व्यासे वारिधिनिहते रूपहृते व्याससागराभिहते ।

त्रिशरादिविषमसंख्याभक्तमृणं स्वं पृथक्क्रमात् कुर्यात् ॥from the diameter multiplied by four and divided by onereduce and add in turn, the diameter multiplied by four, andrespectively divided by the odd numbers three, five and so on.

In the previous article in this series, we made a brief acquaintance with the Madhava series described by this verse.

\( \frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} \)

How though, did Madhava and his successors establish this result, and the others mentioned in the article? Was this a one-off, or did they establish a sufficient body of techniques that deserve to be called Calculus?

### Building up the rationale

Consider a circle inscribed in a square. What the figure above shows is the top-right quadrant, which contains a quarter of the circumference. The top point, where the circle touches the square is marked ** E **(for “East”) and the rightmost point as

**(for “South”). This follows the convention of Indian Mathematics, where figures are oriented with East towards the top and North to the left. Arcs usually start from the East and grow Northward, as we shall see in later reasoning. Here, we follow Yuktibhāṣā, which we have seen in the previous article, and look at the quadrant with the South-East corner (the अग्निकोणः) as in the above figure, as the problem is symmetric.**

*S*Having marked the centre of the circle as *O*, and the corner of the quadrant as *A* (for “Agni”, as above), we draw the diagonal *OA* of the square. If the diameter of the circle is considered to be *d*, we know that the side of the quadrant would be *d/2*, or *r*, the radius. Divide the line segment EA into n equal parts, and mark the intermediate equidistant points as \( A_1, A_2 … A_{n-1} \), each of which we join to the center *O*. Let *OA* meet the circle at *C*, and \( OA_1, OA_2 … OA_{n-1} \) respectively meet the circle at \( C_1, C_2 … C_{n-1} \). We draw \( EB_1, C_1B_2 … C_{n-1}B_n \) perpendicular to \( OA_1, OA_2 … OA_{n-1}, OA \) . respectively. We also draw \( A_1P_2, A_2P_3 … A_{n-1}P_n \), also perpendicular to \( OA_1, OA_2 … OA_{n-1}, OA \) respectively.

Arc ES is a quarter of the circle, and Arc EC is therefore an eighth part of the circumference. \( EC = EC_1 + C_1C_2 + … C_{n-1}C\) . If we can approximate the arcs \( C_{k-1}C_k \), we can approximate the circumference. But how do we do that?

#### A Brief Excursion into Indian Trigonometry

Consider the arc *CS*, centred at *O*. *OC* = *r* is a radius. *CB* is perpendicular to *OS*. We take angle COB as \(\theta\). In terms of modern trigonometry,

\( sin(\theta) = \frac{CB}{OC}\),

\(cos(\theta) = \frac{OB}{OC}\), and

\(tan(\theta) = \frac{CB}{OB}\).

Indian “trigonometry”, or more accurately, “circleometry” refers to the half-chord *CB* as the jyā (“bowstring”) or bhuja (“arm”) of arc *CS*, and OB as the koti*–*jyā (“pointed bowstring”), and BS as the utkrama*–*jyā or śara (“arrow”). *OC,* the radius is the karṇa. Strictly, it is the arc, not the angle that is the independent variable. However, circles are considered to be 21600′ in circumference by convention, which converts the arc into what we would now call an angle. Most calculations can be done in these units, until we need to convert into units of real length, when we use the “rule of three” (त्रैराशिकम्) – or the principle of proportionality. The familiar relation

\( sin^2(\theta) + cos^2(\theta) = 1 \)

becomes the jyā-koti-karṇa-nyāya (ज्याकोटिकर्णन्याय:)

\( jyā^2 + koti^2 = karṇa^2 \)

Each of the karṇas \( OA_i = k_i \) are determined by the jyā-koti-karṇa-nyāya as \( k_i^2 = OE^2 + EA_i^2 = r^2 + (\frac{ir}{n})^2\)

We approximate the arcs with their jyās, and say

\(

\begin{align*} EC &= EC_1 + C_1C_2 + … C_{n-1}C \\

&\approx EB_1 + C_1B_2 + … C_{n-1}N_n

\end{align*}

\)

Triangle \( A_1OE\) can be seen to be similar to Triangle \( A_1EB_1\) and hence \( \frac{EB_1}{EA_1} = \frac{OE}{OA_1}\)

and hence,

\( EB_1 = (\frac{r}{n})\frac{r}{k_1} \)

#### Illustrating similarity

To describe the same figure we are looking at, Yuktibhāṣā uses a very interesting analogy, which may be accessible to those who have seen traditional Kerala Architecture. Consider the roof-frame of a rectangular building, with frame-rods (കഴുക്കോല്). The connector-rod (വള) enters the first (straight) frame-rod perpendicularly, but enters other frame-rods, which are at an angle to the original straight one, at an angle to the perpendicular. To make the point about similarity of triangles \( A_1OE\) and \(A_1EB_1\) , Yuktibhāṣā points out that the frame-rods are analogous to \(OA_i\) and the connector rod analogous to \(EA_i\). By a parallel arms argument, the angle between the connecting rod and the perpendicular to a frame-rod must be the same as the angle between the frame-rod and the “straight” frame rod.

