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On the computation of the arcsin function in the Kerala school of astronomy and mathematics

Dr. V. N. Krishnachandran
Principal (retd), Govt. Victoria College, Palakkad – 678001
(Formerly) Professor of Mathematics,
Govt. Engineering College, Thrissur – 680009
(email: [email protected])
Abstract

This paper examines how the mathematicians and astronomers of the Kerala school tackled the problem of computing the values of the arcsin function. Four different approaches are discussed all of which are found in Nīlakaṇṭha Somayājī’s (1444–1545 CE) Tantrasaṅgraha and the roots of all of which can be traced to ideas originally articulated by Saṅgamagrāma Mādhava (c. 1340–1425 CE): (i) a simple method when the argument is small; (ii) an iterative method when the argument is small; (iii) a method based on a lookup table; (iv) a method when the argument is large. The paper also contains the original Sanskrit verses describing the various methods and English translations thereof. Moreover, there is a presentation of a novel method for computing the circumference of a circle found in Jyeṣṭhadeva’s (c. 1500–1575 CE) Yuktibhāṣā which is based on method (i) for computing the arcsin function. All methods have been illustrated with numerical examples. A surprising by-product of the investigation is a totally unexpected appearance of a core integer sequence, namely, the entry A001764 in the Online Encyclopedia of Integer Sequence, while studying the iterative method for computing the arcsin function.   
2000 Mathematics Subject Classification: 01A32, 33B10, 11B83.
Keywords and phrases: Kerala school of astronomy and mathematics, computation of arcsin, Sangamagrāma Mādhava, Nīlakaṇṭha Somayājī, Tantrasaṅgraha, integer sequence A001764.

1 Introduction

Saṅgamagrāma Mādhava’s (c. 1340–1425 CE) methods for computing the values of the jyā, kojyā and śara functions are well known. His methods to compute the arc length when its tangent is given is also well known. In this paper we examine how the mathematicians and astronomers of the Kerala school tackled the related problem of computing the values of the the arc lengths when the values of the jyā-s are given, or, equivalently, computing the values of the arcsin function. This problem is highly significant and it arises in connection with several astronomical problems especially in connection with problems involving the “equation of centre” of the various planets. In particular, the problem of computing values of the arcsin function arises in the computation of the candravākya-s. Since Mādhava is famous as the author of a more accurate set of candravākya-s than the then available Vararuci’s candravākya-s, he definitely should have addressed the problem. No direct account of how Mādhava had tackled the problem have survived. However, in Nīlakaṇṭha Somayājī’s (1444–1545 CE) Tantrasaṅgraha there is a detailed discussion of the problem and it could be the case that the solution presented by Somayājī might have had its origins in the mind of Mādhava. In support of this surmise, it may be pointed out that Somayājī’s methods for computing small arc lengths make use of the Mādhava-Newton series for the sine function.

Before taking up Nīlakaṇṭha Somayājī’s methods for computing values of the arcsin function, it is illuminating to see how the astronomers and mathematicians of the classical age of Indian astronomy had handled the problem. Āryabhaṭīya has not considered the problem. But Brahmagupta (c. 598–668 CE) and Bhāskara II (c. 1114–1185 CE) have given approximate expressions for the arcsin function. These expressions are derived from Bhāskara I’s (c. 600–680 CE) approximation function for the sine function. Hence, we begin the paper with a discussion on this very interesting approximation function and then consider the associated expression for the arcsin function. This is followed by Somayājī’s discussion of the methods for computing the arcsin function. There are separate methods for computing arc lengths corresponding to small jyā-s and large jyā-s. Both are dealt with in detail. Jyeṣṭhadeva’s (c. 1500–1575 CE) Yuktibhāṣā contains a novel method for the computation of the circumference of a circle which makes use the arcsin function. We have included a discussion of this also in this chapter.

2 Bhāskara I’s approximation to the sine function

In this section, we consider a remarkable rational function approximation to the sine function due to Bhāskara I (c. 600-680). Bhaskara I gave this approximation formula in his Mahābhāskarīya and nearly all subsequent Indian mathematicians and astronomers, except those belonging to the Kerala school, have also given equivalent forms of the formula. What is curious and interesting is the fact that that none of the mathematicians of the Kerala school have cared to mention it in their writings; perhaps they thought they had a more accurate formula in their hands in the form of an infinite series and why bother about an approximation formula! Neither Bhāskara I nor any of his successors, true to their style, have given any rationale for the formula. Historians of mathematics have come up with several candidate rationales that could be the rationale by which Bhāskara I originally arrived at the approximation formula. However no candidate has won universal approval!

2.1 The approximation formula

The formula is stated in verses 17–19 Chapter VII of Mahābhāskarīya. The verses and their English translation are given below (see [4] p. 45 for the verses and p. 207 for the English translation; see also [3]):

[Uncaptioned image]

makhyādirahitaṃ karma vakṣyate tatsamāsataḥ |\scriptstyle|
cakrārdhāṃśakasamūhādviśodhyā ye bhujāṃśakāḥ ||\scriptstyle||
taccheṣuguṇitā dviṣṭhaāḥ śodhyāḥ khābhreṣukhābdhitaṭaḥ |\scriptstyle|
caturthāṃśena śeṣasya dviṣṭhamantyaphalaṃ hatam ||\scriptstyle||
bāhukoṭyoḥ phalaṃ kṛtsnaṃ kramotkramaguṇasya vā |\scriptstyle|
labhyate candratīkṣṇāṃśvostārāṇāṃ vāpi tattvataḥ ||\scriptstyle||

“ I briefly state the rule (for finding the bhujaphala and kotiphala etc.) without making use of the Rsine-differences, 225, etc.

Subtract the degrees of the bhuja (or koti) from the degrees of half a circle (i.e., 180180^{\circ}). Then multiply the remainder by the degrees of the bhuja (or koti) and put down the result at two places. At one place subtract the result from 40500. By one-fourth of the remainder (thus obtained) divide the result at the other place as multiplied by the antyaphala (i.e., the epicyclic radius). Thus is obtained the entire bahuphala (or kotiphala) for the Sun, Moon, or the star-planets. So also are obtained the direct and inverse Rsines.”

In the above verses, the reference to “Rsine-differences, 225, etc.” is a reference to Āryabhaṭa’s sine table for a discussion on this table).

2.2 Rendering in modern notations

Let xx^{\circ} be the angle subtended by the arc whose jyā) is sought and let rr be the radius of the circle. Bhāskara I’s approximation formula can be expressed in these notations as follows:

jyā(x)rx(180x)14[40500x(180x)].\text{jy\={a}}\,(x^{\circ})\approx\frac{r\cdot x(180-x)}{\tfrac{1}{4}[40500-x(180-x)]}. (1)

Since jyā(x)=rsin(x)\text{jy\={a}}\,(x^{\circ})=r\sin(x^{\circ}), we have

sin(x)4x(180x)40500x(180x).\sin(x^{\circ})\approx\frac{4x(180-x)}{40500-x(180-x)}. (2)

This is Bhāskara I’s approximation formula to the sine functions in modern notations. Note that this has to be understood in the following form:

sin(x)4x(180x)59090x(180x).\sin(x^{\circ})\approx\frac{4\cdot x^{\circ}\cdot(180^{\circ}-x^{\circ})}{5\cdot 90^{\circ}\cdot 90^{\circ}-x^{\circ}\cdot(180^{\circ}-x^{\circ})}. (3)

If angles are measured in radians, Eq.(3) can be put in the following form:

sin(θ)=16θ(πθ)5π24θ(πθ).\sin(\theta)=\frac{16\theta(\pi-\theta)}{5\pi^{2}-4\theta(\pi-\theta)}. (4)

Some properties of the formula can be easily observed.

