What's In A Name?

Previously, we saw the nihatya series for jyā and śara and how that enables a very accurate estimate of those functions for any arc, accurate to the fourths. We also saw how this led to the partially precomputed vidvān series as well as the table of पठितज्या accurate to thirds – श्रेष्ठं नाम वरिष्ठानाम् etc.
The jyā series is

निहत्य चापवर्गेण चापं तत्तत्फलानि च । 
हरेत् समूलयुग्वर्गैस्त्रिज्यावर्गाहतैः क्रमात् ॥
चापं फलानि चाधोऽधो न्यस्योपर्युपरित्यजेत् ।
जीवापत्यै संग्रहोऽस्यैव विद्वान् इत्यदिना कृतः ॥

\(
jyā(s) = s – s.\frac{s^2}{R^2.(2^2+2)} + s.\frac{s^2}{R^2.(2^2+2)}\frac{s^2}{R^2.(4^2+4)} – s.\frac{s^2}{R^2.(2^2+2)}.\frac{s^2}{R^2.(4^2+4)}.\frac{s^2}{R^2.(6^2+6)} …
\)

which corresponds to the modern series

\(
sin(\theta) = {\theta} – \frac{\theta^3}{3!} + \frac{\theta^5}{5!} …
\)

How was this derived? Thankfully, Yuktibhāṣā comes to our aid again, and gives us a very detailed derivation

इष्टचापः – Pick and arc, any arc

As is usual with Indian convention, we pick the first quadrant of the standard circle, with East at the top, and North at the left of our figure. Any arc EC is chosen that falls within the quadrant, and we wish to find an saṅkalita for jyā. We chose points \(C_1, C_2, … C_{n-1}\) so that they divide EC into n equal parts, where n is very large. The midpoints of arcs \(C_1C_2, C_2C_3, C_3C_4, …\) are \(M_1, M_2, M_3 …\) respectively. For convenience, only the arc segments \(C_{j-1}C_j\) and \(C_jC_{j+1}\) are shown. The length of each of the arc segments is denoted as \(\alpha\).

Consider the arc segment \(EC_{j+1}\). Since it is the (j+1)th segment, its jyā, koṭijyā, and śara are denoted \(B_{j+1}, K_{j+1}, S_{j+1}\) respectively. Similarly, the jyā, koṭijyā, and śara of arc segment \(EM_{j+1}\) are called \(B_{j+\frac{1}{2}}, K_{j+\frac{1}{2}}, S_{j+\frac{1}{2}}\) respectively.

The differential jyā (खण्डज्या)

The difference \(B_{j+1} – B_j = C_{j+1}P_{j+1} – C_jP_j\) is the differential jyā, or खण्डज्या. We can compute this by noting the similarity of triangles \(C_{j+1}FC_j\) and \(OM_{j+1}Q_{j+1}\) by the perpendicular sides rule, earlier seen in the yukti of व्यासे वारिधिनिहते. That allows us to use the the rule of three (त्रैराशिकम्)‌ to link the differential jyā and koṭijyā/śara with the midpoint koṭijyā and jyā respectively. Taking the karṇas (hypotenuses) \(OM_{j+1} = R\) and \(C_{j+1}C_j = \alpha \) as the pramāṇa and iccā respectively, we get:

\(
\begin{align*}
B_{j+1} – B_j &= C_{j+1}F \\
&= \frac{\alpha}{R}. {OQ_{j+1}} \\
&= \frac{\alpha}{R}.K_{j+\frac{1}{2} }\\
K_j – K_{j+1} &= S_{j+1} – S_j = C_jF \\
&= \frac{\alpha}{R}. {M_{j+1}Q_{j+1}} \\
&= \frac{\alpha}{R}.B_{j+\frac{1}{2}}
\end{align*}
\)

Thus, the differential jyā at an arc segment is the midpoint koṭijyā of the arc segment multiplied by the length of the arc segment divided by the radius (and vice versa, but with a negative sign). Thus is the first step in the yukti, and the first time we see the differential/derivative analog turning up. By similar arguments, we can also prove that

\(
\begin{align*}
B_{j+\frac{1}{2}} – B_{j-\frac{1}{2}} &= \frac{\alpha}{R}.K_j \\
K_{j-\frac{1}{2}} – K_{j+\frac{1}{2}} &= S_{j+\frac{1}{2}} – S_{j-\frac{1}{2}} \\
&= \frac{\alpha}{R}.B_j
\end{align*}
\)

That is to say, the differentials at the midpoint are related the same way to the koṭijyā/jyā at the ends of the arc segments.

Since the number n is large, and the arc segments are very small (अणुपरिमित); in modern notation, we can say \(\frac{\alpha}{R} = d \theta \), and the relations become equivalent to:

\(
\frac{d sin \theta}{d \theta} = cos \theta \\
\frac{d cos \theta}{d \theta} = -sin \theta \\
\)

Second differentials (खण्डज्यान्तरम्)

The second differentials of jyā, or खण्डज्यान्तरम् can be computed using the relations in the previous section as:

\(
\begin{align*}
(B_{j} – B_{j-1}) – (B_{j+1} – B_j) &= \frac{\alpha}{R}.{K_{j-\frac{1}{2}}-K_{j+\frac{1}{2}}} \\
&= (\frac{\alpha}{R})^2. B_j \\
(S_{j+1} – S_{j}) – (S_{j} – S_{j-1}) &= (K_{j} – K_{j+1}) – (K_{j-1} – K_{j}) \\
&= \frac{\alpha}{R}.{B_{j+\frac{1}{2}}-B_{j-\frac{1}{2}}} \\
&= (\frac{\alpha}{R})^2. K_j \\
\end{align*}
\)

Note the implicit negative signs in both equations, appearing in two different ways.

Setting up recurrences

We can use the differentials to set up recurrence relations, which will then enable us to write out the full sankalita.

First, we sum up the first differentials of śaras at arc segment midpoints to get
\(
\begin{align*}
(S_{\frac{3}{2}} – S_{\frac{1}{2}}) + (S_{\frac{5}{2}} – S_{\frac{3}{2}}) + … + (S_{n-\frac{1}{2}} – S_{n-\frac{3}{2}}) & = S_{n-\frac{1}{2}} – S_{\frac{1}{2}} \\
& = \frac{\alpha}{R}. (B_1 + B_2 + … B_{n-1})
\end{align*}
\)

Next, we sum up second differentials of jyā to each intermediate point to get
\(
\begin{align*}
(B_2 – B_1) – (B_1-0) &= -(\frac{\alpha}{R})^2. B_1 \\
B_2 &= 2.B_1 – (\frac{\alpha}{R})^2. B_1 \\
(B_3 – B_2) – (B_2 – B_1) &= -(\frac{\alpha}{R})^2. B_2 \\
B_3 &= 2B_2 – B_1 – (\frac{\alpha}{R})^2. B_2 \\
&= 3.B_1 – (\frac{\alpha}{R})^2. (B_2 + 2 B_1) \\
(B_4- B_3) – (B_3 – B_2) &= -(\frac{\alpha}{R})^2. B_3 \\
B_4 &= 2B_3 – B_2 – (\frac{\alpha}{R})^2. B_3 \\
&= 3.B_1 – (\frac{\alpha}{R})^2. (B_3 + 2.B_2 + 3.B_1) \\
… \\
B_n &= n.B_1 – (\frac{\alpha}{R})^2. (B_{n-1} + 2.B_{n-2} + …(n-2).B_3 + (n-1).B_2 + n.B_1) \\
\end{align*}
\)

Yukti 1 – Approximations by substitution

Yuktibhāṣā provides us two proofs, one employing both recurrences, and one employing only one. We start with the latter.

First, we note that \(n.B_1 = s\) and \(\alpha = \frac{s}{n}\) at the large n limit.

