How Did Madhava Find π Without a Calculator?

Here are the first few totals, each already multiplied by 4:

Terms added	Total × 4	Above or below π?
1	4	above
2	2.667	below
3	3.467	above
4	2.895	below
5	3.340	above

Each new term jumps the total across π, a little less far each time: the next term is 4/11 = 0.364, then 4/13 = 0.308. So π is always trapped between two totals in a row. That already tells you something useful: after any number of terms, π lies between the last total and the one before.

Now look at how far off the total is, as the widget showed:

Terms added	Total × 4	How far from π
10	3.0418396…	0.0998
100	3.1315929…	0.0100
1,000	3.1405926…	0.0010
10,000	3.1414926…	0.0001

The pattern is very clean: after n terms, the total is off by almost exactly 1/n. Ten times as many terms buys one more decimal, no more. So for ten decimals, the gap must be under 0.00000000005, which needs about 20,000,000,000 terms: twenty billion fractions. Adding one a second, day and night, would take more than six hundred years. Madhava found π to eleven decimals in his lifetime, so he cannot have done it this way.

Look again at the zigzag. The totals jump above π, then below, then above, and π sits somewhere between the last two. Where, exactly? If you knew that, you could stop early and correct the total instead of adding billions more terms.

A first guess: π is about halfway between two totals in a row. After 4 terms the total is 2.8952 and the next would be 3.3397; halfway is 3.1175. That is better than either, but only the 3.1 is right.

Madhava’s simplest rule, a step in the Kerala books’ argument, is a little different and much better. In the times-4 form of the sum:

Stop after any odd number, call it p. Take the even number just after it, p + 1. Add 2 ÷ (p + 1) if your last term was taken away, or subtract it if your last term was added.

After 4 terms the last odd number is 7, so p + 1 = 8 and the last term (1/7) was taken away. Add 2/8:

$$2.8952 + 0.25 = 3.1452$$

Two decimals right, from four fractions, as good as a thousand terms without the trick. After 10 terms the same rule gives 3.14184, three decimals; after 100 terms, six.

Why the even number?

Here is an argument, the kind Madhava’s followers may have used (historians aren’t sure). Call the leftover the amount the total still has to move to reach π. Stop at two odd numbers in a row, say just before 7 and just after it. The two leftovers point in opposite directions, and together they make up the one jump between them: the term with 7, 4/7. So

leftover before 7 + leftover after 7 = 4/7.

The leftovers change only slowly from one stopping place to the next, so each is about half of 4/7. Better still, each leftover belongs to a gap between two odd numbers: the gap after 5 is centered on 6, the gap after 7 on 8. Give each gap the value 2 ÷ (its even number), and the two add to 2/6 + 2/8 = 0.583, very close to 4/7 = 0.571. That is Madhava’s rule: the leftover after the odd number p is about 2 ÷ (p + 1).

The idea of correcting the end of a sum is called an end correction, antya-samskara in the Kerala books. Madhava did not stop at one. Three are credited to him, each sharper than the last. Written for the plain sum (without the 4), after n terms:

Correction	After 10 terms, π comes out as	Decimals right
none	3.0418396…	0
$\frac{1}{4n}$ (the simple one)	3.1418396…	3
$\frac{n}{4n^2 + 1}$	3.1415902…	5
$\frac{n^2 + 1}{4n^3 + 5n}$ (the best)	3.1415927…	6

(The first one is the even-number rule again: with n terms the last odd number is 2n − 1, so the even number after it is 2n, and 2 ÷ 2n = 4 × 1/(4n).) With the best correction, 75 terms give π to thirteen decimals. Nobody knows how the third one was found: the books state it, but don’t say how.

Now the real puzzle. Why should odd numbers have anything to do with circles? The Kerala proof, as Jyeshthadeva, three generations of teachers after Madhava, wrote it down in his book Yuktibhasha (between about 1530 and 1550: sources differ), uses only a square, Pythagoras’ theorem, similar triangles and the sum of a geometric series. Here it is.

