courseshub.world · Lives in Mathematics and Physics

Madhava of Sangamagrama

Saṅgamagrāma Mādhava (സംഗമഗ്രാമ മാധവൻ), called Golavid, “master of the sphere”

A course in two strands: the Kerala astronomer of around 1400 whose own mathematical writings are lost but whose results his successors carefully credited to him, and the infinite series for π, the arctangent, the sine and the cosine that those successors preserved and proved, two and a half centuries before they appeared in Europe.

Born c. 1340, Kerala Died c. 1425 Signature result \(\dfrac{\pi}{4}=1-\dfrac13+\dfrac15-\dfrac17+\cdots\), with end corrections Level advanced undergraduate
Partial sums of \(4(1-\tfrac13+\tfrac15-\cdots)\) zigzag slowly toward π (open circles). With Mādhava’s correction term \(n/(4n^2+1)\) added at the end, the same sums sit on the line from the second term (filled).
Lesson 00

How to use this course

Mādhava is the founder of what historians call the Kerala school of astronomy and mathematics. He is credited with the earliest known infinite series for π, the arctangent, the sine and the cosine, together with ingenious corrections that make a slowly converging series fast. None of the works that would have contained these results survives. What we have are a few astronomical texts of his, and the testimony of his intellectual descendants, who quote his verses by name and, in the Malayalam Yuktibhāṣā of about 1530, explain how the series are proved.

The course therefore separates three layers of evidence: Mādhava’s surviving works, later texts that quote verses and attribute them to him, and results of the school whose author is uncertain.

Prerequisites for Part II
  • Calculus: Riemann sums, the integral \(\int_0^1\frac{dx}{1+x^2}\), Taylor series of \(\sin\), \(\cos\), \(\arctan\).
  • Geometric series, alternating series and their error estimates.
  • The companion courses on Āryabhaṭa (sine table, radius 3438) and Brahmagupta are helpful but not required.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations in the spirit of the Yuktibhāṣā, and exercises with folded solutions. The Kerala texts state everything in words, often in verse with numbers encoded in syllables. The formulas here are modern transcriptions.

Reference

Timeline

All dates CE. Dates for Mādhava himself are estimates from internal evidence; see Lesson 01.

499Āryabhaṭa’s Āryabhaṭīya, the foundational text for Kerala astronomers.
c. 1340Mādhava born, by the usual estimate, in a house he names Ilaññippaḷḷi at Saṅgamagrāma.
c. 1400Veṇvāroha, his short work on the true Moon, with a reference date in 1402.
after 1418Agaṇita tables, whose latest reference year corresponds to 1418.
c. 1425Conventional date of death.
c. 1430His pupil Vaṭaśśeri Parameśvara completes the Dṛggaṇita, reforming parameters by observation.
144414 June: birth of Nīlakaṇṭha Somayājī, pupil of Parameśvara’s son Dāmodara.
1500Nīlakaṇṭha’s Tantrasaṅgraha, which quotes series results.
c. 1530Jyeṣṭhadeva’s Yuktibhāṣā in Malayalam gives proofs of the series.
16th c.Śaṅkara Vāriyar’s commentaries, including the Yuktidīpikā, quote Mādhava’s verses on the sine and cosine series by name.
1671James Gregory finds the arctangent series in Scotland.
1673–76Gottfried Wilhelm Leibniz finds \(\pi/4=1-\tfrac13+\tfrac15-\cdots\); Isaac Newton’s sine and cosine series circulate.
1835Charles Whish’s paper on the Hindu quadrature of the circle, Transactions of the Royal Asiatic Society, brings the Kerala series to European notice.
1940s–50sC. T. Rajagopal and collaborators analyse the Yuktibhāṣā proofs.
1972K. V. Sarma’s A History of the Kerala School of Hindu Astronomy.
2008English translation of the Yuktibhāṣā by K. V. Sarma, with explanatory notes by K. Ramasubramanian, M. D. Srinivas and M. S. Sriram.
Part I · Formation
Lesson 01

Saṅgamagrāma and the sources

Almost nothing is known of Mādhava’s life beyond his name, his village and his house. In the Veṇvāroha he describes his birthplace in a Sanskrit phrase meaning a vihāra inhabited by the bakula tree, a learned rendering of the Malayalam house name Ilaññippaḷḷi. Later writers call him Saṅgamagrāma Mādhava, and Nīlakaṇṭha refers to him by the title Golavid, “knower of the sphere”. An epithet, Emprān, has been read as indicating a family of Tuḷu Brahmin origin; this is plausible but not established.

Where was Saṅgamagrāma?

The name means “village at the confluence”. The traditional identification is Irinjalakuda, near Thrissur, whose temple deity is Saṅgameśvara. The grounds are thin. A recent proposal by V. N. Krishnachandran places it at Kūṭallūr, on the Nila (Bhāratappuzha) river where a tributary joins it; the only known manuscript of Mādhava’s Sphuṭacandrāpti came from the Kūṭallūr family collection. The question is open.

Dating

No document gives Mādhava’s birth or death. The estimate c. 1340–1425 comes from three kinds of evidence: the dates in his own astronomical works (about 1400 for the Veṇvāroha, after 1418 for the Agaṇita), the dated works of his pupil Parameśvara (from about 1430), and the chain of teachers recorded by Nīlakaṇṭha, who names Parameśvara as Mādhava’s student and Parameśvara’s son Dāmodara as his own teacher.

