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Most students meet the Kerala school through one result: the series π/4 = 1 − 1/3 + 1/5 − ..., found by Madhava around 1400. Our post on Madhava and the infinite series for pi covers that result and how fast it converges. This post is about the person who explained why it is true. Jyesthadeva's Yuktibhasa is the book that kept the reasoning alive, and two of its steps are simple enough to rebuild with a pencil.
Who was Jyesthadeva?
Very little is known about his life. The MacTutor biography from the University of St Andrews gives his dates as about 1500 to 1575 and says he wrote the Yuktibhasa in Malayalam, the language of Kerala, rather than in Sanskrit. K. Ramasubramanian of IIT Bombay, in his 2022 paper for the International Congress of Mathematicians, dates the book to about 1530 and calls Jyesthadeva a junior contemporary of Nilakantha Somayaji. His pupil Acyuta Pisarati carried the line of teachers on.
MacTutor describes the Yuktibhasa as half astronomy and half mathematics, and notes that, unusually for an Indian mathematical text, it gives proofs and derivations. Most texts of the tradition state rules in compressed verse. A modern edition with translation and notes, Ganita-Yukti-Bhasa of Jyesthadeva, was published in 2008 by K. V. Sarma, K. Ramasubramanian, M. D. Srinivas and M. S. Sriram.
Why does a book of proofs matter so much?
Madhava's own writings on the series have not survived. Ramasubramanian notes that later Kerala authors all credit the series to him, and that the Yuktibhasa holds the most detailed account of the school's work, from Madhava through Parameshvara, Damodara and Nilakantha. Without it, historians would have results but not the reasoning.
The reasoning took a long time to be noticed outside Kerala. Ramasubramanian describes how an early-19th-century report on these series by Charles Whish was largely ignored in Europe for about a century. Only from the 1940s did Indian mathematicians such as C. T. Rajagopal and K. Mukunda Marar publish detailed studies, and Ramavarma Thampuran and A. R. Akhileswara Ayyar edited the mathematics part of the Yuktibhasa.
How do you get an infinite series from a single division?
The Yuktibhasa starts with an ordinary product a × c/b, where b is larger than c. As Ramasubramanian explains it, the text rewrites the division by b as a division by c plus a correction, then corrects the correction, and so on. Writing x = (b − c)/c, the result is
a × c/b = a − ax + ax² − ax³ + ...
which is the series for a/(1 + x).
Worked example. Take a = 1, b = 5, c = 4, so the exact answer is 4/5 = 0.8. Then x = (5 − 4)/4 = 1/4.
1 term: 1
2 terms: 1 − 0.25 = 0.75
3 terms: 0.75 + 0.0625 = 0.8125
4 terms: 0.8125 − 0.015625 = 0.796875
5 terms: 0.796875 + 0.00390625 = 0.80078125
The partial sums swing above and below 0.8 and close in. Ramasubramanian notes that both the Yuktibhasa and Sankara Variyar's Kriyakramakari say there is no logical end to the corrections, and that the Kriyakramakari states (b − c) must be smaller than c so the terms keep shrinking. That is the modern condition |x| < 1. Try b = 9, c = 4: x = 5/4, and the terms 1, 1.25, 1.5625 grow instead of shrinking.
How did the Yuktibhasa estimate sums of powers?
The second tool is called samaghata-sankalita, the summation of powers. The text argues that for large n, 1 + 2 + ... + n is practically n²/2, then uses that to show 1² + 2² + ... + n² is practically n³/3, and continues to the general rule n^(k+1)/(k+1). Ramasubramanian calls the argument "tantamount to providing a proof by induction".
Check it with numbers.
n = 100: 1 + 2 + ... + 100 = 5050, against n²/2 = 5000, about 1% high.
n = 100: the sum of squares is 338,350, against n³/3 = 333,333, about 1.5% high.
n = 1000: the sum of squares is 333,833,500, against 333,333,333, about 0.15% high.
The estimate is never exact, but the relative gap shrinks as n grows. That shrinking gap is what makes the final step legitimate, and it is the idea behind a limit.
How do the two pieces give π/4?
Ramasubramanian sets out the argument. Draw a circle of radius r inside a square and divide one side of the square, from its midpoint to a corner, into n equal parts. Lines from the centre to the division points cut the circle into small arcs. The text argues that for large n each tiny arc can be replaced by its half-chord; in Ramasubramanian's translation, the half-chords of the arc-bits are essentially the arc-bits themselves.
Adding the pieces gives one-eighth of the circumference as a sum of terms like (r/n) × r²/(r² + (ir/n)²). Each term is a division of exactly the kind in the worked example, so it can be expanded as a series. Each column of that expansion is a sum of even powers of 1, 2, ..., n, which the power-sum rule turns into 1, 1/3, 1/5 and so on. The result is π/4 = 1 − 1/3 + 1/5 − 1/7 + ...
What can a student take from Jyesthadeva?
Write down the reason, not only the rule. The Yuktibhasa survived as the explanation of results whose original texts were lost. When you solve a problem, one line on why each step is allowed is worth keeping.
Test an approximation with numbers. Comparing 338,350 with 333,333 takes a minute and tells you how good the estimate is. Our guide to sequences and series for JEE gives you more series to test.
Watch the conditions. The rule from division only works when the terms shrink. Most wrong answers with infinite series come from using a formula outside its range.
If the step from a finite sum to a limit still feels shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit limits on their own before going back to series.
Try it now: expand 3/4 using b = 4, c = 3, add five terms, and check how close you get to 0.75.
