On this page
In JEE mathematics you meet infinite series as finished formulas: the binomial expansion, the sum of a geometric progression, perhaps the series for sin x in a calculus aside. Around 1400, a mathematician in Kerala was working out series like these, and asking a practical question: how many terms do you need before the answer is good enough? That question is where Madhava of Sangamagrama becomes useful to a student today.
Who was Madhava?
The MacTutor biography from the University of St Andrews gives his dates as about 1350 to 1425 and places his birth near Cochin. None of his mathematical writings survive; some of his astronomical texts do. His mathematics is known through later members of what is now called the Kerala school, such as Nilakantha, about a century later, and Jyesthadeva, who wrote the Yuktibhasa in Malayalam around 1550 and explained the methods.
A modern edition and translation, Ganita-Yukti-Bhasa of Jyesthadeva, was published in 2008 by K. V. Sarma, K. Ramasubramanian, M. D. Srinivas and M. S. Sriram. Much of what is known about Madhava has come to light only in recent decades. In a 1978 paper, C. T. Rajagopal and M. S. Rangachari wrote, as quoted by MacTutor, that he "took the decisive step onwards from the finite procedures of ancient mathematics to treat their limit-passage to infinity".
What series did Madhava find?
According to MacTutor, Madhava found, around 1400, series equivalent to the Maclaurin expansions of sin x, cos x and arctan x. Written in modern notation, the arctan series is
arctan x = x − x³/3 + x⁵/5 − x⁷/7 + ...
Putting x = 1, where arctan 1 = π/4, gives
π/4 = 1 − 1/3 + 1/5 − 1/7 + ...
In Europe these results are associated with James Gregory, Isaac Newton and Gottfried Leibniz in the seventeenth century, and the π series is often called the Leibniz series. MacTutor's point is that the Kerala results came first. Our post on Leibniz and calculus covers the European side.
A worked example: why the simple series is not enough
Add the first five terms of the π/4 series.
1 − 1/3 = 0.66667
+ 1/5 = 0.86667
− 1/7 = 0.72381
+ 1/9 = 0.83492
Multiply by 4: 4 × 0.83492 = 3.3397. After five terms you do not even have the first decimal place right. After ten terms the total, times 4, is about 3.0418, still wrong in the first decimal. The series converges, but very slowly, because the terms shrink only like 1/n.
Madhava's answer was a correction term. MacTutor reports that he added a remainder R to the partial sum, subtracting it when the last term used was positive and adding it when negative, and gave three forms, including R = 1/(4n) and R = n/(4n² + 1), where n is the number of terms.
Apply the second form after five terms. R = 5/(4 × 25 + 1) = 5/101 = 0.049505. The last term, +1/9, was positive, so subtract: 0.83492 − 0.04950 = 0.78542. Times 4, that is 3.14166, against the true 3.14159. Five terms plus one correction beat thousands of terms without it.
How did Madhava get eleven decimal places?
MacTutor reports that he gave π = 3.14159265359, correct to eleven decimal places, and that this can be reached with a faster series obtained by putting x = 1/√3, where arctan(1/√3) = π/6:
π = √12 × (1 − 1/(3 × 3) + 1/(5 × 3²) − 1/(7 × 3³) + ...)
Each term is now about three times smaller than the last. Try the first few by hand: one term gives 3.4641, two give 3.0792, three give 3.1562, four give 3.1379 and five give 3.1426. The partial sums swing above and below π and close in fast. Choosing where to evaluate a series, so that its terms shrink quickly, is a technique still used in numerical work.
What can a student take from Madhava's way of working?
Ask how fast, not just whether. In sequences and series, practise estimating how many terms you need, not only whether a series converges. Our guide to sequences and series for JEE gives you the standard series to test.
Estimate the error you are leaving out. Madhava's correction is an estimate of the tail of the series. When you round or truncate, write down roughly how big the neglected part is.
Change the problem to one that is easier to compute. Moving from x = 1 to x = 1/√3 gave the same π with far fewer terms. Many hard problems become easy after a substitution.
Learn from the chain of teachers. His work survived because students wrote it down and explained it. Teaching a method to someone else is a reliable way to keep it.
If the step from a series to its sum still feels like magic, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson, so you can fill that one gap. IIT Bombay's NPTEL course Mathematics in India: From Vedic Period to Modern Times goes much deeper into the Kerala school.
A short practice routine
Compute 4 × (1 − 1/3 + 1/5 − 1/7) and then apply Madhava's correction R = 1/(4n) with n = 4 (add it, since the last term was negative). (Answer: about 2.8952 before the correction and 3.1452 after it.)
Write out five terms of the √12 series and check the partial sums given above.
Explain to a friend why 1 − 1/3 + 1/5 − ... converges slowly, using the size of its terms.
Tonight, pick one series from your textbook and estimate how many terms you would need for three correct decimal places.