What survives

  • Securely his: the Veṇvāroha, on computing the Moon’s true position.
  • Attributed with good reason: the Sphuṭacandrāpti, the Candravākyāni (a list of 248 Moon sentences), the Lagnaprakaraṇa on the ascendant, and the Agaṇita tables.
  • Doubtful: the Mahājyānayanaprakāra (“method of computing great sines”), the only surviving text that sets out the sine series as a procedure; most historians now place it in the sixteenth century, by a follower.
  • Lost: any general treatise containing the series. They are known through verses quoted by Nīlakaṇṭha, Śaṅkara Vāriyar and others with explicit attribution to Mādhava.

Exercise 1.1Mādhava’s pupil was active around 1430; that pupil’s son taught Nīlakaṇṭha, born in 1444. Assuming teachers are typically 20–40 years older than their pupils, what range of birth years for Mādhava is consistent with this chain? Does it support c. 1340?

Show solution
Parameśvara’s dated work from about 1430 and a long period of observation suggest he was born around 1360. A teacher 20–40 years older would be born between about 1320 and 1340. Independently, Mādhava’s own works date from about 1400–1420, which fits a man in his sixties or seventies if born around 1340. Both lines are consistent with c. 1340, but neither excludes a date a decade or two earlier or later.
Lesson 02

Kerala and the Āryabhaṭa tradition

Kerala, the narrow coastal strip of southwestern India, was in Mādhava’s time a land of small kingdoms, prosperous from the pepper trade, with a culture of Sanskrit learning centred on Brahmin households and temples. Astronomy there followed the Āryapakṣa, the school of Āryabhaṭa, and a Kerala calendar system called parahita (“for the benefit of the people”) had been in use since the seventh century.

Household schools

Teaching was carried on within and between families, from teacher to pupil, often over many decades. The Kerala lineage can be traced for two centuries: Mādhava taught Parameśvara; Parameśvara’s son Dāmodara taught Nīlakaṇṭha Somayājī and Jyeṣṭhadeva; Jyeṣṭhadeva taught Acyuta Piṣāraṭi; Śaṅkara Vāriyar was a pupil of Nīlakaṇṭha. Texts were written in Sanskrit verse, sometimes with Malayalam prose commentaries, and copied on palm leaf.

Why series?

The tradition inherited Āryabhaṭa’s sine table, with 24 values in a quadrant and radius 3438′ (see the Āryabhaṭa course), and with it a constant pressure for more accurate sines, more accurate π, and better interpolation. Kerala astronomers also cared about the accuracy of eclipse predictions and corrected parameters by observation. Infinite series answered the demand for arbitrary precision.

Exercise 2.1The Āryabhaṭīya gives π ≈ 3.1416 and the radius 3438. Mādhava’s school used the radius \(3437'44''48'''\) (minutes, seconds, thirds). Compute \(21600/2\pi\) in sexagesimal and show that his value is correct to the nearest third.

Show solution
\(21600/2\pi=3437.746770\ldots\) minutes. The fractional part \(0.746770\times60=44.8062\) seconds, and \(0.8062\times60=48.37\) thirds. Hence \(3437'44''48'''\) to the nearest third, an accuracy of about one part in \(7\times10^8\).
Lesson 03

The astronomer of the Moon

The works that certainly survive from Mādhava are about the Moon. The Moon moves fast and irregularly, so its true position at a given moment is the hardest computation in a traditional calendar, and errors show up quickly in the timing of eclipses and festivals. The Veṇvāroha, “ascent of the bamboo”, gives a method for the true Moon at successive intervals, climbing through a period like the joints of a bamboo stem.

Moon sentences

The Candravākyāni are 248 numbers, each encoded as a short Sanskrit phrase in the kaṭapayādi system, in which consonants stand for digits. They give the Moon’s accumulated true motion day by day over a cycle of 248 days, accurate to seconds of arc. Moon sentences already existed in Kerala in an older, less accurate form attributed to Vararuci; Mādhava’s are a recomputation.

The 248-day cycle

The Moon’s irregularity is governed by its anomaly, which repeats every anomalistic month of about 27.5546 days. Nine anomalistic months are almost exactly 248 days. After 248 days the Moon’s correction for anomaly repeats almost exactly, so a table over one such cycle can be reused, with a small correction for the drift accumulated over many cycles.

A date

The Veṇvāroha starts from a moment when the Moon’s anomaly is zero, expressed as a number of days since the beginning of the Kali Yuga. Krishnachandran’s reckoning gives 1,502,008 days plus 5180 anomalistic months, which falls on 10 March 1402 in the Julian calendar.

Exercise 3.1With the anomalistic month \(27.554550\) days, compute 9 anomalistic months, and the drift after one cycle of 248 days. After how many cycles does the drift reach one day?

Show solution
\(9\times27.554550=247.99095\) days, so one cycle of 248 days overshoots by \(0.00905\) days. The drift reaches one day after \(1/0.00905\approx110\) cycles, about 75 years. A table of 248 Moon sentences therefore needs a correction for epochs a few decades apart, which is why such tables carry a reference date.

Exercise 3.2The Kali Yuga began at Julian day number 588,465.5 (midnight, 17/18 February 3102 BCE, proleptic Julian). Show that \(1{,}502{,}008+5180\times27.554550\) days later is in March 1402.

Show solution
\(5180\times27.554550=142{,}732.57\), so the total is \(1{,}644{,}740.57\) days, and the Julian day is \(2{,}233{,}206.1\). This Julian day number is 18 March 1402 in the proleptic Gregorian calendar. The two calendars then differed by 8 days, so the date is 10 March 1402 in the Julian calendar, as Krishnachandran gives it.
Part II · The mathematics
Lesson 04

Sums of powers

For the general reader. To find the length of a curved arc, the Kerala mathematicians cut it into many small pieces, approximated each piece, and added. The sums that arise are of the form \(1^p+2^p+\cdots+n^p\). They needed to know how such a sum behaves when \(n\) is very large, and the answer they found is the heart of what we now call integration.

