## BR Maths Corner-1

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Amber G.
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### Re: BR Maths Corner-1

Schmidt wrote:

We can use a^3 + b^3 >= a^2b + ab^2
...

Excellent. This concept (Rearrangement Inequality #note 1), is, in my opinion, intuitively more obvious and powerful than AM/GM type concept. (This is what I was thinking when I made the problem)

If there are \$100 bills and \$1 bills, and you have to pick 10 bills of one kind, and 1 bill of the other, even a child will pick 10 bills of \$100 and 1 of \$1 rather than the other way. (In general a*a^2 + b* b^2 >= a * b^2 + b* a^2 === same concept)

So rewriting what you wrote:
given abc=1
a^3+b^3 >= ab (a+b) = (a+b)/c ==> a^3+b^3+1 >= (a+b+c)/c ==> 1/(a^3+b^3+1) <= c/(a+b+c)
Rest is simple.
(Original problem a^5+b^5+ab reduces the same way by noticing a^5+b^5 >= a^2*b^2(a+b))
***
Unfortunately I have not seen this (Rearrangement Inequality) covered in a typical math class. Though it is quite powerful - one can easily derive AM/GM or power-mean or Cauchy–Schwarz Inequality from this.

Intuitively obvious, this inequality is: (/taking lazy way out and not to write a longer post / --> Just look it up in wiki ) .
One more powerful part is, unlike AM/GM the values here don't have to be positive.
Last edited by Amber G. on 25 Jun 2020 00:51, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

Wow. I really like that Rearrangement Inequality. I did not come across it before. The one I know best is Cauchy-Schwartz Inequality and AM > GM.

The problem you have given above was posed in the Facebook group called Inequalities. One proof made the same mistake as above. I pointed out the mistake, but whatever I tried, I couldn't fix it.

Amber G.
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### Re: BR Maths Corner-1

^^^Thanks, glad it was enjoyable. Yes, this inequality is not as well known, but in my coaching this is one concept I *always* cover (along with Power-Mean, AM/GM/HM and Cauchy-Schwartz - and perhaps Holder, Jensen etc.- my approach is more of intuitive understanding rather than formal treatment).

With this, for example a*a+b*b >= a*b +b*a is nothing but AM/GM.

Also it is *very* intuitive - If one has to pick, say (a,b,c) bills from a stack of \$100, \$10, \$5 etc bills, .. and given a>=b>c , best strategy is to pick a \$100 bills, b \$10 bills, c \$5 bills etc..).

Another way to see this is intuitively obvious, as any physicist or a kids can see, is to think of bunch of kids (weights - x_1, x_2, x_3..) sitting on a sew-saw at distances (y_1, y_2,. y_3...)from the fulcrum. The greatest turning moment around the fulcrum will be obtained when the heaviest kid sits the farthest away from the fulcrum .. and so on. This is nothing but the arrangement inequality.
---
Both US and India's olympiads (and JEE type exams) and International Math Olympiads often have such problems of inequalities, so learning these tools are good for exams too .
---

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Formal solution, as I posted before is:

x = ( (3+2 √ 2)^n - (3 - 2 √ 2)^n) / (4 √ 2)
y = ((3+2 √ 2) ^n + (3 - 2 √ 2) ^n - 2)/ 4

Answers are given by choosing n=0,1,2,3,....

Surprising the "practical" solution is *VERY* easy as 3+2 √ 2 ≈ 5.82842712475 all you have to do is to find (5.82842712475^n) .. for example
y=(5.82842712475^n - 2)/4 (Just take the integer part only .)

OK, but how does one arrive at this solution, algebraically?

---
In some of the books, I have seen the formula is credited to Euler, but of course, this was well known to Brahmgupta and *many* others for thousands of years.

