BR Maths Corner1
Re: BR Maths Corner1
^^^^
Very funny . I tried hacking some code together, but it seemed to run for ever. Then I recalled something weird I'd read a while ago (no, I'm not a math major, but I tend to remember odd stuff for some reason).
Brocard's Problem
On the other hand, I did get some practice in hacking lua code before shifting to the ol' standby once the values started getting really large.
Very funny . I tried hacking some code together, but it seemed to run for ever. Then I recalled something weird I'd read a while ago (no, I'm not a math major, but I tend to remember odd stuff for some reason).
Brocard's Problem
On the other hand, I did get some practice in hacking lua code before shifting to the ol' standby once the values started getting really large.
Re: BR Maths Corner1
Dear BR Gurus,
I need some help understanding the zeller's congruence for day of the week problems.
Could someone provide an example? Im having problems getting the correct answer using the reference:
http://en.wikipedia.org/wiki/Zeller's_congruence
Im having case of...... weak brain cells......
Thanks already......
I need some help understanding the zeller's congruence for day of the week problems.
Could someone provide an example? Im having problems getting the correct answer using the reference:
http://en.wikipedia.org/wiki/Zeller's_congruence
Im having case of...... weak brain cells......
Thanks already......
Re: BR Maths Corner1
girish.r wrote:http://en.wikipedia.org/wiki/Zeller's_congruence
Im having case of...... weak brain cells......
Seems straight forward. Can you give an input date for which you are getting a wrong answer?
Re: BR Maths Corner1
Remembering Ramanujan: India Celebrates Its Famous Mathematical Son
notes of Ramanujam
kamments section
India, home of the number zero, ends a yearlong math party in unique fashion
An intuitive mathematical genius, Ramanujan's discoveries have influenced several areas of mathematics, but he is probably most famous for his contributions to number theory and infinite series, among them fascinating formulas (pdf ) that can be used to calculate digits of pi in unusual ways.
Last December Prime Minister Manmohan Singh declared 2012 to be a National Mathematics Year in India in honor of Ramanujan's quasiquicentennial.
India's mathematical heritage extends far beyond Ramanujan's time. The nation is considered home of the concept of zero. Babylonians had used a space as a placeholder (similar to the role of "0" in the number 101), but this space could not stand alone or at the end of a number. (In our number system, as in theirs, this could be problematic; imagine trying to tell the difference between the numbers 1 and 10 by context alone.) In India, however, zero was treated as a number like any other. India is also the home of our decimal numeral system.
ndian government and mathematical societies pursued several projects to celebrate their year of mathematics, from enrichment programs for students and teachers to the "Mathematical Panorama Lectures" that occurred around the country. This series of 20 short lecture courses, which will continue into 2013, brings prominent mathematicians from different fields to Indian universities to deliver five or six lectures. M. S. Raghunathan, president of the Ramanujan Mathematical Society and chair of the organizing committee for the National Mathematics Year, wrote in an email that he hopes the lectures will facilitate an infusion of Indian talent into fields that lack it right now.
Two longerterm projects begun this year could help as well: a documentary on the history of Indian mathematics and a mathematics museum in Chennai. Raghunathan hopes that the documentary will be available in 2014 and the museum will open its doors in 2015.
This yearlong fete is culminating in "The Legacy of Srinivasa Ramanujan," a conference at the University of Delhi from December 17 to 22. Included are technical lectures on mathematics influenced by Ramanujan's work, public presentations on Ramanujan's notebooks, dance performances and a film about Ramanujan's life. The annual SASTRA Ramanujan Prize, which recognizes a mathematician age 32 or younger who works in a field influenced by Ramanujan, will be awarded as well. The awardee this year is Zhiwei Yun of Stanford University, whose work lies at the intersection of geometric representation theory, algebraic geometry and number theory.
notes of Ramanujam
kamments section
Re: BR Maths Corner1
I quickly went through the comments section. There can be a lot speculations as to what would have happened, what could have happened etc. The book "The Man Who Knew Infinity" by Kanigel (which has been recommended in the past by AmberG, Kasturi, and yours truly) is worth reading to understand the circumstances and the British Mathematical environment at that time. According to the book British Math was weak in rigor and formalism (as opposed to the prcatices of their continental peers in France and Germany) and it was due to Hardy's efforts that British Mathematicians turned more rigorous. Since Ramanujan was selftaught from British books he had the same weakness. Hardy could see past the haze as he had already figured out what was ailing the British mathematical community and started a program to clean it up. In to this milieu he brought in Ramanujan. R's ideas were very original but formal proofs are a must for wider acceptance and these techniques he learnt from Hardy. This is where the lost note books (which George Andrews and Bruce Berendt started working on) come in. When R went back to India after he was wasting away from TB, he lived on his death bed for 10 months during which time he dictated several theorems to his wife who wrote them down and sent to Cambridge. These were sitting in the basement gathering dust until G. Andrews was rooting through some stuff and found the note books which are termed the last lost notebooks. Shortly thereafter he was joined by Bruce Berendt. Prof. Berendt and his students worked on these for several years and Prof. BB passed out about 10 doctoral students just based on the 10 month worth of R's notes. What I gathered from the book is that by this time R's writings became quite formal and sophisticated (though some amount of his original mystic still remained). Hardy needs to be given credit for polishing R's genius. That is only one of the things Hardy is famous for.