  1.  (1)

    The formula gives the exact values of sin(x)\sin(x^{\circ}) for x=0,30,90,150x=0^{\circ},30^{\circ},90^{\circ},150^{\circ} and 180180^{\circ}, namely, the values 0,12,1,12,00,\frac{1}{2},1,\frac{1}{2},0.

  2.  (2)

    The formula is symmetrical about 9090^{\circ}, that is, the formula gives the same value for sin(90x)\sin(90^{\circ}-x^{\circ}) and sin(90+x)\sin(90^{\circ}+x^{\circ}).

2.3 Accuracy of the formula

Figure 1 shows the percentage error in the value of sin(x)\sin(x^{\circ}) computed using Bhāskara I’s approximation formula for the sine function. It can be seen that the maximum relative error is less than 1.8%.

Refer to caption
Figure 1: Graph of the percentage relative error in Bhāskara I’s approximation formula for the sine function

2.4 On the rationale of the formula

As already indicated, neither Bhāskara nor his followers have given the rationale of the sine approximation formula. In the literature on the history of Indian mathematics one can see several modern rationales for the formula (see, for example, [3], [9] and [10]). We present below one such rationale which is an adaption of a rationale given in [10]. Our rationale is to look for the simplest function that satisfies the two properties (1) and (2) of Bhāskara I’s approximation formula.

2.5 The rationale

The simplest function which satisfies property (2) and which takes the value 0 when x=0,180x=0,180 is

f(x)=kx(180x).f(x)=kx(180-x).

Now, f(30)=12f(30)=\frac{1}{2} if k=19000k=\frac{1}{9000} and f(90)=1f(90)=1 if k=18100k=\frac{1}{8100}. This suggests that, for f(x)f(x) to satisfy property (1), kk should be of the form 1g(x)\frac{1}{g(x)} with g(x)g(x) satisfying property (2) and with g(30)=9000g(30)=9000 and g(90)=8100g(90)=8100.

The simplest general function which satisfies property (2) is

g(x)=a+bx(180x).g(x)=a+bx(180-x).

Since g(30)=9000g(30)=9000 and g(90)=8100g(90)=8100, we must have

a+4500b\displaystyle a+4500b =9000,\displaystyle=9000,
a+8100b\displaystyle a+8100b =8100.\displaystyle=8100.

Solving these equations we get a=14×40500a=\frac{1}{4}\times 40500 and b=14b=-\frac{1}{4}.

Thus the function

x(180x)14×4050014x(180x)\frac{x(180-x)}{\frac{1}{4}\times 40500-\frac{1}{4}x(180-x)}

satisfies the properties (1) and (2). This is precisely Bhāskara I’s sine approximation formula.

3 Brahmagupta’s formula for arcsin

Brahmagupta in his Brahmasphuṭasiddhānta has given a formula for computing the values of the arcsin function. The verses and an English translation thereof are reproduced below (see verses 24–25 Chapter 14 [8]; see [2] for the English translation):

[Uncaptioned image]

iṣṭjyā saṃguṇitā paṁcakayamalaika śūnyacandramasā |\scriptstyle|
iṣṭajyāpādayutavyāsārddhavibhājitālabdham ||\scriptstyle||
navatikṛteḥ prohyapadaṃ navateḥ saṃśodhya śeṣakalāḥ |\scriptstyle|
evaṃ dhanuriṣṭāyā bhavati jyayā vinā jyābhiḥ ||\scriptstyle||

“Multiply 10125 by the given jyā and divide by the quarter of the given jyā plus the radius; subtracting the quotient from the square of 90, extract the square-root and subtract (the root) from 90; the remainder will be in degrees and minutes; thus will be found the arc of the given jyā without the table of jyā-s.”

3.1 Rendering in modern notations

Let m=jyā(s)m=\text{jy\={a}}\,\,(s), where ss is an arc, measured in degrees, of a circle of radius rr, then:

s=90810010125m(m4+r).s=90-\sqrt{8100-\frac{10125m}{(\frac{m}{4}+r)}}. (5)

3.2 Rationale of the formula

Brahmagupta has not given any rationale for the expression in Eq.(5). It can be seen that it follows from from Eq.(1). Rrom Eq.(1) we have:

mrs(180s)14[40500s(180s)].m\approx\frac{r\cdot s(180-s)}{\tfrac{1}{4}[40500-s(180-s)]}.

This can be written as

s2180s+10125m(m4+r)=0.s^{2}-180s+\frac{10125m}{(\frac{m}{4}+r)}=0.

This is a quadratic equation in ss and solving it for ss we get precisely the formula in Eq.(5). The negative sign of the radical is retained as it is assumed that ss is less than 9090^{\circ}.

4 Nīlakaṇṭha Somayājī’s method for computing arcsin of small jyā-s

4.1 Somayājī’s approximate expression for jyā  (s)(s)

Somyājī, in Tantrasṅgraha, has given an an approximate expression for jyā  (s)(s) in the following verse (see verse 17 Chapter 2 in [6]):

[Uncaptioned image]

śiṣṭacāpaghanaṣaṣṭhabhāgato vistarārdhakṙtibhaktavarjitam |\scriptstyle|
śiṣṭacāpamiha śiñjinī  bhavet spaṣṭatā bhavati cālpatāvaśāt ||\scriptstyle||

“Divide one-sixth of the cube of the remaining arc by the square of the trijyā. This quantity when subtracted from the remaining arc becomes the śiñjinī  (the dorjyā or jyā) corresponding to the remaining arc). The value is accurate because of the smallness [of the arc].”

4.1.1 Rendering in modern notations

Let ss be small arc of a circle of radius rr, then, the verse states that

jyā(s)=ss36r2.\text{jy\={a}}\,\,(s)=s-\frac{s^{3}}{6r^{2}}. (6)

4.1.2 Rationale of the result

Somayājī has not given any rationale for the result. However, it can be seen easily that the result follows from the Mādhava-Newton series for jyā  (s)(s), namely,

jyā(s)=\displaystyle\text{jy\={a}}\,\,(s)= s[ss2(22+2)r2{ss2(22+2)r2s2(42+4)r2\displaystyle s-\Big{[}s\cdot\frac{s^{2}}{(2^{2}+2)r^{2}}-\Big{\{}s\cdot\frac{s^{2}}{(2^{2}+2)r^{2}}\cdot\frac{s^{2}}{(4^{2}+4)r^{2}}-
(ss2(22+2)r2s2(42+4)r2s2(62+6)r2[])}],\displaystyle\phantom{s-}\,\,\Big{(}s\cdot\frac{s^{2}}{(2^{2}+2)r^{2}}\cdot\frac{s^{2}}{(4^{2}+4)r^{2}}\cdot\frac{s^{2}}{(6^{2}+6)r^{2}}-\Big{[}\cdots\Big{]}\Big{)}\Big{\}}\Big{]},

by neglecting terms containing s5s^{5} and higher powers of ss.