Goals
  • State the asymptotic rule for \(\sum k^p\) used in the Yuktibhāṣā.
  • Prove it by the recursive argument of the text.
  • Recognize it as \(\int_0^1x^p\,dx=\frac1{p+1}\).
Proposition (saṅkalita, sums of powers)

For each integer \(p\ge0\),

\[ \lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^p=\frac{1}{p+1}. \]

Argument of the Yuktibhāṣā, in modern form. Write \(S_p(n)=\sum_{k=1}^n k^p\). The text observes that

\[ n\,S_{p}(n)-S_{p+1}(n)=\sum_{k=1}^{n}(n-k)\,k^{p}=\sum_{j=1}^{n-1}S_p(j), \]

since each \(k^p\) with \(k\le j\) appears once in \(S_p(j)\) for \(j=k,\dots,n-1\). If \(S_p(j)\approx j^{p+1}/(p+1)\) for large \(j\), then the right side is approximately \(\frac{1}{p+1}S_{p+1}(n)\). So \(nS_p(n)\approx S_{p+1}(n)\big(1+\tfrac1{p+1}\big)\), and \(S_{p+1}(n)\approx\frac{p+1}{p+2}\,n\cdot\frac{n^{p+1}}{p+1}=\frac{n^{p+2}}{p+2}\). Starting from \(S_0(n)=n\), induction gives the rule. The text reasons with large \(n\) and neglects lower powers, which is the content of the limit.

Read as Riemann sums, the proposition says \(\frac1n\sum_{k=1}^n(k/n)^p\to\int_0^1x^p\,dx\).

Exercise 4.1For \(n=10\), compare \(\sum k^2\) with \(n^3/3\) and \(\sum k^3\) with \(n^4/4\). What is the relative error, and how does it shrink with \(n\)?

Show solution
\(\sum_{k=1}^{10}k^2=385\) against \(333.3\); \(\sum k^3=3025\) against \(2500\). Exact formulas: \(\sum k^2=\frac{n^3}3+\frac{n^2}2+\frac n6\), \(\sum k^3=\frac{n^4}4+\frac{n^3}2+\frac{n^2}4\). The relative errors are about \(\frac{3}{2n}\) and \(\frac{2}{n}\): 15% and 21% at \(n=10\), and they tend to 0 like \(1/n\).

Exercise 4.2Verify the identity \(nS_1(n)-S_2(n)=\sum_{j=1}^{n-1}S_1(j)\) for \(n=4\), and use it to derive the exact formula for \(S_2(n)\) from \(S_1(j)=j(j+1)/2\).

Show solution
\(n=4\): \(4\cdot10-30=10\), and \(S_1(1)+S_1(2)+S_1(3)=1+3+6=10\). In general \(\sum_{j=1}^{n-1}\frac{j(j+1)}2=\frac{(n-1)n(n+1)}{6}\). So \(S_2(n)=n\cdot\frac{n(n+1)}{2}-\frac{(n-1)n(n+1)}6=\frac{n(n+1)}{6}(3n-(n-1))=\frac{n(n+1)(2n+1)}6\).
Lesson 05

The series for π

For the general reader. Add one, subtract a third, add a fifth, subtract a seventh, and so on forever: the result is a quarter of π. This statement, known in Europe as Leibniz’s series, is attributed in Kerala texts to Mādhava. The Yuktibhāṣā derives it by measuring an eighth of a circle with small pieces, exactly in the spirit of integral calculus.

Goals
  • State the series and its attribution.
  • Follow the Yuktibhāṣā derivation: rectification of the arc.
  • Estimate how slowly the bare series converges.
Theorem (Mādhava; Leibniz 1673)

The circumference \(C\) of a circle of diameter \(d\) satisfies

\[ \frac{C}{4d}=\frac{\pi}{4}=1-\frac13+\frac15-\frac17+\cdots=\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}. \]

The Kerala verse, quoted by later authors as Mādhava’s, instructs: multiply the diameter by four, then alternately subtract and add the diameter times four divided by the odd numbers 3, 5, 7, ….

Derivation (after Yuktibhāṣā, chapter 6)

Take a circle of radius 1 centred at \(O\) and the square circumscribing it. On the side tangent at \(A=(1,0)\), mark points \(P_k=(1,k/n)\) for \(k=0,\dots,n\), running from \(A\) to the corner \((1,1)\). The lines \(OP_k\) cut the eighth of the circle from \(A\) to \(45^\circ\) into \(n\) small arcs. The arc between \(OP_{k-1}\) and \(OP_k\) is approximately the perpendicular from \(P_{k-1}\) to \(OP_k\), scaled down to the circle. Similar triangles give its length as approximately

\[ \Delta\theta_k\approx\frac{1/n}{1+(k/n)^2}. \]

Summing, the eighth-circle arc \(\pi/4\) is the limit of \(\sum_{k=1}^n\frac{1}{n}\,\frac{1}{1+(k/n)^2}\). Expand each term as a geometric series, \(\frac1{1+x^2}=1-x^2+x^4-\cdots\), interchange the sums, and apply the rule of Lesson 04 to each \(\sum k^{2j}\):

\[ \frac\pi4=\lim_{n\to\infty}\sum_{j\ge0}(-1)^j\frac{1}{n^{2j+1}}\sum_{k=1}^{n}k^{2j}=\sum_{j\ge0}\frac{(-1)^j}{2j+1}. \]

In modern terms this is \(\int_0^1\frac{dx}{1+x^2}=\arctan1\). The expansion at \(x=1\) is the boundary case of the geometric series, and a rigorous proof needs Abel’s theorem or a direct remainder estimate; the text’s reasoning with “sufficiently large \(n\)” leaves this gap.