\(
\begin{align*}
B_n &= n.B_1 – (\frac{\alpha}{R})^2. (B_{n-1} + 2.B_{n-2} + …(n-2).B_3 + (n-1).B_2 + n.B_1) \\
&\approx s – (\frac{s}{n.R})^2. (B_{n-1} + 2.B_{n-2} + …(n-2).B_3 + (n-1).B_2 + n.B_1) \\
\end{align*}
\)

Our first approximation for jyā sets it equal to the corresponding arc – \( B_j \approx \frac{j.s}{n}\). Substituting this back into the recurrence, and noting that we get a dvitiya-sankalita, which leads us to the second approximation:
\(
\begin{align*}
B &\approx s – (\frac{s}{n.R})^2 (B_{n-1} + 2.B_{n-2} + …(n-2).B_3 + (n-1).B_2 + n.B_1) \\
&\approx s – \frac{1}{R^2}.(\frac{s}{n})^3.SS^{(2)}_{n} \\
&\approx s – \frac{1}{R^2}.\frac{s^3}{1.2.3}
\end{align*}
\)

Now this leads to a better approximation \( B_j \approx \frac{j.s}{n} – \frac{1}{R^2}.\frac{(j.s)^3}{n^3.1.2.3}\). Substituting this into the recurrence, and using the triangle law of sankalitas gets us to the next approximation:

\(
\begin{align*}
B &\approx s – \frac{1}{R^2}.\frac{s^3}{1.2.3} + \frac{1}{R^4}.\frac{s^5}{1.2.3.4.5} \\
B_j &\approx \frac{j.s}{n} – \frac{1}{R^2}.\frac{(\frac{j.s}{n})^3}{1.2.3} + \frac{1}{R^4}.\frac{(\frac{j.s}{n})^5}{1.2.3.4.5} \\
\end{align*}
\)

and so on, for ever, resulting in:

\(
\begin{align*}
B &\approx s – \frac{1}{R^2}.\frac{s^3}{1.2.3} + \frac{1}{R^4}.\frac{s^5}{1.2.3.4.5} – \frac{1}{R^6}.\frac{s^7}{1.2.3.4.5.6.7} … \\
&= s – s.\frac{s^2}{R^2.(2^2+2)} + s.\frac{s^2}{R^2.(2^2+2)}\frac{s^2}{R^2.(4^2+4)} – s.\frac{s^2}{R^2.(2^2+2)}.\frac{s^2}{R^2.(4^2+4)}.\frac{s^2}{R^2.(6^2+6)} …
\end{align*}
\)

Yukti 2 – Bidirectional substitution

To start with, we note that

\(
\begin{align*}
S_{n-\frac{1}{2}} – S_{\frac{1}{2}} & = \frac{\alpha}{R}. (B_1 + B_2 + … B_{n-1}) \\
S_n &\approx \frac{\alpha}{R}. (B_1 + B_2 + … B_{n-1})
\end{align*}
\)

This is a legitimate approximation for large n, as \(S_n\) will turn out to be quadratic, and hence \(S_n \approx S_{n-\frac{1}{2}}\) and \(S_{\frac{1}{2}} \approx 0\) are fine.

Substituting the recurrence for \(S_n, S_{n-1}\) etc. into the recurrence for B,

\(
\begin{align*}
B_n &= n.B_1 – (\frac{\alpha}{R})^2. (B_{n-1} + 2.B_{n-2} + …(n-2).B_3 + (n-1).B_2 + n.B_1) \\
&\approx n.B_1 – \frac{\alpha}{R}. (S_n + S_{n-1} + ….+ S_1)
\end{align*}
\)

Now, we start by noting that \(n.B_1 = s\) and \(\alpha = \frac{s}{n}\). Our intent is to approximate B, compute S using the recurrence, and obtain a better approximation for B by substituting the obtained S back into the recurrence for B. This will lead to a series of improving approximations. We start with the simplest approximation for \(B_j \approx \frac{j.s}{n}\) . By substituting into the recurrence for S, we get an improved estimate for \(B_j\)

\(
\begin{align*}
S_n &\approx \frac{s}{n.R}. (B_1 + B_2 + … B_{n-1}) \\
&\approx \frac{1}{R}.(\frac{s}{n})^2(1+2 + … n-1) \\
&\approx \frac{1}{R}.(\frac{s}{n})^2.\frac{n^2}{2} \\
&\approx \frac{1}{R}.\frac{s^2}{2} \\
B_n &\approx s – \frac{s}{n.R}.\frac{1}{R}.(\frac{s}{n})^2(1^2+2^2 + … (n-1)^2) \\
&\approx s – \frac{1}{R^2} \frac{s^3}{3!} \\
\end{align*}
\)

Repeated substitutions of the result of one recurrence into the other lead us to the infinite series:

\(
\begin{align*}
B &\approx s – \frac{1}{R^2}.\frac{s^3}{1.2.3} + \frac{1}{R^4}.\frac{s^5}{1.2.3.4.5} – \frac{1}{R^6}.\frac{s^7}{1.2.3.4.5.6.7} … \\
&= s – s.\frac{s^2}{R^2.(2^2+2)} + s.\frac{s^2}{R^2.(2^2+2)}\frac{s^2}{R^2.(4^2+4)} – s.\frac{s^2}{R^2.(2^2+2)}.\frac{s^2}{R^2.(4^2+4)}.\frac{s^2}{R^2.(6^2+6)} … \\
S &\approx \frac{1}{R}.\frac{s^2}{1.2} – \frac{1}{R^3}.\frac{s^4}{1.2.3.4} + \frac{1}{R^5}.\frac{s^6}{1.2.3.4.5.6} … \\
&\approx \frac{s^2}{2R} – \frac{s^2}{2R}.\frac{s^2}{R^2.(4^2-4)} + \frac{s^2}{2R}.\frac{s^2}{R^2.(4^2-4)}.\frac{s^2}{R^2.(6^2-6)} …
\end{align*}
\)

Which gives us both the निहत्य चापवर्गेण relations.

These are quite intricate proofs, and are quite rigorous as well. Approximations that are made turn out to be quite legitimate in the limit of large n, which indicates that the Kerala School worked well with small and large n limits. Another point of note here is the introduction of first and second differentials. Sankalitas being the Kerala analogue of integrals, these differentials play the role of derivatives.

What's better than an improved jyā table and an improved method for interpolation? A full jyā series that lets you compute jyā of any arc, to any degree of precision you wish.

निहत्य चापवर्गेण चापं तत्तत्फलानि च । 
हरेत् समूलयुग्वर्गैस्त्रिज्यावर्गाहतैः क्रमात् ॥
चापं फलानि चाधोऽधो न्यस्योपर्युपरित्यजेत् ।
जीवापत्यै संग्रहोऽस्यैव विद्वान् इत्यदिना कृतः ॥

Multiply the arc by the square of the arc. Multiply successive results by the square of the arc again, and divide by the square of the radius (trjyā) multiplied by sum of successive odd numbers and their squares. Set the arc and the other terms up in a vertical line, and subtract the last from the next, and the result from the next higher, and so on, to get the jīva (jyā). The same can be done by the phrases vidvān and so on.

Or:

\(
jyā(s) = s – s.\frac{s^2}{R^2.(2^2+2)} + s.\frac{s^2}{R^2.(2^2+2)}\frac{s^2}{R^2.(4^2+4)} – s.\frac{s^2}{R^2.(2^2+2)}.\frac{s^2}{R^2.(4^2+4)}.\frac{s^2}{R^2.(6^2+6)} …
\)

निहत्य चापवर्गेण रूपं तत्तत्फलानि च ।
हरेद्विमूलयुग्वर्गैस्त्रिज्यावर्गाहतैः क्रमात् ॥
किन्नु व्यासदलेनैव द्विघ्नेनाद्यं विभज्यताम् ।
फलान्यधोऽधः क्रमशो न्यस्योपर्युपरित्यजेत्
शरापत्यै संग्रहोऽस्यैव स्तेनस्त्रीत्यादिना कृतः ॥

Multiply one by the square of the arc. Multiply successive results by the square of the arc again, and divide by the square of the radius (trjyā) multiplied by difference of successive odd numbers from their squares. Let the divisor of the first term be twice the radius. Set the terms up in a vertical line, and subtract the last from the next, and the result from the next higher, and so on, to get the śara. The same can be done by the phrases stenastrī and so on.

Or:
\(
śara(s) = \frac{s^2}{2R} – \frac{s^2}{2R}.\frac{s^2}{R^2.(4^2-4)} + \frac{s^2}{2R}.\frac{s^2}{R^2.(4^2-4)}.\frac{s^2}{R^2.(6^2-6)} …
\)

Since \( jyā(s) = R.sin(\frac{s}{R})\), and \(n.(n+1) = n^2+n=(n+1)^2-(n+1)\) these two series can be seen to be equivalent to the modern series:
\(
sin(\theta) = {\theta} – \frac{\theta^3}{3!} + \frac{\theta^5}{5!} … \\
vers(\theta) = \frac{\theta^2 }{2!} – \frac{\theta^4}{4!} + \frac{\theta^6}{6!} …
\)

These series yield jyā or śara of any arc, correct to thirds of arc, as we will soon see.

Vidvān – Partly Precomputed Series

Having derived these series, we ask, how many terms do we need to compute? We could go as far as we need to to get the accuracy we need. Mādhava has made things even easier for us by giving us a partially pre-computed series that is accurate to the thirds of arc.

विद्वान्स्तुन्नबलः कवीशनिचयः सर्वार्थशीलस्थिरो ।
निर्विद्धाङ्गनरेन्द्ररुङ् निगदितेष्वेषु क्रमात् पञ्चसु ॥
आधस्त्यात् गुणितादभीष्टधनुषः कृत्या विहृत्यान्तिम-
स्याप्तं शोध्यमुपर्युपर्यथ घनेनैवं धनुष्यन्ततः ॥
The phrases vidvān (44″), tunnabala (33″06”'), kavīśanicaya (16'05”41”'), sarvārthaśīlasthiro (273'57”47”') nirviddhāṅganarendraruṅ (2220'39”40”') are successive multipliers, placed in reverse order, successively multiplied by the square of the arc divided by the square of a quarter-circle, and subtracted from the next, finally multiplied by the cube of the desired arc divided by the cube of a quarter circle, then subtracted from the desired arc.

