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.