Exercise 5.1Show that the error after \(n\) terms satisfies \(\big|\tfrac\pi4-\sum_{k=0}^{n-1}\tfrac{(-1)^k}{2k+1}\big|=\int_0^1\frac{x^{2n}}{1+x^2}dx\), lies between \(\frac1{2(2n+1)}\) and \(\frac{1}{2n+1}\), and deduce how many terms give π to 10 decimals.

Show solution
From \(\frac1{1+x^2}=\sum_{k=0}^{n-1}(-x^2)^k+\frac{(-x^2)^n}{1+x^2}\), integrate on \([0,1]\). Since \(\tfrac12\le\frac1{1+x^2}\le1\), the integral of \(\frac{x^{2n}}{1+x^2}\) lies between \(\frac1{2(2n+1)}\) and \(\frac1{2n+1}\). The error in π is 4 times this, about \(1/n\). For error below \(5\times10^{-11}\) one needs \(n\approx2\times10^{10}\) terms. The bare series is useless for computation, which is the problem Lesson 06 solves.

Exercise 5.2Compute the Riemann sums \(\sum_{k=1}^n\frac1n\frac1{1+(k/n)^2}\) and \(\sum_{k=0}^{n-1}\frac1n\frac1{1+(k/n)^2}\) for \(n=10\), and compare with \(\pi/4\approx0.785398\).

Show solution
The right-endpoint sum is \(0.759981\) and the left-endpoint sum \(0.809981\), which bracket \(\pi/4\) because \(\frac1{1+x^2}\) is decreasing. They differ by exactly \(\frac1n(1-\frac12)=0.05\). Their average, \(0.784981\), is the trapezoid rule and is already within \(4\times10^{-4}\).
Lesson 06

End corrections

For the general reader. The series for π creeps toward its limit, overshooting and undershooting by a little less each time. Mādhava saw that one can stop after a modest number of terms and add a single carefully chosen fraction, which cancels most of the remaining error. His three successive correction terms make a few dozen terms as good as billions.

Goals
  • State the three correction terms (antya-saṃskāra).
  • Derive them from the requirement that the corrected sum not depend on where one stops.
  • Quantify the accuracy gained.
Rule (attributed to Mādhava)

Let \(S_n=\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\), stopping after the term with denominator \(2n-1\). Then \(\frac\pi4\approx S_n+(-1)^nR_n\), with successively better choices

\[ R_n=\frac{1}{4n},\qquad R_n=\frac{n}{4n^2+1},\qquad R_n=\frac{n^2+1}{4n^3+5n}. \]

Where they come from

The Yuktibhāṣā asks that the corrected value be the same whether one stops at \(n\) or at \(n+1\). Since \(S_{n+1}=S_n+\frac{(-1)^n}{2n+1}\), this means

\[ R_n+R_{n+1}=\frac{1}{2n+1}. \]

No simple \(R_n\) satisfies this exactly, so one measures the defect. For \(R_n=\frac1{4n}\): \(R_n+R_{n+1}-\frac1{2n+1}=\frac{1}{4n(n+1)(2n+1)}\), of order \(n^{-3}\). For \(R_n=\frac{n}{4n^2+1}\) the defect is \(-\frac{4}{(2n+1)(4n^2+1)(4n^2+8n+5)}\), of order \(n^{-5}\). The text argues that a smaller defect means a better correction, and it finds the third term in the same way.

The three corrections are the successive convergents of the continued fraction

\[ R_n=\cfrac{1}{4n+\cfrac{1}{n+\cfrac{4}{4n+\cfrac{9}{n+\cdots}}}}\,, \]

whose next convergent is \(\frac{4n^3+13n}{16n^4+56n^2+9}\).

Exercise 6.1Take \(n=10\), so \(S_{10}=0.7604599047\). Compute \(4(S_{10}+R_{10})\) for each of the three corrections and compare with π.

Show solution
\(n\) is even, so add. \(R=\frac1{40}\): \(4(0.7604599+0.025)=3.1418396\), error \(2.5\times10^{-4}\). \(R=\frac{10}{401}\): \(3.1415902\), error \(-2.4\times10^{-6}\). \(R=\frac{101}{4050}\): \(3.14159271\), error \(5\times10^{-8}\). Ten terms plus one fraction give seven correct decimals; the bare ten terms give \(3.0418\).

Exercise 6.2Verify that \(\frac{1}{4n}+\frac{1}{4(n+1)}-\frac{1}{2n+1}=\frac{1}{4n(n+1)(2n+1)}\), and explain why the first correction already improves the error from order \(1/n\) to order \(1/n^3\).

Show solution
Common denominator \(4n(n+1)(2n+1)\): the numerator is \((n+1)(2n+1)+n(2n+1)-4n(n+1)=(2n+1)^2-4n^2-4n=1\). If the defects \(\delta_m=R_m+R_{m+1}-\frac1{2m+1}\) are of order \(m^{-3}\), the error of the corrected sum is an alternating sum of the \(\delta_m\) for \(m\ge n\), which is bounded by the first one, of order \(n^{-3}\).
Lesson 07

Faster series for π

For the general reader. Beyond correcting the slow series, the Kerala texts give other series for π that converge quickly from the start. One, attributed to Mādhava, uses the circle of diameter \(d\) and the number \(\sqrt{12}\), and gives π to eleven decimal places with 21 terms. A circumference value quoted as his, 2,827,433,388,233 for a circle of diameter nine hundred billion, is correct in all but its last digit.

