What's In A Name?

Yuktibhāṣā does not explicitly calculate the volumes of pyramids or cones, but it would be instructive to use the method of Sankalita to do so, to demonstrate the generalizability of this method beyond what is explicitly shown.

Volume of a Pyramid

Consider a square pyramid of side a, and height h. We divide it into n slices of equal thickness such that \(\frac{h}{n}\) is the thickness of the slice. The base of each slice will have varying dimensions. Let \(a_k\) be the side of slice number k.

By trairāśika, we have \(a_k = \frac{a.k}{n}\). Thus, the volume of slice number k is

\(
\begin{align*}
V_k &= \frac{h}{n} . a_k^2 \\
&= \frac{h}{n}. {(\frac{a.k}{n})}^2 \\
& = \frac{h.a^2.k^2}{n^3} \\
\end{align*}
\)

The total volume is the sum of all such segments. Taking n to parardhā, and using the value of vargasankalita seen earlier,

\(
\begin{align*}
V &= \sum_{0}^{n} V_k \\
&= \sum_{0}^{n} \frac{h.a^2.k^2}{n^3} \\
&= \frac{h.a^2}{n^3}.S_n^{(2)} \\
&= \frac{h.a^2}{n^3}.\frac{n^3}{3} \\
&= \frac{h.a^2}{3}.
\end{align*}
\)

Which is identical to the modern result. A rectangular pyramid can be handled similarly.

Volume of a Cone

As it turns out, the volume of a cone is also quite similar. Taking the base of the cone to have radius r and using a similar slicing methodology, the volume of slice number k becomes \(V_k = \pi. {(\frac{k.r}{n})}^2\), and the volume of the cone becomes

\(
\begin{align*}
V &= \sum_{0}^{n} V_k \\
&= \frac{\pi . h . r^2}{n^3}.S_n^{(2)} \\
&= \frac{\pi . h . r^2}{n^3}.\frac{n^3}{3} \\
&= \frac{\pi . h . a^2}{3}.
\end{align*}
\)

again, identical to the modern result.

Clearly then, the Sankalita is equivalent to the modern integral, at least for powers (and by extension, polynomials).

Since the 19th century, modern calculus has relied on ideas of limits for its theoretical grounding. The infinitesimal, favoured by Leibniz, was used to describe and derive results in calculus prior to that.

aṇu and parārdha

Kerala Calculus relies on two concepts, aṇu and parārdha, to ground and derive their results. For example, the मूलसङ्कलितम् result for a large number of segments, which we have seen before, was derived for small segments using these.

परार्धः / parārdha is treated as a number that is larger than any other number, while अणु / aṇu is a number smaller than any other. A side of finite length divided into parārdha segments will have each segment of aṇu size.

parārdha and saṅkalitas

\( 1+2+…n=\frac{n.(n+1)}{2} \)

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 saṅkalita, 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 \( 1+2+…n=\frac{n.(n+1)}{2} \).

Now, we keep the side of the square the same, and increase the number of segments to parārdha, so that each segment is aṇu-sized. Denoting parārdha as p,

\( 1+2+…p=\frac{p.(p+1)}{2} = \frac{p^2}{2} \).

since one greater than parārdha is parārdha itself.

Using this property of parārdha, we see the sums/integrals of powers and the sum of sums and higher order sums defined for aṇu-sized segments.

aṇu and saṅkalitas

Another trick that the Kerala school has used is that aṇu-sized parts of saṅkalitas can be ignored, if the rest of the saṅkalita can be rendered into finite form.

For example, in the derivation of the famous pi series, we come to a point where the quarter circumference, divided into parārdha segments is expressed in terms of the chords of the aṇu-sized segments as \( EC \approx (\frac{r}{n})\frac{r^2}{k_0k_1} + (\frac{r}{n}).(\frac{r^2}{k_1.k_2}) + … + (\frac{r}{n}).(\frac{r^2}{k_i.k_{i+1}}) + … (\frac{r}{n}).(\frac{r^2}{k_{n-1}.k_n}) \)

This can be simplified by using \( \frac{1}{k_i.k_{i+1}} \approx (\frac{1}{2}) (\frac{1}{k_i^2}+\frac{1}{k_{i+1}^2})\), which works because the difference between the two becomes
\( (\frac{1}{2}) (\frac{1}{k_i^2}+\frac{1}{k_{i+1}^2}) – \frac{1}{k_i.k_{i+1}} = (\frac{1}{2})(\frac{1}{k_i}-\frac{1}{k_{i+1}})^2\)

which is less than aṇu-sized, and can be ignored compared to the saṅkalita which ends up finite-sized.

Again, we get:

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

where one more asymmetry remains, which we can smoothen out by noting that \( k_n = \sqrt{2}.k_0\), and \(k_n-k_0\) is aṇu-sized.

We then get

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

which we have seen can be transformed into a set of infinite saṅkalitas with finite closed forms.

It must be noted that these aṇu drops only work because the total saṅkalita can be compacted into finite closed form as a result.

