BR Maths Corner1
Re: BR Maths Corner1
^^^
Are you looking for a good preconditioner?
Are you looking for a good preconditioner?
Re: BR Maths Corner1
The sum is π^2/6. The most elegant proof of convergence I have seen involves using a square of 2x2 and putting strips of width (1/2),(1/4)... and putting squares in those strips.
Always been told Euler solved it first.
Always been told Euler solved it first.

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Re: BR Maths Corner1
Najunamar wrote:The sum is π^2/6. The most elegant proof of convergence I have seen involves using a square of 2x2 and putting strips of width (1/2),(1/4)... and putting squares in those strips.
Always been told Euler solved it first.
Yes. It is pi^2/6 . But how do you do it without integral calculus in an "elementary" way ? It can't be that elementary , especially if it took someone with a mind like Euler's to do it.
Re: BR Maths Corner1
vina wrote:Najunamar wrote:The sum is π^2/6. The most elegant proof of convergence I have seen involves using a square of 2x2 and putting strips of width (1/2),(1/4)... and putting squares in those strips.
Always been told Euler solved it first.
Yes. It is pi^2/6 . But how do you do it without integral calculus in an "elementary" way ? It can't be that elementary , especially if it took someone with a mind like Euler's to do it.
Wiki article on Basel Problem gives one elementary proof using Trig. Also a link in the references gives a write up that has 14 other proofs.
Wiki gives an elementary proof for 2 being an upper bound through the inequality
1/n^2 < 1/(n1)  1/n for n >= 2 and the partial sum telescopes to S_n < 2  1/n. So, as n goes to infinity, the inequality holds with S_n < 2.
Q for AmberG  why is Ramanujam connected to this problem?
(Edited to complete properly).
Re: BR Maths Corner1
^^^ Search wiki for zeta function (or "zeta elementary proof" etc..) for details...
For some elementary proofs see (for example I get this from google):
http://math.stackexchange.com/questions/8337/differentmethodstocomputesumlimitsn1inftyfrac1n2
or
http://www.freag.net/de/t/4v0u/references_zeta
or
this is perhaps mentioned above 
http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf
****
I have not seen mentioned it in "elementary" calculus .. (for example, calculus books in Engineering schools in India I have seen, just mentioned the result without proof, just mentioning that proof is not easy)... In my case, first time I saw the proof in a text book was in graduate school (IIT math graduate course), using MittagLefler theorem (wiki it) from complex variable theory using partial fractional representation of cotangent..
that is start with:
K pi cot (k*pi) = 1 + 2k^2(1/(k^21)+1/(k^24)+1/(k^29) + ...)
expand it.. etc..
I have some of my own favorite methods ...I may post in separate posts..
Wrt to Ramanujan ..I may post a few fun comments, but in the meanwhile check out google .. search for "ramanujan and zeta" function..quite interesting.
For some elementary proofs see (for example I get this from google):
http://math.stackexchange.com/questions/8337/differentmethodstocomputesumlimitsn1inftyfrac1n2
or
http://www.freag.net/de/t/4v0u/references_zeta
or
this is perhaps mentioned above 
http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf
****
I have not seen mentioned it in "elementary" calculus .. (for example, calculus books in Engineering schools in India I have seen, just mentioned the result without proof, just mentioning that proof is not easy)... In my case, first time I saw the proof in a text book was in graduate school (IIT math graduate course), using MittagLefler theorem (wiki it) from complex variable theory using partial fractional representation of cotangent..
that is start with:
K pi cot (k*pi) = 1 + 2k^2(1/(k^21)+1/(k^24)+1/(k^29) + ...)
expand it.. etc..
I have some of my own favorite methods ...I may post in separate posts..
Wrt to Ramanujan ..I may post a few fun comments, but in the meanwhile check out google .. search for "ramanujan and zeta" function..quite interesting.
Re: BR Maths Corner1
Here is Euler's proof, quite simple in a way but not rigorous but Euler being Euler, it gives the correct result..
Compare the expression of (sin(x)/x) as a function of (x^2)
sin (x) = x  x^3/6 + x^5/5! .....
if I put x^2 = y
(sin (x) /x) = 1  y/6 + y^2/120  ....
Sum of reciprocals of roots of y = (1/6)
Since the roots of the equations are at {(pi^2), (2pi)^2, (3pi)^2, ...)
we have (1/6) = (1/pi^2) (1+1/2^2+1/3^2 + ......)
so 1+ 1/4+ 1/9 + ... = pi^2/6
Compare the expression of (sin(x)/x) as a function of (x^2)
sin (x) = x  x^3/6 + x^5/5! .....
if I put x^2 = y
(sin (x) /x) = 1  y/6 + y^2/120  ....