The method, as you can see, is fairly well known, easy to prove (extremely easy to prove). Interesting this theme does appear many times in competitive exams.
(Similar problem I asked here in brf before - find a right-angle triangle with integer sides (a,b,c) such that b=a+1. (Examples (3,4,5))
-------
Sudarshanji - Just curious, did you find any other solution to my first problem - how high up did you (or your computer program) go? TIA.

The first problem meaning - the sum of integers up to y being equal to the sum of squares up to x? I don't have a computer program, was doing it by hand in Excel, so only got the first two <x,y> pairs. Even the easier problem was set up in Excel.

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Okay - here is a simple problem (as easy as 1,2,3 ) I posed in a group consisting of mainly Google Engineer's

Find three positive rational numbers such that their sum is 6 and the product is 6.

One answer is 1,2,3 as 1+2+3 = 1*2*3 = 6
(Only fractions - rational numbers which are positives - are allowed.
(Let us see if brfites visiting this forum can find the answer(s) before the other group... computers /books etc are allowed .. googling is discouraged but I can't stop it )

(Request: Please For this problem (only) _ No invoking great mathematicians and writing about fantastic theorems ... Just the answer(s) please.. if one answer appears find a different answer - Imagine your audience just know how to add and multiply ordinary fractions! I want to see how many answers we get here in brf ...)

So this problem is:

abc=6;-------a+b+c=6

For a given a:

b+c=6-a;-------b(6-a-b)=6/a

Working with that last equation:

b^2-(6-a)*b+6/a=0

b=0.5*( (6-a) +/- sqrt( (6-a)^2 - 24/a ) )

For rational b, the discriminant has to be positive:

(6-a)^2 >= 24/a

(6-a)^2 is a parabola with a minimum at a=6; 24/a is a rectangular hyperbola with a discontinuity at a=0. We want all a where the hyperbola is below the parabola.

The hyperbola will be below the parabola for all a<0. There are also three points where the parabola and hyperbola have the same value (since this is a cubic equation). Too lazy to find the exact values of a, but roughly:

Root 1 (call it R1): a is between 0.9 and 1
R2: a is between 3 and 4
R3: a is between 7 and 8

For a<0, the hyperbola is of course below the parabola, but a can't be negative, so this case is out.

For 0<=a<R1, the hyperbola is above the parabola, hence this case is out.

For R1<=a<R2, the hyperbola is below the parabola (which is what we want).

For R2<=a<R3, the hyperbola is above the parabola, hence this case is out.

For R3<=a<Inf, the hyperbola is below the parabola, and a is also positive, but b and c will both turn out negative, hence this case is out.

So any value of a between R1 and R2 will work, except for the condition that a, b, and c all have to be rational. Not sure how to enforce this condition.

For a given a, it is easy to see that b will be one solution to the quadratic, and c will be the other solution - i.e.:

b= 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a) )

c= 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a) )

A further point - a=1, b=2, c=3 is a solution, but this is technically the same solution as a=2, b=1, c=3, or a=2, b=3, c=1, etc. So a further condition could be enforced, that a<=b<=c. This could modify the range of a (restrict it from the current R1<=a<=R2).

sudarshan
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### Re: BR Maths Corner-1

<a, b, c>=<49/15, 25/21, 54/35> (Rearrange in ascending order if so desired).

Check it out: a+b+c=6, abc=6

Obtained by a little logic and then trial and error, will elaborate in a while.

Amber G.
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### Re: BR Maths Corner-1

<a, b, c>=<49/15, 25/21, 54/35> (Rearrange in ascending order if so desired).

Check it out: a+b+c=6, abc=6

Obtained by a little logic and then trial and error, will elaborate in a while.