Some of the comments from both sides in the Scientific American are from people who are uninformed and insecure.
Some of the comments from both sides in the Scientific American are from people who are uninformed and insecure.
Last edited by Vayutuvan on 24 Dec 2012 22:29, edited 1 time in total.
Re: BR Maths Corner1
^^yes, that book is really good. I have had it with me for over a decade iirc. One of the books from my collection that my father has appreciated .
btw, India being home of zero and decimal system is generally acknowledged. Shouldnt the list include algebra and trigonometry?
btw, India being home of zero and decimal system is generally acknowledged. Shouldnt the list include algebra and trigonometry?
Re: BR Maths Corner1
Viv ji,
By "trigonometry", you mean "Pythogoros Theorem"? I would say yes due to Boudhayana's estimate of Pi to be 3.14. Algebra comes from the Arab mathematician al Jabar al Khoworizmi  so does Algorithms. Of course, Number Theory gives rise to Algebra and Number Theory is well developed by Indians  Brahmagupta's "Pell equation".
By "trigonometry", you mean "Pythogoros Theorem"? I would say yes due to Boudhayana's estimate of Pi to be 3.14. Algebra comes from the Arab mathematician al Jabar al Khoworizmi  so does Algorithms. Of course, Number Theory gives rise to Algebra and Number Theory is well developed by Indians  Brahmagupta's "Pell equation".
Re: BR Maths Corner1
ArmenT wrote:^^^^
Very funny . I tried hacking some code together, but it seemed to run for ever. Then I recalled something weird I'd read a while ago (no, I'm not a math major, but I tend to remember odd stuff for some reason).
Brocard's Problem
On the other hand, I did get some practice in hacking lua code before shifting to the ol' standby once the values started getting really large.
Sorry ! Seriously I first saw this from a book/review article which had hundreds of Ramanujan's problems, long ago. Some problems were not very not hard (relatively) and some were really fun. And some were .. "how do I attack this to solve?" ... I too spent lot of computing hours (in old days IBM 9094 and IBM 1620  i am dating myself) for the problem above. .. BTW though the problem may have been attributed to Brocard, Ramujan did not know about it when he posed this problem, it was independently discovered by him.
Another problem sticks in my mind from the same source:
For what values of n ... (2^n  7 ) is a perfect square?
(One can easily check that it is true for n=3,4,5 etc...
I am fortunate enough to have known two people (both of them physicists but very knowledgeable with Ramnujan's work, actually one is very famous and is considered one of the expert about R's math) who have given me years of enjoyment explaining some of Ramanujan's work..
Re: BR Maths Corner1
^^^ BTW, unlike the other problem, this (the above) problem does have an interesting solution...
(For those who want to peek you will find one solution here)
May be of interest to some ... such problems and more..
Notebooks of S. Ramanujan  Collecters Item from TIFR
(For those who want to peek you will find one solution here)
May be of interest to some ... such problems and more..
Notebooks of S. Ramanujan  Collecters Item from TIFR
Re: BR Maths Corner1
A trailer ...
[youtube]zrIY2XRE2B0#![/youtube]
[youtube]zrIY2XRE2B0#![/youtube]
Re: BR Maths Corner1
Amber G. wrote:For what values of n ... (2^n  7 ) is a perfect square?
(One can easily check that it is true for n=3,4,5 etc...
Just a wild guess  if (n is a Mersenne Prime) then (2^n  7) is perfect square.
checked only for a few values. Would working in binary give some insight?
Re: BR Maths Corner1
matrimc wrote:Amber G. wrote:>>For what values of n ... (2^n  7 ) is a perfect square? <<
(One can easily check that it is true for n=3,4,5 etc...
Just a wild guess  if (n is a Mersenne Prime) then (2^n  7) is perfect square.
checked only for a few values. Would working in binary give some insight?
Yes, the problem happens to be equivalent to finding a Mersenne number which is a triangular number. (which is very easy to show)... but it is same as stating the problem in a slightly different form.
The problem is, relatively well known and even wiki has a page for this.
Last edited by Amber G. on 25 Dec 2012 09:26, edited 1 time in total.