Sum of reciprocals of roots of y = (1/6)
Since the roots of the equations are at {(pi^2), (2pi)^2, (3pi)^2, ...)
we have (1/6) = (1/pi^2) (1+1/2^2+1/3^2 + ......)
so 1+ 1/4+ 1/9 + ... = pi^2/6

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Re: BR Maths Corner1
Amber G. wrote:check out google .. search for "ramanujan and zeta" function..quite interesting.
That man was simply simply amazing . Truly the Man Who Knew Infinity and really Saraswati's favorite son.
Re: BR Maths Corner1
^^^ One of the most (if not the most in many people's views) important/famous theorem yet to be solved/proved is Riemann hypothesis.. some what connected with above type of series..One of the most celebrated work by Ramanujan is with zeta functions..
(zeta (2) is above series... zeta(3) = 1 + 1/2^3 + 1/3^3 + ... etc)
There are many interesting and awesome results and connection between this and prime numbers...
For example.. the above series is also equal to
(1/(11/4)) (1/(11/9))(1/(11/5^2)) ...
That is terms are product of factors like (1/(11/p^2)) where p goes (2,3,5,7,..prime number)
Pretty fantastic...
Even more fantastic .. (almost lunatic..) result credited to Ramanujan is
1+2+3+4+5 ..... = 1/12
(Don't believe me, check out
http://en.wikipedia.org/wiki/Ramanujan_summation
(When Ramanujan wrote that famous letter to Hardy about such "methods", he said something to the effect  "If I showed you these without really explaining my methods, you will send me to mental asylum ..."
(zeta (2) is above series... zeta(3) = 1 + 1/2^3 + 1/3^3 + ... etc)
There are many interesting and awesome results and connection between this and prime numbers...
For example.. the above series is also equal to
(1/(11/4)) (1/(11/9))(1/(11/5^2)) ...
That is terms are product of factors like (1/(11/p^2)) where p goes (2,3,5,7,..prime number)
Pretty fantastic...
Even more fantastic .. (almost lunatic..) result credited to Ramanujan is
1+2+3+4+5 ..... = 1/12
(Don't believe me, check out
http://en.wikipedia.org/wiki/Ramanujan_summation
(When Ramanujan wrote that famous letter to Hardy about such "methods", he said something to the effect  "If I showed you these without really explaining my methods, you will send me to mental asylum ..."
Re: BR Maths Corner1
^^^Ramanujan is simply amazing  especially in light of Bruce Berndt's work we now know several things about him (of course Hardy). One of the biggest obstacles he faced is expressing his mathematical thoughts in the modern language of math developed by Russell and Whitehead. It was very difficult for him to communicate. God knows what he would have done if he had formal training in math as we do...
Added later: AmberJi, I was thinking about the problem in terms of convergent series  I knew the answer was pi^2/6 but was unable to derive (without calculus). That's where the genius of Euler stands considering that his productivity did not diminish in spite of going blind during the later part of his life.
Added later: AmberJi, I was thinking about the problem in terms of convergent series  I knew the answer was pi^2/6 but was unable to derive (without calculus). That's where the genius of Euler stands considering that his productivity did not diminish in spite of going blind during the later part of his life.
Re: BR Maths Corner1
Okay, let me post a classical question, I would urge people who know the proof to wait...
Qn: Show that the primes are infinite in number.
The proof is attributed to yet another genius. I simply love the proof as it shows the elegance, beauty and the sublime power of reductio ad absurdum.
Qn: Show that the primes are infinite in number.
The proof is attributed to yet another genius. I simply love the proof as it shows the elegance, beauty and the sublime power of reductio ad absurdum.
Re: BR Maths Corner1
^^^ Here is one of the most "show off" proof of above, which is a favorite of mine..
See above few posts related to zeta function and its relationship with prime numbers..
We know that (for proof just see Ramanujan's note book .. )
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
Now if prime numbers are finite ==> RHS is rational number...
But we know LHS is irrational (because pi is irrational)
Contradiction..(reductio ad absurdum)
Hence there are infinite number of primes..
(BTW this has been discussed on page 1 of this dhaga)
See above few posts related to zeta function and its relationship with prime numbers..
We know that (for proof just see Ramanujan's note book .. )
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
Now if prime numbers are finite ==> RHS is rational number...
But we know LHS is irrational (because pi is irrational)
Contradiction..(reductio ad absurdum)
Hence there are infinite number of primes..