4.2 Formula for computing arcsin function

Somayājī in verse 37 Chapter 2 of Tantrasaṅgraha has indicated that the result in Eq.(6) can be used to compute the arc lengths corresponding to a given jyā provided that the arcs are small. However, Somayājī did not explicitly state how the value of ss is to be computed from this equation. But, fortunately, Śankara Vāriyar (c. 1500–1560 CE), in his Laghuvivṛtī commentary on Tantrasaṅgraha, has explained how exactly this can be done (see [6] pp. 92–93). We rewrite Eq.(6) in the following form

s=jyā(s)+s36r2s=\text{jy\={a}}\,\,(s)+\frac{s^{3}}{6r^{2}} (7)

and then, in the right side, we replace ss by jyā  (s)(s) to get

sjyā(s)+(jyā(s))36r2.s\approx\text{jy\={a}}\,\,(s)+\frac{(\text{jy\={a}}\,\,(s))^{3}}{6r^{2}}. (8)

This is Somayājī’s first method for computing values of arc lengths corresponding to small arcs.

4.3 A polynomial approximation to the arcsin function

Here we show that Somayājī’s expression to compute the arcsin function given by Eq.(8) is equivalent to the power series expression for arcsin(x)\arcsin(x) truncated at the term containing x3x^{3}.

Taking s=rθs=r\theta and jyā  (s)=rsinθ(s)=r\sin\theta, Eq.(8) can be written as

θsinθ+sin3θ6.\theta\approx\sin\theta+\frac{\sin^{3}\theta}{6}.

Let x=sinθx=\sin\theta so that θ=arcsin(x)\theta=\arcsin(x). Then we have:

arcsin(x)x+x36.\arcsin(x)\approx x+\frac{x^{3}}{6}. (9)

Thus Eq.(8) is equivalent to the polynomial approximation to arcsin(x)\arcsin(x) given by Eq.(9). Incidentally, the approximation to arcsin(x)\arcsin(x) given by Eq.(9) is the approximation to arcsin(x)\arcsin(x) obtained by neglecting terms containing x5x^{5} and higher powers of xx in the power series expression for arcsin(x)\arcsin(x), namely,

arcsin(x)=x+x36+3x540+5x7112+35x91152+.\arcsin(x)=x+\frac{{{x}^{3}}}{6}+\frac{3{{x}^{5}}}{40}+\frac{5{{x}^{7}}}{112}+\frac{35{{x}^{9}}}{1152}+\cdots. (10)

5 Śankara Vāriyar’s iterative method for computing approximate values of the arcsin function

Śankara Vāriyar, in his Laghuvivṛtī commentary on Tantrasṅgraha, after stating the formula given in Eq.(8), has given an iterative method for computing the arc lengths more accurately. The iterative process starts from the approximate relation in Eq.(7), which is rewritten as follows:

s=m+Δ,s=m+\Delta, (11)

where m=jyā(s)m=\text{jy\={a}}\,\,(s) is known and ss is unknown, and where

Δ=s36r2.\Delta=\frac{s^{3}}{6r^{2}}. (12)

In Eq.(12), as the first approximation to ss we take ss as mm and compute

Δ1=m36r2\Delta_{1}=\frac{m^{3}}{6r^{2}}

Using this in Eq.(11), we get the first approximation to ss as

s1=m+Δ1.s_{1}=m+\Delta_{1}.

We next use this value of ss in Eq.(12) to get the second approximation to Δ\Delta as

Δ2=(m+Δ1)36r2.\Delta_{2}=\frac{(m+\Delta_{1})^{3}}{6r^{2}}.

Using this in Eq.(11), we get the second approximation to ss as

s2=m+Δ2.s_{2}=m+\Delta_{2}.

The process is continued until two successive approximations to ss are nearly equal.

5.1 Śankara Vāriyar’s Algorithm to compute arc lengths

The above described procedure can be summarised as an algorithm thus. Let mm be the jyā of an arc ss of a circle of radius rr. Given mm, to compute a sequence of approximations sis_{i} to ss and to generate an approximate value of ss:

  1.   Step 1.

    Set s0=ms_{0}=m, Δ0=0\Delta_{0}=0.

  2.   Step 2.

    For i=1,2,3,i=1,2,3,\ldots, repeat the following until successive values of sis_{i}-s are nearly equal:

    1. (a)

      Δi=(m+Δi1)36r2.\Delta_{i}=\frac{(m+\Delta_{i-1})^{3}}{6r^{2}}.

    2. (b)

      si=m+Δi.s_{i}=m+\Delta_{i}.

  3.   Step 3.

    When sisi+1s_{i}\approx s_{i+1}, then ssis\approx s_{i}.

Remarks

The above algorithm to compute arc lengths does not generate successively better approximations to the true value of the arc length. Moreover, the sequence {si}\{s_{i}\} does not converge to the true arc length ss. In fact we have:

s0\displaystyle s_{0} =m,\displaystyle=m,
s1\displaystyle s_{1} =m+m36r2,\displaystyle=m+\frac{m^{3}}{6r^{2}},
s2\displaystyle s_{2} =m+m36r2+m512r4+m772r6+m91296r8,\displaystyle=m+\frac{{{m}^{3}}}{6{{r}^{2}}}+\frac{{{m}^{5}}}{12{{r}^{4}}}+\frac{{{m}^{7}}}{72{{r}^{6}}}+\frac{{{m}^{9}}}{1296{{r}^{8}}},
s3\displaystyle s_{3} =m+m36r2+m512r4+m718r6+7m9324r8+ higher powers of m,\displaystyle=m+\frac{{{m}^{3}}}{6{{r}^{2}}}+\frac{{{m}^{5}}}{12{{r}^{4}}}+\frac{{{m}^{7}}}{18{{r}^{6}}}+\frac{7{{m}^{9}}}{324{{r}^{8}}}+\text{ higher powers of $m$},
s4\displaystyle s_{4} =m+m36r2+m512r4+m718r6+55m91296r8+ higher powers of m.\displaystyle=m+\frac{{{m}^{3}}}{6{{r}^{2}}}+\frac{{{m}^{5}}}{12{{r}^{4}}}+\frac{{{m}^{7}}}{18{{r}^{6}}}+\frac{55{{m}^{9}}}{1296{{r}^{8}}}+\text{ higher powers of $m$}.

These may be compared with the power series expansion of the arcsin function given in Eq.(10).

5.2 Illustrative examples

5.2.1 Illustrative example 1

Let us compute the arc whose jyā is m=224  50′′  22′′′=809422′′′m=224^{\prime}\,\,50^{\prime\prime}\,\,22^{\prime\prime\prime}=809422^{\prime\prime\prime}. This is the first entry in Mādhava’s sine table and it corresponds to jyā  (225)(225^{\prime}). We take the radius of the circle as r=3437  44′′  48′′′=12375888′′′r=3437^{\prime}\,\,44^{\prime\prime}\,\,48^{\prime\prime\prime}=12375888^{\prime\prime\prime}.

  1.   Step 1.

    As initial approximations, we take s0=809422′′′s_{0}=809422^{\prime\prime\prime} and Δ0=0\Delta_{0}=0.

  2.   Step 2.

    Results of computations are taken as rounded to the nearest integer.