Great! Many (most) find (8, -3/2, -1/2) -- numbers which are not all positive, so I put the condition "positive" part.
Let us see if more such numbers are found! Amber G.
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### Re: BR Maths Corner-1

This is a Gem! For Ramanujan's 100th year - a webinar by none other than prof Professor Bruce Berndt. Save the link watch when you can.
Lot of Math, Photographs stories etc..
https://youtu.be/cAjMMvQIY9o

There was just today

A few days ago there was a presentation by Prof Ribet I will post the youtube or video link if there is some interest here. VERY nice lecture about Fermat's Last Theorem. (I do have PP slides of the lecture - will try to share it here). (The background needed to understand is graduate level math but need not be expert -- It is one of the most understandable lectures I have seen for FLT)

(BTW - with those two lectures .. the last few problems I posted would be child's play )

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:Great! Many (most) find (8, -3/2, -1/2) -- numbers which are not all positive, so I put the condition "positive" part.
Let us see if more such numbers are found! I tried:

Let a=x/y where x and y are both integers

b = 0.5 * ( (6-a) - sqrt( (6-a)^2 - 24/a ) )
c = 0.5 * ( (6-a) + sqrt( (6-a)^2 - 24/a ) )

b = ( 1 / ( 2 * y * sqrt(x) ) ) * ( sqrt(x) * (6*y-x) - sqrt( x*(6*y-x)^2 - 24*y^3 ) )
c = ( 1 / ( 2 * y * sqrt(x) ) ) * ( sqrt(x) * (6*y-x) + sqrt( x*(6*y-x)^2 - 24*y^3 ) )

For both numerator and denominator for b and c to be integers, one way could be that x is a perfect square (say k^2), with the additional condition that k * (6*y-x)^2 - 24*y^3 has to be a perfect square. Trying combinations of k and y, I kept getting <1,2,3> (various permutations of that) plus <8,-1.5,-0.5>, plus another solution which had b & c both being negative. But then k=7, y=15 yielded the solution above: <49/15, 25/21, 54/35>.

But of course, this is not the only way to get rational b & c.

Amber G.
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### Re: BR Maths Corner-1

^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

sudarshan
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### Re: BR Maths Corner-1

Amber G. wrote:^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

k=143, y=8023 worked: <a, b, c> (ascending order) = <15123/16159, 25538/10153, 20449/8023>.

Going much higher than that runs into issues with even .NET long (int64) types.

The other set with negatives that I found was:

<a, b, c> (ascending order) = <-361/68, -32/323, 867/76>.

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. wrote:This is a Gem! For Ramanujan's 100th year - a webinar by none other than prof Professor Bruce Berndt. Save the link watch when you can.
...
(BTW - with those two lectures .. the last few problems I posted would be child's play )

That reminds me, Prof. Berndt is retiring. He has several books and a stack of papers he ever published. I am supposed to go and root through the material and take whatever I can. He has biographies of several Indian mathematicians as well as the book "Vedic Mathematics". Our common friend, who is a vegan and yoga practitioner asked him about the book. Prof. Berndt answered thusly: "Hindus had a very advanced mathematical system thousands of years back".

I consider myself lucky to have known him.

Amber G.
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### Re: BR Maths Corner-1

^^^Thanks. I highly recommend the recent talk I posted, it is very good. Prof Bruce Berndt, along with others - many pieces written/edited/compiled with Freeman Dyson and him about Ramanjuan's work are very good (my personal favorites). UVA's Ken Ono, recently has given some good popular lectures about Ramanujan's work too.

Say Hi to prof. (last few problems in brf may amuse him - as they are inspired/compiled from Ramnujan's very popular work. )

For those who may not know - Prof Bruce Berndt's multi-volume books about "Ramanjuan's Papers/notebooks" is *very* good, it gives proofs of many conjunctures of Ramnujans and comments on insights of many of his concepts - It is a *very* good reference for anyone interested in this aspect. (There was an effort - may be still going on - to compile an wikipedia/encyclopedia type writing of all of Ramanujan's work).

Amber G.
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### Re: BR Maths Corner-1

sudarshan wrote:
Amber G. wrote:^^^ Very good!
How about trying some more numbers .. like k=143, and y=8023 or something like that. (Just curious, what other set you found where b,c were negative (other than -3/2, -1/2))..

k=143, y=8023 worked: <a, b, c> (ascending order) = <15123/16159, 25538/10153, 20449/8023>.