\(
jyā(s) = s – \frac{s^3}{5400^3}.(2220'39”40”' – \frac{s^2}{5400^2}.(273'57”47”' – \frac{s^2}{5400^2}.(16'05”41”' – \frac{s^2}{5400^2}.(33″06”' – \frac{s^2}{5400^2}.44”'))))
\)

We also have the convenient encodings नानाज्ञानतपोधरः for \(5400^2 = 29160000\) and अज्ञाननुन्ने नवतत्त्वसंशयः for \(5400^3 = 157464000000\)

Similarly for śara:
स्त्येनस्त्रीपिशुनस्सुगन्धिनगनुद्भद्राङ्गभव्यासनो
मीनाङ्गो नरसिंह ऊनधनकृद्भूरेव षट्स्वेषु तु ।
आधस्त्यात् गुणितादभीष्टधनुषः कृत्या विहृत्यान्तिम-
स्याप्तं शोध्यमुपर्युपर्यथ फलं स्यादुत्क्रमज्यान्त्यजम् ॥

\(
śara(s) = \frac{s^2}{5400^2}.(4241'09”0”' – \frac{s^2}{5400^2}.(872'03”05”' – \frac{s^2}{5400^2}.(71'43”24”' – \frac{s^2}{5400^2}.(3'09”37”' – \frac{s^2}{5400^2}.(5″12”' – \frac{s^2}{5400^2}.6”')))))
\)

Where do we get these coefficients from? It can be seen that these are derived from computing the Mādhava series for an arc of 5400′ (and hence the division by 5400′ in the formula, by त्रैराशिकम् )

The Mādhava Jyā Table

For those un-inclined to compute the series for arbitrary arcs, worry not – the Mādhava jyā table – understood to have been computed using the vidvān method – exists. Similar to the Āryabhaṭa table, Mādhava jyā table is a verse-encoded table of jyās at the standard 24 points, accurate to the thirds, ie: 3600 times more accurate than Āryabhaṭa. This can be used with the इष्टदोःकोटिधनुषोः method of interpolation to obtain intermediate values.

	Verse	Value	Modern
1	श्रेष्ठं नाम वरिष्ठानां	224,50,22,0	224,50,21,49
2	हिमाद्रिर्वेदभावनः	448,42,58,0	448,42,57,35
3	तपनो भानु सूक्तज्ञो	670,40,16,0	670,40,16,2
4	मध्यमं विद्धि दोहनं	889,45,15,0	889,45,15,36
5	धिगाज्योनाशनं कष्टं	1105,1,39,0	1105,1,38,56
6	छन्नभोगा शयाम्बिका	1315,34,7,0	1315,34,7,26
7	मृगाहारो नरेशोयं	1520,28,35,0	1520,28,35,27
8	वीरो रणजयोत्सुकः	1718,52,24,0	1718,52,24,11
9	मूलं विशुद्धं नाळस्य	1909,54,35,0	1909,54,35,11
10	गानेषु विरळा नराः	2092,46,3,0	2092,46,3,29
11	अशुद्धिगुप्ता चोरश्रीः	2266,39,50,0	2266,39,50,12
12	शङ्कुकर्णो नगेश्वरः	2430,51,15,0	2430,51,14,35
13	तनुजो गर्भजो मित्रं	2584,38,6,0	2584,38,5,31
14	श्रीमानत्र सुखी सखे	2727,20,52,0	2727,20,52,22
15	शशी रात्रौ हिमाहारौ	2858,22,55,0	2858,22,55,6
16	वेगज्ञः पथि सिन्धुरः	2977,10,34,0	2977,10,33,43
17	छायालयो गजो नीलो	3083,13,17,0	3083,13,16,56
18	निर्मलो नास्ति सत्कुले	3176,3,50,0	3176,3,49,57
19	रात्रौ दर्पणमभ्राङ्गं	3255,18,22,0	3255,18,21,34
20	नागस्तुङ्गनखो बली	3320,36,30,0	3320,36,30,12
21	धीरो युवा कथालोलः	3371,41,29,0	3371,41,29,8
22	पूज्यो नारीजनैर्भगः	3408,20,11,0	3408,20,10,56
23	कन्यागारे नागवल्ली	3430,23,11,0	3430,23,10,38
24	देवो विश्वस्थली भृगुः	3437,44,48,0	3437,44,48,22

A minor point to note is the use of ळ encoding to 9 in a couple of lines.

Why Three Methods?

Comparing निहत्य चापवर्गेण with the famous व्यासे वारिधिनिहते series, we see that it is tougher to calculate through, and therefore could not have been expected to be used by someone with basic ability. It must be presumed that Mādhava and his followers left multiple options available for those with varying skills in computation.

Yukti for the Mādhava Jyā Series

As usual, we have detailed derivations of these available from Yuktibhāṣā, but we will leave that for the next article. As with the series itself, these proofs are involved, and require a bit of diligence.

Previously, we saw the मूलसङ्कलितम् \( S_n = 1+2+…n=\frac{n.(n+1)}{2} \) and also noted that as n becomes large, \(S_n \approx \frac{n^2}{2}\). We also saw the समघातसङ्कलितानि, sums of higher powers, including the sum of squares – वर्गसङ्कलितम्, and sum of cubes – घनसङ्कलितम्. We noted that in the limit of large n, the kth power sankalita tends to:

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

We also saw सङ्कलितसङ्कलितम्, the second order sankalita which is the sum of first order sankalitas \(
SS^{(2)}_n = S_n + S_{n-1} + …. + 1
\), and also other higher order sankalitas.

The Triangle Law of Sankalitas

To help us with the proof of Madhava's sankalita for jyā, we will note an interesting, but simple result for sankalitas, which I have taken the liberty of calling the Triangle Law

\(
\begin{align*} SS^{(k)}_n &= SS^{(k-1)}_n + SS^{(k-1)}_{n-1} + …. + 1 \\
&= SS^{(k-2)}_n + SS^{(k-2)}_{(n-1)} + SS^{(k-2)}_{(n-2)} + … + 1 \\
&\quad+ SS^{(k-2)}_{(n-1)} + SS^{(k-2)}_{(n-2)} + … + 1 \\
&\quad+ SS^{(k-2)}_{(n-2)} + … + 1 \\
&\quad … \\
&\quad + 1 \\
& = SS^{(k-2)}_n + 2.SS^{(k-2)}_{(n-1)} + 3.SS^{(k-2)}_{(n-2)} + … n.1 \\
& = \sum_{j=1}^{n}j.SS^{(k-2)}_{(n-(j-1))}
\end{align*}
\)

This result allows us to express the kth sankalita-sankalita as a sum of (k-2)th sankalita-sankalitas, in addition to the elementary expression in terms of (k-1)th sankalita-sankalitas, which we will find useful.

Recap

In the previous article in this series, we saw how Āryabhaṭa's jyā table was calculated. It was a remarkably simple recurrence relation, with accuracy to one minute of arc (using the Indian standard circle of circumference 21600′) at 24 points 225′ apart.

प्रथमाच्चापज्यार्धाद्यैरूनं खण्डितं द्वितीयार्धम् ।
तत्प्रथमज्यार्धांशैस्तैस्तैरूनानि शेषाणि ।।

We set both first jyā B₁ and the first difference K₁ as equal to 225. Then, further differences (खण्डज्या) and actual jyās (पिण्डज्या) are calculated using the recurrence relations

\(
K_2 = K_1 – \frac{B_1}{B_1} \\
K_n = K_{n-1} – \frac{B_{n-1}}{B_1} \\
B_n = K_n + B_{n-1}
\)

The resulting खण्डज्या are encoded in the verse

मखि भखि फखि धखि णखि ञखि ङखि हस्झ स्ककि किष्ग श्घकि किघ्व |
घ्लकि किग्र हक्य धकि किच स्ग झश ङ्व क्ल प्त फ छ कलार्धज्याः ||

For intermediate points in between the 24 standard tabular points, we use linear interpolation.

Making the tabular jyā more accurate

The Kerala School came up with two approaches to make tabular sines, or पठितज्या as they called it, more accurate. The first, which we shall see in this article, is a modification of Āryabhaṭa's recurrence, plus a better method for interpolation. The second, which we shall see later, resulted in a full infinite series for jyā allowing any degree of accuracy to be achieved.