We can compute further jyās \(C_1B_2\) thus, using similarity arguments similar to the above:

\( \frac{OE}{OA_2} = \frac{A_1P_2}{A_1A_2}\)

\( \frac{r}{k_2} = \frac{A_1P_2}{\frac{r}{n}}\)

and

\( \frac{C_1B_2}{OC_1} = \frac{A_1P_2}{OA_1}\)

\( \frac{C_1B_2}{r} = \frac{A_1P_2}{k_1}\)

and hence

\( C_1B_2 = (\frac{r}{n}).(\frac{r^2}{k_1.k_2})\)

Thus,

\( \begin{align*}

EC &\approx EC_1 + C_1C_2 + … C_iC_{i+1} + … + C_{n-1}C \\

&= (\frac{r}{n})\frac{r}{k_1} + (\frac{r}{n}).(\frac{r^2}{k_1.k_2}) + … + (\frac{r}{n}).(\frac{r^2}{k_i.k_{i+1}}) + … (\frac{r}{n}).(\frac{r^2}{k_{n-1}.k_n}) \\

&= (\frac{r}{n})\frac{r^2}{k_0k_1} + (\frac{r}{n}).(\frac{r^2}{k_1.k_2}) + … + (\frac{r}{n}).(\frac{r^2}{k_i.k_{i+1}}) + … (\frac{r}{n}).(\frac{r^2}{k_{n-1}.k_n}) \end{align*}\)

since \(k_0 = r\)

This is a formidable equation, but can be simplified further with an approximation. Before we go there, we note that Yuktibhāṣā points out that the approximate equality above tends to a full equality when the number of segments (ie *n*) increases, and each segment is very small.

### Simplifying

Yuktibhāṣā notes here that the divisor is rather inconvenient (ഹാരകന്നാനുരൂപം – हारकन्नानुरूपम्) and proposes a simplification

\( \frac{1}{k_i.k_{i+1}} \approx (\frac{1}{2}) (\frac{1}{k_i^2}+\frac{1}{k_{i+1}^2})\)

which is justified by the argument that two adjacent \( k_i \) will tend to each other.

Thus, we have:

\( EC \approx \frac{r}{n}. (\frac{1}{2}).(\frac{r^2}{k_0^2}+\frac{r^2}{k_1^2}+\frac{r^2}{k_1^2}+\frac{r^2}{k_2^2}+ … +\frac{r^2}{k_{n-2}^2}+\frac{r^2}{k_{n-1}^2} + \frac{r^2}{k_{n-1}^2}+\frac{r^2}{k_n^2}) \)

One more minor asymmetry remains, which an also be approximated away at large *n*

\( k_n = \sqrt{2}.k_0\)

We note that both of these appear only once in the expression, unlike others which appear twice. At the cost of an error \(\frac{r}{4.n}\) which vanishes for large *n*, we replace one instance of \(k_0\) with \(k_n\).

\( EC \approx \frac{r}{n}. (\frac{r^2}{k_1^2}+ … \frac{r^2}{k_{n-1}^2}+\frac{r^2}{k_n^2})\)

Now we have a nice and symmetric expression, and more importantly, one that can be summed with the Sankalita technique** **known since Aryabhata, and improved by Narayana Pandita, combined with a nice sleight-of-hand, called shodyaphala (शेध्यफलम्).

#### Shodyaphala

If we have an expression \(a.\frac{b}{c}\), where *c* is an inconvenient divisor, but b isn't, we can convert this into a series using this trick.

First, we note that \(a.\frac{b}{c} = a – a.\frac{c-b}{c}\),

now, using the same trick on \(\frac{c-b}{c} = \frac{c-b}{b} – \frac{c-b}{b}\frac{c-b}{c}\) repeatedly,

\(a.\frac{b}{c} = a – a.\frac{c-b}{b}+a.(\frac{c-b}{b})^2 – … + (-1)^m.a.(\frac{c-b}{b})^{m-1}.\frac{c-b}{c} \),

If *c-b* is less than *c*, the final term will eventually be negligible. We have elimnated an incovenient divisor, at the cost of converting a single term into an infinite series.

