Madhava series

Madhava's cosine series

In Madhava's own words

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.

Rendering in modern notations

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

• The following numerators are formed first:

${\displaystyle r\cdot s^2,r\cdot s^2\cdot s^2,r\cdot s^2\cdot s^2\cdot s^2,\cdots }$

• These are then divided by quantities specified in the verse.

${\displaystyle r\cdot \frac{s^2}{(2^2-2)r^2},r\cdot \frac{s^2}{(2^2-2)r^2}\cdot \frac{s^2}{(4^2-4)r^2},r\cdot \frac{s^2}{(2^2-2)r^2}\cdot \frac{s^2}{(4^2-4)r^2}\cdot \frac{s^2}{(6^2-6)r^2},\cdots }$

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

${\displaystyle \text{sara}=r\cdot \frac{s^2}{(2^2-2)r^2}-\left[r\cdot \frac{s^2}{(2^2-2)r^2}\cdot \frac{s^2}{(4^2-4)r^2}-\left[r\cdot \frac{s^2}{(2^2-2)r^2}\cdot \frac{s^2}{(4^2-4)r^2}\cdot \frac{s^2}{(6^2-6)r^2}-\cdots \right]\right]}$

Transformation to current notation

Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ and śara = r(1 − cos θ). Substituting these in the last expression and simplifying we get

${\displaystyle 1-\cos \theta =\frac{{\theta }^2}{2!}-\frac{{\theta }^4}{4!}+\frac{{\theta }^6}{6!}+\cdots }$

which gives the infinite power series expansion of the cosine function.

Madhava's reformulation for numerical computation

The last line in the verse ′as collected together in the verse beginning with stena, stri, etc.′ is a reference to a reformulation introduced by Madhava himself to make the series convenient for easy computations for specified values of the arc and the radius. As in the case of the sine series, Madhava considers a circle one quarter of which measures 5400 minutes (say C minutes) and develops a scheme for the easy computations of the śara′s of the various arcs of such a circle. Let R be the radius of a circle one quarter of which measures C. Then, as in the case of the sine series, Madhava gets R = 3437′ 44′′ 48′′′.

Madhava's expression for śara corresponding to any arc s of a circle of radius R is equivalent to the following:

${\displaystyle \begin{array}{cc}\text{jiva }& =R\cdot \frac{s^2}{R^2(2^2-2)}-R\cdot \frac{s^4}{R^4(2^2-2)(4^2-4)}-\cdots \\ & ={\left(\frac{s}{C}\right)}^2\left[\frac{R{\left(\frac{\pi }{2}\right)}^2}{2!}-{\left(\frac{s}{C}\right)}^2\left[\frac{R{\left(\frac{\pi }{2}\right)}^4}{4!}-{\left(\frac{s}{C}\right)}^2\left[\frac{R{\left(\frac{\pi }{2}\right)}^6}{6!}-\cdots \right]\right]\right]\end{array}}$

Madhava now computes the following values:

The śara can now be computed using the following scheme:

$\text{sara }={\left(\frac{s}{C}\right)}^2\left[\left(\frac{R(\frac{\pi }{2}{)}^2}{2!}\right)-{\left(\frac{s}{C}\right)}^2\left[\left(\frac{R(\frac{\pi }{2}{)}^4}{4!}\right)-{\left(\frac{s}{C}\right)}^2\left[\left(\frac{R(\frac{\pi }{2}{)}^6}{6!}\right)-{\left(\frac{s}{C}\right)}^2\left[\left(\frac{R(\frac{\pi }{2}{)}^8}{8!}\right)-{\left(\frac{s}{C}\right)}^2\left[\left(\frac{R(\frac{\pi }{2}{)}^{10}}{10!}\right)-{\left(\frac{s}{C}\right)}^2\left(\frac{R(\frac{\pi }{2}{)}^{12}}{12!}\right)\right]\right]\right]\right]\right]$

This gives an approximation of śara by its Taylor polynomial of the 12'th order. This also involves one division, six multiplications and five subtractions only. Madhava prescribes this numerically efficient computational scheme in the following words (translation of verse 2.438 in Yukti-dipika):

The six stena, strīpiśuna, sugandhinaganud, bhadrāngabhavyāsana, mīnāngonarasimha, unadhanakrtbhureva. Multiply by the square of the arc divided by the quarter of the circumference and subtract from the next number. (Continue with the result and the next number.) Final result will be utkrama-jya (R versed sign).

Madhava's arctangent series

In Madhava's own words

Madhava's arctangent series is stated in verses 2.206 – 2.209 in Yukti-dipika commentary (Tantrasamgraha-vyakhya) by Sankara Variar. A translation of the verses is given below. Jyesthadeva has also given a description of this series in Yuktibhasa.

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).

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.

Rendering in modern notations

Let s be the arc of the desired sine (jya or jiva) y. Let r be the radius and x be the cosine (kotijya).

• The first result is ${\displaystyle \frac{y\cdot r}{x}}$.
• Form the multiplier and divisor ${\displaystyle \frac{y^2}{x^2}}$.
• Form the group of results:

${\displaystyle \frac{y\cdot r}{x}\cdot \frac{y^2}{x^2},\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2},\cdots }$

• These are divided in order by the numbers 1, 3, and so forth:

${\displaystyle \frac{1}{1}\frac{y\cdot r}{x},\frac{1}{3}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2},\frac{1}{5}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2},\cdots }$

• Sum of odd-numbered results:

${\displaystyle \frac{1}{1}\frac{y\cdot r}{x}+\frac{1}{5}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}+\cdots }$

• Sum of even-numbered results:

${\displaystyle \frac{1}{3}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}+\frac{1}{7}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}+\cdots }$

• The arc is now given by

${\displaystyle s=\left(\frac{1}{1}\frac{y\cdot r}{x}+\frac{1}{5}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}+\cdots \right)-\left(\frac{1}{3}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}+\frac{1}{7}\frac{y\cdot r}{x}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}\cdot \frac{y^2}{x^2}+\cdots \right)}$

Transformation to current notation

Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get

• ${\displaystyle \theta =\tan \theta -\frac{{\tan }^3\theta }{3}+\frac{{\tan }^5\theta }{5}-\frac{{\tan }^7\theta }{7}+\cdots }$.

Letting tan θ = q we finally have

• ${\displaystyle {\tan }^{-1}q=q-\frac{q^3}{3}+\frac{q^5}{5}-\frac{q^7}{7}+\cdots }$

Another formula for the circumference of a circle

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

${\displaystyle c=\sqrt{12d^2}-\frac{\sqrt{12d^2}}{3\cdot 3}+\frac{\sqrt{12d^2}}{3^2\cdot 5}-\frac{\sqrt{12d^2}}{3^3\cdot 7}+\cdots }$

Since c = π d this can be reformulated as a formula to compute π as follows.

${\displaystyle \pi =\sqrt{12}\left(1-\frac{1}{3\cdot 3}+\frac{1}{3^2\cdot 5}-\frac{1}{3^3\cdot 7}+\cdots \right)}$

This is obtained by substituting q = ${\displaystyle 1/\sqrt{3}}$ (therefore θ = π / 6) in the power series expansion for tan⁻¹ q above.

Comparison of convergence of various infinite series for π