What we have seen so far of the Kerala School is quite interesting, to say the least. We have seen:

• A line integral (circumference of a circle).
• Areas of a circle and sphere and Volume of a sphere.
• Integrals (sankalitas) for powers.
• Power series approximation for what is called \(\pi\) today.
• Trigonometry similar to the modern version, with jyā instead of sine.
• Power series for jyā, kojyā and sara functions.
• The idea of limits.
• The differential (khanḍa-jyā) taken to the limit of small arcs, which is equivalent to the derivative.
  • We have also seen the khanḍajyāntara, the second derivative of the jyā.

Was all of this merely one-offs relevant only in context, albeit substantial, or did the Kerala School produce a general system of calculus?

Interestingly, in the Tantrasaṃgraha, we come across an remarkable verse not explained in the Yuktibhāṣā:

चन्द्रबाहुफलवर्गशोधितत्रिज्यकाकृतिपदेन संहरेत्
तत्र कोटिफललिप्तिकाहतां केन्द्रभुक्तिरिह यच्च लभ्यते ॥
तद्विशोद्य मृगादिकेः गतेः क्षिप्यतामिह तु कर्कटादिके
तद्भवेत्स्फुटतरागतिर्विधोरस्य तत्समयजा रवेरपि ॥

The question at hand here is how to calculate the instantaneous (तत्समयजा) velocity of the moon (and sun), given their mean velocities and the equations of position. Since we have not delved into these astronomical models, we will merely note that the equation of position of the moon is given by (in modern terms)

\( \theta = \theta_0 – arcsin(\frac{r_o}{R}.sin(\theta – \theta_0)) \)

The instantaneous velocity, therefore, is

\( \frac{d{\theta}}{dt} = \frac{d{\theta_0}}{dt} – \frac{{r_o}.cos(\theta – \theta_0).\frac{d{\theta_0}}{dt}}{\sqrt{R^2 – r_0^2.sin^2(\theta-\theta_0)}} \)

And this is precisely what the verse gives. To generate this expression, one would need to know the derivative of the arcsin function, as well as the chain rule of differentiation, both of which were apparently known to Nilakaṇṭha Somayāji, though never explicitly stated. Also, the idea that the instantaneous velocity is the derivative of the instantaneous position had to have been appreciated.

Evidently, the calculus that we have seen earlier had been taken to some degree of generalization behind the scenes by the Kerala School, which is unfortunately not available to us. Knowing this gives us the impetus to reconstruct a general calculus using the means available to the Kerala School with a good amount of assurance that a similar path was tread by them.

सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः ।
मूलमस्फुटफलं चतुर्भुजे स्पष्टमेवमुदितं त्रिबाहुके ॥
Take the sum of all sides, and divide it into half. Multiply the result by the four differences of itself with each side. The square root of the result is is the approximate area of a quadrilateral, and precise for a triangle. (This is exact for a cyclic quadrilateral)

\(
\begin{align*}
A &= \sqrt{s.(s-a)(s-b)(s-c)(s-d)} &\textit{cyclic quadrilateral} \\
A &= \sqrt{s.(s-a)(s-b)(s-c)} &\textit{triangle} \\
\end{align*}
\)

In the previous article, we proved this relationship to be exact, and gave an intuitive proof for triangles. We will now examine the full proof for triangles in Yuktibhāṣā of Jyeṣṭhadeva.

Consider triangle ABC, with BC being the base (भूमि:). The perpendicular from AP to BC (लम्बः) divides BC into two segments (आबाधौ) \(d_1\) and \(d_2\). Following modern convention, we denote the length of the side opposite A (BC) as a, AC as b and AB as c.

The Area of the triangle ABC

\(
Ar = \frac{1}{2}a.h \\
Ar^2 = \frac{1}{4}a^2.h = (\frac{a}{2})^2.h
\)

Our aim is to show that this is equivalent to \( Ar^2 = s.(s-a).(s-b).(s-c)\)

Do do this, we first note that
\(
s.(s-a) = (\frac{b+c}{2})^2-(\frac{a}{2})^2 = (\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2 \\
h^2 = c^2 – d_1^2 = b^2- d_2^2 = \frac{b^2+c^2}{2} – \frac{d_1^2+d_2^2}{2} \\
s.(s-a) \approx h^2 \\
(s-b)(s-c) = (\frac{a}{2})^2 – (\frac{b}{2} – \frac{c}{2})^2 \\
(s-b)(s-c) \approx (\frac{a}{2})^2 \\
s.(s-a)(s-b)(s-c) \approx (\frac{a}{2})^2.h^2 = Ar^2
\)

This proves that the two expressions are at least approximately equal. To prove exact equality, we must prove that the differences between the factors are in such proportion that their product evens out.

\(
\begin{align*}
h^2 – s.(s-a) &= (\frac{b^2+c^2}{2} – \frac{d_1^2+d_2^2}{2}) – ((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2) \\
&= (\frac{b-c}{2})^2-(\frac{d_2-d_1}{2})^2 \\
s.(s-a) &= h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2) \\
\end{align*}
\)
and therefore
\(s.(s-a).(s-b).(s-c) = (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)).((\frac{a}{2})^2 – (\frac{b}{2} – \frac{c}{2})^2)\)
with one factor a bit more than \(h^2\), and another a bit less than \(a^2\).

To prove that everything evens out on multiplication, we use the relationship \( p.q = (p+r).(q – q.\frac{r}{p+r})\), which is one of the multiplication tricks used by the Kerala School. Therefore,
\(
\begin{align*}
Ar^2 &= h^2.(\frac{a}{2})^2 \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2)}) \\
\end{align*}
\)

To make further progress, we use the ābādha relationship proven previously
\(
d_2^2 – d_1^2 = b^2 – c^2 \\
\frac{b-c}{d_2+d_1} = \frac{d_2-d_1}{b+c} \\
\frac{(\frac{b-c}{2})}{(\frac{d_2+d_1}{2})} = \frac{(\frac{d_2-d_1}{2})}{(\frac{b+c}{2})} \\
\frac{(\frac{b-c}{2})^2}{(\frac{d_2+d_1}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2}{(\frac{b+c}{2})^2} \\
\frac{(\frac{b-c}{2})^2}{(\frac{d_2+d_1}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2-(\frac{b-c}{2})^2}{(\frac{b+c}{2})^2-(\frac{d_2+d_1}{2})^2} \\
\frac{(\frac{b-c}{2})^2}{(\frac{a}{2})^2} = \frac{(\frac{d_2-d_1}{2})^2-(\frac{b-c}{2})^2}{(\frac{b+c}{2})^2-(\frac{d_2+d_1}{2})^2} \\
\)

Thus
\(
\begin{align*}
Ar^2 &= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2))((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)}{((\frac{b+c}{2})^2-(\frac{d_1+d_2}{2})^2)}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)) ((\frac{a}{2})^2 – (\frac{a}{2})^2.\frac{(\frac{b-c}{2})^2}{(\frac{a}{2})^2}) \\
&= (h^2 + ((\frac{d_2-d_1}{2})^2 – (\frac{b-c}{2})^2)) ((\frac{a}{2})^2 – (\frac{b-c}{2})^2) \\
&= s.(s-a).(s-b).(s-c)
\end{align*}
\)
Which is what we set out to prove.

And so, we have yuktis to prove सर्वदोर्युतिदलं as exact for areas of cyclic quadrilaterals as well as triangles.

With the jyā recurrence without the standard radius having been demonstrated in the previous article, we return to cyclic quadrilaterals, and the proof of Bhāskarācārya's verse on the areas of cyclic quadrilaterals and triangles:

सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः ।
मूलमस्फुटफलं चतुर्भुजे स्पष्टमेवमुदितं त्रिबाहुके ॥
Take the sum of all sides, and divide it into half. Multiply the result by the four differences of itself with each side. The square root of the result is is the approximate area of a quadrilateral, and precise for a triangle. (This is exact for a cyclic quadrilateral)

\(
\begin{align*}
A &= \sqrt{s.(s-a)(s-b)(s-c)(s-d)} &\textit{cyclic quadrilateral} \\
A &= \sqrt{s.(s-a)(s-b)(s-c)} &\textit{triangle} \\
\end{align*}
\)

Yukti

Consider a cyclic quadrilateral ABCD, with sides AB=a (भूमिः – longest side), BC=b, DA=d (भुजे), and CD=c (मुखम्) (not to be confused with the cyclic diameter, which we will not need for this yukti). Here, we assume \(c > b \), and will consider the opposite case a bit later. Consider the primary diagonal BD, with midpoint at M. Draw perpendiculars (लम्बौ) AE and CF to BD. Extend AE to P and CF to Q such that AP = CQ = AE+CF.

We compute the area of ABCD (denoted as Ar) as the sum of areas of ABD and BCD, which in turn is computed as \(\frac{1}{2}.BD.(AE+CF) = \frac{1}{2}.BD.AP \) by the triangle area law (लम्बगुणं भूम्यर्धं स्पष्टं त्रिभुजे फलं भवति). Now if we can compute AP, we have what we need. Examinig the rectange APCQ, we see that

\(
\begin{align*}
AP^2 &= AC^2-PC^2 \\
&= AC^2-EF^2 \\
Ar^2 &= \frac{1}{4}.BD^2.(AC^2-EF^2) \\
\end{align*}
\)

This gets us a step further. We already know BD and AC in terms of the sides from a previous article as the square root of the ratio of the associated कर्णाश्रितभुजाघातैक्यम् with that of the other diagonal, multiplied by the भुजाप्रतिभुजाघातैक्यम्. Therefore

\(
BD^2 = \frac{(cd+ab)}{(ad+bc)}. (ac+bd) \\
AC^2 = \frac{(ad+bc)}{(cd+ab)}. (ac+bd)
\)

Now all we need is to find EF, which we can see from the figure as the distance between the perpendiculars AE and CF on diagonal BD, or the लम्बनिपातान्तरम् .

Computing लम्बनिपातान्तरम्

From the figure, we can see that EF = ME + MF, and

\(
\begin{align*}
ME &= MD – DE \\
&= \frac{1}{2}(BE+ED) – ED \\
&= \frac{1}{2}(BE-ED) \\
MF &= MB – FB \\
&= \frac{1}{2}(DF+FB) – FB \\
&= \frac{1}{2}(DF – FB) \\
\end{align*}
\)

Now BE and ED are the आबाधौ of triangle ABD and DF and FB are the आबाधौ of triangle BCD. This means

\(
(BE-ED) = \frac{AB^2-DA^2}{BD} \\
(DF-FB) = \frac{CD^2-BC^2}{BD} \\
EF = \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD}
\)

AB > AD, since we've assumed AB as the longest side. We've also assumed c > b, or CD > BC. What if BC > CD? Then, the perpendiculars CF and AE will fall on the same side of midpoint M on BD as below.

