Srinivasa Ramanujan (1887 - 1920)

Ramanujan was born in Southern India in 1887, and although he was only 33 on his death, he was certainly one of India’s greatest mathematician.

His genius was such that he also ranks as a world class mathematician as well. And yet his story is bizarre - a self taught mathematician who failed to gain a degree, and when his genius was finally recognised by other world class mathematicians, he was working as a clerk.

Ramanujan’s early education showed that he was able and motivated, but it did not highlight his phenomenal mathematical talent. Only other mathematicians would have recognised that.

He had the talent of asking the ‘right question’. Here’s an incident that occurred at Ramanujan’s schoool.

Teacher: ‘If you divide any number by itself, you get 1.’

Ramanujan:’Is zero divided by zero also equal to one?’

If there was a single turning point, it seems to have come from a book he came across at High School. This was a book by G S Carr called Synopsis of elementary results in pure mathematics.

This book became Ramanujan’s tutor and while it undoubtedly motivated him it also had an unfortunate side effect. This book provided the only model that he had of written mathematical arguments. The book was filled with results that Ramanujan devoured - but it contained only short proofs and notes about these results. At the same time, the book, published in 1856, was out of date by the time Ramanujan used it. This had a profound consequence on his later work.

After High School he was awarded a scholarship to study at the local government college. However, his illness and interest in mathematics took so much time away from the rest of his study that he failed his examination which would have enabled him to go on to the University of Madras. Typically, he passed in mathematics, but failed in everything else. He was forced to take a job and obtained a clerk’s post in Madras. He did not give up mathematics.

Ramanujan was recognised now by those around him as a promising mathematician, and they did all they could to support him. He, in turn, continued his work in mathematics. In a fateful decision, he decided to send off some of his results to mathematicians in England.

He wrote a polite letter together with a list of results and awaited the outcome. Two of his letters were ignored but the one addressed to Professor Hardy at Cambridge provoked a response.

Hardy was impressed with Ramanujan’s work and soon arranged for him to come to Cambridge to pursue mathematics. The journey was long (in those days people travelled by boat and it took about a month), and as a strict Brahmin, Ramanujan’s stringent dietary habits were often a problem. With the outbreak of World War I, and the subsequent food shortages, Ramanujan soon developed dietary problems. The long cold English winters also had dire effects on Ramanujan’s health, and he became so ill that he was unable to work. His health deteriorated so much over the next few years that in 1917 doctors feared for his life.

The 1917 illness in England was probably the result of an incurable disease. He stayed in a nursing home for well over a year and his spirits must have been lifted when, in early 1918, he became the first Indian Fellow of the Royal Society. Early the next year he returned to India hoping that the climate would aid his recovery. But his health only declined and he died on 26 April 1920 - tragically young, tragically unfulfilled.

Ramanujan’s mathematics

Ramanujan had an intimate familiarity with numbers. During his illness Hardy visited Ramanujan in hospital. When Hardy remarked that he had taken taxi number 1729, a singularly unexceptional number, Ramanujan immediately contradicted Hardy. He explained that this number was actually quite remarkable, since it was the smallest integer that can be represented in two ways by the sum of two cubes.

1729=1^3+12^3

\qquad\,=9^3+10^3

An area of mathematics that Hardy and Ramanujan worked together on was a formula for the number p(n) of partitions of a number n. A partition of a positive integer n is just an expression for n as a sum of positive integers, regardless of order.

So,

p(4)=5

because 4 can be written as the partitions,

1+1+1+1,\quad 1+1+2,\quad 2+2,\quad 1+3\quad \mbox{and}\quad 4.

The problem of finding p(n) was studied by Euler who found a formula for an infinite expression that gave the values for the partition function for each value of n. Unfortunately, as the expression was infinite, it couldn’t be evaluated directly. This is the expression,

{1\over{(1-x)(1-x^2)(1-x^3)(1-x^4)\cdots}}

Hardy and Ramanujan came up with an approximate formula for the partition function; it is:

p(n)\approx{1\over{4\sqrt{3}}}{\left({e^{\pi\sqrt{2n\over{3}}}\over{n}}\right)}

and then refined this to make it more accurate. Their final result is a formula which is exceedingly complicated but exact.

Ramanujan’s years in England were mathematically productive, and he gained the recognition he hoped for when Cambridge granted him a Bachelor of Science degree “by research” in 1916. Here is another piece of his output during this period. It concerns work in an area of mathematics called elliptic functions. These are functions, like the partition function, that have amazing symmetries, and their study leads to equally amazing results about number properties. If three sequences of numbers, \left\{a_n\right\},\left\{b_n\right\} and \left\{c_n\right\} are such that,

\sum\limits_{n\ge0}a_nx^n={1+53x+9x^2\over{1-82x-82x^2+x^3}}

\sum\limits_{n\ge0}b_nx^n={2-26x-12x^2\over{1-82x-82x^2+x^3}}

\sum\limits_{n\ge0}c_nx^n={2+8x-10x^2\over{1-82x-82x^2+x^3}}

then they obey the identity,

a_n^3+b_n^3+c_n^3=(-1)^n.

Here is a simpler problem, equally dazzling: what is the value of,

\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+\cdots}}}?

Ramanujan showed that it was just, x+1. What an amazing result; what an amazing man.