Re: BR Maths Corner1
Amber G. wrote:NO! Why should it? Even for the very next value of n (next Mersenne prime) your assertion is does not seem to be correct. (2^77 = 57 is not a perfect square).
2^7  7 = 121 = 11^2
but in general my assertion doesn't hold for n and that was a wild guess.
Code: Select all
... deleted as exact arbitrary precision integer arithmetic is needed ...
OK. There seems to be some rounding off happening. That is the problem with doing Number Theory problems using a computer . I will retry with AmberG's hints above (not look at Wiki page, of course. No fun in that, is there?).
Re: BR Maths Corner1
Ok sorry..my mistake ... 2^7  7 was a typo! I meant 2^6  7 .. (I wasn't talking about prime perse as prime to me did not make much sense... for example, 2^157 is perfect square while 15 is not prime).. etc..
FWIW (hope people don't mind)  putting my "teachers hat" , the above algorithm is not what I will do. ..
first  taking a sqrt is quite expensive in computing power ..(Unnecessary using real numbers and introducing rounding off error etc)...the following part is edited as the original post is edited ..
For example the following is much better (using integer math only ..)
take y, find y*y, add 7 and just keep dividing it by 2 (to see if it is a power of 2)... (BTW dividing by 2 is quite fast in binary math)... (You use only multiply and add and divide by 2  which is much cheaper in terms of computing cycles)
Hth...
FWIW (hope people don't mind)  putting my "teachers hat" , the above algorithm is not what I will do. ..
first  taking a sqrt is quite expensive in computing power ..(Unnecessary using real numbers and introducing rounding off error etc)...the following part is edited as the original post is edited ..
take y, find y*y, add 7 and just keep dividing it by 2 (to see if it is a power of 2)... (BTW dividing by 2 is quite fast in binary math)... (You use only multiply and add and divide by 2  which is much cheaper in terms of computing cycles)
Hth...
Re: BR Maths Corner1
Easy enough to hack. I actually had the function for computing an integer sqrt already done thanks to my attempt at AmberG's last puzzle. Code for that function was filched from here:
http://www.codecodex.com/wiki/index.php?title=Calculate_an_integer_square_root#Python
and I had to modify it slightly for python 3 (using // instead of / to do integer divisions. The same code should work in python 2.x with either / or //)
I didn't need to use pow(), since python already has the ** operator from the very beginning. Python's long integer will work with large numbers, so ** is exact. With python 3, the difference between integer and long integer is completely invisible.
For all numbers in the range of (3, 1000], it prints solutions of:
I guess I should try it in scheme next, just for a programming exercise
http://www.codecodex.com/wiki/index.php?title=Calculate_an_integer_square_root#Python
and I had to modify it slightly for python 3 (using // instead of / to do integer divisions. The same code should work in python 2.x with either / or //)
I didn't need to use pow(), since python already has the ** operator from the very beginning. Python's long integer will work with large numbers, so ** is exact. With python 3, the difference between integer and long integer is completely invisible.
Code: Select all
#!/usr/bin/python
"""Ramanujan sqrt problem"""
def isqrt(n):
"""Calculate the nearest integer square root using Newton's method"""
xn = 1
xn1 = (xn + n//xn)//2
while abs(xn1  xn) > 1:
xn = xn1
xn1 = (xn + n//xn)//2
while xn1*xn1 > n:
xn1 = 1
return xn1
def is_a_square(n):
"""Returns a square root if n is a square, 1 if not"""
foo = isqrt(n)
if foo ** 2 == n:
return foo
return 1
for n in range(3, 1000):
x = 2 ** n  7
y = is_a_square(x)
if y > 1:
print(n, x, y)
For all numbers in the range of (3, 1000], it prints solutions of:
Code: Select all
3 1 1
4 9 3
5 25 5
7 121 11
15 32761 181
I guess I should try it in scheme next, just for a programming exercise
Last edited by ArmenT on 25 Dec 2012 09:36, edited 1 time in total.
Re: BR Maths Corner1
ArmenT  my point was, it is much easier to calculate square (than square root), and much easier to see if a number is 2^n or not ..so I will start in reverse .. get a number, square it, add 7 to it and see if you keep dividing it by 2 to reach 1 or some other odd number...
BTW, 3,4,5,7,15 are the only possible values.. proof of this is not that obvious.. That's where genius of Ramanujan comes in.. Ramanujan stated it in his book without any proof, one of the first proof of this wasn't published till 30 years after his death.
Here is one of the recent paper on this:
http://www.math.tifr.res.in/~saradha/debrecen.pdf
or
http://www.math.tifr.res.in/~saradha/saradharev.pdf
BTW, 3,4,5,7,15 are the only possible values.. proof of this is not that obvious.. That's where genius of Ramanujan comes in.. Ramanujan stated it in his book without any proof, one of the first proof of this wasn't published till 30 years after his death.