(BTW this has been discussed on page 1 of this dhaga)
Re: BR Maths Corner1
kasthuri wrote:...
Added later: AmberJi, I was thinking about the problem in terms of convergent series  I knew the answer was pi^2/6 but was unable to derive (without calculus).
Actually even with calculus, it is not that easy to derive. This is what fascinates us that with such a simple series (where you have 1,2,3...) you get pi in the answer.
That's where the genius of Euler stands considering that his productivity did not diminish in spite of going blind during the later part of his life.
He was also a devoted son, took care of his mother..
His outlook of life was extremely positive. Never uttered any unkind word to others. He lost sight in one eye.. when he lost the sight in the other eye (and went completely blind) ... his comment was something like "There goes distraction".. no bitterness.. he continued to do math , science etc...
(He lost sight in his right eye due to fever .. friends used to call him cyclops, he never minded..)
BTW he was one of the first person (this was in 1700's) to have successful cataract operation.
His productivity actually increased after he went completely blind.. His memory was perfect.. he could remember what he wrote on each page and each line of every paper he wrote .. (he would tell his scribes to see page xxx .. line yyy of zzz paper ... simply amazing memory)
He wrote some thing like 50 important papers in one year alone after he lost his sight.. (most of us have not written that even in our lifetime.)...
On his death one of the famous lines written were:
He ceased to calculate and to live
Re: BR Maths Corner1
Amber G. wrote:^^^ Here is one of the most "show off" proof of above, which is a favorite of mine..
See above few posts related to zeta function and its relationship with prime numbers..
We know that (for proof just see Ramanujan's note book .. )
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
Now if prime numbers are finite ==> RHS is rational number...
But we know LHS is irrational (because pi is irrational)
Contradiction..(reductio ad absurdum)
Hence there are infinite number of primes..
(BTW this has been discussed on page 1 of this dhaga)
Very nice! I never imagined you would relate this with the previous problem you posed. Doesn't the relation,
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
assume infinite number of primes? Otherwise, how would the product be convergent?
Anyway, sorry that I didn't see this problem was posted on the first page...so lets forget it.
Re: BR Maths Corner1
Amber G. wrote: This is what fascinates us that with such a simple series (where you have 1,2,3...) you get pi in the answer.
There is another beauty to this series. As you may know, sum (1/n) is divergent, whereas sum (1/n^2) is not...isn't it amazing that an order increase in denominator makes a whole lot of difference?
Re: BR Maths Corner1
kasthuri wrote:Okay, let me post a classical question, I would urge people who know the proof to wait...
Qn: Show that the primes are infinite in number.
The proof is attributed to yet another genius. I simply love the proof as it shows the elegance, beauty and the sublime power of reductio ad absurdum.
We've already done this one on this very thread. A geezer named Euclid supplied a very elegant proof.
Re: BR Maths Corner1
kasthuri wrote:<edited >
Very nice! I never imagined you would relate this with the previous problem you posed. Doesn't the relation,
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
assume infinite number of primes? Otherwise, how would the product be convergent?
Anyway, sorry that I didn't see this problem was posted on the first page...so lets forget it.
If the series is convergent than the product is finite .. if prime numbers are finite, the product is finite anyway. (And no, the proof does not assume that there are infinite primes)
Key point is, if pi is irrational, then RHS can not have finite number of terms.
(BTW, proof that pi is irrational was not produced till late (around 1800) and is much harder than, say proving that e is irrational.)
Re: BR Maths Corner1
Amber G. wrote:If the series is convergent than the product is finite .. if prime numbers are finite, the product is finite anyway. (And no, the proof does not assume that there are infinite primes)
Key point is, if pi is irrational, then RHS can not have finite number of terms.
(BTW, proof that pi is irrational was not produced till late (around 1800) and is much harder than, say proving that e is irrational.)
Ji, I noted the irrationality of pi argument (and the difficulty of the proving that pi is irrational). However, there is another key point  the RHS is equal to LHS. How do we derive that assuming the finiteness of primes?
Added later:
Sorry, I didn't read the proof that 6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
But, I guess that proof (in Ramanujan's notebook) would involve the infiniteness of the primes. And this is precisely the problem that your proof would *potentially* assume that there are infinite primes...
Re: BR Maths Corner1
kasthuri wrote:Ji, I noted the irrationality of pi argument (and the difficulty of the proving that pi is irrational). However, there is another key point  the RHS is equal to LHS. How do we derive that assuming the finiteness of primes?