Improved recurrence

This verse is from Tantrasaṅgraha of Nīlakaṇṭha Somayāji and is also seen in Yuktibhāṣā

विलिप्तदशकोना ज्या राश्यष्टांशधनुः कलाः ।
आद्यज्यार्धात्ततो भक्ते सार्धदेवाश्विभिस्ततः ।।
त्यक्ते द्वितीयखण्डज्या द्वितीयज्या च तद्युतिः ।
ततस्तेनैव हारेण लब्धं शोद्यं द्वितीयतः ।।
खण्डात्तृतीयखण्डस्स्यात् द्वितीयस्तद्युतो गुणः ।
तृतीयस्स्यात्ततश्चैवं चतुर्थाद्याः क्रमात्गुणाः ।।

The jyā of an eighth of a rāśi is its length minus ten seconds
From the first jyā is subtracted itself divided by 233′ 30″ to get the second jya difference and the second jyā is obtained by adding that to the first jyā
Further subtrahends are found by dividing successive jyās with the same divisor, and are subtracted from jyā differences to find successive differences.
Further jyās are obtained by adding successive differences to prior jyās

The Kerala school replaced the Āryabhaṭa recurrence with the precise relation

\(
K_n = K_{n-1} – {(\frac{a}{R})}^2.B_{n-1} \\
B_n = K_n + B_{n-1}
\)

where R is the radius, and a is the full-chord of the quarter circumference divided by 24, (ie 225′). So far, this is perfectly precise, as we shall see in the next article where we will look at the saṅkalita for this. To make this computable, they use the approximation \(a = 225′\). This is more accurate than the Āryabhaṭa approximation which sets the corresponding jyā to 225′, since the full-chord is between the jyā and the corresponding arc in length. This leads to the approximations
\({(\frac{a}{R})}^2 \approx 233′ 30″\) (नीलोबालारिः or सार्धदेवाश्वि in bhūtasaṅkhyā).
This was later improved by Śankaravāriyar to \({(\frac{a}{R})}^2 \approx 233′ 32″\) (रङ्गेबालास्त्री)

Also,
\(K_1 \approx 224′ 50″\)
This again was improved by Śankaravāriyar as,
\(K_1 \approx 224′ 50″22‴\)

What we gain here is better accuracy, at the cost of

1. A more complex divisor – 233′ 32″ in the place of 225′
2. A more complex first sine – 224′ 50″ 22”' in the place of 225′

Comparsion of tabular jyā accuracy

Comparing the accuracy of tabular jyā by these methods, with the modern values, we see that Śankaravāriyar is accurate to a few seconds of arc, while Nīlakaṇṭha's table is off by upto about 20 seconds. Āryabhaṭa, as we have noted is accurate to about a minute of arc.

	Āryabhaṭa	Nīlakaṇṭha	Śankaravāriyar	Modern
1	225	224,50,0,0	224,50,22,0	224,50,21,49
2	449	448,42,13,37	448,42,58,1	448,42,57,35
3	671	670,39,9,19	670,40,16,55	670,40,16,2
4	890	889,43,45,11	889,45,17,10	889,45,15,36
5	1105	1104,59,43,37	1105,1,41,33	1105,1,38,56
6	1315	1315,31,45,42	1315,34,11,31	1315,34,7,26
7	1520	1520,25,45,33	1520,28,41,30	1520,28,35,27
8	1719	1718,49,4,4	1718,52,32,46	1718,52,24,11
9	1910	1909,50,42,38	1909,54,46,57	1909,54,35,11
10	2093	2092,41,36,2	2092,46,19,8	2092,46,3,29
11	2267	2266,34,45,13	2266,40,10,30	2266,39,50,12
12	2431	2430,45,29,17	2430,51,40,20	2430,51,14,35
13	2585	2584,31,36,59	2584,38,37,34	2584,38,5,31
14	2728	2727,13,37,36	2727,21,31,36	2727,20,52,22
15	2859	2858,14,51,2	2858,23,42,25	2858,22,55,6
16	2978	2977,1,37,15	2977,11,30,3	2977,10,33,43
17	3084	3083,3,25,0	3083,14,23,12	3083,13,16,56
18	3177	3175,52,59,31	3176,5,7,7	3176,3,49,57
19	3256	3255,6,29,41	3255,19,50,32	3255,18,21,34
20	3321	3320,23,34,1	3320,38,11,52	3320,36,30,12
21	3372	3371,27,26,0	3371,43,24,23	3371,41,29,8
22	3409	3408,4,58,21	3408,22,20,35	3408,20,10,56
23	3431	3430,6,46,22	3430,25,35,30	3430,23,10,38
24	3438	3437,27,10,26	3437,47,29,10	3437,44,48,22

Improved Interpolation

Rather than basic linear interpolation, the Kerala School introduced a more complex interpolation equation to improve accuracy:

इष्टदोःकोटिधनुषोः स्वसमीपसमीरिते
ज्ये द्वे सावयवे न्यस्य कुर्यादूनाधिकं धनुः
द्विघ्नतल्लिप्तिकाप्तैकशरशैलशिखीन्दवः
न्यस्याच्छोधाय च मिथः तत्संस्कारविधित्सया
छित्वैकां प्राक् क्षिपेज्जह्यात् तद्धनुष्याधिकोनके
अन्यस्यामथ तां द्विघ्नां तथा स्यामिति संस्कृतिः‌
इति ते कृतसंस्कारे स्वगुणौ धनुषोस्तयोः
तत्राल्पीयः कृतिं त्यक्त्वा पदं त्रिज्याकृते परः‌
For a desired arc, find the tabular jyā and kojyā below and above it
compute the differences of arc length to the tabular points above and below,
Divide 13751 by these and keep aside as divisors for mutual correction
Divide each one (ie jyā below and kojyā above) by this, and subtract from other (jyā above and kojyā below). Mutliply the results by two, and add to the original (ie jyā below and kojyā above) to get the interpolated jyā and kojyā.
Of the two, take the smaller, square it, subtract from the square of the radius and take the root to get the other

Again, this verse is from Tantrasaṅgraha and is also seen in Yuktibhāṣā. The idea is better explained mathematically thus:

\(
D = \frac{13751}{2.d} \\
\begin{align*}
jyā(B+d) &= jyā(B) + \frac{2}{D}(kojyā(B) – \frac{jyā(B)}{D}) \\
jyā(B-d) &= jyā(B) – \frac{2}{D}(kojyā(B) + \frac{jyā(B)}{D}) \\
kojyā(B-d) &= kojyā(B) + \frac{2}{D}(jyā(B) – \frac{kojyā(B)}{D})\\
kojyā(B+d) &= kojyā(B) – \frac{2}{D}(jyā(B) + \frac{kojyā(B)}{D})\\
\end{align*}
\)

How does this work? Where does the mysterious 13751 come from? It turns out that it is a decent approximation for four times the radius of the standard circle. Using the जीवे परस्परन्यायः (which we will see in a later article), which modern readers will know as the equation
\( sin(A+B) = sin(A).cos(B) + cos(A).sin(B)\),
and the approximation for small t
\(
sin(t) = t \\
cos(t) = \sqrt{1-t^2}
\)
and recasting to the jyā framework allows us to derive the equations above.

We aren't done yet

Granted, this gets us better accuracy than Āryabhaṭa, but knowing the Kerala school, the gains from this are clearly not sufficient to satisfy their rather exacting standards. So, this improved method is extended further to generate an infinite series for jyā, just as we saw for the circumference. We will see that the jyā series produces much improved accuracy than the recurrence relation, and will lead us to the famous Madhava jyā table and its interesting mnemonics.

Having calculated the ratio of the circumference and diameter of a circle to any accuracy we want with the Madhava series, we move onto the question of calculating ज्या/jyā or the half-chord of any arc. ज्या is the Indian counterpart of the modern sine function, and a backgrounder on how Indians thought of it, and the entire subject of what we call “trigonometry” would be interesting. We have seen a brief summary of this in an earlier post, but this would be a good time for a more detailed picture.

Bows, strings and arrows

Think of an arc of a circle as a bow – depicted in red in the figure. The chord of the arc represents the bowstring, and is therefore called समस्तज्या – full bowstring – and the half-chord (green in the figure) correspondingly becomes अर्धज्या, or just ज्या. Āryabhaṭa is often fastidious about calling the half-chord अर्धज्या, but later mathematicians merely say ज्या. The radius of the circle (purple) represents half of the bowstring in stretched position. The कोटीज्या or कोटिः (in yellow) is the horizontal difference in positions between the normal and stretched bowstrings. The difference of radius and कोटीज्या is the arrow or शरः (in blue) placed on the bow. The शरः is not commonly seen in modern trigonometry, but turns out to be quite useful in Indian Trigonometry.

A more formal picture

Consider the circle centered on O, oriented with East upwards, as is the convention. The North South diameter is NOS, and the East West diameter EOW. Consider Arc EA of the circle. The ज्या of this arc is segment AJ, and the corresponding कोटीज्या (or कोटिः) is AK. EJ is the शरः.

OE/OA/ON/OS are all equal to the radius (व्यासार्धः, also called कर्णः in this context). कर्णः means radius in this context, but can also mean diagonal, or hypotenuse of a right triangle. OA fulfils all these meanings at the same time.