Here is one of the recent paper on this:
http://www.math.tifr.res.in/~saradha/debrecen.pdf
or
http://www.math.tifr.res.in/~saradha/saradharev.pdf
Re: BR Maths Corner1
Amber G. wrote:ArmenT  my point was, it is much easier to calculate square (than square root), and much easier to see if a number is 2^n or not ..so I will start in reverse .. get a number, square it, add 7 to it and see if you keep dividing it by 2 to reach 1 or some other odd number...
Didn't see your post until after I'd posted my previous reply .
By the way, how does repeated dividing by 2 prove if a number is square or not? As far as I can see, it only proves that a number is a multiple of 2.
Say I have 36. Divide by 2, I have 18. Divide by 2, I have 9. Divide by 2, I have 4. Divide by 2, I have 2. Divide by 2, I have 1.
Say I have 11. Divide by 2, I have 5. Divide by 2, I have 2. Divide by 2, I have 1.
But 36 is a square and 11 isn't... Am I misunderstanding your algorithm?
One more quick way to find out if a number is a multiple of 2 or not is to examine its representation in binary. A power of 2 will have only one bit set. If it has more than one bit set, then it is not a power of 2 .
Also if you're integerdividing by 2, fastest way is to bit shift to the right (for languages that support the >> operator). In decimal arithmetic, we learn to divide by 10 really fast by simply lopping off the last digit (if doing integer division. For decimal division, we simply move the decimal point to the left for each division by 10). The same concept applies when the number is represented in binary  to integer divide by 2, simply lop off the last digit (i.e. shift all the binary digits of the number one place to the right)
Re: BR Maths Corner1
AmberG and ArmenT
If factoring can be reduced to detecting whether an integer is a perfect square, then computational experiments are doomed. Is there such a result? It doesn't matter but just academic interest only. In any case, I am going to catch Leno and go to bed. Merry Xmas and hope all of you get some nice xmas presents.
Added Later
Newton's method posted by ArmenT has quadratic convergence. So finding integer square root is algorithmically not difficult.
More in a later post.
If factoring can be reduced to detecting whether an integer is a perfect square, then computational experiments are doomed. Is there such a result? It doesn't matter but just academic interest only. In any case, I am going to catch Leno and go to bed. Merry Xmas and hope all of you get some nice xmas presents.
Added Later
Newton's method posted by ArmenT has quadratic convergence. So finding integer square root is algorithmically not difficult.
More in a later post.
Last edited by Vayutuvan on 26 Dec 2012 01:27, edited 2 times in total.
Re: BR Maths Corner1
^^ArmenT
Working in binary,
2^n  7 = (2^n  1)  6
= (111....11)  (110) < In binary
\/
n ones
That's why I asked whether working in binary will get me somewhere. Looks like that is a dead end.
AmberG, a request when posting a problem  can you rate it similar to Knuth's book? That way, all of us know how much effort to be expended. Of course, this is giving away information
Working in binary,
2^n  7 = (2^n  1)  6
= (111....11)  (110) < In binary
\/
n ones
That's why I asked whether working in binary will get me somewhere. Looks like that is a dead end.
AmberG, a request when posting a problem  can you rate it similar to Knuth's book? That way, all of us know how much effort to be expended. Of course, this is giving away information
Re: BR Maths Corner1
matrimc wrote:Viv ji,
By "trigonometry", you mean "Pythogoros Theorem"? I would say yes due to Boudhayana's estimate of Pi to be 3.14. Algebra comes from the Arab mathematician al Jabar al Khoworizmi  so does Algorithms. Of course, Number Theory gives rise to Algebra and Number Theory is well developed by Indians  Brahmagupta's "Pell equation".
It is beyond Pythagoras theorem. The basic concepts of the relation of the angle to the chord, the definitions and use of sine/cosine etc, as well as tables of sine values are all from Indian maths.
I remember reading somewhere (Scientific American?) long back that Sine also comes from the Arabs having been incorrectly written down as jba from the original jya (in Sanskrit). This mistransliteration was taken to mean a 'bay' by the Europeans and they therefore translated it to sine as a result. The kojya became cosine.
Re: BR Maths Corner1
viv wrote:I remember reading somewhere (Scientific American?) long back that Sine also comes from the Arabs having been incorrectly written down as jba from the original jya (in Sanskrit). This mistransliteration was taken to mean a 'bay' by the Europeans and they therefore translated it to sine as a result. The kojya became cosine.
Also
Etymologically, the word sine derives from the Sanskrit word for chord, jiva*(jya being its more popular synonym). This was transliterated in Arabic as jiba جــيــب, abbreviated jb جــــب . Since Arabic is written without short vowels, "jb" was interpreted as the word jaib جــيــب, which means "bosom" (or pocket like shirt pocket AmberG), when the Arabic text was translated in the 12th century into Latin by Gerard of Cremona. The translator used the Latin equivalent for "bosom", sinus (which means "bosom" or "bay" or "fold")..The English form sine was introduced in the 1590s
From Victor J Katx, A history of mathematics.