Goals
  • Derive the \(\sqrt{12}\) series from the arctangent.
  • Derive the rapidly converging series obtained by absorbing end corrections.
  • Check the 13-digit circumference value.
Theorem (the \(\sqrt{12}\) series) \[ \pi=\sqrt{12}\left(1-\frac{1}{3\cdot3}+\frac{1}{5\cdot3^2}-\frac{1}{7\cdot3^3}+\cdots\right)=\sqrt{12}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)\,3^k}. \]

Put \(x=1/\sqrt3\) in \(\arctan x=x-\frac{x^3}3+\frac{x^5}5-\cdots\) (Lesson 08): \(\frac\pi6=\frac1{\sqrt3}\sum_k\frac{(-1)^k}{(2k+1)3^k}\), and \(6/\sqrt3=\sqrt{12}\). The terms shrink by a factor 3 at each step.

Theorem (transformed series) \[ \frac{\pi}{4}=\frac34+\frac{1}{3^3-3}-\frac{1}{5^3-5}+\frac{1}{7^3-7}-\cdots,\qquad \frac{\pi}{16}=\frac{1}{1^5+4\cdot1}-\frac{1}{3^5+4\cdot3}+\frac{1}{5^5+4\cdot5}-\cdots \]

These arise by writing the slow series with the end correction \(R_n\) applied at every step. With \(R_n=\frac1{4n}\), each pair of consecutive corrected sums differs by \(\pm\frac{1}{(2n+1)^3-(2n+1)}\), which gives the first; with \(R_n=\frac{n}{4n^2+1}\) one gets the second, whose terms fall off like \(k^{-5}\). Both appear in the Kerala texts; their attribution to Mādhava personally is less certain than for the \(\sqrt{12}\) series.

Exercise 7.1Sum 21 terms of the \(\sqrt{12}\) series (by computer). How many decimals of π are correct? Estimate the error from the first omitted term.

Show solution
The sum is \(3.1415926535956\ldots\), which exceeds π by about \(5.8\times10^{-12}\); the first 11 decimals, 3.14159265359, agree with π. The first omitted term is \(\sqrt{12}/(43\cdot3^{21})\approx7.7\times10^{-12}\), and for an alternating series with decreasing terms the error is smaller than it, as observed.

Exercise 7.2The verse attributed to Mādhava gives circumference 2,827,433,388,233 for diameter \(9\times10^{11}\). Compute \(9\times10^{11}\pi\) to the nearest integer. Which digit is wrong?

Show solution
\(9\times10^{11}\times3.14159265358979=2{,}827{,}433{,}388{,}230.8\), which rounds to …231. The quoted value ends in 233, two units too large: the ratio \(2827433388233/(9\times10^{11})=3.141592653592\ldots\) is correct to 11 decimals. Krishnachandran suggests a small computational slip; how Mādhava obtained the value is not known.

Exercise 7.3Prove \(\frac{1}{(2n+1)^3-(2n+1)}=\frac{1}{4n(n+1)(2n+1)}\), and so connect the first transformed series to Exercise 6.2.

Show solution
\((2n+1)^3-(2n+1)=(2n+1)\big((2n+1)^2-1\big)=(2n+1)\cdot4n(n+1)\). So the terms of the first transformed series are exactly the defects of the correction \(R_n=\frac1{4n}\). Let \(T_n=S_n+(-1)^n\frac1{4n}\) be the corrected sum. Then \(T_{n+1}-T_n=(-1)^n\big(\frac1{2n+1}-\frac1{4n}-\frac1{4(n+1)}\big)=\frac{(-1)^{n+1}}{4n(n+1)(2n+1)}\). Starting from \(T_1=1-\frac14=\frac34\) and summing these increments, \(T_n\to\frac\pi4\) gives the series.
Lesson 08

The arctangent series

For the general reader. The series for π is one case of a more general rule: from the sine and cosine of an angle, compute the angle itself. In Europe this is Gregory’s series of 1671. In Kerala it was stated in verse and attributed to Mādhava, in the language of sines and cosines of an arc in a circle of radius \(R\).

Goals
  • State the series in Mādhava’s form and in modern form.
  • Derive it as in Lesson 05 with a general endpoint.
  • Understand the restriction to arcs of at most \(45^\circ\).
Theorem (Mādhava; Gregory 1671)

For an arc \(s\) in a circle of radius \(R\), with sine \(y=R\sin(s/R)\) and cosine \(x=R\cos(s/R)\), and \(|y|\le x\):

\[ s=\frac{Ry}{x}-\frac{R}{3}\frac{y^3}{x^3}+\frac{R}{5}\frac{y^5}{x^5}-\cdots. \]

Equivalently, \(\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\cdots\) for \(|t|\le1\).

The verse, preserved in Śaṅkara Vāriyar’s Yuktidīpikā, describes a recursion: the first result is the sine times the radius divided by the cosine; each subsequent result is the previous one multiplied by the square of the sine and divided by the square of the cosine; these are divided by 1, 3, 5, …, and the arc is the sum of the odd-placed quotients minus the sum of the even-placed ones. It adds that the sine should be smaller than the cosine.

Derivation. Repeat Lesson 05 with the tangent segment from \(A\) to \((1,t)\) cut into \(n\) parts: \(\arctan t=\lim\sum_{k=1}^n\frac{t/n}{1+(kt/n)^2}=\int_0^t\frac{du}{1+u^2}\). Expanding \(\frac1{1+u^2}\) in a geometric series is valid for \(|u|<1\), and the sums of powers now give \(\frac{t^{2j+1}}{2j+1}\). The condition \(|t|\le1\) is the condition sine ≤ cosine in the verse.

Exercise 8.1Use six terms of the series at \(t=1/\sqrt3\) to approximate π, and compare with six terms of the series at \(t=1\).