Using the notation \(EA = s \text{,} \quad \theta = \frac{s}{R} \), we get:

\(\begin{align*}
jyā(s) &= AJ = R.sin(\theta) \\
koṭi(s) &= AK = R.cos(\theta) \\
śara(s) &= EJ = R.versine(\theta) = R.(1-cos(\theta)) \\
\end{align*}\)

Quite evidently, we can derive the ज्याकोटिकर्णन्यायः \( jyā^2 + koṭi^2 = karṇa^2 \), by noting that \( AJ^2 + AK^2 = OA^2 \).

In modern trigonometry, we talk of sines and cosines of angles (arcs divided by the radius), whereas ज्या etc. are functions of arcs. Again, the ज्या etc. are lengths, not ratios as the sine and cosine functions are. Therefore, ज्या etc. are dependent on the actual radius/circumference of the circle, which can make life complicated if we use them as is. Therefore, a convention of all circles having radius 21600′ was adopted.

These names and conventions were first used by Āryabhaṭa, and adopted by later Indian mathematicians.

Building a table of jyā

Āryabhaṭa's first step towards being able to calculate the jyā of any arc was to build a table of a fixed number of them. This construction later became standard among all of his followers, thanks to his ingenious method of calculating those tabular jyās. Āryabhaṭa divided the NE quadrant of the circle into 24 arcs of equal length, EC₁, C₁C₂, … C₂₂C₂₃, C₂₃N. The respective jyās are C₁B₁, C₂B₂, … C₂₃B₂₃ and their respective koṭis are C₁K₁, C₂K₂, … C₂₃K₂₃. How are these calculated?

There is the hard, but accurate way. Through geometrical reasoning, Āryabhaṭa proved that \(jyā(30) = \frac{r}{2}\), \(jyā(90) = r \) and \( jyā(\frac{s}{2}) = \frac{1}{2}. \sqrt{jyā(s)^2 + śara(s)^2}\). Using these and the ज्याकोटिकर्णन्यायः all of tabular jyā can be calculated accurately. However, a whole lot of square roots need to be calculated, which limits practical calculation accuracy. This problem sets the stage for Āryabhaṭa's ingenious method.

The Āryabhaṭa Recurrence Relation

प्रथमाच्चापज्यार्धाद्यैरूनं खण्डितं द्वितीयार्धम् ।
तत्प्रथमज्यार्धांशैस्तैस्तैरूनानि शेषाणि ।।
From the first jyā. subtract its ratio with the first jya (ie itself, therefore 1) to get the second difference
Further differences are computed from the previous by subtracting the ratio of the previous sine with the first

Each of the equal arcs EC₁, C₁C₂, … C₂₂C₂₃, C₂₃N has length equal to 21600/(4*24) =225, which was one reason exactly 24 divisions were chosen rather than 23 or 25. Using a small arc approximation, Āryabhaṭa set jyā(225′) = 225′. Further jyās are calculated using recurrence relations.

Let the jyā of EC₁, C₁C₂, … C₂₂C₂₃, C₂₃N be denoted as B₁, B₂, … B₂₄. We denote their differences as K₁, K₂, … K₂₄, such that K₁ = B₁, K₂ = B₂ – B₁, K₃ = B₃ – B₂ etc. The Āryabhaṭa recurrence relation is

\(
K_2 = K_1 – \frac{B_1}{B_1} \\
K_n = K_{n-1} – \frac{B_{n-1}}{B_1}
\)

Since we already know K₁ = B₁ = 225′, this lets us calculate the full table of jyās. Āryabhaṭa calls the K_n as खण्डज्या and the B_n as पिण्डज्या. Calculating the K_n leads to the verse of all 24 खण्डज्या (encoded in his peculiar encoding)

मखि भखि फखि धखि णखि ञखि ङखि हस्झ स्ककि किष्ग श्घकि किघ्व |
घ्लकि किग्र हक्य धकि किच स्ग झश ङ्व क्ल प्त फ छ कलार्धज्याः ||

S.No	Angle	खण्डज्या	पिण्डज्या	3438′ * Modern Sine
1	03° 45′	225	225′	224.8560
2	07°   30′	224	449′	448.7490
3	11°   15′	222	671′	670.7205
4	15°   00′	219	890′	889.8199
5	18°   45′	215	1105′	1105.1089
6	22°   30′	210	1315′	1315.6656
7	26°   15′	205	1520′	1520.5885
8	30°   00′	199	1719′	1719.0000
9	33°   45′	191	1910′	1910.0505
10	37°   30′	183	2093′	2092.9218
11	41°   15′	174	2267′	2266.8309
12	45°   00′	164	2431′	2431.0331
13	48°   45′	154	2585′	2584.8253
14	52°   30′	143	2728′	2727.5488
15	56°   15′	131	2859′	2858.5925
16	60°   00′	119	2978′	2977.3953
17	63°   45′	106	3084′	3083.4485
18	67°   30′	93	3177′	3176.2978
19	71°   15′	79	3256′	3255.5458
20	75°   00′	65	3321′	3320.8530
21	78°   45′	51	3372′	3371.9398
22	82°   30′	37	3409′	3408.5874
23	86°   15′	22	3431′	3430.6390
24	90°   00′	7	3438′	3438.0000

These are accurate to the minute (= 1/21600 of the circumference), which is remarkable for something that's that simple to remember. Incidentally, the Āryabhaṭa recurrence can be considered a discrete approximation of the modern relation \( \frac{d}{d\theta}sin \theta = – sin \theta\).

Reversing this table gives the values of koṭi for the same arcs, since koṭi(s) = jyā(90-s). Using śara(s) = R- koṭi(s) lets us calculate those as well.

Calculation of jyā at points between the points of the table is done through simple linear interpolation. Thus, we're home and dry to a decent accuracy for all three – jyā, koṭi, and śara, with nothing more than basic arithmetic!

Going beyond 90 degrees

Consider the four points A₁, A₂, A₃, A₄ . By convention, we measure angles in Indian astronomy from the East point (first point of Aries, or मेषादिः), and therefore the points on the circle are marked E, N, W, S, respectively. For a point in the NW quadrant (ie, arc greater than quarter but less than half the circumference), we take the complementary arc WA₂, and read out (or interpolate) its tabular jyā, to get A₂B₂, the jyā of EA₂. For A₃, we use WA₃ to calculate A₃B₃ with a negative sign, and for A₄, we use the complementary arc EA₄ to calculate A₄B₄, again with a negative sign. This corresponds to the modern relationships:

\(\begin{align*}
sin(90+\theta) &= cos(\theta) = sin(90-\theta) \\
sin(180+\theta) &= -sin(\theta) \\
sin(270+\theta) &= -cos(\theta) = -sin(90-\theta) \\
\end{align*}\)

Getting even better

Is this the end of the story then? Not quite. While these values were “good enough” for a long time, errors tended to accumulate, and just as we saw with calculating circumference, the need for more accurate values was felt. Also, linear interpolation is clearly not quite good enough for a non-linear function, as we would instantly note in a modern context.

The Kerala School came up with solutions for both of these.

• First, they discovered much improved recurrence relations, and a proper infinite series for these functions which improved accuracy much further.
• Secondly, they found a much more accurate interpolation, using the जीवे परस्परन्यायः relation for sin(A+B).

All this will be explored in later articles.

31415926536

A famous verse in a later work from the Kerala school, with a कटपयादि mnemonic of pi to ten decimal places goes:

आनूननून्नाननुनुन्ननित्यैस्  समाहताश्चक्रकलाविभक्ताः‌ ।
चण्डांशुचन्द्राधमकुम्भिपालैर्व्यासस्तदर्धं त्रिभमौर्विका स्यात् ।।

करणपद्धतिः – पुतुमनसोमयाजी

The circumference of a circle in minutes of arc is multiplied by आनूननून्नाननुनुन्ननित्यम् (=10000000000) and divided by चण्डांशुचन्द्राधमकुम्भिपालः = (31415926536) to yield the diameter ….

This is accurate to 11 digits including the leading 3.

How was चण्डांशुचन्द्राधमकुम्भिपालः calculated?

In the previous article in the series, we saw how the व्यासे वारिधिनिहते series converged too slowly to be of practical use, and how it could be made useful by adding a संस्कार, or correction factor. We saw two correction factors, the first order \(\frac{4d}{2(p+1)}\) and the second order \(\frac{\frac{p+1}{2}}{(p+1)^2+1}\), and how they made the series converge well. Neither, however, is good enough to calculate this result in less than 30 terms, which is what would be manually feasible.