Re: BR Maths Corner1
ArmenT wrote:Amber G. wrote:ArmenT  my point was, it is much easier to calculate square (than square root), and much easier to see if a number is 2^n or not ..so I will start in reverse .. get a number, square it, add 7 to it and see if you keep dividing it by 2 to reach 1 or some other odd number...
Didn't see your post until after I'd posted my previous reply .
By the way, how does repeated dividing by 2 prove if a number is square or not? As far as I can see, it only proves that a number is a multiple of 2.
Sorry for being sloppy and not very clear. What I meant was, that it is easy to check if a number
2^n. (As you said, in binary it is 1 and all zeros.. or what I was trying to say that you can keep it dividing by 2  and will keep getting an even number till you reach 1..).
What I was also trying to say, that instead of taking 2^n  7 and taking a square root, it is much easier , to start with an integer (say y) and square it, add 7 and see if it is 2^n .
(It is not easy, but if you think about it, the choice for y is very restrictive.. more of in later post)
BTW:
Can you prove (it is not really very hard) that apart from 4, there is no other even value of n.
Re: BR Maths Corner1
Amber G. wrote:BTW:
Can you prove (it is not really very hard) that apart from 4, there is no other even value of n.
If you like to peek, here is a very simple proof of that...
If n is even, say n=2m then we want to find ..
2^n  7 = y^2
or 4^m7 = y^2
Let 2^m = x then
x^2  7 = y^2 or x^2 = Y^2 +7
Obviously we must have x>y,
also if y>3, then (y+1)^2 = y^2+2y+1 > y^2+7
so y+1>x>y which is not possible if both x and y are integers
Re: BR Maths Corner1
Amber G. wrote:viv wrote:I remember reading somewhere (Scientific American?) long back that Sine also comes from the Arabs having been incorrectly written down as jba from the original jya (in Sanskrit). This mistransliteration was taken to mean a 'bay' by the Europeans and they therefore translated it to sine as a result. The kojya became cosine.
AlsoEtymologically, the word sine derives from the Sanskrit word for chord, jiva*(jya being its more popular synonym). This was transliterated in Arabic as jiba جــيــب, abbreviated jb جــــب . Since Arabic is written without short vowels, "jb" was interpreted as the word jaib جــيــب, which means "bosom" (or pocket like shirt pocket AmberG), when the Arabic text was translated in the 12th century into Latin by Gerard of Cremona. The translator used the Latin equivalent for "bosom", sinus (which means "bosom" or "bay" or "fold")..The English form sine was introduced in the 1590s
From Victor J Katx, A history of mathematics.
Thanks Amber G.
This is another aspect that must be acknowledged more widely. Not just 'zero', but decimal system, trigonometry and much more of course but we can start with going beyond just zero in the popular mind. Will be a good boost to the confidence (and pride) of young students of maths who otherwise see only European names associated with theorems or methods.
Re: BR Maths Corner1
AmberG wrote: (It is not easy, but if you think about it, the choice for y is very restrictive.. more of in later post)
Is it something along these lines?
2^n  7 = 2^n  8 +1 = 8(2^(n3)  1) +1
Let 2^(n3)  1 (which is the n3 rd
If x is m'th Triangular number, i.e. x = m(m+1)/2
2^n  7 = (2m+1)^2
But x is also 2^(n3)  1
M(n) can be computed following the recurrence M(0) = 0; M(n) = 2*M(n1)+1
T(m) = m(m+1)/2
one can generate M(n) and T(m) and whenever they are equal, output n+3
But computation wise since M(n) grows exponentially, one cannot afford to compute all the triangular numbers that fall between M(n) and M(n+1).
EDIT: I edited the post above after looking at the definition of Mersenne Numbers which are same as Mersenne Primes. I was of the impression that any number of the form 2^n1 is a Mersenne Number which is also a prime is Mersenne Prime. But the definition seems to be Mersenne Prime == Mersenne Number.
Last edited by Vayutuvan on 26 Dec 2012 23:51, edited 2 times in total.
Re: BR Maths Corner1
May be of interest to some .. Tribute to R is also going on at ..
The newest museum of math opened in New York..
NYTimes article:
Opening the Doors to the Life of Pi
or
20 Fascinating Things We Saw At Manhattan's New Museum Of Mathematics
Here is the site's website:
http://momath.org/
The newest museum of math opened in New York..