Going much higher than that runs into issues with even .NET long (int64) types.

....

As some may have guessed, there is a fairly simple method to attack these kind of problems.. one can get generate these numbers, as many as you want, fairly easily - once the method is known - the fun part is - the numbers go big quite fast so without theory brute force calculation can only produce a few numbers. (There are infinite set of such numbers)

Some math theory here is quite interesting, and gets interesting and complicated. When I have some time (and if there is interest) I may put some comments here, but standard math courses in number-theory cover this.

Amber G.
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### Re: BR Maths Corner-1

Another interesting old problem, which just have been solved (lot of buzz in math world) is "square in a loop" problem. Basically this is an old problem, very easy to describe:

Draw a close loop. (Imagine you started from point A, took a trip, went to many different places and came back to point A): The conjecture was that one can *always* find 4 points on that closed loop / path such that these four points will make a square.

(People are claiming that it has been proven - I have not looked at the proof but was interested in that problem over the years).

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. wrote:As some may have guessed, there is a fairly simple method to attack these kind of problems.. one can get generate these numbers, as many as you want, fairly easily - once the method is known - the fun part is - the numbers go big quite fast so without theory brute force calculation can only produce a few numbers. (There are infinite set of such numbers)

I am thinking more on the lines of start with (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) and search intheir vicinity.

Let us assume that all three are equal to 2. Then their product is larger than 6. If one is close to 6, the other two have two are bounded fro below. If two are equal say close to 3, then the other is bounded from below by 2/3.

Let us assume that the three rational numbers are a/x, b/y, and c/z and also (a,x), (b,y), and (c,z) are pairs of relative primes. We need x, y, and z to be greater than 1 as well as a > x, b> y, c > z.

One equation we can get is cyc+(1/((b/y)(c/z)) = 1

where cyc+ denotes the cyclic addition.

...
?

Amber G.
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### Re: BR Maths Corner-1

Ronald Lewis Graham - One of the principal architects of the rapid development worldwide of discrete mathematics in recent years died today.

Vayutuvan
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### Re: BR Maths Corner-1

That is sad. Great man. I have the book "Concrete Mathematics" Knuth co-authored with Graham and Patashnik. I read large parts of it. but need to attmpt problems.
Last edited by Vayutuvan on 08 Jul 2020 23:00, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

https://writings.stephenwolfram.com/2016/04/who-was-ramanujan/

I highly recommend this blog article by Wolfram. He has great insights. I never met Wolfram even I passWolfram Corp. HQ on my way to work and back.

Amber G.
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### Re: BR Maths Corner-1

Vayutuvan wrote:That is sad. Great man. I have the book "Concrete Mathematics" Knuth co-authored with Graham and Patashnik.

He was also known for fun math - like magical tricks, and kind of problems yours truly puts in this thread. I think he contributed a lot for Martin Gardener's classic column (Mathematica Games which ran for decades and was very popular).. He will be missed by many.
Here is a picture from a few years ago: (With his Magical Mathematics Book) Amber G.
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### Re: BR Maths Corner-1

Speaking of Ramanujan .. These slides from a recent lecture by Prof Henri Darmon puts my April 1, post from some time ago in a whole new perspective.. . (BTW Darmon's lecture, as usual was excellent - you can find the slides and video which will be posted fairly soon on the university's cloud platform so just search for it)

Old post: Here:
(This was an April Fool Joke, BTW ..)

Amber G. wrote:On this day let me bring you an exciting story....

Anyway I got this email from a math friend who was working on some old Ramajnujan problem regarding Heegner theorem. She found an interesting result. The result is not published yet so you are reading it first hand here.

Let me present it here - she found that amazingly

(ln(12^3(231^2-1)^3+744)/pi)^2 = 163

Here ln = natural logarithm (that is base e)
Pi = 3.14159265....
etc..