Now:

\(
EF = ME – MF \\
ME = \frac{1}{2}(BE-ED) \\
MF = \frac{1}{2}(DF – FB) \\
(BE-ED) = \frac{AB^2-DA^2}{BD} \\
(DF-FB) = \frac{BC^2-CD^2}{BD} \\
EF = \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD}\\
\)

Thus, we end up with the same expression for EF, irrespective of whether c > b or b > c.
\(
\begin{align*}
EF &= \frac{1}{2}.\frac{(AB^2+CD^2)-(DA^2+BC^2)}{BD} \\
&= \frac{1}{2}.\frac{(a^2+c^2)-(d^2+b^2)}{BD}
\end{align*}
\)

Putting it all together

\(
\begin {align*}
Ar^2 &= \frac{1}{4}.BD^2.(AC^2-EF^2) \\
&= \frac{1}{4}.(BD^2.AC^2 – BD^2.EF^2) \\
BD^2.AC^2 &= \frac{(cd+ab)}{(ad+bc)}. (ac+bd) . \frac{(ad+bc)}{(cd+ab)}. (ac+bd) \\
&= (ac+bd)^2 \\
BD^2.EF^2 &= BD^2 .\frac{1}{4}.\frac{((a^2+c^2)-(d^2+b^2))^2}{BD^2} \\
&= \frac{1}{4}.((a^2+c^2)-(d^2+b^2))^2 \\
Ar^2 &= \frac{1}{4}.((ac+bd)^2 – \frac{1}{4}.((a^2+c^2)-(d^2+b^2))^2) \\
&= (\frac{ac+bd}{2})^2 – (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))^2 \\
&= (\frac{ac+bd}{2} + (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))) . (\frac{ac+bd}{2} – (((\frac{a}{2})^2+(\frac{c}{2})^2)-((\frac{d}{2})^2+(\frac{b}{2})^2)))) \\
&= ((\frac{a}{2} + \frac{c}{2})^2 – (\frac{b}{2}-\frac{d}{2})^2). ((\frac{d}{2} + \frac{b}{2})^2 – (\frac{a}{2}-\frac{c}{2})^2) \\
&= \frac{a+c+b-d}{2} . \frac{a+c+d-b}{2} . \frac{d+c+b-a}{2}. \frac{d+c+a-b}{2}
\end{align*}
\)

To simplify further, we note that \(s = \frac{a+b+c+d}{2}\)

\(
\begin {align*}
Ar^2 &= \frac{a+c+b-d}{2} . \frac{a+c+d-b}{2} . \frac{d+c+b-a}{2}. \frac{d+c+a-b}{2} \\
&= (s-d) . (s-b). (s-a). (s-c) \\
Ar &= \sqrt{(s-a).(s-b).(s-c).(s-d)} \\
\end{align*}
\)

Which is सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः मूलम्, which is where we need to be.

Extending this yukti to a triangle

The simplest way to extend this result to a triangle, is to consider a triangle (त्र्यश्रम्) as a cyclic quadrilateral with the smallest side d = 0. This will lead directly to

\(
Ar = \sqrt{s.(s-a).(s-b).(s-c)}
\)

We also have a direct yukti for this expression, which we will see in a later article.

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.

\(
B_3 = \frac{B_2^2 – B_1^2}{B_1} \\
B_4 = \frac{B_3^2 – B_1^2}{B_2} \\
… \\
B_{k+1} = \frac{B_{k}^2 – B_1^2}{B_{k-1}} \\
\)

In the previous article in this series, we began a search for a jyā recurrence without using the standard radius R, and went off into triangles and cyclic quadrilaterals. In this article, we will dive deeper into the latter.

The sampurṇajyā relationship

First, we consider an relationship about sampurṇajyā and associated arcs (चाप) in a circle. This will be very useful for us in the following discussion.

Consider a वृत्तान्तर्गतचतुरश्रम् (cyclic quadrilateral). By convention, we consider AB, the longest side on the west (ie, the bottom of the picture). This is called the भूमिः or base. The sides AD and BC are called भुजे (singular – भुजा), and the side DC is the मुखम् . We draw DM as the लम्बः‌ (perpendicular) to AC, a कर्णः (diagonal). EP is parallel to DM and of the same length, so AE = DC. Therefore, arc ED = Arc AD – Arc AE = Arc AD – Arc DC. Also, ED = PM = AM-AP = AM-MC (since arcs AE and DC are the same)

In the previous article, we had noted the relationship between the आबाधौ (basal segments) of a triangle, which in the case of triangle ADC leads us to \(AD^2-DC^2 = AM^2 – MC^2 = AC.(AM-MC)\). Segments AD and DC are the full-chords (सम्पूर्णज्या)‌ of arcs AD and DC. Similarly, AC is the full-chord of arc AC the sum of arcs AM and MC, and AM-MC = ED = PM is the full-chord corresponding to the difference of arcs AM and MC.
Thus, setting

\(
s_1 = arc(AD) \\
s_2 = arc(DC)
\)

leads to the first sampurṇajyā relationship

\(
[sampurṇajyā(s1)]^2 – [sampurṇajyā(s2)]^2 = sampurṇajyā(s1).sampurṇajyā(s2)
\)