Show solution
At \(t=1/\sqrt3\): \(6\sum_{k=0}^{5}\frac{(-1)^k}{(2k+1)\sqrt3\,3^k}\approx3.14131\), error \(3\times10^{-4}\). At \(t=1\): \(4(1-\tfrac13+\tfrac15-\tfrac17+\tfrac19-\tfrac1{11})\approx2.9760\), error \(0.17\). A smaller argument converges geometrically faster.

Exercise 8.2John Machin (1706) used \(\frac\pi4=4\arctan\frac15-\arctan\frac1{239}\). Prove the identity using \(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\), and explain why Mādhava’s series combined with it is so much faster than the \(\sqrt{12}\) series.

Show solution
\(\tan2\alpha\) with \(\tan\alpha=\frac15\) is \(\frac{2/5}{24/25}=\frac5{12}\); \(\tan4\alpha=\frac{10/12}{1-25/144}=\frac{120}{119}\). Then \(\tan(4\alpha-\frac\pi4)=\frac{120/119-1}{1+120/119}=\frac1{239}\). The series at \(t=\frac15\) gains a factor 25 per term, and at \(\frac1{239}\) a factor 57,121, against a factor 3 for \(t=1/\sqrt3\).
Lesson 09

The sine and cosine series

For the general reader. The sine of an angle can be computed exactly from the angle by an infinite sum of powers. In Europe these series are due to Isaac Newton, around 1669, and are usually written with factorials. In Kerala, verses attributed to Mādhava give them in a form designed for hand calculation: a short list of constants, each encoded as a word, and a nested rule that uses only a few multiplications.

Goals
  • State the series in modern form and in Mādhava’s nested form.
  • Recompute his coefficients.
  • See how the Yuktibhāṣā derives the series by iterating an integral equation.
Theorem (Mādhava; Newton c. 1669) \[ \sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots,\qquad 1-\cos\theta=\frac{\theta^2}{2!}-\frac{\theta^4}{4!}+\cdots. \]

Mādhava’s form

Work in a circle with \(R=3437'44''48'''\) and quarter-circumference \(q=5400'\). For an arc \(s\) (in minutes) the sine is \(R\sin(s/R)\). Since \(s/R=\frac{\pi}{2}\cdot\frac{s}{q}\), the cubic term is \(R\frac{(\pi/2)^3}{3!}\big(\frac sq\big)^3\), and so on. Mādhava precomputes the constants \(c_{2k+1}=R\frac{(\pi/2)^{2k+1}}{(2k+1)!}\) and nests:

\[ R\sin\frac sR=s-\Big(\frac sq\Big)^3\bigg(c_3-\Big(\frac sq\Big)^2\Big(c_5-\Big(\frac sq\Big)^2\big(c_7-(\tfrac sq)^2(c_9-(\tfrac sq)^2c_{11})\big)\Big)\bigg). \]

The verse, quoted in Śaṅkara Vāriyar’s Yuktidīpikā, encodes the constants as words in the kaṭapayādi system, the last being vidvān (\(44'''\)) and the first nirviddhāṅganarendrarung (\(2220'39''40'''\)):

\(c_3\)\(2220'\,39''\,40'''\)
\(c_5\)\(273'\,57''\,47'''\)
\(c_7\)\(16'\,5''\,41'''\)
\(c_9\)\(33''\,6'''\)
\(c_{11}\)\(44'''\)

A companion verse does the same for the versine \(R(1-\cos)\), with constants from \(4241'9''0'''\) down to \(6'''\).

Derivation (after Yuktibhāṣā, chapter 7)

Divide an arc into \(n\) equal small arcs and consider the sines and cosines at the division points. The second difference of the sines is proportional to the sine itself (the rule seen in the Āryabhaṭa course): \(\Delta^2 J_k=-2(1-\cos\alpha)J_k\). Summing twice, as a discrete double integral, gives the integral equation \(\sin\theta=\theta-\int_0^\theta\!\int_0^\phi\sin\psi\,d\psi\,d\phi\). Start from the crude approximation \(\sin\psi\approx\psi\) and substitute it back repeatedly; each round uses the sums of powers of Lesson 04 and adds one more term \(\pm\theta^{2k+1}/(2k+1)!\).

Exercise 9.1Recompute \(c_3\) and \(c_{11}\) from \(R=21600/2\pi\) and check them against the verse.

Show solution
\(c_3=R(\pi/2)^3/6=3437.7468\times3.8758/6\approx2220.661'\), and \(0.661\times60=39.66''\), \(0.66\times60\approx39.6'''\), so \(2220'39''40'''\) to the nearest third. \(c_{11}=R(\pi/2)^{11}/11!\approx0.012373'=0.7424''=44.5'''\), matching vidvān = 44 thirds.

Exercise 9.2Use the nested rule with \(s=1800'\) (an arc of \(30^\circ\)) to compute \(R\sin30^\circ\), and compare with \(R/2\).

Show solution
\(s/q=\tfrac13\), \((s/q)^2=\tfrac19\). In minutes, \(c_{11}=0.012222\), \(c_9=0.551667\), \(c_7=16.094722\), \(c_5=273.963056\), \(c_3=2220.661111\). Inside out: \(c_9-c_{11}/9=0.550309\); \(c_7-0.550309/9=16.033577\); \(c_5-16.033577/9=272.181547\); \(c_3-272.181547/9=2190.418717\); finally \(1800-2190.418717/27=1800-81.126619=1718.873381'\). And \(R/2=1718.873385'\). The rule is correct to about a third of a third.

Exercise 9.3Show by iterating \(\sin\theta=\theta-\int_0^\theta\int_0^\phi\sin\psi\,d\psi\,d\phi\) twice, starting from \(\sin\psi\approx\psi\), that one gets \(\theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}\).