    Iteration 1:

    Δ1=(m+Δ0)36r2=(809422′′′+0)36×(12375888′′′)2=577′′′\Delta_{1}=\frac{(m+\Delta_{0})^{3}}{6r^{2}}=\frac{(809422^{\prime\prime\prime}+0)^{3}}{6\times(12375888^{\prime\prime\prime})^{2}}=577^{\prime\prime\prime}

    s1=m+Δ1=809422′′′+577′′′=809999′′′s_{1}=m+\Delta_{1}=809422^{\prime\prime\prime}+577^{\prime\prime\prime}=809999^{\prime\prime\prime}

    Iteration 2:

    Δ2=(m+Δ1)36r2=(809422′′′+577′′′)36×(12375888′′′)2=578′′′\Delta_{2}=\frac{(m+\Delta_{1})^{3}}{6r^{2}}=\frac{(809422^{\prime\prime\prime}+577^{\prime\prime\prime})^{3}}{6\times(12375888^{\prime\prime\prime})^{2}}=578^{\prime\prime\prime}

    s2=m+Δ2=809422′′′+578′′′=810000′′′s_{2}=m+\Delta_{2}=809422^{\prime\prime\prime}+578^{\prime\prime\prime}=810000^{\prime\prime\prime}

    Iteration 3:

    Δ3=(m+Δ2)36r2=(809422′′′+578′′′)36×(12375888′′′)2=578′′′\Delta_{3}=\frac{(m+\Delta_{2})^{3}}{6r^{2}}=\frac{(809422^{\prime\prime\prime}+578^{\prime\prime\prime})^{3}}{6\times(12375888^{\prime\prime\prime})^{2}}=578^{\prime\prime\prime}

    s3=m+Δ3=809422′′′+578′′′=810000′′′s_{3}=m+\Delta_{3}=809422^{\prime\prime\prime}+578^{\prime\prime\prime}=810000^{\prime\prime\prime}

  3.   Step 3.

    s2=s3=810000′′′s_{2}=s_{3}=810000^{\prime\prime\prime} and so s=810000′′′=225s=810000^{\prime\prime\prime}=225^{\prime}. This agrees with the true value of ss.

5.2.2 Illustrative example 2

Let us compute the arc whose jyā is m=448  42′′  58′′′=1615378′′′m=448^{\prime}\,\,42^{\prime\prime}\,\,58^{\prime\prime\prime}=1615378^{\prime\prime\prime}. This is the second entry in Mādhava’s sine table and it corresponds to jyā  (450)(450^{\prime}).

  1.   Step 1.

    As initial approximations, we take s0=1615378′′′s_{0}=1615378^{\prime\prime\prime} and Δ0=0\Delta_{0}=0.

  2.   Step 2.

    Results of computations are taken as rounded to the nearest integer.

    Iteration 1:

    Δ1=(m+Δ0)36r2=4587′′′\Delta_{1}=\frac{(m+\Delta_{0})^{3}}{6r^{2}}=4587^{\prime\prime\prime}

    s1=m+Δ1=1619965′′′s_{1}=m+\Delta_{1}=1619965^{\prime\prime\prime}

    Iteration 2:

    Δ2=(m+Δ1)36r2=4626′′′\Delta_{2}=\frac{(m+\Delta_{1})^{3}}{6r^{2}}=4626^{\prime\prime\prime}

    s2=m+Δ2=1620004′′′s_{2}=m+\Delta_{2}=1620004^{\prime\prime\prime}

    Iteration 3:

    Δ3=(m+Δ2)36r2=4626′′′\Delta_{3}=\frac{(m+\Delta_{2})^{3}}{6r^{2}}=4626^{\prime\prime\prime}

    s3=m+Δ3=1620004′′′s_{3}=m+\Delta_{3}=1620004^{\prime\prime\prime}

  3.   Step 3.

    s2=s3=1620004′′′s_{2}=s_{3}=1620004^{\prime\prime\prime} and so s=1620004′′′=450  0′′  4′′′s=1620004^{\prime\prime\prime}=450^{\prime}\,\,0^{\prime\prime}\,\,4^{\prime\prime\prime}. In this case, there is a difference of 4′′′4^{\prime\prime\prime} from the true value of ss.

5.3 Śankara Vāriyar’s algorithm throws up a core integer sequence!

The algorithm described in Section 5.1 has completely unexpectedly thrown up an important integer sequence which appears in a large number of combinatorial problems. The Online Encyclopedia of Integer Sequences has characterised this as a “core integer sequence”. To see how the particular integer sequence emerges via Śankara Vāriyar’s algorithm, let us reformulate the algorithm in the following form: we write m=xm=x, t=16r2t=\frac{1}{6r^{2}}, Δi=yi\Delta_{i}=y_{i}, s0=xs_{0}=x and y0=0y_{0}=0. Then we have, for i=1,2,i=1,2,\ldots:

yi\displaystyle y_{i} =t(x+yi1)3\displaystyle=t(x+y_{i-1})^{3}
si\displaystyle s_{i} =x+yi1.\displaystyle=x+y_{i-1}.

Let us compute a few values of sis_{i}:

s0\displaystyle s_{0} =(x),\displaystyle=(x),
s1\displaystyle s_{1} =(x+tx3),\displaystyle=(x+t{{x}^{3}}),
s2\displaystyle s_{2} =(x+tx3+3t2x5)+3t3x7+t4x9,\displaystyle=(x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}})\;+3{{t}^{3}}{{x}^{7}}+{{t}^{4}}{{x}^{9}},
s3\displaystyle s_{3} =(x+tx3+3t2x5+12t3x7)+28t4x9+57t5x11+96t6x13+ higher powers of x,\displaystyle=(x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}}+12{{t}^{3}}{{x}^{7}})\;+28{{t}^{4}}{{x}^{9}}+57{{t}^{5}}{{x}^{11}}+96{{t}^{6}}{{x}^{13}}+\text{ higher powers of $x$},
s4\displaystyle s_{4} =(x+tx3+3t2x5+12t3x7+55t4x9)+192t5x11+618t6x13+ higher powers of x,\displaystyle=(x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}}+12{{t}^{3}}{{x}^{7}}+55{{t}^{4}}{{x}^{9}})\;+192{{t}^{5}}{{x}^{11}}+618{{t}^{6}}{{x}^{13}}+\text{ higher powers of $x$},
s5\displaystyle s_{5} =(x+tx3+3t2x5+12t3x7+55t4x9+273t5x11)+1185t6x13+ higher powers of x,\displaystyle=(x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}}+12{{t}^{3}}{{x}^{7}}+55{{t}^{4}}{{x}^{9}}+273{{t}^{5}}{{x}^{11}})\;+1185{{t}^{6}}{{x}^{13}}+\text{ higher powers of $x$},
s6\displaystyle s_{6} =(x+tx3+3t2x5+12t3x7+55t4x9+273t5x11+1428t6x13)+ higher powers of x.\displaystyle=(x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}}+12{{t}^{3}}{{x}^{7}}+55{{t}^{4}}{{x}^{9}}+273{{t}^{5}}{{x}^{11}}+1428{{t}^{6}}{{x}^{13}})\;+\text{ higher powers of $x$}.

Consider the integers that appear as coefficients in the parenthesised expressions in the sis_{i}-s:

1,1,3,12,55,273,1428,.1,1,3,12,55,273,1428,\ldots.

This is precisely the sequence A001764 in the Online Encyclopedia of Integer Sequences (OIES) (see [5]). This sequence, which appears in a large number of combinatorial problems, has been characterised as a core sequence in the Encyclopedia. As per the Encyclopedia, the jj-th entry aja_{j} in the above integer sequence is given by

aj=(3j)!j!(2j+1)!.a_{j}=\frac{(3j)!}{j!(2j+1)!}.

Thus we have:

sn=(a0x+a1tx3+a2t2x5++antnx2n+1)+ higher powers of x.s_{n}=(a_{0}x+a_{1}tx^{3}+a_{2}t^{2}x^{5}+\cdots+a_{n}t^{n}x^{2n+1})\;+\text{ higher powers of $x$}.

