Ramanujam !

Blazing goIITian

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14 Mar 2012 17:41:01 IST
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14 Mar 2012 17:41:01 IST
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Ramanujam !
Engineering Entrance , JEE Main , BITSAT , JEE Main & Advanced , Mathematics , Algebra , Trigonometry , academic , Motivational


By age 11, he had exhausted the mathematical knowledge of two college students who were lodgers at his home. He was later lent a book on advanced trigonometry written by S. L. Loney

 He met deputy collector V. Ramaswamy Aiyer, who had recently founded the Indian Mathematical Society. Ramanujan, wishing for a job at the revenue department where Ramaswamy Aiyer worked, showed him his mathematics notebooks. As Ramaswamy Aiyer later recalled:


One of the first problems he posed in the journal was: sqrt{1+2sqrt{1+3 sqrt{1+cdots}}}.


He waited for a solution to be offered in three issues, over six months, but failed to receive any. At the end, Ramanujan supplied the solution to the problem himself. On page 105 of his first notebook, he formulated an equation that could be used to solve the infinitely nested radicals problem.

x+n+a = sqrt{ax+(n+a)^2 +xsqrt{a(x+n)+(n+a)^2+(x+n) sqrt{cdots}}}

Using this equation, the answer to the question posed in the Journal was simply 3. Ramanujan wrote his first formal paper for the Journal on the properties of Bernoulli numbers. One property he discovered was that the denominators (sequence A027642 in OEIS) of the fractions of Bernoulli numbers were always divisible by six. He also devised a method of calculating Bn based on previous Bernoulli numbers. One of these methods went as follows:

It will be observed that if n is even but not equal to zero,

(i) Bn is a fraction and the numerator of {B_n over n} in its lowest terms is a prime number,

(ii) the denominator of Bn contains each of the factors 2 and 3 once and only once,

(iii) 2^n(2^n-1){b_n over n} is an integer and 2(2^n-1)B_n, consequently is an odd integer.


In his 17-page paper, "Some Properties of Bernoulli's Numbers", Ramanujan gave three proofs, two corollaries and three conjectures.Ramanujan's writing initially had many flaws. As Journal editor M. T. Narayana Iyengar noted.

Mr. Ramanujan's methods were so terse and novel and his presentation so lacking in clearness and precision, that the ordinary [mathematical reader], unaccustomed to such intellectual gymnastics, could hardly follow him.


Some more:


In mathematics, there is a distinction between having an insight and having a proof. Ramanujan's talent suggested a plethora of formulae that could then be investigated in depth later.

It is said that Ramanujan's discoveries are unusually rich and that there is often more to them than initially meets the eye.As a by-product, new directions of research were opened up.


Examples of the most interesting of these formulae include the intriguing infinite series for π, one of which is given below:  rac{1}{pi} = rac{2sqrt{2}}{9801} sum^infty_{k=0} rac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}.

This result is based on the negative fundamental discriminant d = −4×58 with class number h(d) = 2 (note that 5×7×13×58 = 26390 and that 9801=99×99; 396=4×99) and is related to the fact that



 e^{pi sqrt{58}} = 396^4 - 104.000000177dots.

Compare to Heegner numbers, which have class number 1 and yield similar formulae. Ramanujan's series for π converges extraordinarily rapidly (exponentially) and forms the basis of some of the fastest algorithms currently used to calculate π. Truncating the sum to the first term also gives the approximation 9801sqrt{2}/4412 for π, which is correct to six decimal places.

One of his remarkable capabilities was the rapid solution for problems. He was sharing a room with P. C. Mahalanobis who had a problem,

"Imagine that you are on a street with houses marked 1 through n. There is a house in between (x) such that the sum of the house numbers to left of it equals the sum of the house numbers to its right. If n is between 50 and 500, what are n and x?"

This is a bivariate problem with multiple solutions. Ramanujan thought about it and gave the answer with a twist: He gave a continued fraction. The unusual part was that it was the solution to the whole class of problems. Mahalanobis was astounded and asked how he did it.


"It is simple. The minute I heard the problem, I knew that the answer was a continued fraction. Which continued fraction, I asked myself. Then the answer came to my mind", Ramanujan replied.

His intuition also led him to derive some previously unknown identities, such as:

 left [ 1+2sum_{n=1}^infty rac{cos(n	heta)}{cosh(npi)}ight ]^{-2} + left [1+2sum_{n=1}^infty rac{cosh(n	heta)}{cosh(npi)}ight ]^{-2} = rac {2 Gamma^4 left ( rac{3}{4}ight )}{pi}

For all 	heta, where Gamma(z) is the gamma function. Expanding into series of

powers and equating coefficients of 	heta^0	heta^4, and 	heta^8 gives some deep identities for the hyperbolic secant.




Comments (5)

Blazing goIITian

Joined: 30 Aug 2011 22:55:38 IST
Posts: 359
15 Mar 2012 08:29:19 IST
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 ramanujam's number 

sum of 2 cubes

Cool goIITian

Joined: 17 Jan 2009 04:17:41 IST
Posts: 58
15 Mar 2012 22:31:46 IST
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very nice

Cool goIITian

Joined: 17 Jan 2009 04:17:41 IST
Posts: 58
15 Mar 2012 22:32:15 IST
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nice one

Blazing goIITian

Joined: 25 Jul 2011 21:59:21 IST
Posts: 322
18 Mar 2012 18:21:01 IST
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Rate it if u find it useful!

New kid on the Block

Joined: 6 Nov 2012 19:08:38 IST
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6 Nov 2012 19:08:49 IST
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hai! my name is hanani iam giving aquestion please answer this if you can. how many times does 9 greater than 8?

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