Added later:
Sorry, I didn't read the proof that 6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
But, I guess that proof (in Ramanujan's notebook) would involve the infiniteness of the primes. And this is precisely the problem that your proof would *potentially* assume that there are infinite primes...
The proof (6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
) does NOT depend on infiniteness (or finiteness) of the prime numbers..(of course the proof is much harder.. requires good knowledge of number theory etc..).. it is like using a cannon to swat a fly but the proof is rigorous and valid. (Ramanujan had quite a few "methods" like that which were later got solid rigorousness behind them)
Re: BR Maths Corner1
Amber G. wrote:The proof (6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
) does NOT depend on infiniteness (or finiteness) of the prime numbers..
Amberji,
Just consider the RHS  (11/2^2)(11/3^2)(11/5^2)(11/7^2)...= Product(11/pn^2), where pn is a sequence of prime numbers.
The moment we write the above product, we *implicitly* assume the infiniteness of the prime numbers. The prime numbers are a sequence, which by definition has the domain the set of natural numbers which is infinite. Ramanujan's or other proof *need not* explicitly use the infiniteness of the primes. By writing primes as a sequence there is an implicit assumption. Even if we consider the above infinite product as limit of partial products, there is an assumption that given any n0 > N, the product (11/2^2)(11/3^2)...(11/pn0^2) exists which means the primes are essentially infinite.
Re: BR Maths Corner1
Product(11/pn^2), where pn is a sequence of prime numbers.
The moment we write the above product, we *implicitly* assume the infiniteness of the prime numbers.
... No, it is not a (necessary infinite) sequence, ALL that is assumed is that *all* prime numbers are used (finite or infinite).
Becomes clear if you look at the proof in any number theory book.
(Proofs are pretty rigorous)
****
Here is a similar argument, which may make it clear...
One can prove, for any M ... if all prime numbers less than M are .. p1,p2,p3....pn
Then 1+1/2+1/3+1/4+ .... 1/M < 1/((11/p1)(11/p2)(11/p3)..... (11/pn))
(Again proof is little hard but any number theory book will have it)
Again from this, if all prime numbers are finite ( n is finite) the RHS is finite, and for LHS M can be as large as one wants.. but we know LHS diverges... hence there can't be finite number of primes..
Re: BR Maths Corner1
kasthuri wrote:
There is another beauty to this series. As you may know, sum (1/n) is divergent, whereas sum (1/n^2) is not...isn't it amazing that an order increase in denominator makes a whole lot of difference?
As said above we do know that
1+1/2+1/3+1/4+.... diverges ...
How about the series
1/2+1/3+1/5+1/7+1/11+1/13 .... (reciprocals of all prime numbers only)
Is it convergent of divergent? (This was in the news when a famous bug in intel chip was discovered)
How about the series:
1+1/2+1/3+1/4+...1/7+1/8+1/10+.... 1/18+1/20+...
All digit excepts "9" appears in the sum.. all numbers which have "9" in them are ignored.
Re: BR Maths Corner1
Nice. I did some googling and found that the sum of reciprocal primes diverges (proved by Euler):
i.e. 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13..... diverges
but the sum of reciprocal twin primes (i.e. pairs of primes that only differ by 2 (e.g.) 3 and 5, 5 and 7, 11 and 13 etc.) converges to Brun's constant.
i.e. (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + (1/29 + 1/31) + .... converges
It was by attempting to calculate a more accurate value of Brun's constant using the newly introduced Pentium chip that the Intel FDIV bug was discovered.
@AmberG: Would you please mind looking at my post on previous page. I'm not sure if I follow your reasoning. Hope it doesn't cause too much trouble and thanks in advance for any clues .
i.e. 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13..... diverges
but the sum of reciprocal twin primes (i.e. pairs of primes that only differ by 2 (e.g.) 3 and 5, 5 and 7, 11 and 13 etc.) converges to Brun's constant.
i.e. (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + (1/29 + 1/31) + .... converges
It was by attempting to calculate a more accurate value of Brun's constant using the newly introduced Pentium chip that the Intel FDIV bug was discovered.
@AmberG: Would you please mind looking at my post on previous page. I'm not sure if I follow your reasoning. Hope it doesn't cause too much trouble and thanks in advance for any clues .
Re: BR Maths Corner1
^^^
1. Problem's origin is from old Hindu mathematicians, hence vanar and Jamun etc.. (The problem has appeared in various versions where jamuns become coconuts or something like that)
2. Since division by 3 happens 4 times, for any given solution one can add 3^4 to find another solution. (That is if 79 works, so will 79+3^4=160 and other higher numbers.. ( 160+81=241 etc).