Moreover, there is a closed form expression for the infinite series n=0antnx2n+1\sum_{n=0}^{\infty}a_{n}t^{n}x^{2n+1} (for example, see [1]), namely,

23tsin(13arcsin(33t2x))=x+tx3+3t2x5+12t3x7+55t4x9+273t5x11+.\frac{2}{\sqrt{3t}}\sin{\Big{(}\frac{1}{3}\operatorname{arcsin}\Big{(}\frac{3\sqrt{3t}}{2}x\Big{)}\Big{)}}=x+t{{x}^{3}}+3{{t}^{2}}{{x}^{5}}+12{{t}^{3}}{{x}^{7}}+55{{t}^{4}}{{x}^{9}}+273{{t}^{5}}{{x}^{11}}+\cdots.

6 Computing arcsin function using a lookup table

Nīlakaṇṭha Somayājī has suggested that a lookup table can be used to find the arc lengths corresponding to certain pre-defined jyā values. This method has a severe limitation: it cannot be used to find the arc lengths corresponding to any arbitrary jyā values. The table lists only 24 jyā values the minimum being 105  43′′105^{\prime}\,\,43^{\prime\prime} and the maximum being 304  36′′304^{\prime}\,\,36^{\prime\prime}. The difference between a pair successive jyā values is not a constant.

6.1 The rationale of the method

To begin with, we assume that the arcs and the corresponding jyā-s are all small. Let ss be a small arc and let jyā  (s)=m(s)=m. We are given mm and we have to find ss. From Eq.(7), we have

smm36r2.s-m\approx\frac{m^{3}}{6r^{2}}. (13)

Somayājī assumes that mm and rr are in minutes so that sms-m given by Eq.(13) is also in minutes. If

sm=k′′=(k60),s-m=k^{\prime\prime}=(\tfrac{k}{60})^{\prime}, (14)

then we have

k60=m36r2.\frac{k}{60}=\frac{m^{3}}{6r^{2}}. (15)

From this, we get

m=(kr210)13minutes.m=\Big{(}\frac{kr^{2}}{10}\Big{)}^{\frac{1}{3}}\,\,\text{minutes}. (16)

Using this in Eq.(14), we get

s=[k60+(kr210)13]minutes.s=\Big{[}\frac{k}{60}+\Big{(}\frac{kr^{2}}{10}\Big{)}^{\frac{1}{3}}\Big{]}\,\,\text{minutes}. (17)

This means that, if the difference between arc length and jyā is k′′k^{\prime\prime}, then the arc length is given Eq.(17). The lookup table gives the the values of ss for k=1,2,,24k=1,2,\ldots,24.

6.2 The table

Table 1 lists the values of the arc lengths for various values of the difference between the arc lengths and the corresponding jyā-s as given in the Laghuvivṛtī commentary of Tantrasaṅgraha (see [6] pp. 94–95). The Laghuvivṛtī commentary lists the values of m=jyā(s)m=\text{jy\={a}}\,\,(s) given by Eq.(16) for k=1,2,,24k=1,2,\ldots,24 whereas Table 1 lists the values of jyā(s)\text{jy\={a}}\,\,(s) as well as the values of ss given by Eq.(17). In the table, we have also given the kaṭapayādi encodings of the values of jyā  (s)(s)-s. These encodings are taken from the following verses given in the Laghuvivṛtī commentary.

[Uncaptioned image]

lavaṇaṃ nindyaṃ kapilā gopī cararāśayastavārthitayā |\scriptstyle|
laghunoddiṣṭo rājñaḥ praḷayo dhāmnāṃ trinetra narakapuram ||\scriptstyle||
savadhūṭīndro jalasūradrīhimavān gurustriśaṅkuvaraḥ |\scriptstyle|
varado vajrī tilabhūrmeruḥ kālena tatra nṛpaticaraḥ ||\scriptstyle||
tilakaṃ sāndraṃ dhāvatisarit na me kuñjaro nivṛttajaraḥ |\scriptstyle|
śreṣṭhakaḷatramamāśādhātrī dhūpo’gnīnāmbutilavanagaḥ ||\scriptstyle||

kk jyā (s)(s) jyā (s)(s) ss
\prime\prime (in kaṭapayādi) \prime \prime\prime \prime \prime\prime
1 lavaṇaṃ nindyaṃ 105 43 105 44
2 kapilā gopī 133 11 133 13
3 cararāśaya 152 26 152 29
4 stavārthitayā 167 46 167 50
5 laghunoddiṣṭo 180 43 180 48
6 rājñaḥ praḷayo 192 02 192 08
7 dhāmnāṃ trinetra 202 08 202 15
8 narakapuram 211 20 211 28
9 savadhūṭīndro 219 47 219 56
10 jalasūradrī 227 38 227 48
11 himavān guru 234 58 235 09
12 striśaṅkuvaraḥ 241 52 242 04
13 varado vajrī 248 24 248 37
14 tilabhūrmeruḥ 254 36 254 50
15 kālena tatra 260 31 260 46
16 nṛpaticaraḥ 266 10 266 26
17 tilakaṃ sāndraṃ 271 36 271 53
18 dhāvatisarit 276 48 277 06
19 na me kuñjaro 281 50 282 09
20 nivṛttajaraḥ 286 40 287 00
21 śreṣṭhakaḷatra 291 22 291 43
22 mamāśādhātrī 295 55 296 17
23 dhūpo’gnīnā 300 18 300 41
24 mbutilavanagaḥ 304 36 305 00
Table 1: Lookup table for computing values of the arcsin function

The values given by the Laghuvivṛtī commentary are remarkably accurate. For example, for k=24k=24, the commentary gives s=305  0′′s=305^{\prime}\,\,0^{\prime\prime} whereas a modern computation using Maxima software, with r=216002πr=\frac{21600}{2\pi}, yielded s=304  58.03′′s=304^{\prime}\,\,58.03^{\prime\prime}, a difference of nearly 2′′2^{\prime\prime} only.

6.3 How to use the lookup table

The lookup table gives the values of the arc lengths corresponding to a predefined set of jyā values. If the jyā whose arc length is to be determined happens to be very close to one of the values listed in the table, then the corresponding arc length can be read off from the table. For example, if jyā(s)=200\text{jy\={a}}\,\,(s)=200^{\prime} which is very close to the value 202  08′′202^{\prime}\,\,08^{\prime\prime} listed in the table, then we may take ss as the corresponding arc length given in the table, namely, 202  15′′202^{\prime}\,\,15^{\prime\prime}.

It is interesting to speculate why the table contains only values up to k=24′′k=24^{\prime\prime}. One important application of the table might have been the computation of the true longitudes of the Moon which involved the computation of an expression of the form (see [6] p. 90)

arcsin(780sin(θ0θm)).\arcsin\big{(}\tfrac{7}{80}\sin(\theta_{0}-\theta_{m})\big{)}.

The maximum value of 780sin(θ0θm)\tfrac{7}{80}\sin(\theta_{0}-\theta_{m}) is

780×(3437  44′′  48′′′)=300  48′′  10′′′.\tfrac{7}{80}\times(3437^{\prime}\,\,44^{\prime\prime}\,\,48^{\prime\prime\prime})=300^{\prime}\,\,48^{\prime\prime}\,\,10^{\prime\prime\prime}.

Hence in the computation of the true longitudes of the Moon, one would be required to compute arc lengths corresponding to jyā-s less than this maximum value only. Note that the value of jyā corresponding to k=23′′k=23^{\prime\prime} is 300  18′′300^{\prime}\,\,18^{\prime\prime} and that corresponding to k=24′′k=24^{\prime\prime} is 304  36′′304^{\prime}\,\,36^{\prime\prime}, and that the maximum value lies between these two values.