Take a circle of radius 1, centered at O. Its circumference is 2π, so an eighth of it, the arc from straight across to the diagonal of a square, is 2π ÷ 8 = π/4. That arc is what we will measure.

Draw the square OABC with side 1, O at one corner and A at the end of the radius. Cut the side AB into n equal steps, each 1/n long, and draw lines from O to every cut point. They slice the eighth of a circle into n pieces, one piece per step.

One piece, measured

Every step on the side is the same length, but the pieces of arc are not: the steps near A make long pieces, those near B, farther from O, make short ones. How short? Take one step, from the cut point P₁ down to P₂, and call its length δ (it is 1/n). The figure shows it.

Two pairs of similar triangles do all the work.

1. Triangles P₁P₂D and OP₂A both have a right angle (at D and at A) and share the angle at P₂, so they are similar. Matching sides: P₁D ÷ P₁P₂ = OA ÷ OP₂, so $P_1D = \delta \div OP_2$ (since OA = 1).
2. Triangles OEF and OP₁D both have a right angle (at F and at D) and share the angle at O, so they are similar. Matching sides: EF ÷ OE = P₁D ÷ OP₁, and OE = 1, so EF = P₁D ÷ OP₁.

Put them together: $EF = \delta \div (OP_1 \times OP_2)$. When the steps are thin, EF is almost the piece of arc EG, and OP₁ and OP₂ are almost equal, so their product is almost $OP_2^{\,2}$. By Pythagoras, $OP_2^{\,2} = 1 + t^2$, where t is the distance from A down to P₂. So each piece is about

$$\text{piece} \approx \frac{1/n}{1 + t^2}, \qquad t = \frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}.$$

That is what the widget’s estimate does. With 4 slices, the four estimates add up to 0.7203, and times 4 that is 2.881; with 40 slices, 3.116. The estimates fall a little short, because each piece uses the farther cut point, where 1 + t² is biggest. But the shortfall shrinks as the slices get thinner, and the total heads for π/4. So far this is just a way of measuring the arc very finely. The next step turns it into the sum of odd fractions.

A geometric series inside every piece

Dividing by 1 + t² is awkward. The Kerala trick is to turn it into adding and subtracting powers of t. Look at

$$1 - t^2 + t^4 - t^6 + \cdots$$

It is a geometric series: first term 1, each term the last one times $-t^2$. You met its sum in Grade 10: the first k terms add up to

$$\frac{1 - (-t^2)^k}{1 - (-t^2)} = \frac{1 - (-t^2)^k}{1 + t^2}.$$

When t is less than 1, $(-t^2)^k$ shrinks towards 0 as k grows, and the sum settles at exactly $\frac{1}{1 + t^2}$. Check with t = 1/2: 1 ÷ 1.25 = 0.8, and 1 − 0.25 + 0.0625 − 0.015625 = 0.796875, closing in on 0.8. Jyeshthadeva gets the same series a different way, by writing 1/(1 + x) inside itself again and again: $\frac{1}{1+x} = 1 - x \cdot \frac{1}{1+x} = 1 - x\left(1 - x \cdot \frac{1}{1+x}\right) = \cdots$

So each piece is about $\frac{1}{n}\left(1 - t^2 + t^4 - \cdots\right)$. Now add up all n pieces, one column at a time.

• The 1s: n of them, times 1/n, give 1.
• The t²s: $\frac{1}{n}\left(\frac{1^2}{n^2} + \frac{2^2}{n^2} + \cdots + \frac{n^2}{n^2}\right) = \frac{1^2 + 2^2 + \cdots + n^2}{n^3}$.
• The t⁴s: $\frac{1^4 + 2^4 + \cdots + n^4}{n^5}$, and so on.

