of the product of two irrational numbers (!!) e & pi, can be exactly equal to anything

[sudarshan]

I have a feeling that much of the gains from AI simply comes from the fact that computers are millions or billions or times faster than humans at performing basic operations, so they notice patterns or converge to solutions much faster, even with lower efficiency, even with random searches. For example - Genetic Algorithms. I've implemented this for some small-scale problems. The strength of the method is supposedly in its strategy of mutations and crossovers. But when I disabled those and implemented a pure random search (but ensuring that the best solution was preserved for future generations at each step), the method performed just as well! Of course, this is for smaller and less intensive problems, but most use-cases of GA are also for relatively small problems (for example - when a company downloads Python source code and tries out the technique on one of their data-mining requirements). However, like I said, I'm pretty far from the "cutting edge" of AI.

Yes, the successes are bragged about, but I myself am aware of instances where certain real-world problems proved to be too much for these algorithms. So the researcher tries out one technique after the other, based on recommendations from others. "Why are you trying that classifier? Try a self-organizing map, those things are great, I've had good results with those!" Those techniques work for toy problems, the nature of certain real-world situations makes them unsuitable, or maybe the researcher was just applying them wrong. That's when improvisation comes in, and new techniques are occasionally born, which again work great for that particular problem.

Maybe I'm just rambling here....

[sudarshan]

It's not exactly equal to anything, it's a series . Maybe taking the series expansions of those two irrationals and multiplying them out term-by-term would work. But the result would need to group the terms just right. And I still can't figure out those continued fractions. How would one evaluate those? One would need to start at the bottom, so truncate to a specific number of terms? Then increase the number of terms and see if it was sufficiently converged? Also seems like the convergence of those continued fractions would be slow.

VT ji, I think you're reading me wrong . I have no formal training in number theory or math, I just try out my high-school or college techniques on the problems which AmberG posts here, sometimes they work. Then there's Excel, I just set things up there brute-force and again, sometimes they work. I can take a stab at the above problem this week, haven't got any free time yet.

[sudarshan]

There's certainly bandwagoning, no doubt. But after decades in the doldrums, thanks to new algorithmic breakthroughs and computing tipping points, we have a situation where some real AI applications are possible (example: look how good Google Translate is). Sooner or later we will hit the limit and maybe get stuck there for God knows how long.

Where I think there might be *new insights* from AI/ML and in general the ability to crunch massive amounts of data to look for patterns, will be areas like Feigenbaum's constants in chaos theory. There might be hidden constants that such number crunching might uncover. Lets see....

Yes, that kind of massive number crunching was what I was talking about, but would a random search work just as well as AI there? Especially in optimization algos, the stochastic techniques just seem to show a marginal improvement over pure random search. This is again based on my limited experience with smaller problems.

[SriKumar]

It's not exactly equal to anything, it's a series . Maybe taking the series expansions of those two irrationals and multiplying them out term-by-term would work. But the result would need to group the terms just right. And I still can't figure out those continued fractions. How would one evaluate those? One would need to start at the bottom, so truncate to a specific number of terms? Then increase the number of terms and see if it was sufficiently converged? Also seems like the convergence of those continued fractions would be slow. .

True, it is a series and therefore the final solution (i.e. after summation of terms) will vary depending on how terms are taken into consideration. What I meant was that since there is a series on the RHS AND there is (it seems like) a series on the LHS, one could make a case that they are exact, IF you take a similar number (maybe not exactly the same) of terms on both sides. To me, the = symbol suggests that they both converge similarly, depending on how many terms are taken. I am guilty of proposing something without actually trying it out on Excel or Matlab, but please excuse me today .

I did a bit of googling around and found that 'e' can be expressed as a series which has almost the same form as the 2nd term on the RHS. https://en.wikipedia.org/wiki/List_of_r ... 20sequence.

Similarly, pi also can be expressed as a series and this page gives several series approximations; though none look very similar to the first term on RHS in the equation posted by AmberG. https://en.wikipedia.org/wiki/List_of_f ... ing_%CF%80

So, it seems that perhaps the LHS is some sort of a series (and the RHS obviously so). I have not, unfortunately, progressed beyond A.P., G.P.; and H.P. was always a headache ....atleast A.P. and G.P. series appear have some symmetry to them and were more tolerable ).

Added later: I'll walk back a statement of mine, above, atleast partially. It will be interesting to see if it is indeed true that the LHS and RHS are exact for a (any) given number of terms in the series expansion. I am not quite sure now if this will be the case.

[SriKumar]

Yes, the successes are bragged about, but I myself am aware of instances where certain real-world problems proved to be too much for these algorithms. So the researcher tries out one technique after the other, based on recommendations from others. "Why are you trying that classifier? Try a self-organizing map, those things are great, I've had good results with those!" Those techniques work for toy problems, the nature of certain real-world situations makes them unsuitable, or maybe the researcher was just applying them wrong. That's when improvisation comes in, and new techniques are occasionally born, which again work great for that particular problem.

Maybe I'm just rambling here....

Definitely useful observations, from someone who has tried out the algorithms first hand. There are (in my mind anyway) two categories of problems- one where nature's 'rules' or behavior is known and codified through some equations, and in the second category where this is not the case. In the first category, would be problems that we know a lot about, for example speeds or rockets or spacecraft after 200 days past launch. This can be worked out purely from the forces it experiences- rocket thrust, gravity of earth, moon, Mars (for Mangalyaan). It is all Newton's laws and one would not need to use 'A.I.' to solve this. Even weather prediction is actually 'old-school' physics- a combination of thermodynamics and fluid mechaincs (over-simplifying it here) solved using powerful computers. To your point, if powerful computers did not exist, weather prediction would not have been possible (or would have been very inaccurate for anything more than a few hours out).

It is the second category of problems where there is no specific (known) equation describing the behavior of the system (e.g. behavior of humans under various circumstances- online, offline, 1-1 interactions etc) that, it appears, people are trying various 'A.I.' algorithms (Neural networks for example) that they are having some success with. THese class of problems may not have any underlying physics that one needs to keep track of, or reconcile with (unlike in the second category where physics-based theory has been developed based on experimentation and formalization). Big ramble here....but this has been gnawing at me for a while.

[Amber G.]

Meanwhile if anyone wants to try one's hand on using AI to prove something fun.. here it is on a lazy Sunday ...
( Question #541 posed by one of my favorite mathematician ( ) in Journal of the Indian Mathematical Society".



(Believe it or not, Ramanujan's this problem I saw *very* recently* in a Physics professor's lecture .. the method used and insight given is quite deep -- there are two sub plots - a) an interesting and powerful method to do differential equations using series methods, and b) using a continued fraction - which has lots of practical applications now..

The continued fraction technique, as it turns out was developed by Jacobi to treat certain integrals or differential equations and it seems utterly magical! (Ramanujan's problem is a particular solution) -- Idea being ..a function at , say f(x), obeys a differential equation. He uses this to compute f in terms of f/f', then f'/f'' in terms of f''/f''', and so on FOREVER until he gets the formula like that..)

[Vayutuvan]

I will come back to AI/Experimental Math/Computational Science (all three are different from each other with some overlap) in a bit.

Meanwhile, here is a nice introduction to Continued Fractions. I did not know about CFs till a few years back as I never had used it in my work. I found that it is a very easy read - 15 minutes tops.

https://perl.plover.com/classes/cftalk/ ... de001.html

[Amber G.]

Speaking of AI - International Math Olympiad 2020 has just started. Next year there will be AI program taking part in the contest - will be interesting to see how these programs compare with the best young minds.

[Vayutuvan]

AI programs will be hard-pressed in the kinds of problems posed in IMO, IMHO.

[Amber G.]

I don't know all the results.. will be interesting to see how AI did in this IMO...
Meanwhile here is one problem from this contest - try your favorite AI program (or do it yourself / get help from others etc.

