Question

srinivas ramujan achivements

Answer 1

he ivent hadi ramanujan number

Answer 2

mathematics, there is a distinction between insight and formulating or working through a proof. Ramanujan proposed an abundance of formulae that could be investigated later in depth. G. H. Hardy said that Ramanujan's discoveries are unusually rich and that there is often more to them than initially meets the eye. As a byproduct of his work, 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:

{\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}.} {\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}.}
This result is based on the negative fundamental discriminant d = −4 × 58 = −232 with class number h(d) = 2. 26390 = 5 × 7 × 13 × 58 and 16 × 9801 = 3962 and is related to the fact that

{\textstyle e^{\pi {\sqrt {58}}}=396^{4}-104.000000177\dots .} {\textstyle e^{\pi {\sqrt {58}}}=396^{4}-104.000000177\dots .}
This might be compared 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
9801√2
/
4412
for π, which is correct to six decimal places. See also the more general Ramanujan–Sato series.

One of Ramanujan's remarkable capabilities was the rapid solution of problems, illustrated by the following anecdote about an incident in which P. C. Mahalanobis posed a problem: