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:
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).
Madhava's Series for चापीकरणम् (arctangent)
इष्टज्यात्रिज्ययोर्घातात् कोट्याप्तं प्रथमं फलं
ज्यावर्गं गुणकं कृत्वा कोटिवर्गं च हारकम् ।
प्रथमादिफलेभ्योऽथ नेया फलततिर्महुः
एकत्र्याद्योजसङ्ख्याभिर्भक्तेष्वेतेष्वनुक्रमात् ।
ओजानां संयुतेस्त्यकक्त्वा युग्मयोगं धनुर्भवेत् ॥
Multiply the jyā by trijyā (radius), and divide by koṭi. This is your first term. To get successive terms, multiply the previous term by the square of the jyā divided by the square of the koṭi. Divide the terms by the odd numbers 1,3,5,7 etc in turn. From the sum of the odd numbered terms, subtract the sum of the even numbered terms, to get the arc.
This means: given a ज्या/कोटिः pair j, k of a circle of radius r, the associated arc is given by:
which is equivalent to:
\( arctan(t) = t – \frac{t^3}{3} + \frac{t^5}{5} – \frac{t^7}{7} … \)
दोः कोट्योरल्पमेवेष्टं कल्पनीयमिह स्मृतम् ।
लब्धीनां नावसानं स्यादन्यताऽपि मुहुः कृते ।।
For the above procedure to work, we must have \(j \lt k\).
If that's not true, swap j and k, calculate the complementary arc using the same procedure, and subtract from a quarter of the circumference to get the desired result.