(If you don't believe it, just use a calculator / google to check it out)

Of course, the result is rather surprising. Pi is an irrational number. And natural logarithms of even ordinary numbers are almost always irrational so finding an integer result of this rather complicated expression is VERY remarkable. Naturally it has baffled all those mathematicians who were on this email list. Once the result is published I think it will cause a major revolution in number theory.

I am sure, my math friends know that 163 is largest Heegner number - that is this is the largest number where quadratic imaginary number fields admit unique factorisation in their ring of integers.

Slides from the recent lecture, which many may find interesting or at least be amused by it -- and impressed by genius of Ramanujan. <... and.. > (For more, see Ramanujan's Constant say here: https://mathworld.wolfram.com/RamanujanConstant.html#:~:text=Numbers%20such%20as%20the%20Ramanujan,this%20sort%20of%20near%2Didentity.

Vayutuvan
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### Re: BR Maths Corner-1

One quick question @Amber G. Would Prof. Bruce Berendt recognize you by Amber G.?

Amber G.
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### Re: BR Maths Corner-1

^^^Vayutuvanji No. (But even if someone comes to know about more private information than I have shared here, please do not post. Thanks.
(He will certainly be familiar with the type of last few problems I posted and it will be *very* easy to give the answers or provide some very easy way to solve, so it will be very interesting to see his reaction)

Amber G.
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### Re: BR Maths Corner-1

Another, *very* interesting popular lecture from Prof Maryna Viazovska ( Invited lectures series shared by UVA on zoom.
How one can pack spheres in n-dimensional space with lot of interesting number-theory results and it's relation to physics.
(I think many here will like it)

https://virginia.zoom.us/rec/play/tcJ4dumtrjg3SYaR5QSDUf5wW420ev-s1iEZ_6ZcnkewUSEFOlX3YbVDZbd2uArG2Ba1c2_kp8JPdYOk?fbclid=IwAR1b1oEAZ_nXvkGhiJYTjGq6w-rMIfU3lLzlRheXuOhQGXtUSqbPZs1GjeU

Vayutuvan
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### Re: BR Maths Corner-1

Amber G. ji, no problem. I won't.

Does the "add/multiply to 6 problem" has anything to do with continued fraction expansion of sqrt(19)?

√19 = [4;2,1,3,1,2,8,2,1,3,1,2,8,...] (sequence A010124 in the OEIS). The pattern repeats indefinitely with a period of 6.

Amber G.
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### Re: BR Maths Corner-1

^^^ No, the continued fraction is related to Diophantine equations of second degree. Integer solutions for first or second degree equations are relatively easy and there are many standard methods.
For Cubic equations, the calculations are a little more involved.

For example, the sum/product = 6 equation is basically standard cubic equation y^2 = x^3 - 9x + 9 ;

Every integer solution x, can give three numbers (a,b,c) where a = 6/(3-x), b = (6 - 3x + y )/(3-x) and c= (6 - 3x - y)/(3-x)
(Check out, that a+b+c =6 and abc =6) so every rational x, will also give rational (a,b,c). (x != 3).
(In addition if x<2 all three values will be positive)
***
Easy to see x=0, (y^=9), x=1 (y^2=1), gives rational solutions - As said before there are standard methods to find out, if there are finite or infinite values in integer (or rational) domain and if there are infinite values, one can find as many as one wishes .. again standard methods (unfortunately not taught that much in ordinary math classes -

***
Some math is not that hard (that's why I posted here in brf, but theory and some concepts here - like modular functions etc are quite important in modern math and do get quite involved. BTW Fermat's Last Theorem depends in reducing x^n+y^n=z^n into a form very similar to
above equation (y^2 = x^3+....) and then proving it has no solutions in rational domain)... basically proving by contradiction that if FLT had one solution then t can write it the form of above equation etc... (Okay I am simplifying 600 page proof into a post . ).

****