Using the sampurṇajyā relationship we saw in the previous article, we can now achieve what we set out to – find a jyā recurrence without using the standard radius. First, recollect the sampurṇajyā relationship.
\(
[sampurṇajyā(s1)]^2 – [sampurṇajyā(s2)]^2 = sampurṇajyā(s1).sampurṇajyā(s2)
\)
Since \( jyā(s) = sampurṇajyā(2s)/2\), denoting the पठितज्याः (tabular jyā) as \(B_1, B_2 … B_{24}\) as usual, we can say:
\(
\begin{align*}
B_2^2 – B_1^2 &= B_3.B_1 \\
B_3^2 – B_1^2 &= B_4.B_3 \\
B_k^2 – B_1^2 &= B_{k+1}.B_{k-1}\\
..
\end{align*}
\)
Thus, knowing the first two tabular jyā, we can compute the rest.
B_3 = \frac{B_2^2 – B_1^2}{B_1} \\
B_4 = \frac{B_3^2 – B_1^2}{B_2} \\
… \\
B_{k+1} = \frac{B_{k}^2 – B_1^2}{B_{k-1}} \\
\)
This is summarized in the verse
तत्तज्ज्यावर्गमाद्यज्यावर्गहीनं हरेत्पुनः ।
आसन्नाधस्थशिञ्जिन्या लब्धस्स्यादुत्तरोत्तराः ।।
From the square of a known jyā, subtract the square of the first, and divide
by the jyā below, to get the succeeding jyā, and so on
B_{k+1} = \frac{B_{k}^2 – B_1^2}{B_{k-1}} \
\)