Show solution
First: \(\int_0^\phi\psi\,d\psi=\phi^2/2\), \(\int_0^\theta\phi^2/2\,d\phi=\theta^3/6\), so \(\sin\theta\approx\theta-\theta^3/6\). Second: substitute \(\psi-\psi^3/6\): \(\int_0^\phi=\phi^2/2-\phi^4/24\), \(\int_0^\theta=\theta^3/6-\theta^5/120\). Hence \(\sin\theta\approx\theta-\theta^3/6+\theta^5/120\).
Lesson 10

The sine table to thirds

For the general reader. Āryabhaṭa’s sine table of 499 gave 24 values rounded to whole minutes. The table attributed to Mādhava has the same 24 arcs but gives each sine to the nearest third, a sixtieth of a second. Converted to a circle of radius 1, its values are accurate to about seven or eight decimal places.

Goals
  • Read the table and check entries against modern values.
  • Understand how accuracy in thirds translates into decimals.
  • Use the table with a second-order interpolation rule.

The table is preserved in Nīlakaṇṭha’s commentary on the Āryabhaṭīya and in Śaṅkara Vāriyar’s works, where it is attributed to Mādhava. Each value is a kaṭapayādi phrase whose digits are read in reverse. A selection, with the modern value \(R\sin\theta\), \(R=21600/2\pi\):

\(3^\circ45'\)\(224'\,50''\,22'''\)
\(15^\circ\)\(889'\,45''\,15'''\) (modern \(889'45''15.6'''\))
\(30^\circ\)\(1718'\,52''\,24'''\) (modern \(1718'52''24.2'''\))
\(45^\circ\)\(2430'\,51''\,15'''\) (modern \(2430'51''14.6'''\))
\(60^\circ\)\(2977'\,10''\,34'''\)
\(90^\circ\)\(3437'\,44''\,48'''\)

How the table was computed is not recorded. The series of Lesson 09, which reproduce \(R\sin30^\circ\) to a fraction of a third, would serve, and most historians think Mādhava used them.

Proposition (second-order interpolation)

For small \(h\), \(\sin(\theta+h)=\sin\theta+h\cos\theta-\frac{h^2}{2}\sin\theta+O(h^3)\).

Kerala texts give rules of this type for finding sines between table entries; with the cosine available from the table at the complementary arc, they use only tabulated values.

Exercise 10.1One third is \(1/3600\) of a minute. Show that rounding to the nearest third in a circle of radius about 3438′ gives an accuracy of about \(4\times10^{-8}\) relative to the radius, and check the claim that the table gives \(\sin45^\circ\) to about 7 decimals.

Show solution
The rounding error is at most \(\tfrac12\cdot\frac1{3600}\)′ \(\approx1.4\times10^{-4}\)′, and \(1.4\times10^{-4}/3437.75\approx4\times10^{-8}\). For \(45^\circ\): \(2430'51''15'''=2430.854167'\), divided by \(3437.746771\) gives \(0.70710681\), against \(\sin45^\circ=0.70710678\): seven correct decimals.

Exercise 10.2Use the table values at \(30^\circ\) and \(60^\circ\) (the cosine of \(30^\circ\) is the sine of \(60^\circ\)) and the second-order rule to compute \(R\sin31^\circ\), with \(h=60'\). Compare with \(R\sin31^\circ\approx1770.5705'\).

Show solution
In the circle of radius \(R\), \(h\) in radians is \(60/R=0.0174533\). \(R\sin31^\circ\approx J+hK-\tfrac{h^2}2J\) with \(J=1718.8734'\), \(K=2977.1760'\): \(1718.8734+51.9612-0.2618=1770.5731'\). The modern value is \(1770.5705'\). The error, \(0.0026'\), is exactly the neglected third-order term \(-\tfrac{h^3}6K\approx-0.0026'\).
Part III · Life and legacy
Lesson 11

The school after Mādhava

Mādhava’s pupil Vaṭaśśeri Parameśvara, whose family came from Ālattiyūr near the Nila, made observation the basis of reform. Over more than fifty years he recorded eclipses and planetary positions, and his Dṛggaṇita (“computation agreeing with observation”), completed around 1430, revised the Kerala parameters. His son Dāmodara taught the next generation.

Nīlakaṇṭha

Nīlakaṇṭha Somayājī, born on 14 June 1444 by his own statement of the Kali day, was the school’s great synthesizer. His Tantrasaṅgraha of 1500 revised the planetary models, with a more accurate equation of centre for Mercury and Venus, and his long commentary on the Āryabhaṭīya discussed the reasons behind the rules, including the series. In it he proposed a geometrical model in which Mercury, Venus, Mars, Jupiter and Saturn move around the mean Sun, which in turn moves around the Earth, a scheme resembling the one Tycho Brahe proposed in 1588.

Jyeṣṭhadeva and Śaṅkara Vāriyar

Around 1530 Jyeṣṭhadeva, a pupil of Dāmodara, wrote the Yuktibhāṣā (“exposition of reasons”) in Malayalam prose. It is the earliest surviving text to give derivations of the series for π, the arctangent, the sine and the cosine, and of the end corrections. Śaṅkara Vāriyar, a pupil of Nīlakaṇṭha, wrote commentaries on the Tantrasaṅgraha and on Bhāskara II’s Līlāvatī that quote Mādhava’s verses by name. The line continued through Acyuta Piṣāraṭi to Mēlpattūr Nārāyaṇa Bhaṭṭatiri (1559–1632), after whom original work faded.

Exercise 11.1The Kerala authors were careful to attribute verses to Mādhava by name. Using this course, list which results rest on direct quotation with attribution, which on later exposition, and which are of uncertain authorship.

