BR Maths Corner-1

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Vayutuvan
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Re: BR Maths Corner-1

Postby Vayutuvan » 26 Sep 2020 01:49

Amber G. wrote:I don't know all the results.. will be interesting to see how AI did in this IMO...


One question: How do they describe the problem to the AI? AI has to understand the problem just like human participants or the problem is specified in some formal language?

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Re: BR Maths Corner-1

Postby Amber G. » 26 Sep 2020 02:34

Vayutuvan wrote:
Amber G. wrote:I don't know all the results.. will be interesting to see how AI did in this IMO...


One question: How do they describe the problem to the AI? AI has to understand the problem just like human participants or the problem is specified in some formal language?


From what I have seen - AI at present - best programs quite routinely - can fairly easily understand the problems as written. At present except
for IMO level problems, many other contest (even JEE type problems, from what I have seen) problems are no challenge to best AI programs. (They can crack JEE math type questions :)..

Heck even brf problems given by me - at least for most parts I tried :) - good or even semi-decent AI programs I tried solved them..(A iphone image of the written problem is often good enough)

For example we spent a few pages here for x^2 - 61y^2 = 1, I just typed "solve in integers x^2-61y^2=1 and I got the complete result. :) :) with algebraic form as well as all those large integers which makes it easy for me to copy and paste :)..

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Re: BR Maths Corner-1

Postby Amber G. » 27 Sep 2020 00:38

sudarshan wrote:
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.

Will be nice to see interesting discussion it generates.

My solution, which I thought earlier, was a little too complicated and some solution I have seen are also long.. but I have found a method which is very nice.. will put when I get some time.

Reducing all even numbers (so that only case you have to consider are all odd numbers) can be thought in a much simpler way. If the set contains all the even numbers say 2a , 2b, 2c, 2d, 2e, 2f, 2g ....and it works then the set (a,b,c,d,e,f,g) will work too. (for example if AM of 2a and 2b is GM of (2c, 2d and 2e) then AM of a and b will be GM of (c,d,e).

So one can consider only odd numbers without loss of generality.

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Re: BR Maths Corner-1

Postby sudarshan » 28 Sep 2020 09:32

Just thinking out loud.

Let's say that we have n odd numbers with the property, that the AM of any pair, can be represented as the GM of a set of 1 or more numbers from the same set. If all n numbers are the same number, we have no problem. We want to see if we can find a set of n numbers, where all numbers are not the same value, which will satisfy the condition. So in such a set, by definition, one number will be the largest, and one number will be the second largest (there could be multiple instances of the largest and/ or second largest number in the set, of course).

Let's say the largest value in the set is y, and the second largest value is x. If there are other unique numbers in the set, then they are all smaller than x.

1. There is no number in the set larger than y.

2. There is also no number in the set, which lies between x and y (since x is the second largest number).

3. The arithmetic mean of x and y is a = (x+y)/2.

3.a. Since x < y, that means x < a < y (i.e., x is strictly less than a, and a is strictly less than y, meaning, a lies between x and y).

3.b. Since x is at least 1, y cannot be a multiple of a (1*a < y, and 2*a = x+y, which is already larger than y).

4. There has to be some combination of one or more numbers in the set, whose GM equals a = (x+y)/2.

4.a. The GM of any k numbers g1, g2, ... gk is (g1*g2*...gk)^(1/k).

5. If the set <g1, g2, ... gk> (which is a subset of the full set of n numbers) consists of just one number g1 - then by definition, its GM is g1 itself. So by 4., g1 will have to be the same as a, which lies between x and y. But g1 is a member of the full set of n numbers, and by 2., there is no number in the set between x and y.

So there have to be at least two numbers in the set, such that their GM equals the value a = (x+y)/2.

But x is the second largest value in the set, and y is the largest. If the two numbers are y and y (assuming multiple occurrences of y in the set), the GM will be y itself, which is too large (3.a). If the two numbers are x and x (assuming multiple occurrences of x in the set), the GM will be x itself, which is too small (3.a).

How about x and y? Recall, the AM of two unequal positive numbers will *always* be larger than its GM. So even x and y won't work. And the other numbers in the set of n are smaller than x (x is the second largest, remember?) so they definitely won't work.

6. We need a way to "pull up" the value of the GM to be a = (x+y)/2. So we need to include more numbers in the GM calculation, which are large enough to pull the GM up. x, however, is already smaller than a, and will pull *down* the GM value. So these additional numbers can only be further occurrences of y (since y is the only number in the set of n, which is larger than a). Thus, it is necessary to have multiple occurrences of the largest number, y in the set.

Of course, if we include many instances of y in the GM set, we can include a few x's (or even numbers smaller than x) since the many instances of y will compensate for these smaller numbers.

