You
can see Madhava's
cosine series 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 the poetic 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.
|
|
You can find the inverse tangent series of Madhava 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).
When rendered in modern terms, it becomes
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
Change 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