Since the ADC can be arbitrary, we can also set

\(
s_1 = arc(AC) \\
s_2 = arc(AM-MC) = arc(ED)
\)

which leads to the second sampurṇajyā relationship

\(
sampurṇajyā(s_1).sampurṇajyā(s_2) = [sampurṇajyā(\frac{s_1+s_2}{2})]^2 – [sampurṇajyā(\frac{s_1-s_2}{2})]^2
\)

Diagonals of a वृत्तान्तर्गतचतुरश्रम् (cyclic quadrilateral)

Consider the rather imposing figure of a cyclic quadrilateral ABCD, as above. AB is the भूमिः (longest side) and sides AD, BC are the भुजे.
What about all the other stuff then? Points E and F on the circle are marked such that AE = DC and AF = BC (and so are the corresponding arcs). We connect the midpoints of Arcs ED and FB, called H and G respectively, to create a diameter HG. (Why is HG a diameter? Ponder the symmetry of AEDCBF).

त्रयः कर्णाः – Three diagonals

BD is the first diagonal, corresponding to the sampurṇajyā of arc(BD) (which is equal to arc(BC) + arc(CD)). Similarly AB, the second diagonal is the sampurṇajyā of arc(AB), which is the sum of arc(AC) + arc(CB).

Let us now consider DF. DF is sampurṇajyā of \(arc(DF) = arc(AD) + arc(AF) = arc(AD) + arc(AC)\), ie – the sampurṇajyā of the sum of the arcs of opposite sides! (what about EB then? By symmetry, EB=DF)

Thus, we conceptualize DF (=EB) as the third diagonal of the वृत्तान्तर्गतचतुरश्रम्, corresponding to the sampurṇajyā of the sum of arcs AD and BC (or arcs AB and CD). We think of sides, as well as diagonals (all three!) as sampurṇajyā of arcs corresponding to sides or sum of adjacent or opposite sides. This is an interesting way of thinking about cyclic quadrilaterals.

कर्णाश्रितभुजाघातैक्यम्

That imposing, yet compact phrase means sum of products of corresponding sides associated with a diagonal. For the first diagonal DB, the pairs of corresponding sides are (AD, DC) and (AB, BC). Thus the कर्णाश्रितभुजाघातैक्यम् is \(AD.DC + AB.BC\). Using the sampurṇajyā relations, we can say
\(
AD.DC = [sampurṇajyā(\frac{arc(AD)+arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 \\
AB.BC = [sampurṇajyā(\frac{arc(AB)+arc(BC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2
\)

Consider the sum of first terms of the two equations. We know that
\(
arc(AD)+arc(DC) = 360 – (arc(AB)+arc(BC))\\
\frac{arc(AD)+arc(DC)}{2} = 180 – \frac{arc(AB)+arc(BC)}{2}
\)

This makes sampurṇajyā of \(\frac{arc(AD)+arc(DC)}{2}\) and \(\frac{arc(AB)+arc(BC)}{2}\) jyā and koṭi of a right triangle, which means the square of their sum will be equal to the square of the diameter.
\(
[sampurṇajyā(\frac{arc(AD)+arc(DC)}{2})]^2 + [sampurṇajyā(\frac{arc(AB)+arc(BC)}{2})]^2 = d^2 \\
AD.DC + AB.BC = d^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2
\)

We have simplified a bit, but have some more way to go. To simplify further, we note that
\(
\frac{arc(AD)-arc(DC)}{2} = \frac{arc(DE)}{2} = arc(DH)\\
\frac{arc(AB)-arc(BC)}{2} = \frac{arc(BF)}{2} = arc(GB) \\
\)
\(
\begin{align*}
AD.DC + AB.BC &= d^2 – [sampurṇajyā(\frac{arc(AD)-arc(DC)}{2})]^2 – [sampurṇajyā(\frac{arc(AB)-arc(BC)}{2})]^2 \\
&= d^2 – [sampurṇajyā(arc(DH)]^2 – [sampurṇajyā(arc(GB))]^2 \\
&= d^2 – DH^2 – GB^2 \\
&= GD^2 – GB^2
\end{align*}
\)
since GDH is inscribed in a semicircle, and hence a right triangle.

Again, by the sampurṇajyā relation
\(
\begin{align*}
AD.DC + AB.BC &= GD^2 – GB^2 \\
&= sampurṇajyā(arc(GD)+arc(GB)).sampurṇajyā(arc(GD)-arc(GB)) \\
&= sampurṇajyā(arc(GD)+arc(GF)).sampurṇajyā(arc(GD)-arc(GB)) \\
&= sampurṇajyā(arc(GF)).sampurṇajyā(arc(DB)) \\
&= DF . DB
\end{align*}
\)
Thus, the कर्णाश्रितभुजाघातैक्यम् – the sum of products of sides associated with a diagonal DB is the product of the diagonal (DB) with the third diagonal (DF).