3. If you think out of the box, and (instead of restricting yourself to positive numbers only), include negative numbers then it is quite obvious that (2) will work as a solution. (2 jamuns ... every brother gets 1 jamun and vanar gets 1 jamun leaves you back with 2 jamuns... and the process can repeat.)
4. 2 would be a good answer for the problem, but if one does not like the concept of "negative 2 jamuns".. one can always add 81 to 2 and get the next solution 79.
Problem is a good example where if one extends the domain it becomes easier to solve a problem.
I'm still not certain about how you derived 3^4 or the 2. What's that got to do with the word "vanar" anyway? ..
1. Problem's origin is from old Hindu mathematicians, hence vanar and Jamun etc.. (The problem has appeared in various versions where jamuns become coconuts or something like that)
2. Since division by 3 happens 4 times, for any given solution one can add 3^4 to find another solution. (That is if 79 works, so will 79+3^4=160 and other higher numbers.. ( 160+81=241 etc).
3. If you think out of the box, and (instead of restricting yourself to positive numbers only), include negative numbers then it is quite obvious that (2) will work as a solution. (2 jamuns ... every brother gets 1 jamun and vanar gets 1 jamun leaves you back with 2 jamuns... and the process can repeat.)
4. 2 would be a good answer for the problem, but if one does not like the concept of "negative 2 jamuns".. one can always add 81 to 2 and get the next solution 79.
Problem is a good example where if one extends the domain it becomes easier to solve a problem.
Re: BR Maths Corner1
ArmenT wrote:Nice. I did some googling and found that the sum of reciprocal primes diverges (proved by Euler):
i.e. 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13..... diverges
but the sum of reciprocal twin primes (i.e. pairs of primes that only differ by 2 (e.g.) 3 and 5, 5 and 7, 11 and 13 etc.) converges to Brun's constant.
.
Thanks. (posts above + a little more logic .. and one can have the proof that the first series indeed diverges..).. It diverges so slowly that even if one takes 10^100 terms the sum would still be less 10)
Re: BR Maths Corner1
I believer this is the Indian team for the 53rd International Mathematical Olympiad (2012). ( July 416 in Mar del Plata, Argentina)
Good luck to The Indian Team:
(1) Debdyuti Banerjee
(2) Akashdeep Dey
(3) Prafulla Sushil Dhariwal
(4) Shubham Bhakta
(5) Rijul Saini
(6) Mrudul Madhav Thatte
Good luck to The Indian Team:
(1) Debdyuti Banerjee
(2) Akashdeep Dey
(3) Prafulla Sushil Dhariwal
(4) Shubham Bhakta
(5) Rijul Saini
(6) Mrudul Madhav Thatte
Re: BR Maths Corner1
^^^ This is (I believe) US team:
The 2012 USA team includes:
Xiaoyu He (gold medals at the 2010 and 2011 IMO)
Ravi Jagadeesan ( sophmore at Phillips Exeter Academy, Exeter, NH) (Jagadeesan has taken part in the US Physics Olympiad and the North American Computational Linguistics Olympiad)
Mitchell Lee, (gold medal 2011 IMO).
Bobby Shen
David Yang (gold medal 2011 IMO)
The 2012 USA team includes:
Xiaoyu He (gold medals at the 2010 and 2011 IMO)
Ravi Jagadeesan ( sophmore at Phillips Exeter Academy, Exeter, NH) (Jagadeesan has taken part in the US Physics Olympiad and the North American Computational Linguistics Olympiad)
Mitchell Lee, (gold medal 2011 IMO).
Bobby Shen
David Yang (gold medal 2011 IMO)
Re: BR Maths Corner1
Some may enjoy this MAAproduced documentary "Hard Problems" (Which follows U.S. team during the 2006 IMO) ...
Link: The Road to the World's Toughest Math Contest
Link: The Road to the World's Toughest Math Contest
Re: BR Maths Corner1
Amber G. wrote:^^^ Here is one of the most "show off" proof of above, which is a favorite of mine..
Haven't seen this efore. Know the Euclid's, of course. The above is really very short and elegant proof.
Re: BR Maths Corner1
This is a little out of the box. But I would like to know the best methods to create deep abiding interest (and/or groom talent) in mathematics/Physics in young kids. Noting that we live in Mumbai
We may be starting a family soon and I would like to ensure that kids growing up in the house have access to all resources so that they can develop strong mathematical skills. I am only marginally competent in higher mathematics (that is to say most of problems in this dhaaga are beyond me at this stage) but I religiously visit it to understand problems being posed.
I have already noted a correlation with classical musical training and mathematical aptitude. So the kids when they show up will get some instrumental music training. I would like to know what books, puzzles, other games etc need to inculcated at an early age to get the kids to develop mathematics.
We may be starting a family soon and I would like to ensure that kids growing up in the house have access to all resources so that they can develop strong mathematical skills. I am only marginally competent in higher mathematics (that is to say most of problems in this dhaaga are beyond me at this stage) but I religiously visit it to understand problems being posed.
I have already noted a correlation with classical musical training and mathematical aptitude. So the kids when they show up will get some instrumental music training. I would like to know what books, puzzles, other games etc need to inculcated at an early age to get the kids to develop mathematics.
Re: BR Maths Corner1
Vriksh wrote:I would like to know what books, puzzles, other games etc need to inculcated at an early age to get the kids to develop mathematics.
Vrikshji,
Just out of curiosity, why only math/physics?
Re: BR Maths Corner1
Amber G. wrote:Product(11/pn^2), where pn is a sequence of prime numbers.
The moment we write the above product, we *implicitly* assume the infiniteness of the prime numbers.
... No, it is not a (necessary infinite) sequence, ALL that is assumed is that *all* prime numbers are used (finite or infinite).
Becomes clear if you look at the proof in any number theory book.
(Proofs are pretty rigorous)
****
Here is a similar argument, which may make it clear...
One can prove, for any M ... if all prime numbers less than M are .. p1,p2,p3....pn
Then 1+1/2+1/3+1/4+ .... 1/M < 1/((11/p1)(11/p2)(11/p3)..... (11/pn))
(Again proof is little hard but any number theory book will have it)
Again from this, if all prime numbers are finite ( n is finite) the RHS is finite, and for LHS M can be as large as one wants.. but we know LHS diverges... hence there can't be finite number of primes..
Amber Ji,
While I really appreciate the proofs (especially, the last one) I would like to point out that most number theory books out there take the concept of limit (and convergence/divergence) for granted (which is okay btw as these are well established notions and a number theory book cannot be a course in calculus). The notion of limit and continuity has had a tough ride when we go through the history of mathematics (thanks to the monumental works of Karl Weierstrass and Cauchy). The beauty and the simplicity of the Euclid's original proof is that it doesn't involve any notion of infinity in its proof. When one says "All" in the proof above, there is an inherent assumption that the "All" is "infinite".
In other words, can one prove this statement
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
assuming the primes are finite?
I mean a rigorous mathematical proof. As a mathematician in my previous avatar I should be able to follow any rigor...
Let me point out the assumption of infiniteness of primes in the last proof. We one says "1+1/2+1/3+1/4+ .... 1/M" diverges, mathematically it means that given any natural number N, for all numbers M > N, the sequence of partial sums given by,
P_M = 1+1/2+1/3+1/4+ .... 1/M
diverges. If p1,p2,p3....pn are prime numbers less than M, and say pn > pn1>...> p1, then for sufficiently large M, pn>N for any given natural number N, and this is precisely the infinite assumption of the primes.
Again, calculus dictates here and infiniteness of primes *is* an assumption unless I see a formal proof assuming the finiteness of primes (which I don't think exists).
Re: BR Maths Corner1
@kasthuri:
Math/Physics because to me that is the epitome of beauty. Everything else is subjective. Can't really explain it but I have seen too many bright people with tremendous mathematical gifts and intuitiveness both in IITB madarassa and Khan. Grew up with incredible stories of mathematical/physics elegance of people like Cauchy, Gauss, Maxwell, Newton, Ramanujam, Chandrasekhar, Raman, Salam, Bose etc and from a very young age wanted to emulate such worthies, but realize that one can only gaze in awe and hope that if not myself some kid around me could have similar talent.
Math/Physics because to me that is the epitome of beauty. Everything else is subjective. Can't really explain it but I have seen too many bright people with tremendous mathematical gifts and intuitiveness both in IITB madarassa and Khan. Grew up with incredible stories of mathematical/physics elegance of people like Cauchy, Gauss, Maxwell, Newton, Ramanujam, Chandrasekhar, Raman, Salam, Bose etc and from a very young age wanted to emulate such worthies, but realize that one can only gaze in awe and hope that if not myself some kid around me could have similar talent.
Re: BR Maths Corner1
kasthuri wrote:
In other words, can one prove this statement
6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
assuming the primes are finite?
I mean a rigorous mathematical proof. As a mathematician in my previous avatar I should be able to follow any rigor...
Thanks for the comments, I appreciate it.
Short answer is Yes (that is the proof is rigorous, and one does NOT have to assume that primes are infinite to prove they are infinite).