Third order संस्कारः (correction)

अन्ते समसङ्ख्यादलवर्गः सैको गुणः सैव पुनः
युगगुणितो रूपयुतः समसङ्ख्यादलहतः भवेद्धारः

\( C_p = (-1)^{\frac{p+1}{2}}.4d.\frac{(\frac{p+1}{2})^2+1}{(4(\frac{p+1}{2})^2+5).\frac{p+1}{2}}\)

Terms	Result with third order correction	Accuracy (places after decimal point)
10	3.141592705349	6
20	3.141592654020	9
28	3.141592653631	10
30	3.141592653615	10
40	3.141592653593	11
50	3.141592653591	11
100	3.141592653590	11

By calculating to 28 terms, we achieve our desired result of चण्डांशुचन्द्राधमकुम्भिपालः = 3145926536.

Alternate series

By incorporating the correction term into the series itself, Yuktibhāṣā derives quicker-converging series as alternatives to व्यासे वारिधिनिहते. There are equivalent to using the correction terms, so are more of a curiosity than providing any further benefit.

We have seen the स्थौल्य (error term) before: \( E_p = \frac{1}{a_{p-1}} + \frac{1}{a_{p}} – \frac{1}{p} \) . We fold it into the series thus

\(\begin{align*}
C &= 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … \pm \frac{4.d}{p} \mp \frac{4d}{a_p} \\
&= 4.d [1 – \frac{1}{a_1} + (\frac{1}{a_1} + \frac{1}{a_3} – \frac{1}{3}) – (\frac{1}{a_3} + \frac{1}{a_5} – \frac{1}{5}) … \mp (\frac{1}{a_{p-1}}+\frac{1}{a_p}-\frac{1}{p}) \\
&= 4.d [ 1 – \frac{1}{a_1}] + 4d [E_3-E_5+…\mp E_p]
\end{align*}\)

By simple rearrangement, we have transformed the original series with संस्कारः into alternate series using the respective स्थौल्य. We have calculated the स्थौल्य for first and second order corrections previously. Unfortunately, the स्थौल्य for the most accurate correction is hard to use. Thus, we get the two alternate series incorporating संस्कारः

The Madhava series is slow and boring, says … Madhava

यत्सङ्ख्ययात्र हरणे कृते निवृत्ता हृतिस्तु जामितया
तस्या ऊर्ध्वगता या समसङ्ख्या तद्दलं गुणोऽन्ते स्यात् ||
तद्वर्गो रूपयुतो हारो व्यासाब्धिघातात् प्राग्वत् 
ताभ्यामाप्तं स्वमृणे कृते धने क्षेप एव करणीयः ||
लभ्धः परिधिः सूक्ष्मो बहुकृत्वो हरणतोऽतिसूक्ष्मश्च ||
When you stop division out of boredom, remember the last divisor
take the next even number, and halve it. That is your numerator.
The square of the numerator, plus one is your divisor. Multiply by four times the diameter as before.
The correction you thus calculate, add to the result if the last operation you did was subtraction, and vice versa.
The corrected diameter you get is accurate, much more so that by a lot of division (ie: by taking several more terms!)

What exactly does this mean?

We are back with व्यासे वारिधिनिहते, the Madhava Pi (or more accurately circumference) series. We saw in an earlier article how it was demonstrated to be correct by Yuktibhāṣā. To jog the reader's memory, the verse is :

व्यासे वारिधिनिहते रूपहृते व्याससागराभिहते ।
त्रिशरादिविषमसंख्याभक्तमृणं स्वं पृथक्क्रमात् कुर्यात् ॥

which is equivalent to

\( C = 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … \)
\( \frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} \)

The one problem with this series is that it converges extremely slowly. This is precisely why the verse says निवृत्ता हृतिस्तु जामितया (“when you stop dividing out of boredom”). Remember that all this calculation is happening by hand. How many terms would we calculate? 10? 20? 50?

3.14159265358979

These are the first 15 digits of Pi – this is our target.

Madhava's series converges really really slowly towards this number. Setting the diameter to 1, we add more and more terms to the series to calculate the circumference as

Terms	Result
10	3.0418396189294024
20	3.0916238066678385
30	3.1082685666989462
40	3.1165965567938323
50	3.1215946525910105
100	3.131592903558553

50 terms is a lot to calculate by hand, and we haven't got anywhere close to the answer we want. Even after a 100 terms, Aryabhata's 3.1416 is still much better. This series as is, is a theoretical curiosity, and is not much use for practical calculation. Or is it?
(We have conveniently ignored that Madhava himself would have calculated in fractions, not decimals, and would've probably set d = 100000000000, but that doesn't change our point)

Terms	Result	Error	1/Terms
10	3.0418396189294024	0.09975303466039076	0.1
20	3.0916238066678385	0.04996884692195458	0.05
30	3.1082685666989462	0.03332408689084687	0.03333..
40	3.1165965567938323	0.02499609679596082	0.025
50	3.1215946525910105	0.019998000998782572	0.02
100	3.131592903558553	0.009999750031240318	0.01

We notice that the difference between Pi and our approximation is almost the same, but a little less than (and grows increasingly close to) the reciprocal of the number of terms we have calculated. If only we could add a correction term …

संस्कारः – The correction term.

Let's start with the presumption that we want a correction term, to be added to the end of the series, depending on the number of terms in the series. The correction term must

1. Be dependent on the number of terms already calculated (not a constant!)
2. Approximate the rest of the series
3. Converge faster than the series itself.

Let's assume we have computed n terms in the series. The last denominator we computed was \(p = 2n-1\). We are looking for a correction term of the form \(1/a_p\) to add to the series, such that
\( C \approx 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … (-1)^{\frac{p-1}{2}}.\frac{4.d}{p} + (-1)^{\frac{p+1}{2}}.\frac{4.d}{a_p} \)
As n (and p) grow larger, we want the series to converge faster with the correction term added. ie, we want:
\( 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … (-1)^{\frac{p-3}{2}}.\frac{4.d}{p-2} + (-1)^{\frac{p-1}{2}}.\frac{4.d}{a_{p-1}} \approx 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … (-1)^{\frac{p-1}{2}}.\frac{4.d}{p} + (-1)^{\frac{p+1}{2}}.\frac{4.d}{a_p} \) which, after accounting for signs, simplifies to:
\( \frac{1}{a_{p-1}} + \frac{1}{a_{p}} \approx \frac{1}{p}\)

We define the स्थौल्य (error term) \( E_p = \frac{1}{a_{p-1}} + \frac{1}{a_{p}} – \frac{1}{p} \), which we want to keep as small as we can, and want it to go to zero with n much faster than \(\frac{1}{n}\) does.

First order correction term

Our first attempt at a correction term is \( a_p = 2(p+1) = 4n \). This is easy to calculate. Whatever your last denominator was, add one to it, and double your result to get the denominator of your correction term. This is a good start, as we can see :

Terms	Result with first order correction
10	3.141839618929402
20	3.141623806667839
30	3.1416019000322795
40	3.141596556793832
50	3.1415946525910106
100	3.1415929035585526

We are as good as Aryabhata after 20 terms. In 40 terms, we are ahead of him by one digit. By term 100, we have one more digit of accuracy. Good, but we need to do better.

We note that the स्थौल्य is now
\( \begin{align*}
E_p &= \frac{1}{a_{p-1}} + \frac{1}{a_{p}} – \frac{1}{p} \\
&= \frac{1}{2(p-1)} + \frac{1}{2(p+1)} – \frac{1}{p} \\
&= \frac{1}{p^3-p}
\end{align*}\)

We note that a positive स्थौल्य indicates an over-correction, which we can verify from the table. It goes down as the third power of p, indicating that we now converge much faster with this correction term, as we can see from the table.

Second order correction

To improve the correction term, we therefore have to increase its denominator. Can we increase it by a constant 1? That will give us

\( \begin{align*}
E_p &= \frac{1}{a_{p-1}} + \frac{1}{a_{p}} – \frac{1}{p} \
&= \frac{1}{2p-1} + \frac{1}{2p+3} – \frac{1}{p} \
&= \frac{-2p+3}{4p^3+4p^2-3p}
\end{align*}\)

We have now ended up with a term proportional to p in the error numerator. Therefore, Yuktibhāṣā reasons that the denominator of the end-correction is better increased by a factor of 1/p rather than 1, and arrives at the optimal value as \( a_p = 2p+2 + \frac{4}{2p+2}\) That gives us a स्थौल्य of

\( \begin{align*}
E_p &= \frac{1}{a_{p-1}} + \frac{1}{a_{p}} – \frac{1}{p} \
&= \frac{1}{2p-2+\frac{4}{2p-2}} + \frac{1}{2p+2+\frac{4}{2p+2}} – \frac{1}{p} \
&= \frac{-4}{p^5+4p}
\end{align*}\)

This is a good स्थौल्य – negative, implying a slight under-correction, and reducing as the fifth power of p, which means we converge fast with this correction term. Let's see if that really happens

Terms	Result with second order correction	Accuracy (decimal places)
10	3.141590242370799	5
20	3.141592576186889	7
30	3.1415926433443224	7
40	3.1415926511540886	8
50	3.1415926527909903	9
100	3.1415926535648024	11

We have nine decimal places of accuracy with 50 terms, and 11 with 100, which is quite impressive. So, our correction term corresponds to
\( a_p = 2p+2 + \frac{4}{2p+2}\)
The term itself will be proportional to

\( \frac{1}{a_p} = \frac{1}{2p+2 + \frac{4}{2p+2}} = \frac{\frac{p+1}{2}}{(p+1)^2+1}\)

(ie: तस्या ऊर्ध्वगता या समसङ्ख्या तद्दलं गुणोऽन्ते स्यात् || तद्वर्गो रूपयुतो हारो …). Adding this correction term, our series now becomes:

\( C \approx 4.d – \frac{4.d}{3} + \frac{4.d}{5} – \frac{4.d}{7} + … (-1)^{\frac{p-1}{2}}.\frac{4.d}{p} + (-1)^{\frac{p+1}{2}}.\frac{4.d.\frac{p+1}{2}}{(p+1)^2+1} \)

The correction term has the opposite sign to the last term in the series – we add if we have subtracted, and vice versa (स्वमृणे कृते धने क्षेप एव करणीयः).