If the GM calculation set <g1, g2, ... gk> has "k" numbers, then, a^k has to be the product of all the numbers in the GM calculation set. This means the numbers in the GM calculation set can only be composed of the factors of a, and can have no other factors. From 3.b., y is also not a multiple of a.

So y is a number larger than a, all of whose factors are also factors of a, and the set <g1, g2, ... gk> also has multiple occurrences of y in it (i.e., many of the values of g1, g2, ... are actually y itself). <g1, g2, ... gk> also consists of at least 3 numbers, of which at least two are equal to y (see 6.).

Still thinking from here on....

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Re: BR Maths Corner-1

Postby Amber G. » 28 Sep 2020 20:46

^^^ Very good. Agree with point 1 to 6.
Big Hint: Going along the line, I can think of *simpler* point 7 (and possibly 8 ) which finishes your proof.
(Even bigger hint: read *carefully* my previous post (just above yours) :).

Leave it at that, to see someone else or you write down the final argument.
(The problem was posed by an Estonian professor - I liked the problem)

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Re: BR Maths Corner-1

Postby Amber G. » 28 Sep 2020 20:55

^^ I like the problem because, at first site it looks like a combinatorial problem with algebra but it is a little deeper than that.
For example if the statement was changed from positive integers to rational numbers, it will still work, but if you allow all numbers, it does not. (except for the case of only 2 numbers).

One can have, ( (√5 -2), 1, 1) where GM of all three numbers is indeed AM of first two. But in *all* such cases, at least one number has to come out irrational.

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Re: BR Maths Corner-1

Postby Amber G. » 28 Sep 2020 21:04

Okay here is just a pure algebra problem, simple looking, from the same contest - posting here because it is so similar to all the thinking above.

Given a ≥ b ≥ c≥ d>0 and a+b+c+d=1.
prove:
(a+2b+3c+4d) (a^a b^b c^c d^d ) < 1.

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Re: BR Maths Corner-1

Postby Vayutuvan » 29 Sep 2020 02:07

SriKumar wrote: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 :D


Let us modify the problem a little.

Instead of "Prove that" <equality>, make it "Prove or disprove" or "It is conjectured that" and replace the <equality> with the language of real analysis.

What will you do first? You will start by evaluating both RHS and LHS in double precision, double double precision, and so on. You might stop at some reasonable error say like 1.0e-16 (which is considered to be "numerical zero" in 64 bit double computations) or may be even go down 1.0e-312 (which is close to the "numerical zero" in quad precision).

If that works out, then you start looking to prove it using real analysis methods, i.e. convergence of the sequence given in the LHS to a sequence that represents the RHS. Much harder. @Amber G. said "Prove that" which lets you skip the step of experimentation with large numbers and high precision arithmetic. :D

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Re: BR Maths Corner-1

Postby Amber G. » 29 Sep 2020 11:28

FWIW: Some tidbits of this Olympiad.
Rank - China (1), Russia (2), USA (3) India did not take part.
The problem I posted - All USA kids scored 100% ..
(Was curious so looked up score for Pakistan - 0 people solved the problem)

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Re: BR Maths Corner-1

Postby sudarshan » 03 Oct 2020 22:42

Not able to finish that solution, a simple end is still eluding me. I got another couple of steps further, but it's getting more complex all the time.

VT, I verified (numerically) that the series on the RHS plus that continued fraction does come to the LHS on that e/ pi problem. Now what? Expand the series for e and pi, multiply terms, mix and match?

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Re: BR Maths Corner-1

Postby Amber G. » 04 Oct 2020 21:24

sudarshan wrote:Not able to finish that solution, a simple end is still eluding me. I got another couple of steps further, but it's getting more complex all the time.

Here it is:
Assuming we have already agreed that we have to only to consider the case when all the numbers are odd.
(If all numbers are even - or divisible by any one number for that matter - can be factored out. This was a big hint).
We already proved the case for two so let us assume the total number in the set is >2 :) .

- Let L be the largest number in the set.
- It has to be an odd number greater than 1 (else all numbers will be equal :) )
- let p an odd prime ( greater than or equal to 3) divides L
- Let L2 be the next largest number (which is equal or less than L).

Now if L2 is not divisible by p
Then AM (= (L+L2)/2) of L and L2 is also not divisible by p
In this case (since L2 and L are obviously not equal - as L2 is not divisible by P) the AM is greater than L2.
But it is GM of some numbers - out of which L can not be there as one of the numbers . (otherwise p will divide this GM).
But GM of numbers (where *every* possible number remaining is equal or Less than L2) can not be greater than L2.
This is impossible . Contradiction!

==> L2 must be divisible by p.