Some one just showed me Number Theory By GE Andrews, it actually has the same proof which I gave (except it comes after devoting pages to prove that zeta(2) is pi^2/6 etc..).. for details see that book or any other reputable text book.
Here is simple google search .. (the top result  more suitable references are likely to be there)
http://mathoverflow.net/questions/16991/whataretheconnectionsbetweenpiandprimenumbers
You can use the fact that zeta(2)=pi^2/6 to prove the infinitude of primes. If there were finitely many, then the Euler product for zeta(2) would be a rational number, contradicting the irrationality of pi. ... <see the references given there>
(BTW, the problem posted there.. the probability that two integers are prime to each other is another classic problem.. I showed a very elementary and beautiful proof of that to my son  which used no calculus)
Let me point out the assumption of infiniteness of primes in the last proof. We one says "1+1/2+1/3+1/4+ .... 1/M" diverges, mathematically it means that given any natural number N, for all numbers M > N, the sequence of partial sums given by,
P_M = 1+1/2+1/3+1/4+ .... 1/M
diverges. If p1,p2,p3....pn are prime numbers less than M, and say pn > pn1>...> p1, then for sufficiently large M, pn>N for any given natural number N, and this is precisely the infinite assumption of the primes.
I simply put that as if "S_n = 1+1/2+1/3+ ....1/n"
(No infinity involved here, hence it is all valid)
given any number x (however large)
, I can always find n so that S_n > x .
****
Another valid proof for infinitude of primes could be (is in many books)
Let p_n be the number of prime less than n.
One can prove (technique is very similar to above... it is hard but is given in standard number theory books.)
p_n > k n / (ln n) (where k is a known constant)
(This BTW is based on one of Ramanujan's famous work)
Now since you can make RHS as large as you want ==> p_n can be as large as one wants by simply selecting a suitable n.
That completes the proof
*** Added later:
You may also like to see wiki: http://en.wikipedia.org/wiki/Euler_product (pay special attention when "infinity" symbol is used. and when they use product "over primes"  it does not assume that primes are infinite..)
or http://www.physicsforums.com/showthread.php?t=487354
or Worlfram http://mathworld.wolfram.com/PrimeProducts.html
or http://mathworld.wolfram.com/RiemannZet ... Zeta2.html
Re: BR Maths Corner1
^^^I am trying to reason out the difference in the view points. Btw, I am aware of the prime number theorem (the last proof) stated above. It is one of the classical results (also attributed to Guass).
Re: BR Maths Corner1
I think that proof is a valid proof by contradiction. A set is either finite or infinite (Law of excluded middle). Kasthuri ji, are you objecting to the kind of infinity?
Re: BR Maths Corner1
matrimc wrote:I think that proof is a valid proof by contradiction. A set is either finite or infinite (Law of excluded middle). Kasthuri ji, are you objecting to the kind of infinity?
matrimc ji,
This is precisely my point. A set is either finite or infinite. When we say, the set of all primes in
1. 6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
or in
2. for any M ... if all prime numbers less than M are .. p1,p2,p3....pn
Then 1+1/2+1/3+1/4+ .... 1/M < 1/((11/p1)(11/p2)(11/p3)..... (11/pn))
we *have* to assume they are infinite. If we assume they are finite, then the above two relations won't hold.
Re: BR Maths Corner1
kasthuri wrote:^^^I am trying to reason out the difference in the view points. Btw, I am aware of the prime number theorem (the last proof) stated above. It is one of the classical results (also attributed to Guass).
About Prime number theorem Yes it is classical and is attributed to Gauss. To add  Gauss mentioned and studied it, but it was not proven till almost 1900 (Poussin ? ) (using very sophisticated/complex methods) Erdos (my hero as mentioned in previous pages) was the first one to give without using complex variables and, in my opinion, is the only proof I have seen which is easy to understand..
Ramanujan, of course, is famous for tackling some of the most complex results/improvement of this.. (His original claim to find the precise formula did not come out to be correct but he did refined these results quite a bit)
But we may be getting to much out of the general discussion..here ...
BTW have you read the book "Men who knew Infinity" .. if not, strongly recommend it. (This is a book about Ramanujan).
Last edited by Amber G. on 02 Jun 2012 23:42, edited 1 time in total.