This form, as the table above attests, is good enough for calculation (लभ्धः परिधिः सूक्ष्मो), already to a degree of accuracy that was not known anywhere in Madhava's time. Comparing this table with the first, we see that it is a lot more useful to compute this correction term than to compute many many more terms of the series (बहुकृत्वो हरणतोऽतिसूक्ष्मश्च).

By improving the end correction term further, we can do even better than this, but we will leave that for a later article.

After our detour to take a detailed look at sankalitas, we go back to the व्यासे वारिधिनिहते series and – following the reasoning in Yuktibhāṣā – see how we can modify it to convert any ज्या / कोटिः pair to its चापः (arc). In modern terms, this corresponds to finding an infinite series for the arctangent function.

व्यासे वारिधिनिहते … again

When we derived the व्यासे वारिधिनिहते series, we used a quadrant of a square with an inscribed circle, and started off by dividing the eastern (top) side EA into n equal parts. We denote the intermediate karṇas (hypotenuses) as\(k_1, k_2\) etc, and we get:

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

Using \(k_i^2 = r^2 + (\frac{i.r}{n})^2\) and using śodhyaphalas and the large-n-sankalita \( (1^k+2^k…+n^k) \approx \frac{n^{k+1}}{k+1} \), we found that:

\(\begin{align*}
EC &\approx \frac{r}{n} (1+1+ …+1) \\
&\quad -\frac{r}{n}\frac{(\frac{r}{n})^2}{r^2} (1^2 + 2^2 + … n^2) \\
&\quad +\frac{r}{n}\frac{(\frac{r}{n})^4}{r^4} (1^4 + 2^4 + … n^4) \\
&\quad -\frac{r}{n}\frac{(\frac{r}{n})^6}{r^6} (1^6 + 2^6 + … n^6) \\
&\quad …\\
&\approx r – \frac{r}{3} + \frac{r}{5} – \frac{r}{7} + …
\end{align*}\)

One small change

Instead of EC being an eighth of the circumference, we consider an EC less than that. We extend OC to meet the top side at T, and divide ET, whose length we denote as t, into n parts and perform the calculation for EC as we did before. As usual, we denote the intermediate karṇas (hypotenuses) as \(k_1, k_2\) etc, and we get:

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

Using \(k_i^2 = r^2 + (\frac{i.t}{n})^2\) and using śodhyaphalas and sankalita before, we get:

\(\begin{align*}
EC &\approx \frac{t}{n} (1+1+ …+1) \\
&\quad -\frac{r}{n}\frac{(\frac{t}{n})^2}{r^2} (1^2 + 2^2 + … n^2) \\
&\quad +\frac{r}{n}\frac{(\frac{t}{n})^4}{r^4} (1^4 + 2^4 + … n^4) \\
&\quad -\frac{r}{n}\frac{(\frac{t}{n})^6}{r^6} (1^6 + 2^6 + … n^6) \\
&\quad …\\
&\approx r. \frac{t}{r} – \frac{r}{3}\frac{t^3}{r^3} + \frac{r}{5}.\frac{t^5}{r^5} – \frac{r}{7}.\frac{t^7}{r^7} + … \\
&= r. \frac{j}{k} – \frac{r}{3}\frac{j^3}{k^3} + \frac{r}{5}.\frac{j^5}{k^5} – \frac{r}{7}.\frac{j^7}{k^7} + …
\end{align*}\)

For the last step above, we use the fact that the jyā and koṭi of EC are CP and OP respectively, and by त्रैराशिकम्,
\( \frac{t}{r} = \frac{CP}{OP} = \frac{jyā}{koṭi} = \frac{j}{k} \)

And thus, we obtain Madhava's series for चापीकरणम् (the arctangent function, in modern terms).

In the previous article in this series, we saw how sankalitas are the Kerala calculus analogue to integrals in modern calculus, and how they were computed in the limit of large n. In this article, we can take a look at how these were used to compute the volume of a sphere.

Area of a circle

To derive the volume of a sphere, we must first derive the area of a circle of radius r and circumference C.

To find the area of such a circle, we first divide it into radial segments as in the figure, and then reassemble them as below. As the number of segments grows more and more, the assembled figure moves closer to a rectangle. At the limit, we get a true rectangle, and the result \( A = \frac{C}{2}.r \). We can see that this is equivalent to the modern result \( A = \pi . r^2 \). Sankalita doesn't appear directly in this derivation, but is implied in the expression for circumference, which we calculate using (for example), व्यासे वारिधिनिहते …

Volume of a sphere

To compute the volume of a sphere of radius r, diameter \(d = 2.r\) and great-circle circumference C, we divide it into thin horizontal circular segments, slicing parallel to the horizontal axis. Each slice has a finite thickness. We add up the slice volumes to approximate the volume of the sphere, which becomes accurate at the limit of atomically thin slices.

We divide the upper hemisphere of the sphere (of radius r and great-circle circumference C) into n slices of equal thickness \(r/n\). We denote the vertical distance from the centre of the jth slice (numbering from the pole) by \(K_j\), and the radius of the jth slice as \(J_j\) (for कोटिः and ज्या, which we have seen before). The figure represents a vertical great-circle slice through the sphere, showing the radius, \(J_j\) and \(K_j\). The volume of the jth slice \(V_j\) is related to its circumference, using the area formula derived in the previous section.
\( \begin{align*}
C_j &= \frac{C}{r}.J_j && \text{(}C_j\text{ is the circumference of the jth slice, and trairāśikam)} \\
V_j &= \frac{C_j}{2} . J_j . \frac{r}{n} && \text{(using}\quad A = \frac{C}{2}.r \quad\text{)} \\
&= \frac{C}{2.r}. J_j^2. \frac{r}{n} \\
& = \frac{C}{2.n}. J_j^2.
\end{align*}\)

Summing up the upper hemisphere and considering n to be very large, we get
\( \begin{align*}
\frac{V}{2} &= \frac{C}{2.n}. (J_1^2+J_2^2+…J_n^2) \\
V &= \frac{C}{n} (J_1^2+J_2^2+…J_n^2) \\
&= \frac{C}{n} . (r^2+r^2+…r^2) \\
&\quad-\frac{C}{n} . (K_1^2+K_2^2+…K_n^2) && \text{(using jya-koti-karna nyaya)} \\
&= C.r^2 – \frac{C}{n} .\frac{r^2}{n^2} (1^2+2^2+…n^2) \\
&= C.r^2 – \frac{C}{n} .\frac{r^2}{n^2}.\frac{n^3}{3} && \text{(varga-sankalita)} \\
&= C.r^2 – C .\frac{r^2}{3} \\
&= \frac{2}{3}.C.r^2 &&(=\frac{4}{3}.\pi.r^3)\\
& = \frac{C}{6}.d^2
\end{align*}\)
And we see that the same result that is derived in modern textbooks using integration can be obtained by the large-n sankalita method.

This is a slightly modified version of the yukti in Yuktibhāṣā. Instead of the ज्याकोटिकर्णन्यायः (aka “Pythagoras Theorem”) which is known to it, it uses the intersecting-chords theorem (वृत्ते समवर्गो …) to similar effect. We use the former, since it is more familiar to modern readers.

In the previous article in the series, we saw how the famous Madhava circumference (Pi) series was justified using an argument based on the geometry of Kerala roofs, and the mathematical techniques of shodyaphala and sankalita (सङकलितम्). Since the idea of sankalita is central to the Calculus of the Kerala school, it makes sense to spend some time exploring it.