A similar argument for diagonal AC will produce a corresponding result
\( AB.AD + BC.DC = AC. DF \)

Also, by a similar argument, we can prove that the भुजाप्रतिभुजाघातैक्यम् (sums of products of opposite sides) is the product of the first two diagonals
\( AB.DC + BC.AD = AC. BD \)
(This can also be seen by considering the quadrilateral DAFC and noting that AF = BC and FC = AB. We can see that BD is the third diagonal of this quadrilateral DAFC. Considering the कर्णाश्रितभुजाघातैक्यम् of its second diagonal AC as above will quickly lead to the above result. )

कर्णाः

This lets us write down the diagonals in terms of the sides thus:
\(
DB = \sqrt(\frac{(AD.DC + AB.BC).(AB.DC + BC.AD)}{AB.AD + BC.DC})) \\
AC = \sqrt(\frac{(AB.AD + BC.DC).(AB.DC + BC.AD)}{AD.DC + AB.BC})) \\
DF = \sqrt(\frac{(AB.AD + BC.DC).(AD.DC + AB.BC)}{AB.DC + BC.AD})) \\
\)

The first and second diagonals are the square roots product of their कर्णाश्रितभुजाघातैक्यम् divided by that of the other and multiplied by the भुजाप्रतिभुजाघातैक्यम्. The third diagonal is the product of the two कर्णाश्रितभुजाघातैक्ये divided by the भुजाप्रतिभुजाघातैक्यम्.

Area of the वृत्तान्तर्गतचतुरश्रम्

We can now come up with an expression for the area of a cyclic quadrilateral. First we note that adding the areas of ABC and ADC will get us what we want. That gives us the expression
\(
\begin{align*}
A &= \frac{1}{2}(AC.DM + AC.MN) \\
&= \frac{1}{2}(AC.DO)
\end{align*}
\)
(O is obtained by extending DM and BF till they intersect. Since DM is perpendicular to AC which is parallel to BF, DO Is perpendicular to BF)
Using the relationship that the product of two sides of a triangle is equivalent to the third multiplied by its circumdiameter (which we will prove later), we see that:

\(
\begin{align*}
DO &= DB.DF/d \\
A &= \frac{1}{2}(AC.DO) \\
&= \frac{AC.DB.DF}{2d}
\end{align*}
\)

The area of a cyclic quadrilateral is the product of its three diagonals divided by twice the circumdiameter. This is an interesting result, and a good place to stop for now, but is not totally satisfactory, since we are yet to determine the circumdiameter as a function of the sides. We have some more work to do till we prove the सर्वदोर्युतिदलं formula, which is our goal.

The Kerala School, as we saw in past articles, were big on options for jyā computations. First, they improved Aryabhata's jyā recurrence, then added on an improved interpolation method. Later, we saw the Mādhava jyā series, plus a partially pre-computed version of it. Finally, we have the श्रेष्ठं नाम वरिष्ठानां verse, which provides a precomputed jyā table with much better accuracy. What more could one ask for?

What if, they ask, we could find a recurrence relation without that pesky R – the radius of standard circle? We recollect that the improved recurrence relationship

विलिप्तदशकोना ज्या राश्यष्टांशधनुः कलाः ।
आद्यज्यार्धात्ततो भक्ते सार्धदेवाश्विभिस्ततः
etc.

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

where R is the radius, and a is the full-chord of the quarter circumference divided by 24, (ie 225′). 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 \({(\frac{a}{R})}^2 \approx 233′ 32″\).

Can we remove the need for R here? It turns out we can, and this will lead us into an interesting diversion through triangles and cyclic quadrilaterals, which is a journey of interest in itself.

Ābādhas and Area of a Triangle

लम्बगुणं भूम्यर्धं स्पष्टं त्रिभुजे फलं भवति
Half of the base multiplied by the perpendicular to it is the area of a triangle

The first stop in our journey is the elementary expression for area of a triangle. Consider a विषमत्र्यश्रम् (scalene triangle) ABC, with the longest side BC called the भूमिः (base), and AB as the second longest side.

AD is the perpendicular from the भूमिः to the vertex A. We call it the लम्बः (perpendicular). D divides the भूमिः into two segments‌ (आबाधः॒) BD and DC. We draw TR and US, of equal length to and parallel to AD such that they bisect BD and DC. It is evident from this figure that rectangle RSTU has the same area as triangle ABC. (Triangle ATP and PBR are congruent etc.). Therefore,

\(
\begin{align*}
A &= RS. AD \\
&= \frac{1}{2}. BC. AD \\
&= \frac{1}{2}. b. h
\end{align*}
\)

We also see the interesting relationships:

\(
\begin{align*}
BD^2 – CD^2 &= (AB^2 – AD^2)- (AC^2-AD^2) \\
& = AB^2 – AC^2
\end{align*}
\)

Since BC = BD + CD, some elementary manipulation gets us to

\(\begin{align*}
BD &= \frac{1}{2} . (BC + (\frac{AB^2 – AC^2}{BC})) \\
DC &= \frac{1}{2} . (BC – (\frac{AB^2 – AC^2}{BC})) \\
\end{align*}
\)

Cyclic Quadrilaterals

The next thing we need to look at are cyclic quadrilaterals – quadrilaterals that are inscribed in circles – and their diagonals and areas. Bhāskarācārya had given the following formula for the area of a cyclic quadrilateral (approximate for non-cyclic quadrilaterals).