Now same logic for L3- the next largest number.
AM of L and L3 will be larger than L3, and since L2 and L can not be included (as both are divisible by p).
L3 must also be divisible by p (remember p is an odd prime)

Rinse and repeat!
Next largest number must also be divisible by p .. and so it the next largest number ...

So all numbers must be divisible by p !!

Rest is simple - Fermat's infinite descend - Divide all numbers by p and repeat the process on this new set of numbers !!!!
Only way it can end is when the largest number now becomes 1 but in that case all the numbers now have to be equal to 1.)

QED!
(My first solution which I thought was quite a bit more complex but I like the above as it is fairly easy to understand and does not use complicated math)

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Re: BR Maths Corner-1

Postby sudarshan » 04 Oct 2020 22:40

.
.
.
1. But it is GM of some numbers - out of which L can not be there as one of the numbers . (otherwise p will divide this GM).
.
.
.
2. Now same logic for L3- the next largest number.
...


This is what I was missing! I got that L and L2 should both be divisible by a prime factor of the AM, but I arrived at it in a different way, and missed the first statement above.

For the second statement, I figured that all the lower numbers had to be divisible by the same prime factor, but didn't think of repeating the logic with the next largest number. So I wasn't able to prove that intuition.

So close!

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Re: BR Maths Corner-1

Postby Amber G. » 05 Oct 2020 01:55

^^^ Yes...
(The critical part was to see that every prime factor of L must also be a factor of L2 (and thus all other numbers)..The problem becomes a number-theory problem rather than combinatorial which was a interesting part)

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Re: BR Maths Corner-1

Postby Vayutuvan » 05 Oct 2020 05:56

sudarshan wrote:VT, I verified (numerically) that the series on the RHS plus that continued fraction does come to the LHS on that e/ pi problem. Now what? Expand the series for e and pi, multiply terms, mix and match?


I saw this. I will answer after collecting my thoughts.

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Re: BR Maths Corner-1

Postby Amber G. » 05 Oct 2020 06:44

^^^ Trouble or interesting part is there are many examples in math which sometimes boggle even humans. For me, for example one the following from Ramanujan's time --
If Pi(x) is number of primes less than x and Li(x) = integral of 1/log (t) from 0 to x.
Then Pi(x) - Li(x) > 0 is true for x =1, 2, 3, .... 1000, 1000,000 .....10000000000000, ...
It is true for 1 billion .. 1 billion billion .. i billion billion billion but not NOT for all numbers ..
You have to go about 10^300 or more to find that the number becomes -ive

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Re: BR Maths Corner-1

Postby Amber G. » 05 Oct 2020 21:11

For those who are interested in Ramanujan's work here is a current talk - from none other than Prof Ono ( Kyoto RIMS) in case you want to watch - the link will work for next 7 days..
quantum modular forms and representation theory

(In my humble opinion - this talk is excellent - happens to deal with some of the points/theory behind some of the topics we were talking about (eg elliptical functions/cubic equations) - it is understandable too for people like me as it puts many complex items in simpler terms)

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Re: BR Maths Corner-1

Postby Amber G. » 16 Oct 2020 00:25

Thursday (In Hindi = गुरुवार - A day to honor Gurus).
Pictured here are two gurus - great mathematicians - who were also great human beings. One of them received Nobel Prize and the other was honored by "Bharat Ratn "- (the highest civilian award given by India).
Image

Can you identify them?

Here they are getting the honors:
Image

Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 21 Oct 2020 02:23

Ben Sparks explains (and codes) the so-called SIR Model being used to predict the spread of cornavirus (COVID-19).

More links (see in the youtube)
https://youtu.be/k6nLfCbAzgo

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Re: BR Maths Corner-1

Postby Yayavar » 22 Oct 2020 12:53

Amber G. wrote:Thursday (In Hindi = गुरुवार - A day to honor Gurus).
Pictured here are two gurus - great mathematicians - who were also great human beings. One of them received Nobel Prize and the other was honored by "Bharat Ratn "- (the highest civilian award given by India).
Image

Can you identify them?

Here they are getting the honors:
Image


Maharshi Karve.

Amber G.
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Re: BR Maths Corner-1

Postby Amber G. » 22 Oct 2020 19:15

^^^ Yes, left of Maharshi Prof Karve is Physicist Albert Einstein. Photo (if I am not mistaken) was taken in Germany.

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Re: BR Maths Corner-1

Postby sudarshan » 23 Oct 2020 00:19

Amber G ji, maybe you could suggest to Dr. Vidyasagar et al, that the bijaganita (tr.: algebraic; literally: "seed mathematics") terms in that supermodel could use Sanskrit aksharas instead of Greek letters. There's more than enough letters in Indian languages, and after all, the concept of bijaganita did originate in India.


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