NYTimes article:
Opening the Doors to the Life of Pi
or
20 Fascinating Things We Saw At Manhattan's New Museum Of Mathematics
Here is the site's website:
http://momath.org/
Re: BR Maths Corner1
Manjul Bhargava was mentioned in brf before, here is tribute to R from one of the prestigious Mathematician: (From http://www.deccanchronicle.com)
(Please note that I also talked about Panini, Pingala and its relationship to Fibonacci (or more correctly Hemchandra) numbers in brf  so people may be familiar here)
(Please note that I also talked about Panini, Pingala and its relationship to Fibonacci (or more correctly Hemchandra) numbers in brf  so people may be familiar here)
Like poetry and music
Manjul Bhargava, R. Brandon Fradd Professor of Mathematics at Princeton University
I have always found mathematics very similar to poetry and music. In fact, I find that I think about all three in very similar ways. This is true, to a large extent, for all pure mathematicians. In school, mathematics is generally grouped in the ‘science’ category.
But for pure mathematicians, mathematics — like music, poetry, or painting — is a creative art. All these arts involve, and indeed require, a certain creative fire. They all strive to express truths that cannot be expressed in ordinary, everyday language. And they all strive towards beauty.
The connection between music/poetry and mathematics is not only an abstract one. While growing up, I learned from my grandfather how so much incredible mathematics was discovered in ancient times by Indian scholars, who considered themselves not mathematicians but poets (or linguists). Linguists such as Panini, Pingala, Hemachandra and Narayana discovered some wonderful and deep (and now very important and ubiquitous) mathematical concepts while studying poetry.
Here is an example, from before 200 BC, that has particularly fascinated me. In the rhythms of Sanskrit poetry, there are two kinds of syllables — long and short.
A long syllable lasts two beats, and a short syllable lasts one beat. A question that naturally arose for ancient poets was, how many rhythms can one construct with exactly 8 beats, consisting of long and short syllables? For instance, one can take longlonglonglong, or shortshortshortlonglongshort, and so on.
The answer was discovered by the ancients, and is contained in Pingala's classical work Chandashastra, which dates back to between 500 and 200 BC.
Here is the elegant solution. We write down a sequence of numbers as follows. We first write down the numbers 1 and 2, and then each subsequent number is obtained by adding up the two previous numbers.
So, for example, we start with 1 and 2, and then 1+2 is 3, so we have so far 1 2 3. The next number is obtained by adding up the last two numbers 2 and 3, which is 5. So we have so far 1 2 3 5. The next number written is then 3+5 which is 8. In this way, we get a sequence of numbers 1 2 3 5 8 13 21 34 55 89 ....The nth number written tells you the total number of rhythms, consisting of long and short syllables, having ‘n’ beats. So, for 8 beats, the answer is that there are 34 such rhythms in total.
This sequence of numbers is now ubiquitous in mathematics, as well as in a number of other arts and sciences! The numbers are known as the Hemachandra numbers, after the 11th century linguist who first documented and proved their method of generation, called a ‘recurrence relation’ in modern mathematics.
The numbers are also known as the Fibonacci numbers in the West, after the famous Italian mathematician who wrote about them in the 12th century.
These numbers play an important role now in so many areas of mathematics (there is even an entire mathematical journal, the Fibonacci Quarterly, devoted to them!) They also arise in botany and biology.
For example, the number of petals on a daisy tends to be one of these Hemachandra numbers, and similarly for the number of spirals on a pine cone.
One of my favorite photographs, which I keep in my office, is of a vast field of daisies in which every daisy has 34 petals! (Recall that 34 is the same number that appeared as the answer to our question about 8 beat rhythms, revealing a hidden connection that mathematicians now understand.)
This story inspired me as a youngster because it is a wonderful example of how simple ideas grew into concepts so omnipresent, important, and deep, unraveling surprising and beautiful connections between different realms of thought. There are many examples of this phenomenon in mathematics and its sister areas.
I grew up learning many examples from Sanskrit poetry, but it happens throughout the subject of mathematics, and is one of those things that always makes a mathematician's eyes light up.
In some sense, this is the kind of mathematics that still inspires me, and that I still strive for when I do research in number theory.
I think the same is true of all mathematicians. It's about finding the simple questions and ideas that lead one to unexpected, unexplored realms — and to deep, elegant, and lasting mathematics.
There are widespread misconceptions regarding the applications of mathematics. History has shown that much of the greatest and most applicable mathematics has been discovered not by thinking about applications, but by following one's nose and looking for what is most beautiful and elegant.
The area that I work in personally has had many recent applications in the areas of cryptography and coding theory, although I don't think about these applications directly when thinking about mathematics.
My advice to young people is to pursue whatever your passion is — that is where you will do your best work. If that passion is mathematics, pursue mathematics. The country needs you! And, remember to always think of mathematics not just as a science, but also as a creative art.