सर्वदोर्युतिदलं चतुःस्थितं बाहुभिर्विरहितं च तद्धतेः ।
मूलमस्फुटफलं चतुर्भुजे स्पष्टमेवमुदितं त्रिबाहुके ॥

Take the sum of all sides, and divide it into half. Multiply the result by the four differences of itself with each side. The square root of the result is is the approximate area of a quadrilateral, and precise for a triangle. (This is exact for a cyclic quadrilateral)

\(
\begin{align*}
A &= \sqrt{s.(s-a)(s-b)(s-c)(s-d)} &\textit{cyclic quadrilateral} \\
A &= \sqrt{s.(s-a)(s-b)(s-c)} &\textit{triangle} \\
\end{align*}
\)

How do we prove these, and how does this help us come up with a jyā recurrence without using the standard radius R? That will be the subject of the next article, where we will look at the three diagonals of a cyclic quadrilateral. Yes, three. You read that right.

In a previous article, we saw how the area of a circle, and the volume of a sphere are computed using Kerala School methods. We had not looked at the surface area of a sphere at that point, since we had not looked at jyā yet. Now that that is familiar to us, it is time to see how we use saṅkalita to compute the surface area of a sphere.

Segmenting a Sphere

As we did when we computed the volume of a sphere, we divide it into segments. However, instead of dividing it into segments of equal thickness, which is convenient for volume calculation, we divide it into segments of equal arc lengths \(\alpha\), as seen from a vertical slice through the sphere. This makes it more convenient to compute surface area, as each roughly trapezoidal segment on the surface ends up with constant width \(\alpha\). In the limit of large number of segments, each trapezoid becomes a rectangle of area \(C_j.\alpha\) where \(C_j\) is the circumference of the jth segment. If r is the radius of the sphere, and C the circumference of any of its great circles (eg: the equator of the sphere), we can use त्रैराशिकम् to compute the circumference of the jth segment as \(C_j = \frac{C}{r}.B_j\) where \(B_j\) is the jth jyā. Thus, the surface area of the jth segment is \( S_j = C_j.\alpha = \frac{C}{r}.B_j.\alpha\).

Summing up the area of the upper hemisphere to get half the surface area of the sphere, we have

\(\frac{S}{2} = \frac{C}{r}.(B_1 + B_2 + … B_n).\alpha \)

Recollecting the second differential of the jyā from an earlier article, we note that \((B_j – B_{j-1}) – (B_{j+1}-B_j) = (\frac{\alpha}{r})^2.B_j\)

Thus:

\(
\begin{align*}
\frac{S}{2} &= \frac{C}{r}.(B_1 + B_2 + … B_n).\alpha \\
&= \frac{C}{r}.(\frac{r}{\alpha})^2((B_1-0) – (B_2 -B_1) + (B_2-B_1) – (B_3 – B_2) + (B_3 – B_2) – … – (B_{n-1}-B_{n-2}) + (B_{n-1}-B_{n-2}) – (B_n – B_{n-1})).\alpha \\
& = \frac{C}{r}.(\frac{r}{\alpha})^2 (B_1 – (B_n – B_{n-1})) .\alpha \\
&= C.r \\
S &= 2.C.r = C.d && (=4.\pi.r^2)
\end{align*}
\)

Using \(B_1 \approx \alpha\) and \((B_n – B_{n-1}) \approx 0 \) for large n.

Alongside calculus, the Kerala School also came up with some interesting results in Trigonometry, including the now commonly known method to calculate the sines (ज्या) of the sums are difference of two angles (arcs):

जीवे परस्परनिजेतरमौर्विकाभ्याम्
अभ्यस्य विस्तृतिदलेन विभज्यमाने ।।
अन्योन्ययोगविरहानुगुणे भवेतां
यद्वास्वलम्बकृतिभेदपदीकृते द्वे ।।

1) The two jyā (of the two arcs), multiplied by the kojyā of the other, and divided by half the diameter, by adding and subtracting, become the jyā of the similar operations (ie sum or difference of arcs).
2) Alternately, the sums and differences of the square roots of the differences between squares of the two jyā and the lamba can be used.

That is
\(
\begin{align*}
jyā(a \pm b) &= \frac{1}{R}.(jyā(a). kojyā(b) \pm jyā(b). kojyā(a)) \\
sin(A \pm B) &= sin(A).cos(B) \pm sin(B).cos(A)
\end{align*}
\)

The “extra” factor of 1/R , for those used to the modern form, comes from the fact that jyā and kojyā run from 0 to R, while modern sines and cosines run from 0 to 1.

What about the second part of the verse, what exactly is a lamba?

ज्यायोः परस्परं घातात् त्रिज्याप्तं लम्ब इष्यते
The product of the two jyā divided by the radius is the lamba

\(
\begin{align*}
jyā(a \pm b) &= \sqrt{jyā(a)^2 – lamba^2} \pm \sqrt{jyā(b)^2 – lamba^2} \\
lamba &= \frac{jyā(a) . jyā(b)}{R}
\end{align*}
\)
What the “lamba” is geometrically can be seen from the yukti below.

Yukti

How do we justify these results? Yuktibhāṣā provides two proofs, one for either part of the verse.