7 Nīlakaṇṭha Somayājī’s method for computing arcsin of large jyā-s

Nīlakaṇṭha Somayājī in his tantrasaṅgraha has given an approximation formula for the difference between two arcs in terms of the jyā-s and kojyā-s of the arcs. This formula is then used to compute the value of the arcsin function corresponding to a large value of jyā.

7.1 Somayājī’s formula

Nīlakaṇṭha Somayājī’s formula appears in verse 14 Chapter 2 of tantrasaṅgraha (see [6] p. 68).

[Uncaptioned image]

jyayorāsannayorbhedabhaktastatkoṭiyogataḥ |\scriptstyle|
chedastena hṛtādvighnā trijyā taddhanurantaram ||\scriptstyle||

“The sum of the kojyā-s divided by the difference of those two jyā-s, which are close to each other, forms the cheda (divisor). Twice the trijyā divided by this is the difference between the corresponding arcs.”

7.1.1 Rendering in modern notations

Let s1s_{1} and s2s_{2} be two arcs, with s2>s1s_{2}>s_{1}, of a circle of radius rr. Then the quoted verse says:

s2s1=2r(kojyā(s2)+kojyā(s1)jyā(s2)jyā(s1)).s_{2}-s_{1}=\frac{2r}{\left(\dfrac{\text{kojy\={a}}\,\,(s_{2})+\text{kojy\={a}}\,\,(s_{1})}{\text{jy\={a}}\,\,(s_{2})-\text{jy\={a}}\,\,(s_{1})}\right)}. (18)

7.1.2 Equivalent mathematical result

Let the arcs s1s_{1} and s2s_{2} subtend angles θ1\theta_{1} and θ2\theta_{2} at the centre of the circle. Then:

jyā(s1)=rsinθ1,jyā(s2)=rsinθ2,\displaystyle\text{jy\={a}}\,\,(s_{1})=r\sin\theta_{1},\quad\text{jy\={a}}\,\,(s_{2})=r\sin\theta_{2},
kojyā(s1)=rcosθ1,kojyā(s2)=rcosθ2.\displaystyle\text{kojy\={a}}\,\,(s_{1})=r\cos\theta_{1},\quad\text{kojy\={a}}\,\,(s_{2})=r\cos\theta_{2}.

Using these in Eq.(18), we have:

s2s1=2r(cosθ2+cosθ1sinθ2sinθ1).s_{2}-s_{1}=\frac{2r}{\left(\dfrac{\cos\theta_{2}+\cos\theta_{1}}{\sin\theta_{2}-\sin\theta_{1}}\right)}. (19)

7.1.3 Remarks

The mathematical basis of Eq.(18) is a well-known result of elementary trigonometry which states that

tanθθ when θ is small.\tan\theta\approx\theta\text{ when $\theta$ is small}. (20)

This can be seen by expressing the right side of Eq.(19) in the following form:

2r(cosθ2+cosθ1sinθ2sinθ1)\displaystyle\frac{2r}{\left(\dfrac{\cos\theta_{2}+\cos\theta_{1}}{\sin\theta_{2}-\sin\theta_{1}}\right)} =2r(sinθ2sinθ1)cosθ2+cosθ1\displaystyle=2r\frac{(\sin\theta_{2}-\sin\theta_{1})}{\cos\theta_{2}+\cos\theta_{1}}
=2r2cosθ2+θ12sinθ2θ122cosθ2+θ12cosθ2θ12\displaystyle=2r\dfrac{2\cos\frac{\theta_{2}+\theta_{1}}{2}\sin\frac{\theta_{2}-\theta_{1}}{2}}{2\cos\frac{\theta_{2}+\theta_{1}}{2}\cos\frac{\theta_{2}-\theta_{1}}{2}}
=2rtan(θ2θ12)\displaystyle=2r\tan\left(\frac{\theta_{2}-\theta_{1}}{2}\right)
=2rtan(s2s12r).\displaystyle=2r\tan\left(\frac{s_{2}-s_{1}}{2r}\right). (21)

Now, using Eq.(20), we have:

2rtan(s2s12r)2r(s2s12r)=s2s1.2r\tan\left(\frac{s_{2}-s_{1}}{2r}\right)\approx 2r\Big{(}\frac{s_{2}-s_{1}}{2r}\Big{)}=s_{2}-s_{1}.

Thus Eq.(19), and consequently Eq.(18) also, follows from Eq.(20).

7.2 Rationale of the formula

Tantrasaṅgraha does not give any rationale for Eq.(18). The editors of a modern edition of Tantrasaṅgraha have suggested that a geometrical argument could be the method by which Indian astronomers arrived at the result (see [6] pp. 69–70). We present such an argument below which is a slight variant of the argument given in [6].

Refer to caption
Figure 2: Diagram to derive the formula for computing arc lengths

In Figure 2, we assume that s2s1= ¿ PQ s_{2}-s_{1}=\mathchoice{\vbox{ \hbox{\leavevmode\resizebox{15.71458pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\displaystyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{15.71458pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\textstyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{11.0002pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\scriptstyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{7.85728pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\scriptscriptstyle PQ$}}} is small and so we take

s2s1= ¿ PQ =PQ.s_{2}-s_{1}=\mathchoice{\vbox{ \hbox{\leavevmode\resizebox{15.71458pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\displaystyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{15.71458pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\textstyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{11.0002pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\scriptstyle PQ$}}}{\vbox{ \hbox{\leavevmode\resizebox{7.85728pt}{0.0pt}{{\char 62\relax}}} \nointerlineskip\hbox{$\scriptscriptstyle PQ$}}}=PQ. (22)

We shall now compute the length of the chord PQPQ.

Let JJ be the midpoint of the chord PQPQ. We drop perpendiculars PL,QM,JKPL,QM,JK to OAOA and PDPD to QMQM. KK is the midpoint of MLML. From the figure we have:

QD\displaystyle QD =QMPL\displaystyle=QM-PL
=jyā(s2)jyā(s1),\displaystyle=\text{jy\={a}}\,\,(s_{2})-\text{jy\={a}}\,\,(s_{1}),
OK\displaystyle OK =12(OL+OM)\displaystyle=\tfrac{1}{2}(OL+OM)
=12(kojyā(s2)+kojyā(s1)).\displaystyle=\tfrac{1}{2}(\text{kojy\={a}}\,\,(s_{2})+\text{kojy\={a}}\,\,(s_{1})).

Since JJ is the midpoint of PQPQ, OJOJ is perpendicular to PQPQ. It follows that the triangles PQDPQD and JOKJOK are similar. hence:

PQJO=QDOK.\frac{PQ}{JO}=\frac{QD}{OK}.

Using the expressions for QDQD and OKOK, we have:

PQ=2JO(kojyā(s2)+kojyā(s1)jyā(s2)jyā(s1)).PQ=\frac{2\cdot JO}{\left(\dfrac{\text{kojy\={a}}\,\,(s_{2})+\text{kojy\={a}}\,\,(s_{1})}{\text{jy\={a}}\,\,(s_{2})-\text{jy\={a}}\,\,(s_{1})}\right)}.

