The Infinite Power Series Of The Cosine And The Arctangent

By Author: G Kumar
Total Articles: 37
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Madhava's cosine series is stated in verses 2.442 and 2.443 in Yukti-dipika commentary (Tantrasamgraha-vyakhya) by Sankara Variar. A translation of the verses follows.

Multiply the square of the arc by the unit (i.e. the radius) and take the result of repeating that (any number of times). Divide (each of the above numerators) by the square of the successive even numbers decreased by that number and multiplied by the square of the radius. But the first term is (now)(the one which is) divided by twice the radius. Place the successive results so obtained one below the other and subtract each from the one above. These together give the śara as collected together in the verse beginning with stena, stri, etc.

Let r denote the radius of the circle and s the arc-length.

The following numerators are formed first:

s.s^2,
s.s^2.s^2
s.s^2.s^2.s^2

These are then divided by quantities specified in the verse.

1)s.s^2/(2^2-2)r^2,
2)s. s^2/(2^2-2)r^2. s^2/4^2-4)r^2
3)s.s^2/(2^2-2)r^2.s^2/(4^2-4)r^2. s^2/(6^2-6)r^2

As per verse,

sara ...
... or versine = r.(1-2-3)

Let x be the angle subtended by the arc s at the center of the Circle. Then s = rx and sara or versine = r(1-cosx)

Simplifying we get the current notation

1-cosx = x^2/2! -x^4/4!+ x^6/6!......

which gives the infinite power series of the cosine function.

The inverse tangent series of Madhava is given in verse 2.206 – 2.209 in Yukti-dipika commentary (Tantrasamgraha-vyakhya) by Sankara Variar . It is also given by Jyeshtadeva in Yuktibhasha and a translation of the verses is given below.

Now, by just the same argument, the determination of the arc of a desired sine can be (made). That is as follows: The first result is the product of the desired sine and the radius divided by the cosine of the arc. When one has made the square of the sine the multiplier and the square of the cosine the divisor, now a group of results is to be determined from the (previous) results beginning from the first. When these are divided in order by the odd numbers 1, 3, and so forth, and when one has subtracted the sum of the even(-numbered) results from the sum of the odd (ones), that should be the arc. Here the smaller of the sine and cosine is required to be considered as the desired (sine). Otherwise, there would be no termination of results even if repeatedly (computed).

Rendering in modern notations

Let s be the arc of the desired sine, bhujajya, y. Let r be the radius and x be the cosine (kotijya).

The first result is y.r/x
From the divisor and multiplier y^2/x^2
From the group of results y.r/x.y^2/x^2, y.r/x. y^2/x^2.y^2/x^2
Divide in order by number 1,3 etc
1 y.r/1x, 1y.r/3x y^2/x^2, 1y.r/5x.y^2/x^2.Y^2/x^2

a = (Sum of odd numbered results) 1 y.r/1x + 1y.r/5x.y^2/x^2.y^2/x^2+......

b= ( Sum of even numbered results) 1y.r/3x.y^2/x^2 + 1 y.r/7x.y^2/x^2.y^2/x^.y^2/x^2+.....

The arc is now given by
s = a - b

Transformation to current notation

If x is the angle subtended by the arc s at the Center of the Circle, then s = rx and kotijya = r cos x and bhujajya = r sin x. And sparshajya = tan x

Simplifying we get

x = tan x - tan^3x/'3 + tan^5x/5 - tan^7x/7 + .....

Let tan x = z, we have

arctan ( z ) = z - z^3/3 + z^5/5 - z^7/7

Article by G Kumar, astrologer, academician & programmer of
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