The first yukti

Consider arc \( EA_3 = EA_2 + A_2A_3\). We seek to find the jyā in terms of the jyā of the constituent arcs. We take the radius OA₂, and draw A₃L perpendicular to it, extending it to meet the circle at A₁ such that A₁A₂ = A₂A3. Jyā and kojyā of EA₁, EA₂, and EA₃ are A₁B₁, A₂B₂, A₃B₃, and A₁S₁, A₂S₂, A₃S₃ respectively. We draw perpendiculars from L to the East and North axes as LC and LD respectively. LC meets A₃S₃ at Q and LD meets A₃B₃ at P. A₁Q is the lamba, also equal to LP

A₃PL and A₂S₂O are perpendicular triangles, while OLC and OA₂B₂, and OLD and OA₂S₂ are parallel. Hence, each set can be related with त्रैराशिकम्. This gets us:

\(
\begin{align*}
PL &= \frac{S_2O . A_3L}{A_2O} \\
&= \frac{A_2B_2.A_3L}{A_2O} \\
& = \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \\
PA_3 &= \frac{A_2S_2 . A_3L}{A_2O} \\
&= \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3) \\
LC &= \frac{OL.A_2B_2}{A_2O} \\
&= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) \\
LD & = \frac{OL. A_2S_2}{ A_2O}\\
&= \frac{1}{R}. kojyā(EA_2) . kojyā(A_2A_3)
\end{align*}
\)

Using these, we can see that

\(
\begin{align*}
jyā(EA_3) &= A_3B_3 = LC + A_3P \\
jyā(EA_2+A_2A_3) &= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3) \\
jyā(EA_1) &= A_1B_1 = LC – A_3P \\
jyā(EA_2-A_2A_3) &= \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3) – \frac{1}{R}. kojyā(EA_2) . jyā(A_2A_3)
\end{align*}
\)

and also, as a bonus:

\(
\begin{align*}
kojyā(EA_3) &= A_3S_3 = LD- PL \\
kojyā(EA_2+A_2A_3) &= \frac{1}{R}. kojyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \\
kojyā(EA_1) &= A_1S_1 = S_1Q + A_1Q = LD + PL \\
kojyā(EA_2-A_2A_3) &= \frac{1}{R}. kjyā(EA_2) . kojyā(A_2A_3) + \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3)
\end{align*}
\)

This demonstrates the first method of the verse. What about the second, which directly uses the lamba (PL) ?

The second yukti

We have seen that the lamba \( A_1Q = PL= \frac{1}{R}. jyā(EA_2) . jyā(A_2A_3) \).

\(
\begin{align*}
A_3B_3 &= A_3P+PB_3 \\
A_3P &= \sqrt{A_3L^2 – LP^2}\\
PB_3 &= \sqrt{B_3L^2 – LP^2}
\end{align*}
\)

Now, A₃L is evidently jyā(A₂A₃). What is B₃L? Reasoning backwards, we can infer that it must be jyā(EA₂). How do we demonstrate that? We already know that \( PB_3 = LC = \frac{1}{R}. jyā(EA_2) . kojyā(A_2A_3)\), so we can calculate

\(
\begin{align*}
B_3L^2 &= LP^2 + LC^2 \\
&= \frac{1}{R}. jyā(EA_2)^2. (jyā(A_2A_3)^2 + kojyā(A_2A_3)^2) \\
&= jyā(EA_2)^2
\end{align*}
\)

And therefore:

\(
\begin{align*}
jyā(a \pm b) &= A_3B_3 \\
&= \sqrt{jyā(a)^2 – lamba^2} \pm \sqrt{jyā(b)^2 – lamba^2}
\end{align*}
\)

In this case, we used the value of LC already inferred in the previous yukti. Can we make the two yuktis fully independent?

As it turns out, yes we can. Consider a circle centered at O', with triangle A₃LB₃ circumscribed. Once such circle is always possible. In this circle A₃B₃ and A₃L would both be samastajyā (fullchords), and LB₃ would be the samastajyā of their difference (by definition). In a circle of twice the radius of that construction, A₃B₃ and A₃L would both be jyā (halfchords), and on the same circle, LB₃ would have to be the jyā of their difference. Since only one such circumscribing circle can exist, therefore one such circle with A₃B₃ and A₃L as jyā can exist. Therefore, it must be the current one, where A₃B₃ = jyā(EA₃) and A₃L = jyā(A₂A₃). Therefore LB₃ = jyā(EA₂).

Using जीवे परस्परन्यायः

This allows us to use both the Madhava jyā table (श्रेष्ठं नाम वरिष्ठानाम्) , and the Madhava jyā series (निहत्य चापवर्गेण …) to compute accurate jyā. We note that it is much easier to use the series for small arcs. Since we know the jyā at the tabular points accurately, and can calculate jyā and kojyā of small arcs accurately with the series, we use the following technique to calculate the jyā of any arbitrary arc:

• Find the nearest tabular arc, and read off its jyā and kojyā from the table
• Calculate jyā and kojyā of the difference of the desired arc and the nearest tabular arc using the series.
• These are combined using जीवे परस्पर to get an accurate jyā of the desired arc.

This gives us the superior accuracy of the निहत्य चापवर्गेण series, but with faster convergence that we see for smaller arcs. The श्रेष्ठं नाम वरिष्ठानाम् table is accurate to the thirds, and using this method we have a quick and accurate method of calculating jyā for arbitrary arcs that maintains similar accuracy.