The mulasankalita (मूलसङ्कलितम्)

\( 1+2+…n=\frac{n.(n+1)}{2} \)

The मूलसङ्कलितम् has been known in Indian Mathematics since Aryabhata. Yuktibhāṣā proves this known result in two ways, and then extends it to the limiting case of very small segments thus:

Graphical yukti (proof)

Consider the मूलसङ्कलितम् arranged as a two-dimensional figure as here, with the shaded squares expressing successive numbers. The shaded area in this particular figure represents the sum of the first 10 numbers \( 1+2+3+…+10\). If we invert the same shape (the unshaded part of the figure), and place it next to the original uninverted sankalita, we get a rectangle of height 10 and width one more than 10 = 11 units, whose area is therefore 10*11 = 110. The area of the shaded part is therefore half of it, which is \(110/2=55\). Since this argument works for any number, we see that \( 1+2+…n=\frac{n.(n+1)}{2} \). This relationship is the basic one used to prove higher order sankalitas and limits.

Algebraic yukti

Yuktibhāṣā considers the case of a very large n, which occurs in the sankalita form that turns up often in Kerala Calculus. Consider the case where a finite length r is divided into a very large number of segments n.

We change our scale such that each segment is of unit length (and therefore, r = n units in our new scale), and the sum of the segments is expressed as:
\( S_n = (1+2+…n) \)
This can be seen to be equal to:
\(
\begin{align*}
S_n &= ((n-(n-1))+(n-(n-2))+ …n-1+n) \\
&= n*n – ((n-1)+(n-2)+…1) \\
&= n^2 – S_{n-1} \\
\end{align*}
\)

Subcase: regular n

For the regular case:
\( \begin{align*}
S_n&= n^2 – (S_n – n) &&(\text{substituting} \quad S_{n-1} = S_n -n) \\
2. S_n &= n^2 + n \\
S_n &= \frac{n(n+1)}{2}
\end{align*}\)
Which is the same result as before, but derived purely algebraically. But the more interesting case for us in the context of Kerala Calculus is that of a very large n

Subcase: Large n

If n is very large – which means the unit is very small, close to atomic (अणुः), we see that
\( \begin{align*}
S_n&= n^2 – S_n &&(\text{substituting} \quad S_{n-1} \approx S_n) \\
2.S_n &\approx n^2 \\
S_n &\approx \frac{n^2}{2}
\end{align*}\)
In this scale, the length r is n units, and therefore the sum of these segments is equivalent to \(r^2/2\) in this scale.

Extending the Sankalitas – power sums for large n

Extending the mulasankalita, we move on to samaghatasankalitas (समघातसङ्कलितानि) or power sums. Yuktibhāṣā derives their expressions for the limit of large n

(If you find too many details not to your taste, you can skip directly to the summary of results)

Vargasankalita (वर्गसङ्कलितम्)- sums of squares for large n

Using some interesting sleight of hand, and the earlier expression for \(S_n\), the मूलसङ्कलितम् for large n, we can calculate the sum of squares, वर्गसङ्कलितम्, thus:
\(
\begin{align*}
S_n^{(2)} &= 1^2+2^2+3^2+…n^2 \\
&= 1.1 + 2.2 + 3.3 + …n.n \\
&= n.1 + n.2 + …n.n \\
&\quad-((n-1).1 + (n-2).2 + ….1.(n-1)) \\
&= n.S_n \\
&\quad- ((n-1)+(n-2)+(n-3)….+1) \\
&\quad- ((n-2)+(n-3)….+1) \\
&\quad- ((n-3)…..+1) \\
&\quad- … \\
&\quad- 1
\end{align*}
\)
(the triangular sum above is interesting, and we will see it again later)
We can simplify this using the approximation for S_n for large n we derived earlier.
\(
\begin{align*}
S_n^{(2)} &= n.S_n – S_{n-1} – S_{n-2} – … S_1 \\
&\approx \frac{n.n^2}{2} – \frac{(n-1)^2}{2} – \frac{(n-2)^2}{2} – \frac{(n-3)^2}{2} …. \\
&= \frac{n.n^2}{2} – \frac{1}{2}.((n-1)^2 + (n-2)^2 + …. 1) \\
&= \frac{n.n^2}{2} – \frac{1}{2}. S_{n-1}^{(2)}
\end{align*}
\)

Since for large n \( S_n^{(2)} \approx S_{n-1}^{(2)} \), we have
\(
S_n^{(2)} \approx \frac{n.n^2}{2} – \frac{1}{2}. S_{n}^{(2)} \\
S_n^{(2)} \approx \frac{n^3}{3}
\)
for large n

Ghanasankalita (घनसङ्कलितम्) – sums of cubes for large n

Using a very similar argument, we can derive an approximation for the sum of cubes (घनसङ्कलितम्) and all higher power sums (समघातसङ्कलितानि).
\( \begin{align*}
S_n^{(3)} &= 1^3+2^3+3^3+…n^3 \\
&= 1.1^2 + 2.1^2 + … n.n^2 \\
&= n.(1^2+2^2 + ….+n^2) \\
&\quad- ((n-1).1^2 + (n-2).2^2 + ….1.(n-1)^2)
\end{align*}\)
Which simplifies, analogously to the case of the वर्गसङ्कलितम्, to
\(
S_n^{(3)} \approx n . S_n^{(2)} – \frac{1}{3}.S_n^{(3)} \\
S_n^{(3)} \approx \frac{n^4}{4}
\)
for large n

Samaghatasankalita (समघातसङ्कलितम्) – sums of higher powers for large n

Using a very similar argument, we can derive an approximation for the sum of higher powers (समघातसङ्कलितम्)
\( \begin{align*}
S_n^{(k)} &= 1^k+2^k+3^k+…n^k \\
&= 1.1^{k-1} + 2.1^{k-1} + … n.n^{k-1} \\
&= n.(1^{k-1}+2^{k-1} + ….+n^{k-1}) \\
&\quad- ((n-1).1^{k-1}+ (n-2).2^{k-1} + ….1.(n-1)^{k-1})
\end{align*}\)
Which simplifies to
\(
S_n^{(k)} \approx n . S_n^{(k-1)} – \frac{1}{k}.S_n^{(k)} \\
S_n^{(k)} \approx \frac{n^{k+1}}{k+1}
\)
for large n
The astute reader would have noted the analogy with the modern \( \int x^k.dx = \frac{x^{k+1}}{k+1} \). As we would expect from this analogy, we can calculate areas, surface areas and volumes using these sankalitas, as we shall see soon enough. Before that, though, we take a small detour to the concept of the sankalita-sankalita, also called the sum of sums, or second order sum and its higher level analogs.

Extending the Sankalita – the Sankalita-sankalita (सङ्कलितसङ्कलितम्)

We can extend the mulasankalitas in another way too. Instead of adding higher powers of numbers, we can add mulasankalitas themselves! The सङ्कलितसङ्कलितम् is defined as the sum of mulasankalitas up to a given number
\(
\begin{align*}
SS^{(2)}_n &= S_n + S_{n-1} + …. + 1 \\
&= n + (n-1) + (n-2) + … + 1 \\
&\quad+ (n-1) + (n-2) + … + 1 \\
&\quad+ (n-2) + … + 1 \\
&\quad … \\
&\quad + 1
\end{align*}
\)
We saw this triangular sum in the expression for vargasankalita, but did not explicitly name it then.
We can also define higher order sankalita-sankalitas. For example the sankalita-sankalita-sankalita
\(
SS^{(3)}_n = SS^{(2)}_n + SS^{(2)}_{n-1} + SS^{(2)}_{n-2} + …+SS^{(2)}_1
\)
and the general k-th order sankalita-sankalita as
\(
SS^{(k)}_n = SS^{(k-1)}_n + S^{(k-1)}_{n-1} + S^{(k-1)}_{n-1} + … + S^{(k-1)}_{1}
\)

Narayana Pandita (c. 1350 CE) has derived the general expression for the sankalita-sankalitas as
\(
\begin{align*}
SS^{(2)}_n &= \frac{n.(n+1)(n+2)}{1.2.3}\\
SS^{(3)}_n &= \frac{n.(n+1)(n+2)(n+3)}{1.2.3.4}\\
… \\
SS^{(k)}_n &= \frac{n.(n+1)….(n+k)}{1.2.3..(k+1)}
\end{align*}
\)

Sankalita-sankalitas for large n

By directly considering the case of large n in Narayana Pandita's formula, we can see that \(SS^{(k)}_n \approx \frac{n^{k+1}}{(k+1)!} \), where ! denotes the factorial \(k! = 1.2.3..k\)

Yuktibhāṣā calculates this explicitly thus:
\(
\begin{align*}
SS^{(2)}_n &= S_n + S_{n-1} + …. + 1 \\
&\approx \frac{n^2}{2}+ \frac{(n-1)^2}{2} + … 1 &&\text{(for large n)}\\
&= \frac{1}{2}.S^{(2)}_n \\
&= \frac{n^3}{2.3}
\end{align*}
\)
and works up to: \(SS^{(k)}_n \approx \frac{n^{k+1}}{(k+1)!} \) for large n.

In the previous article, we saw how वर्गसङ्कलितम् was used to derive the Madhava Pi series. In the next article, we will see how the area of a circle, and the volume of a sphere are calculated using these.