Concepts of aṇu and parārdha

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.