Now, JOrJO\approx r and so we have

PQ2r(kojyā(s2)+kojyā(s1)jyā(s2)jyā(s1)).PQ\approx\frac{2r}{\left(\dfrac{\text{kojy\={a}}\,\,(s_{2})+\text{kojy\={a}}\,\,(s_{1})}{\text{jy\={a}}\,\,(s_{2})-\text{jy\={a}}\,\,(s_{1})}\right)}. (23)

Combining Eq.(22) and Eq.(23) we get

s2s12r(kojyā(s2)+kojyā(s1)jyā(s2)jyā(s1)).s_{2}-s_{1}\approx\frac{2r}{\left(\dfrac{\text{kojy\={a}}\,\,(s_{2})+\text{kojy\={a}}\,\,(s_{1})}{\text{jy\={a}}\,\,(s_{2})-\text{jy\={a}}\,\,(s_{1})}\right)}.

7.3 Computation of arcsin of large jyā-s

Let ss be an unknown arc whose jyā is known, say, jyā  (s)=m(s)=m. Using the value of jyā  (s)(s), the value of kojyā  (s)(s) can be calculated using the the following result:

kojyā(s)=r2(jyā(s))2=r2m2.\text{kojy\={a}}\,\,(s)=\sqrt{r^{2}-(\text{jy\={a}}\,\,(s))^{2}}=\sqrt{r^{2}-m^{2}}.

Now, let mm be between two consecutive jyā values, say jyā  (s1)(s_{1}) and jyā  (s2)(s_{2}), listed in a table of jyā-s, say Mādhava’s sine table, and let s1<s2s_{1}<s_{2}. Let mm be closer to jyā  (s1)(s_{1}). Using Eq.(18), we have:

ss1\displaystyle s-s_{1} =2r(kojyā (s) + kojyā(s1)jyā(s)jyā(s1))\displaystyle=\frac{2r}{\left(\dfrac{\text{\text{kojy\={a}}\,\,(s) + kojy\={a}}\,\,(s_{1})}{\text{jy\={a}}\,\,(s)-\text{jy\={a}}\,\,(s_{1})}\right)}
=2r(r2m2+kojyā(s1)mjyā(s1))\displaystyle=\frac{2r}{\left(\dfrac{\sqrt{r^{2}-m^{2}}+\text{kojy\={a}}\,\,(s_{1})}{m-\text{jy\={a}}\,\,(s_{1})}\right)}
=p, say,\displaystyle=p,\text{ say},

from which we get s=s1+ps=s_{1}+p. If mm is closer to jyā  (s2)(s_{2}), we have to calculate s2ss_{2}-s and subtract it from s2s_{2} to get ss.

7.3.1 Error estimate

Let the true value of ss1s-s_{1} be pp^{\ast}. Then, by Eq.(21), the approximate value of ss1s-s_{1} is

p=2rtan(p2r).p=2r\tan\big{(}\tfrac{p^{\ast}}{2r}\big{)}.

The error in the approximate value is

|pp|=|2rtan(p2r)p|.|p-p^{\ast}|=\big{|}2r\tan\big{(}\tfrac{p^{\ast}}{2r}\big{)}-p^{\ast}\big{|}.

If we are using Mādhava’s sine table, the maximum value of pp^{\ast} is 225225^{\prime}. So, taking r=3437  44′′  48′′′r=3437^{\prime}\,\,44^{\prime\prime}\,\,48^{\prime\prime\prime}, the maximum error in the computed value of pp^{\ast} is

|2rtan(2252r)225|=4′′  49′′′.\big{|}2r\tan\big{(}\tfrac{225^{\prime}}{2r}\big{)}-225^{\prime}\big{|}=4^{\prime\prime}\,\,49^{\prime\prime\prime}.

7.4 Illustrative example

To illustrate the method with a numerical example, let us calculate arcsin  (3000)=s(3000^{\prime})=s. So, here m=jyā(s)=3000m=\text{jy\={a}}\,\,(s)=3000^{\prime}. We perform all computations in the units of arc-thirds and hence the radius rr will be taken as

r=(34376060+4460+48)′′′=12375888′′′r=(3437\cdot 60\cdot 60+44\cdot 60+48)^{\prime\prime\prime}=12375888^{\prime\prime\prime}

and mm as

m=(30006060)′′′=10800000′′′m=(3000\cdot 60\cdot 60)^{\prime\prime\prime}=10800000^{\prime\prime\prime}

We first calculate kojā  (s)(s):

kojā(s)\displaystyle\text{koj\={a}}\,\,(s) =r2m2\displaystyle=\sqrt{r^{2}-m^{2}}
=123758882108000002\displaystyle=\sqrt{12375888^{2}-10800000^{2}}
=36522603788544\displaystyle=\sqrt{36522603788544}
=6043393′′′ rounded to nearest integer.\displaystyle=6043393^{\prime\prime\prime}\text{ rounded to nearest integer}.

In Mādhava’s sine table, the jyā value 30003000^{\prime} lies between jyā(3600)=2977  10′′  34′′′\text{jy\={a}}\,\,(3600^{\prime})=2977^{\prime}\,\,10^{\prime\prime}\,\,34^{\prime\prime\prime} and jyā(2825)=3083  13′′  17′′′\text{jy\={a}}\,\,(2825^{\prime})=3083^{\prime}\,\,13^{\prime\prime}\,\,17^{\prime\prime\prime} and is closer to jyā(3600)\text{jy\={a}}\,\,(3600^{\prime}). Hence we take s1=3600s_{1}=3600^{\prime} and

jyā(s1)=(29776060+1060+34)′′′=10717834′′′.\text{jy\={a}}\,\,(s_{1})=(2977\cdot 60\cdot 60+10\cdot 60+34)^{\prime\prime\prime}=10717834^{\prime\prime\prime}.

Further, we have:

kojyā(s1)\displaystyle\text{kojy\={a}}\,\,(s_{1}) =kojyā(3600)\displaystyle=\text{kojy\={a}}\,\,(3600^{\prime})
=jyā(54003600)\displaystyle=\text{jy\={a}}\,\,(5400^{\prime}-3600^{\prime})
=jyā(1800)\displaystyle=\text{jy\={a}}\,\,(1800^{\prime})
=(17186060+5260+24)′′′\displaystyle=(1718\cdot 60\cdot 60+52\cdot 60+24)^{\prime\prime\prime}
=6187944′′′.\displaystyle=6187944^{\prime\prime\prime}.

Therefore, we have

p\displaystyle p =ss1\displaystyle=s-s_{1}
=212375888(6043393+6187944900000010717834)\displaystyle=\frac{2\cdot 12375888}{\left(\frac{6043393+6187944}{9000000-10717834}\right)}
=166274′′′,\displaystyle=166274^{\prime\prime\prime},

and hence

s\displaystyle s =s1+p\displaystyle=s_{1}+p
=3600+166274′′′\displaystyle=3600^{\prime}+166274^{\prime\prime\prime}
=3646  11′′  14′′′.\displaystyle=3646^{\prime}\,\,11^{\prime\prime}\,\,14^{\prime\prime\prime}.

This is amazingly accurate because by a modern computation we see that

jyā(3646  11′′  14′′′)\displaystyle\text{jy\={a}}\,\,(3646^{\prime}\,\,11^{\prime\prime}\,\,14^{\prime\prime\prime}) =rsin(3646  11′′  14′′′)\displaystyle=r\sin(3646^{\prime}\,\,11^{\prime\prime}\,\,14^{\prime\prime\prime})
=12375888′′′sin(3646  11′′  14′′′)\displaystyle=12375888^{\prime\prime\prime}\sin(3646^{\prime}\,\,11^{\prime\prime}\,\,14^{\prime\prime\prime})
=10800001′′′\displaystyle=10800001^{\prime\prime\prime}

and the difference of this from the value of mm is only 1′′′1^{\prime\prime\prime}.

