The man who was loved by formulas R. Ramanujam, The Institute of Mathematical Sciences, Chennai This month we enter the 125th birth anniversary of one of the greatest mathematicians of the 20th century, and perhaps one of the greatest ever. We are fortunate to bask in the reflection from his glory, since he was born in India, and spent much of his life in Tamil Nadu. Srinivasa Ramanujan shone brightly as a star in an all too short life, but left a legacy of mathematical gems that sparkle even today. Most of us know his story, but love to hear the tale yet again. Ramanujan was born on 22 December 1887 in Erode. He showed an early penchant for mathematics, mastering trigonometry and proving his own theorems by the age of 12. By the age of 17, Ramanujan conducted his own mathematical research on Bernoulli numbers and the Euler-Mascheroni constant. But then Ramanujan was bad at other subjects, failed exams and lost his scholarship. He worked as a clerk in the Accountant-General's office at the Madras Port Trust Office to support himself while continuing his independent mathematical research. In 1912-13 Ramanujan sent samples of his work to three mathematicians at the University of Cambridge, England. One of them, G. H. Hardy, a great mathematician in his own right, was intrigued. He consulted Littlewood, another eminent mathematician, and they were both fascinated by Ramanujan's theorems. Ramanujan went to England and worked with them, eventually becoming a Fellow of the Royal Society and a Fellow of Trinity College, Cambridge. But cold England took its toll on Ramanujan: at the early age of 32, he died of illness caused by malnutrition and possibly liver infection. He left behind several notebooks containing densely packed formulae written by hand. Box 1 There is a famous story about Ramanujan. When he was ill, Hardy arrived at his residence to visit him. Being generally awkward, Hardy did not know what to say and told Ramanujan that the number of the taxicab he had come in was 1729, a number that seemed to be uninteresting. Ramanujan replied, on the spot, that it was actually a very interesting number mathematically, being the smallest natural number representable in two different ways as a sum of two cubes: 10^3 + 9^3 = 1729 = 12^3 + 1^3. Mathematicians have now generalized this concept to so-called "taxicab numbers". The nth taxicab number, typically denoted Ta(n), is defined as the smallest number that can be expressed as a sum of two positive algebraic cubes in n distinct ways. Ta(1) = 2 = 1^3 + 1^3. Ta(2) is the Ramanujan number above. Ta(3) = 87539319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3. In 1954, Hardy and Wright proved that Ta(n) exists for every n, and the proof can be transformed into an algorithm that can be run on a computer. BOX 2 Ramanujan was remarkably adept at problem solving, and could carry out long and complicated manipulations of formulas inside his head. He was sharing a room with P. C. Mahalanobis who had a problem to solve. He had found the problem in the English magazine 'Strand' in December 1914. Ramanujan was stirring vegetables in a frying pan over the kitchen fire when Mahalanobis read this problem to him. After listening to this problem, still stirring vegetables, Ramanujan asked Mahalanobis to take down the solution, and gave the general solution to the problem, not just the one with the given constraints. The 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 the 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? Ramanujan did not merely answer the problem but gave the answer with a twist: he gave a continued fraction (see definition below). 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. If m is the number of houses in that street and n is the number of the particular house, then: 1 + 2 + ... +(n-1) = (n+1) + (n+2) + ... + m. Summing both sides, n (n-1) / 2 = (m (m+1) / 2) - (n (n+1) / 2). Therefore: (2m+1)^2 - 2(2n)^2 = 1. This is of the general form x^2 - 2y^2 = 1, the Brahmagupta-Bhaskara equation, also called Pell's equation. This can be solved by using continued fraction, as Ramanujan did, or by other (more boring) methods. The answer, incidentally, is: m = 204, n = 288. -------- What is a continued fraction ? A simple infinite continued fraction, for example, is: 1 + 1/(2 +1/(3 + 1/(4 +1/(5 + ... ))) ....). If you truncate it at each step, you get what are called the successive convergents of the continued fraction, thus: 1 ; 1 + 1/2 = 3/2 ; 1 + 1/(2 + 1/3) = 1 + 3/7 = 10/7 ; 1 + 1/(2 + 1/(3 + 1/4)) = 1 + 1 /(2 + 4/13) = 1 + 13/30 = 43/30 ; and so on, endlessly. Let us consider a simpler version of the problem. We are to solve the equation: x2 - 10 y2 = +1 or -1. Mahalanobis tries it and comes out with the answer x = 3 and y = 1 in no time. Now Ramanujan, as soon as he hears the statement of the problem, dictates the following continued fraction: 3 + 1/(6 + 1/(6 + 1/(6 + 1/(6 + ... ))) ... ) Here 3/1 is the first convergent. The numbers 3 and 1 are the first answers to x and y which Mahalanobis has himself discovered by trial and error. Now Ramanujan's continued fraction answer not only gives this first elementary answer but gives an infinity of answers on the assumption that the street may have an infinite number of houses in it! Thus the second convergent, 3 + 1/6 = 19/6 gives the pair, 19 and 6 for x and y. This can be verified to be a true answer: 192 - 10 x 62 = + 1. The third convergent, 3 + 1/(6+1/6) = 3 + 6/37 = 117 / 37 gives the answer x = 117 and y = 37 which also satisfies the relation specified by the problem. The fourth convergent can be calculated to be 721/228 and this pair x = 721 and y = 228 can also be verified to satisfy the relation. And so on it goes. Every convergent gives a pair of numbers, which constitute an answer. The greatness of Ramanujan was that as soon as he heard of the problem, without effort he decided that the answer could be given in the form of a continued fraction and immediately gave the continued fraction, which not only solved the problem but gave an infinity of solutions to it!