Show solution
Quoted with attribution: the sine and cosine series verses (Śaṅkara Vāriyar’s Yuktidīpikā), the arctangent series, the π series and end corrections, the 13-digit circumference value, the sine table. Exposition with proofs: the Yuktibhāṣā, which presents results of the school without always naming the discoverer. Uncertain: the transformed fast series, the Mahājyānayanaprakāra, and the second-order interpolation rule in its various forms.

DiscussionMādhava’s mathematics survives only because his students and their students quoted him. Compare the Kerala household lineage with a modern research group. What are the strengths and the risks of knowledge passed down this way?

Lesson 12

Rediscovery and the transmission debate

Europe

In Europe the same series appeared in the second half of the seventeenth century: James Gregory found the arctangent series in 1671, Gottfried Wilhelm Leibniz the series for \(\pi/4\) in 1673–74, and Isaac Newton the sine and cosine series around 1669. There is no evidence that any of them knew of the Kerala work.

Whish and after

Charles Whish, a British civil servant in the Malabar district, described the Kerala series in a paper published in 1835 in the Transactions of the Royal Asiatic Society, where he wrote that the Kerala authors had laid the foundation for a complete system of fluxions. The paper was largely ignored for a century. From the 1940s C. T. Rajagopal and his collaborators studied the Yuktibhāṣā proofs in detail, and K. V. Sarma’s editions and his history of the Kerala school (1972) made the sources available. The annotated translation of the Yuktibhāṣā by Sarma, Ramasubramanian, Srinivas and Sriram (2008) is now the standard reference.

Transmission?

Jesuit missionaries were active in Kerala from the sixteenth century, and some historians, including A. K. Bag and George Gheverghese Joseph, have suggested that Kerala results might have reached Europe through them. No manuscript, letter or translation documenting such a transfer has been found. David Bressoud and others conclude that the Indian series were not known outside India before the nineteenth century. The question remains open, and the burden of proof lies with documentary evidence.

Assessment

  • Analysis: the first infinite series for trigonometric functions and π, with error-correcting terms of remarkable sophistication.
  • Method: summation of powers as a discrete integration, used to rectify arcs; an informal but correct use of limits.
  • Computation: π to 11 decimals and a sine table to thirds.
  • Astronomy: accurate lunar tables, and a school that made observation the test of theory.
  • Limits: no general concept of function or derivative, results tied to the circle, and proofs that would need modern tools to be made rigorous.

Exercise 12.1Gregory and Leibniz found the arctangent series by integrating a geometric series term by term; the Yuktibhāṣā sums small arcs and uses sums of powers. Write the two derivations side by side in modern notation and identify the single step that corresponds to “interchange of limit and sum”.

Show solution
Gregory–Leibniz: \(\arctan t=\int_0^t\sum_j(-u^2)^j\,du=\sum_j(-1)^j\int_0^tu^{2j}du\). Kerala: \(\arctan t=\lim_n\sum_k\frac tn\sum_j(-1)^j(kt/n)^{2j}=\sum_j(-1)^j\lim_n\frac{t^{2j+1}}{n^{2j+1}}\sum_kk^{2j}\). In both, the critical step is moving the infinite sum over \(j\) outside the integral or the limit in \(n\). It is justified for \(|t|<1\) by uniform convergence, and at \(t=1\) by Abel’s theorem.

Final projectChoose one of: the π series with end corrections, the arctangent series, or the sine series in nested form. Write two pages for a strong undergraduate that paraphrase the Kerala rule, give the Yuktibhāṣā-style derivation in modern notation, carry out a numerical example by hand, and state precisely what a modern rigorous proof requires.

Reference

Sources and further reading

Texts and translations

  1. K. V. Sarma (trans.), with notes by K. Ramasubramanian, M. D. Srinivas and M. S. Sriram, Gaṇita-Yukti-Bhāṣā of Jyeṣṭhadeva, 2 vols., Hindustan Book Agency and Springer, 2008.
  2. K. Ramasubramanian and M. S. Sriram, Tantrasaṅgraha of Nīlakaṇṭha Somayājī, Hindustan Book Agency and Springer, 2011.
  3. K. V. Sarma (ed.), Veṇvāroha of Mādhava of Saṅgamagrāma, Sanskrit College, Tripunithura, 1956.

Biography and history

  1. Madhava of Sangamagramma, MacTutor History of Mathematics.
  2. Madhava of Sangamagrama and Kerala school of astronomy and mathematics, Wikipedia.
  3. V. N. Krishnachandran, Saṃgamagrāma Mādhava: an updated biography, arXiv:2408.12096.
  4. K. V. Sarma, A History of the Kerala School of Hindu Astronomy, Vishveshvaranand Institute, Hoshiarpur, 1972.
  5. K. Plofker, Mathematics in India, Princeton University Press, 2009, ch. 7.
  6. P. P. Divakaran, The Mathematics of India: Concepts, Methods, Connections, Hindustan Book Agency and Springer, 2018.

Mathematics

  1. Madhava series and Madhava’s sine table, Wikipedia.
  2. V. N. Krishnachandran, On Śaṅkara Varman’s (correct) and Mādhava’s (incorrect) values for the circumferences of circles, arXiv:2405.11144.
  3. C. M. Whish, On the Hindú quadrature of the circle, and the infinite series of the proportion of the circumference to the diameter, Transactions of the Royal Asiatic Society of Great Britain and Ireland 3 (1835) 509–523.
  4. R. Roy, The discovery of the series formula for π by Leibniz, Gregory and Nilakantha, Mathematics Magazine 63 (1990) 291–306.
  5. D. Bressoud, Was calculus invented in India?, College Mathematics Journal 33 (2002) 2–13.
  6. G. G. Joseph, The Crest of the Peacock: Non-European Roots of Mathematics, 3rd ed., Princeton, 2011.