1 Ramanujan and π

As every child in middle school knows,

π =Crcmƒerence oƒ the crce

Dmeter oƒ the crce

The ancients where interested in accurately calculating its value because of its applications inastronomical calculations which helped in predicting seasons. It was important to know whenthe rains will come so that they can sow their crops at the proper time.

Many approximations are known to π, the most well known being 227 . Note that, π is not a

rational number. It is not a number likep2 which is a root of the quadratic equation 2−2= 0.

For another example, take the numberp

1+p5. Let us call this α. Then

α =

q

1+p

5

Squaring both sides α2 = 1+p

5

∴ α2−1=p

5

Squaring both sides again,�

α2−1�2= 5

α4−2α2−1= 25

or α4−2α2−26= 0

or 1×α4+0×α3+(−2)×α2+0×α+(−26)×1= 0

π is a transcendental number, meaning there is no polynomial of any degree with rationalcoefficients which is satisfied by this number. Rational coefficients means that the coefficientsare rational number like 1, −2, 3/4 etc. Another way of putting it is to say that if you take anyn rational numbers 0, 1, . . ., n, not all of them zero, then 0+1π+2π2+nπn is neverzero.

Ramanujan had a great memory and he would entertain his friends by reciting things likesanskrit roots and the value of π to several digits. His interest in π led him to do someinteresting work. This is something simple enough to explain. We will also explain someprogress in calculating π using a formula found by Ramanujan.

First, we will explain two geometric constructions for π due to Ramanujan. A classicalquestion asked by the Greeks is, ‘Given a circle, is it possible to construct a square which hasthe same area as the circle using only ruler and compass?’. That the answer this question was‘No’ was found only in 19th century.

Suppose you have a circle of unit radius. Then, its area π×12 = π square units. If a squareof side has the same area, then π = 2 or =

pπ. So, if you can construct the length

pπ

using ruler and compass, then you can construct a square whose area is equal to the area ofthe circle. The point is that any number that you can construct using a ruler and compass willbe the root of an equation whose degree is a power of 2. For example, we can construct a line

segment of lengthp

1+p5. Since this is not possible in the case of π, it is not possible.

But, can we construct a square whose area is approximately equal to the area of the givencircle? Two such constructions were given by Ramanujan. If the circle has unit radius, the sideof the square, say , will be approximately

pπ, so it square, 2, will be an approximation of π.

The first one appeared in the research article ‘Squaring the circle’. ’We will look at the secondconstruction, which appeared in his research paper ‘Modular equations and approximations to

1

π’.

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