Ramanujan – 22 Dec 188726 April 1920
Ramanujan had no formal training in pure mathematics, but from the age of 10, he mastered the works of other mathematicians, such as S.L. Loney’s trigonometry, and the works of Bernoulli and Euler, to whom the Cambridge mathematician and Ramanujan’s mentorcollaborator G.H. Hardy compared him.
In 1904, Ramanujan received a scholarship to study at Government College, Kumbakonam, but lost it the next year when he failed in other subjects because he wanted to do only Mathematics. He ran away from home, landed at Pachaiyappa’s College, Madras, which he left without a degree and for a while lived in extreme povery and on the brink of starvation.
Ramanujan then met deputy collector V. Ramaswamy Iyer, who had recently founded the Indian Mathematical Society, and asked him for a job as a clerk in the revenue department. Iyer, instead, gave him letters of recommendation to meet other mathematicians in Madras. Iyer later said, “I was struck by the extraordinary mathematical results contained in it [the notebooks]. I had no mind to smother his genius by an appointment in the lowest rungs of the revenue department.
Eventually, through his mathematician friends, both Indian and British, Ramanujan sent some of his work to Prof. Hardy at Cambridge University in 1913. The next year, on Prof. Hardy’s persistent invitation, he sailed to England. The two collaborated for over five years. But as Ramanujan’s health failed, he returned to India where he died in 1920.
Ramanujan developed much of his mathematics in isolation while still in India, and his works – mostly only the results, because he never wrote down the proofs to his conjectures on paper, but is said to have worked on them with chalk and slate because paper was expensive – are contained in three notebooks, totalling about 640 pages in all. A fourth notebook of 87 unorganised pages, called the ‘Lost Notebook’ was rediscovered in 1976. His works have inspired many, including Prof. Hardy himself, to explore and create new branches of mathematics.
Ramanujan’s Notebooks 1, 2 and 3 were published as a two volume set, as a photocopy edition of the original manuscripts, in 1957 by TIFR, Mumbai.
An international publication, The Ramanujan Journal, was launched to publish work in all areas of mathematics influenced by his work.
Next: The other side of the mathematics genius

 BRF Oldie
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Re: BR Maths Corner1
Amber G. wrote:In honor of Ramanujan, Here is one problem posed by him... A *VERY* simple looking problem, see if we have a solution here...
(1+n!) is a perfect square when n=4 or 5 or 7, can you find another number for which this is true.
(n! means 1x2x3...n, for example 4! = 1x2x3x4 = 24
5! = 1x2x3x4x5 = 120
and 7! = 120x6x7 etc...)
Ok computer gurus .. or math gurus ... find another n...
It is true for 20, 21, and 22 as well. I can more numbers if you want (and if they exist).
Re: BR Maths Corner1
abhischekcc wrote:Amber G. wrote:In honor of Ramanujan, Here is one problem posed by him... A *VERY* simple looking problem, see if we have a solution here...
(1+n!) is a perfect square when n=4 or 5 or 7, can you find another number for which this is true.
(n! means 1x2x3...n, for example 4! = 1x2x3x4 = 24
5! = 1x2x3x4x5 = 120
and 7! = 120x6x7 etc...)
Ok computer gurus .. or math gurus ... find another n...
It is true for 20, 21, and 22 as well. I can more numbers if you want (and if they exist).
Really?? What whole number is the square root of (1 + 20!) then? I think you might have a mistake in your calculations.
Re: BR Maths Corner1
In light of AmbreG's small print proof, here is a small (may be easy) interesting problem.
Prove that there is no integer between 0 and 1.
(Hint: Use well ordering principle).
Prove that there is no integer between 0 and 1.
(Hint: Use well ordering principle).
Re: BR Maths Corner1
AmberG
I finally googled the 2^n  7 problem. This is what I found.
The link below gives several more interesting problems for hacking
Beeler, M., Gosper, R.W., and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972. Retyped and converted to html ('Web browser format) by Henry Baker, April, 1995.
NUMBER THEORY, PRIMES, PROBABILITY
I also found the above and N's and also tested up to N of 45 (only). Since you said there is a Wiki page, do you know by who andwhen the proof was given?
Thanks
EDIT: Looks like it was proved in the late 60's early 70's.
Some more info. Here is the reference to the proof
Skolem, Thoralf; Chowla, S.; Lewis, D. J.
=> The diophantine equation 2^{n+2}7=x^2 and related problems.
=> Proc. Amer. Math. Soc. 10 1959 663669.
Chowla, S. is the famous Indian mathematician Sarvadaman Chowla
I finally googled the 2^n  7 problem. This is what I found.
The link below gives several more interesting problems for hacking
Beeler, M., Gosper, R.W., and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972. Retyped and converted to html ('Web browser format) by Henry Baker, April, 1995.
NUMBER THEORY, PRIMES, PROBABILITY
ITEM 31 (Schroeppel):
Ramanujan's problem of solutions to
N 2
2  7 = X
was searched to about N = 10^40; only his solutions (N = 3, 4, 5, 7, 15) were found. It has recently been proven that these are the only ones. Another Ramanujan problem: Find all solutions of n! + 1 = x^2.
I also found the above and N's and also tested up to N of 45 (only). Since you said there is a Wiki page, do you know by who and
Thanks
EDIT: Looks like it was proved in the late 60's early 70's.
Some more info. Here is the reference to the proof
Skolem, Thoralf; Chowla, S.; Lewis, D. J.
=> The diophantine equation 2^{n+2}7=x^2 and related problems.
=> Proc. Amer. Math. Soc. 10 1959 663669.
Chowla, S. is the famous Indian mathematician Sarvadaman Chowla
Last edited by Vayutuvan on 27 Dec 2012 02:34, edited 1 time in total.
Re: BR Maths Corner1
In fact a Mersenne Number is defined to be 2^n1 and Mersenne Prime is a Mersenne Number that is prime. Wikipedia directs to Mersenne Prime for Mersenne Number (since Mersenne Number definition is trivial). May be there should be a separate page for "Mersenne Number" on Wikipedia so that 2^n  7 problem can be found from there and crosslinked from Ramanujan page. Fast computation of large Mersenne Numbers is of interest to find large primes as well as parallelization of the same.
Re: BR Maths Corner1
ArmenT wrote:It is true for 20, 21, and 22 as well. I can more numbers if you want (and if they exist).
Really?? What whole number is the square root of (1 + 20!) then? I think you might have a mistake in your calculations.
20!+1 is nothing but 20639383 * 117876683047 and each of them is prime, let alone a square.
22!+1 is divisible by 23 ... (Why ? .. easy proof with a little bit knowledge of math) but not 23^2 so it is not a square.
Abhisekh  If you are just using a computer (or a calculator) be aware that the rounding off error is what giving you wrong results. (Most ordinary calculators or computers, unless specifically programmed for more digits  carry calculations to only 910 digits.. 20! is about 1819 digit number)
Last edited by Amber G. on 27 Dec 2012 03:03, edited 1 time in total.
Re: BR Maths Corner1
matrimc wrote:
I also found the above and N's and also tested up to N of 45 (only). Since you said there is a Wiki page, do you know by who andwhenthe proof was given?
Thanks
EDIT: Looks like it was proved in the late 60's early 70's.
Wiki has this ..PAGE
And according to wiki the first proof was published by Nagell in 1948, (that 3,4,5,7, and 15 are the only solutions) The proof requires a little higher abstract algebra.
http://link.springer.com/article/10.1007%2FBF02592006
Re: BR Maths Corner1
matrimc wrote:In fact a Mersenne Number is defined to be 2^n1 and Mersenne Prime is a Mersenne Number that is prime. Wikipedia directs to Mersenne Prime for Mersenne Number (since Mersenne Number definition is trivial). May be there should be a separate page for "Mersenne Number" on Wikipedia so that 2^n  7 problem can be found from there and crosslinked from Ramanujan page. Fast computation of large Mersenne Numbers is of interest to find large primes as well as parallelization of the same.
See the link I gave above... for link with 2^n type problem..
For Mersenne prime in general .. a very nice link is,,,
http://primes.utm.edu/mersenne/
Re: BR Maths Corner1
]Of course the very famous http://www.mersenne.org/..(Great Internet Mersenne Prime Search) may be known to some..
For those computer gurus (or others, who have computers and can donate a few cpu cycles ) this may be fun..
(For those who do not know, this is a organized search of large primes helped by millions of ordinary folks who donate the use of their computer while the computers are idle)
For those computer gurus (or others, who have computers and can donate a few cpu cycles ) this may be fun..
(For those who do not know, this is a organized search of large primes helped by millions of ordinary folks who donate the use of their computer while the computers are idle)
Re: BR Maths Corner1
Okay, one more question from Ramanujan's book. (This, for a change is relatively not very hard..)
(It is famous, of course, and some may have seen it before .. )
Here goes...
what is the value of (this infinite sequence..
sqrt(1+sqrt(1+2*sqrt(1+3*sqrt(1+4*sqrt( ................................
(It is famous, of course, and some may have seen it before .. )
Here goes...
what is the value of (this infinite sequence..
sqrt(1+sqrt(1+2*sqrt(1+3*sqrt(1+4*sqrt( ................................
Re: BR Maths Corner1
AmberG, thanks. As for the search for large primes, usually when people are searching for primes they search for large Mersenne primes as Mersenne Numbers are relatively easy to compute (shift left and add 1 in binary)  of course, primaility testing is the real difficult part.
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