A deck of n>1 cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.
For which n does it follow that the numbers on the cards are all equal?

[sudarshan]

A deck of n>1 cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.
For which n does it follow that the numbers on the cards are all equal?

n=2 is an obvious candidate. Because if the two numbers on the cards are x and y:

AM=(x+y)/2

GM of one card is either x or y--------Case 1
GM of two cards is sqrt(xy)------------Case 2

For case 1, we get x+y=2x or x+y=2y, both of which mean x=y
For case 2, we get (x+y)^2=4xy, which means x^2-2xy+y^2=0, which means (x-y)^2=0, which means x=y

Will have to think about higher values of n.

[sudarshan]

Ah I see....

Let there be n cards. The arithmetic mean of any pair xp and xq (p and q are indices going from 1 to n) is (xp+xq)/2. The GM of any k cards (with indices k1, k2, k3...kk) is (xk1*xk2*xk3*...xkk)^(1/k).

For example, n might be 17, p might be 3, q might be 11, k might be 3, k1=4, k2=9, k3=14 (whatever) - so we take the arithmetic mean of cards 3 and 11, and the geometric mean of cards 4, 9, and 14 is the same as this arithmetic mean (just to illustrate what I mean by "indexing").

Now the fun starts.

[ (xp+xq)/2 ]^k = xk1*xk2*xk3...xkk

The RHS is an integer. So the LHS has to be an integer. This automatically implies that either both xp and xq are odd, or both xp and xq are even. Because if one is odd and the other is even, then we have a fraction ending in ".5" (like 3.5 or 5.5 or 6.5) to the power of an integer k, which cannot be an integer for any integer value of k.

By extension, if one of the cards has an odd number on it, then they all have to do so, and if one of the cards has an even number on it, then they all have to do so.

Ergo, the n cards have either all odd numbers, or all even numbers on them.

[sudarshan]

We need to be able to find a set of k cards, whose GM is an integer (equal to the AM), for any given pair of cards.

If all the cards have odd numbers on them, the GM of any set of k cards, if at all it is an integer, will also be odd. And vice versa.

So for all odd number cards, the AM also has to be odd, and for all even number cards, the AM also has to be even.

Take the card with the smallest number on it (if there are several such cards, just pick one of them). If the cards have odd numbers - the AM of any pair has to be odd, which means if the smallest number is 1, we can't have any cards with 3, 7, 11, ... on them. The other numbers are restricted to 1, 5, 9, 13, ....

This also applies for the "all even numbers" case - for example, if the smallest number is 2, then the other cards can only have numbers 2, 6, 10, 14, ....

So if the smallest number is x1, then x2, x3, etc. have to be restricted to the range x1 + 4*y where y is a whole number (0, 1, 2, ... ).

Am I on the right track so far?

[Amber G.]

^^^
Nice work.

If all numbers are even, then divide each number by 2 .. a moments thought will tell us that this new set will also have the same property. .. One can repeat the process. As you said that some even and some odd case is not possible so all you have to consider the case for all odds only.

(I can't find, yet an easy way but was able to prove that for all cases the *all* numbers have to be equal. -- but proof is a little involved) (So if smallest number is 1 others have to be 1,1,1,1.... no 1, 5 etc..))


(Interestingly this problem is supposed to be harder one - appearing in the middle)
----
Was a little disappointed not to see India this time which skipped this contest due to Covid.

[Amber G.]

^^^Found a solution of above which is quite elegant and easy to explain.

[sudarshan]

(I can't find, yet an easy way but was able to prove that for all cases the *all* numbers have to be equal. -- but proof is a little involved) (So if smallest number is 1 others have to be 1,1,1,1.... no 1, 5 etc..))

I suspected that was the case, let me see if I can get there. That insight about "if the numbers are all even, then divide them by 2" is a good one, of course it only works because of that "x1 + 4*y" property, else it would yield mixes of odd/ even again.

[Vayutuvan]

Interesting news for encryption and MCMC simulations

https://www.hpcwire.com/2020/09/21/ibm- ... eneration/