The sums of powers

What do those fractions come to when n is large? Try them:

n	$\frac{1 + 2 + \cdots + n}{n^2}$	$\frac{1^2 + 2^2 + \cdots + n^2}{n^3}$	$\frac{1^4 + 2^4 + \cdots + n^4}{n^5}$
10	0.55	0.385	0.25333
100	0.505	0.33835	0.20503
1,000	0.5005	0.33383	0.20050

They head for 1/2, 1/3 and 1/5. In general, **1ᵐ + 2ᵐ + ⋯ + nᵐ is close to $\frac{n^{m+1}}{m+1}$ when n is large.** For squares you can see it exactly from the formula $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$: divided by n³ it is $\frac{1}{6}\left(1 + \frac{1}{n}\right)\left(2 + \frac{1}{n}\right)$, which goes to 2/6 = 1/3. The Yuktibhasha proves the general rule one power at a time, each from the one before; here we check only the squares.

Put them in the columns, and as the slices get thinner and thinner:

$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$$

The odd numbers are the 1, 3, 5, 7 under the powers: the sum of the m-th powers brings a 1/(m + 1), and only even powers of t appear. This is the Madhava–Leibniz series. It is an infinite series: a sum with endless terms whose totals settle on a definite number.

The sum for π/4 is slow because it uses the arc up to t = 1, where the powers of t don’t shrink at all. Madhava’s second series uses a shorter arc where they do.

Measure a twelfth of the circle instead, the arc of 30∘°. For 30∘°, the side of the triangle opposite the angle is $\tan 30^\circ = \frac{1}{\sqrt 3}$, so the same slicing runs only from t = 0 to t = 1/√3, and each power of t² is a third of the one before. The same steps give an arc of π/6, and multiplying by 6 (and tidying $\frac{6}{\sqrt 3}$ into $\sqrt{12}$):

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

Each term is less than a third of the one before, so each one gains about half a decimal:

Terms	Total	Decimals right
1	3.4641016…	0
2	3.0792014…	0
5	3.1426047…	2
10	3.1415905…	5
21	3.14159265359563…	10

Twenty-one terms give ten decimals, against twenty billion for the first series. Madhava’s value of π, as his followers recorded it, is 3.14159265359: π correctly rounded to eleven decimals, more than anyone before him (in 1424, al-Kashi in Samarkand reached sixteen). (The √12 at the front has to be worked out to as many decimals first, by hand, which is itself a long job.)

Twist: a slow sum with a hidden pattern

The first series is slow, but it hides something strange. Add up exactly 5,000,000 terms, multiply by 4, and compare with π, to 40 decimals:

π	3.1415926535 8979323846 2643383279 5028841971
5,000,000 terms	3.1415924535 8979323846 4643383279 5027841971

The gap rule says it should be off by 1/5,000,000, and it is: the 7th decimal is wrong. But then the digits come back right for thirteen places, then one is wrong again, then twelve more are right. Why? Because the leftover is almost exactly Madhava’s simple correction, 4 × 1/(4n) = 1/5,000,000 = 0.0000002, a number with a single non-zero digit, and what is left after that is tinier still and also “round”. The wrong digits can be predicted exactly, from a famous list of numbers called the Euler numbers; in 1992 Jonathan Borwein and Mark Limber used them to get π to 5,263 decimals out of this slow sum.

• Can You Square a Circle? Why no ruler-and-compass drawing can make a length of exactly π, however close you get.
• What Is e, and Why Does Growth Keep Finding It? Another famous number reached by adding more and more terms.
• Can You Add Halves Forever and Still Get One? (Grades 6–8) The simplest endless sum, the geometric series, as a picture.
• How Did Bhaskara Solve Fermat’s Challenge 500 Years Early? Another Indian method found centuries before Europe: the chakravala.
• How Can Throwing Rice Find π? (Computing, Going Further, Grades 6–8) A completely different way to π, by chance.
• How Does a Calculator Find a Square Root? (Computing, Going Further, Grades 9–12) How a machine works out a number like √12 to many decimals.
• Grade 10 Math: Sum of a geometric series The formula used inside every piece of Madhava’s proof.