Noting that \(k_i\) is an inconvenient denominator, and applying shodyaphala and the expression \(k_i^2 = r^2+(\frac{i.r}{n})^2\) for a single term in the series, we get

\( \frac{r}{n}. \frac{r^2}{k_i^2} = \frac{r}{n} – \frac{r}{n}\frac{(\frac{ir}{n})^2}{r^2}+\frac{r}{n}\frac{(\frac{ir}{n})^3}{r^3}- …\)

Writing out the entire series this way (this is elaborate, but spare a thought for those working this out before 1400CE!):

\( \begin{align*}

EC &\approx (\frac{r}{n} – \frac{r}{n}\frac{(\frac{r}{n})^2}{r^2}+\frac{r}{n}\frac{(\frac{r}{n})^4}{r^4}- …) \\

&\quad-(\frac{r}{n} – \frac{r}{n}\frac{(\frac{2r}{n})^2}{r^2}+\frac{r}{n}\frac{(\frac{2r}{n})^4}{r^4}- …) \\

&\quad+(\frac{r}{n} – \frac{r}{n}\frac{(\frac{3r}{n})^2}{r^2}+\frac{r}{n}\frac{(\frac{3r}{n})^4}{r^4}- …) \\

&\quad… \\

&\quad-(\frac{r}{n} – \frac{r}{n}\frac{(\frac{nr}{n})^2}{r^2}+\frac{r}{n}\frac{(\frac{nr}{n})^4}{r^4}- …)

\end{align*}\)

Each line in the above equation corresponds to one term in \( EC \approx \frac{r}{n}. (\frac{r^2}{k_1^2}+ … \frac{r^2}{k_{n-1}^2}+\frac{r^2}{k_n^2})\)

Now we group similar powers, and rewrite as

\(\begin{align*}

EC &\approx \frac{r}{n} (1+1+ …+1) \\

&\quad- \frac{r}{n}\frac{(\frac{r}{n})^2}{r^2} (1^2 + 2^2 + … n^2) \\

&\quad+ \frac{r}{n}\frac{(\frac{r}{n})^4}{r^4} (1^4 + 2^4 + … n^4) \\

&\quad- \frac{r}{n}\frac{(\frac{r}{n})^6}{r^6} (1^6 + 2^6 + … n^6) \\

&\quad …

\end{align*}\)

Whereas we formerly had a finite number of terms (n), but with infinite length, we now have an infinite number of terms with finite length (n) – not that it makes a difference, since we want to make n very large anyway. We are now ready to use sankalita (सङ्कलितम्) compact these terms

#### Sankalita – Summation of series.

Aryabhata shows how to compute finite sums,

\(\begin{align*}

1+2+…n&=\frac{n.(n+1)}{2} &\text{(sankalita)}\\

1^2+2^2+…n^2&=\frac{n.(n+1)(2n+1)}{6} &\text{(vargasankalita)}\\

1^3+2^3+…n^3&=(\frac{n.(n+1)}{2})^2 &\text{(ghanasankalita)}

\end{align*}\)

He also shows how to compute

\(\begin{align*}

&\quad1\\

&+(1+2)\\

&+(1+2+3)\\

&+…\\

&+(1+2+3+..n)=\frac{n.(n+1)(n+2)}{6}&&\text{(sankalita-sankalita)}\end{align*}\)

This was later extended by Narayana Pandita (~1350CE) to show the general kth-order sum-of-sums of n integers as

\(\begin{align*}SS^{(k)}_n &= SS^{(k-1)}_1+SS^{(k-1)}_2…+SS^{(k-1)}_n \\

&= \frac{n(n+1)..(n+k)}{1.2…(k+1)}\end{align*}\)

(Superscript indicates order, not exponentiation)

We will see a lot more of this in later articles, as this is the basis of all Kerala School Calculus.

Yuktibhāṣā notes the following for the power-series sum (समघातसङ्कलितम्) of segments of a fixed length divided into n very small parts

\((1^k+2^k…+n^k) \approx \frac{n^{k+1}}{k+1} \)

This is shown by geometrical arguments relating the (k-1)th such power sum with the kth one, and noting that the pieces are atomic (अणुपरिमितम्). We shall explore this further in a later article. For now, we note that this is the analog to integration in the Kerala School.

Using the sama-ghata-sankalita approximation for small units

\( (1^k+2^k…+n^k) \approx \frac{n^{k+1}}{k+1} \), we have

\(

EC \approx \frac{r}{n} (1+1+ …+1) – \\

\frac{r}{n}\frac{(\frac{r}{n})^2}{r^2} (1^2 + 2^2 + … n^2) + \\

\frac{r}{n}\frac{(\frac{r}{n})^4}{r^4} (1^4 + 2^4 + … n^4) – \\

\frac{r}{n}\frac{(\frac{r}{n})^6}{r^6} (1^6 + 2^6 + … n^6) + \\

…\\

\approx r – \frac{r}{3} + \frac{r}{5} – \frac{r}{7} + …

\)

Now since EC is an eighth of the circumference C

\( \frac{C}{8} = r – \frac{r}{3} + \frac{r}{5} – \frac{r}{7} + … \\

C = 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + …\) (using *d = 2.r*)

and therefore,

\( \frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} \)

Which is the verse व्यासे वारिधिनिहते

### In Conclusion

Examining Yuktibhāṣā's argument, we can see the Kerala School built their Calculus around the ideas of

- Sankalita of small units
- Shodyaphala to get rid of inconvienient denominators in sums
- Approximations of sums by dropping insignificant terms.

We shall see a lot more of this in further articles as we examine further refinements of this series, and series for trigonometric functions, and dive into some of the mathematics in more detail.