The Infinite Pi series of Madhava

By means of the same argument, the circumference can be computed in another way too. That is as (follows): The first result should by the square root of the square of the diameter multiplied by twelve. From then on, the result should be divided by three (in) each successive (case). When these are divided in order by the odd numbers, beginning with 1, and when one has subtracted the (even) results from the sum of the odd, (that) should be the circumference. ( Yukti deepika commentary )


This quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows.


c = SQRT(12 d^2 - SQRT(12 d^2/3.3 + sqrt(12 d^2)/3^2.5 - sqrt(12d^2)/3^3.7 +.......


As c = Pi d , this equation can be rewritten as


Pi = Sqrt(12( 1 - 1/3.3 + 1/3^2.5 -1/3^3.7 +......


This is obtained by substituting z = Pi/ 6 in the power series expansion for arctan (z).


Pi/4 = 1 - 1/3 +1/5 -1/7+.....


This is Madhava's formula for Pi, and this was discovered in the West by Gregory and Liebniz.

Madhava's Pi series

By means of the same argument, the circumference can be computed in another way too. That is as (follows): The first result should by the square root of the square of the diameter multiplied by twelve. From then on, the result should be divided by three (in) each successive (case). When these are divided in order by the odd numbers, beginning with 1, and when one has subtracted the (even) results from the sum of the odd, (that) should be the circumference. ( Yukti deepika commentary )


This quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows.


c = SQRT(12 d^2 - SQRT(12 d^2/3.3 + sqrt(12 d^2)/3^2.5 - sqrt(12d^2)/3^3.7 +.......


As c = Pi d , this equation can be rewritten as


Pi = Sqrt(12( 1 - 1/3.3 + 1/3^2.5 -1/3^3.7 +......


This is obtained by substituting z = Pi/ 6 in the power series expansion for arctan (z).


Pi/4 = 1 - 1/3 +1/5 -1/7+.....


This is Madhava's formula for Pi, and this was discovered in the West by Gregory and Liebniz.

Madhava's Pi series

By means of the same argument, the circumference can be computed in another way too. That is as (follows): The first result should by the square root of the square of the diameter multiplied by twelve. From then on, the result should be divided by three (in) each successive (case). When these are divided in order by the odd numbers, beginning with 1, and when one has subtracted the (even) results from the sum of the odd, (that) should be the circumference. ( Yukti deepika commentary )


This quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows.


c = SQRT(12 d^2 - SQRT(12 d^2/3.3 + sqrt(12 d^2)/3^2.5 - sqrt(12d^2)/3^3.7 +.......


As c = Pi d , this equation can be rewritten as


Pi = Sqrt(12( 1 - 1/3.3 + 1/3^2.5 -1/3^3.7 +......


This is obtained by substituting z = Pi/ 6 in the power series expansion for arctan (z).


Pi/4 = 1 - 1/3 +1/5 -1/7+.....


This is Madhava's formula for Pi, and this was discovered in the West by Gregory and Liebniz.

Arctangent series of Madhava, Gregory and Liebniz

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

We thank www.wikipedia.org for publishing this on their site.

Arctangent series of Madhava, Gregory and Liebniz

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

We thank www.wikipedia.org for publishing this on their site.

The Madhava cosine series

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 Madhava Trignometric Series

The Madhava Trignometric series is one one of a series in a collection of infinite series expressions discovered by Madhava of Sangramagrama ( 1350-1425 ACE ), the founder of the Kerala School of Astronomy and Mathematics. These are the infinite series expansions of the Sine, Cosine and the ArcTangent functions and Pi. The power series expansions of sine and cosine functions are called the Madhava sine series and the Madhava cosine series.

The power series expansion of the arctangent function is called the Madhava- Gregory series.

The power series are collectively called as Madhava Taylor series. The formula for Pi is called the Madhava Newton series.

One of his disciples, Sankara Variar had translated his verse in his Yuktideepika commentary on Tantrasamgraha-vyakhya, in verses 2.440 and 2.441


Multiply the arc by the square of the arc, and take the result of repeating that (any number of times). Divide (each of the above numerators) by the squares of the successive even numbers increased by that number and multiplied by the square of the radius. Place the arc and the successive results so obtained one below the other, and subtract each from the one above. These together give the jiva, as collected together in the verse beginning with "vidvan" etc.


Rendering in modern notations

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

Place the arc and the successive results so obtained one below the other, and subtract each from the one above to get jiva:


Jiva = s-(1-2-3)

When we transform it to the current notation

If x is the angle subtended by the arc s at the center of the Circle, then s = rx and jiva = r sin x.


Sin x = x - x^3/3! + x^5/5! - x^7/7!...., which is the infinite power series of the sine function.


By courtesy www.wikipedia.org and we thank Wikipedia for publishing this on their site.