Re: BR Maths Corner1
kasthuri wrote:matrimc wrote:I think that proof is a valid proof by contradiction. A set is either finite or infinite (Law of excluded middle). Kasthuri ji, are you objecting to the kind of infinity?
matrimc ji,
This is precisely my point. A set is either finite or infinite. When we say, the set of all primes in
1. 6/pi^2 = (11/2^2)(11/3^2)(11/5^2)(11/7^2) ... (all prime numbers)
or in
2. for any M ... if all prime numbers less than M are .. p1,p2,p3....pn
Then 1+1/2+1/3+1/4+ .... 1/M < 1/((11/p1)(11/p2)(11/p3)..... (11/pn))
we *have* to assume they are infinite. If we assume they are finite, then the above two relations won't hold.
No, we don't have to assume they are infinite...
In 2, n is explicitly finite, (all prime numbers less than M) one can prove that for any given M (finite, however large) the relationship holds..
Now if one assumes that there are only finite primes, (say k primes only), then RHS is (always) finite ( even when you take k = n)
But I can always choose M large enough, so that LHS can be as large as I want, if I make it larger than RHS .. there is a contradiction..
QED
***
In 1 
Again one can use the same logic (in the proof to derive the zeta(2) formula) The RHS here just says *all* the primes. The proof DOES not require finiteness.
(The critical part of proof is some what related to the Fundamental theorem of arithmetic (click here for details) ..
(That is, if finite number of primes can "generate" all the numbers then pi has to rational..(which is not true))
***
A trivial example to exhibit the point that one can use "..." but it does not mean infinite numbers..
For example I can say
1+1/2^2+1/3^2 + ...... = (product over all factors of 10)( pi^2/600)
Does not mean there are infinite factors of 10 .. though, if I wanted, I can have written them as (1,2,5....).
hth
Last edited by Amber G. on 02 Jun 2012 23:41, edited 2 times in total.
Re: BR Maths Corner1
Amber G. wrote:kasthuri wrote:^^^I am trying to reason out the difference in the view points. Btw, I am aware of the prime number theorem (the last proof) stated above. It is one of the classical results (also attributed to Guass).
To add about Prime number theorem  Gauss mentioned and studied it, but it was not proven till almost 1900 (Poussin ? ) (using very sophisticated/complex methods) Erdos (my hero as mentioned in previous pages) was the first one to give without using complex variables and, in my opinion, is the only proof I have seen which is easy to understand..
Ramanujan, of course, is famous for tackling some of the most complex results/improvement of this.. (His original claim to find the precise formula did not come out to be correct but he did refined these results quite a bit)
But we may be getting to much out of the general discussion..here ...
BTW have you read the book "Men who knew Infinity" .. if not, strongly recommend it. (This is a book about Ramanujan).
Yes, I have read "Men who knew Infinity" and "A mathematician's apology" both of which are gems. If you like Erdos, you should watch the documentary "N is a number". Erdos used to call children as "epsilons" and he said "A mathematician is a machine that turns coffee into theorems". Erdos whenever he used to visit Texas A&M, he used to stay in my wife's PhD. mentor's house. His name is Dr. Hobbs, a graph theorist.
Re: BR Maths Corner1
Also "The Man Who Loved Only Numbers." It's by Paul Hoffman.
^^^ Thanks about mentioning Dr Hobbs. Small world. He was a good friend of Erdos and liked to tell stories about him in his class ..there is interesting story told by him (I am sure you have heard it) of becoming Erods number of 1 (those who wrote a paper with him)
Here is the story..(Thanks to google I found a reference so I can just cut and paste)
^^^ Thanks about mentioning Dr Hobbs. Small world. He was a good friend of Erdos and liked to tell stories about him in his class ..there is interesting story told by him (I am sure you have heard it) of becoming Erods number of 1 (those who wrote a paper with him)
Here is the story..(Thanks to google I found a reference so I can just cut and paste)
Dr. Hobbs got accepted to Waterloo, Paul Erdos also visited Waterloo since he dabbled in graph theory (as with everything) himself. Hobbs found out that Erdos was going to some conference and made sure to get on the same bus. He went up to Erdos and expressed his desire to publish a paper with him (he wanted an Erdos number of 1). So, Erdos asked him if he had any idea for what they could write about. Hobbs replied, "I was hoping you did." Erdos sighed and asked Hobbs what he did. "I did my thesis on Hamiltonian cycles." Okay... is every cubic graph with girth less than 4 Hamiltonian? Hobbs thought he would try the Dirac formula on the question, but if Erdos hadn't solved it yet then that probably didn't work. Sure enough, the Dirac formula didn't work. He woke up Erdos (Erdos tended to sleep when people bored him) and told him what he found. Erdos nodded, gave him an idea of a proof, and where the hard spots were. Hobbs went back to work, solved the problem, and got his Erdos number
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