8 A novel method for computing the circumference of a circle

Two different methods, both due to Mādhava, for the computation of the circumference of a given circle are well known. One of the methods makes use of the several infinite series expressions for π\pi and the correction terms and the other is an iterative procedure using geometrical ideas. In this section we present a third method for the computation of the circumference. Even though the present method produces only an approximate value of the circumference and is not of much practical significance, we have taken it for presentation for the beauty and elegance of the mathematics involved. (Does this indicate that the astronomers and mathematicians of the Kerala school enjoyed “doing mathematics” for its own sake?) The method and its rationale appears in Yuktibhāṣā (see Gaṇita-Yukti-Bhāṣā Section 7.6; [7] pp. 233-234) and is presented there as an application of the Mādhava-Newton series for jyā. It is also an application of the jīve-paraspara nyāya-s due to Mādhava, the addition and subtraction rules for the sine and cosine functions. Though we are sure that this method is a product of the Kerala school of mathematics, we are not sure who was the original inventor of the method.

8.1 Computation of the circumference of a circle

Refer to caption
Figure 3: Diagram for the description of the third method for the computation of the circumference of a circle

Let CC be the circumference of a circle of diameter DD. Let CC be unknown but an approximation CC^{\ast} to CC be known. The procedure described in Yuktibhāṣā yields a better approximation to CC.

We consider a circle of radius DD and an arc of length C4\frac{C^{\ast}}{4} on this circle (see Figure 3). We calculate a=jyā(C4)a^{\ast}=\text{jy\={a}}\,(\frac{C^{\ast}}{4}) using the Mādhava-Newton series for jyā:

a=(C4)(1D)21123(C4)3+(1D)4112345(C4)5.a^{\ast}=\Big{(}\frac{C^{\ast}}{4}\Big{)}-\Big{(}\frac{1}{D}\Big{)}^{2}\frac{1}{1\cdot 2\cdot 3}\Big{(}\frac{C^{\ast}}{4}\Big{)}^{3}+\Big{(}\frac{1}{D}\Big{)}^{4}\frac{1}{1\cdot 2\cdot 3\cdot 4\cdot 5}\Big{(}\frac{C^{\ast}}{4}\Big{)}^{5}-\cdots. (24)

We next calculate b=kojā(a4)b^{\ast}=\text{koj\={a}}\,(\frac{a^{\ast}}{4}) as follows:

b=D2(a)2.b^{\ast}=\sqrt{D^{2}-(a^{\ast})^{2}}.

The arc C4\frac{C}{4} is one-eighth of the circle of radius DD. Hence we have:111Yuktibhāṣā has given the rationale of these results in an earlier section (see Gaṇita-Yukti-Bhāṣā Section 7.2.2; [7] p. 211).

jyā(C4)=D22,kojyā(C4)=D22.\text{jy\={a}}\,\Big{(}\frac{C}{4}\Big{)}=\sqrt{\frac{D^{2}}{2}},\quad\text{kojy\={a}}\,\Big{(}\frac{C}{4}\Big{)}=\sqrt{\frac{D^{2}}{2}}.

We assume that b>ab^{\ast}>a^{\ast} so that C>CC>C^{\ast}. By jīve-parspara nyāya, we have

Δ\displaystyle\Delta =jyā(C4C4)\displaystyle=\text{jy\={a}}\,\Big{(}\frac{C}{4}-\frac{C^{\ast}}{4}\Big{)}
=1D[jyā(C4)kojyā(C4)kojyā(C4)jyā(C4)]\displaystyle=\frac{1}{D}\left[\text{jy\={a}}\,\Big{(}\frac{C}{4}\Big{)}\text{kojy\={a}}\,\Big{(}\frac{C^{\ast}}{4}\Big{)}-\text{kojy\={a}}\,\Big{(}\frac{C}{4}\Big{)}\text{jy\={a}}\,\Big{(}\frac{C^{\ast}}{4}\Big{)}\right]
=1D[D22bD22a]\displaystyle=\frac{1}{D}\left[\sqrt{\frac{D^{2}}{2}}b^{\ast}-\sqrt{\frac{D^{2}}{2}}a^{\ast}\right]
=(b)22(a)22.\displaystyle=\sqrt{\frac{(b^{\ast})^{2}}{2}}-\sqrt{\frac{(a^{\ast})^{2}}{2}}.

Now, by Eq.(8), we have

C4C4Δ+1D2Δ36=δ(say).\frac{C}{4}-\frac{C^{\ast}}{4}\approx\Delta+\frac{1}{D^{2}}\frac{\Delta^{3}}{6}\\ =\delta\quad\text{(say)}.

Then

CC+4δ.C\approx C^{\ast}+4\delta.

This gives a better approximation to CC than CC^{\ast}. If b<ab^{\ast}<a^{\ast}, then C<CC<C^{\ast} and we will have to consider the difference C4C4\frac{C^{\ast}}{4}-\frac{C}{4} in the above described procedure and CC will be given by C=C4δC=C^{\ast}-4\delta.

DD = 14001400^{\prime}
CC^{\ast} = D×227=4400D\times\frac{22}{7}=4400^{\prime}
C4\frac{C^{\ast}}{4} = 11001100^{\prime}
First term = +C4+\tfrac{C^{\ast}}{4} = +(1100)+(1100^{\prime})
Second term = (1100)(1D)2123(C4)2-(1100^{\prime})(\tfrac{1}{D})^{2}\tfrac{1}{2\cdot 3}(\frac{C^{\ast}}{4})^{2} = (113  10′′  49′′′)-(113^{\prime}\,\,10^{\prime\prime}\,\,49^{\prime\prime\prime})
Third term = +(113  10′′  49′′′)(1D)2145(C4)2+(113^{\prime}\,\,10^{\prime\prime}\,\,49^{\prime\prime\prime})(\tfrac{1}{D})^{2}\tfrac{1}{4\cdot 5}(\frac{C^{\ast}}{4})^{2} = +(3  29′′  37′′′)+(3^{\prime}\,\,29^{\prime\prime}\,\,37^{\prime\prime\prime})
Fourth term = (3  29′′  37′′′)(1D)2167(C4)2-(3^{\prime}\,\,29^{\prime\prime}\,\,37^{\prime\prime\prime})(\tfrac{1}{D})^{2}\tfrac{1}{6\cdot 7}(\tfrac{C^{\ast}}{4})^{2} = (0  3′′  5′′′)-(0^{\prime}\,\,3^{\prime\prime}\,\,5^{\prime\prime\prime})
Fifth term = +(0  3′′  5′′′)(1D)2189(C4)2+(0^{\prime}\,\,3^{\prime\prime}\,\,5^{\prime\prime\prime})(\tfrac{1}{D})^{2}\tfrac{1}{8\cdot 9}(\tfrac{C^{\ast}}{4})^{2} = +(0  0′′  2′′′)+(0^{\prime}\,\,0^{\prime\prime}\,\,2^{\prime\prime\prime})
aa^{\ast} using Eq.(24) = 990  15′′  45′′′990^{\prime}\,\,15^{\prime\prime}\,\,45^{\prime\prime\prime}
(a)2(a^{\ast})^{2} = 980619  49′′  8′′′980619^{\prime}\,\,49^{\prime\prime}\,\,8^{\prime\prime\prime}
(b)2=D2(a)2(b^{\ast})^{2}=D^{2}-(a^{\ast})^{2} = 979380  10′′  52′′′979380^{\prime}\,\,10^{\prime\prime}\,\,52^{\prime\prime\prime}
(a)22\sqrt{\frac{(a^{\ast})^{2}}{2}} = 700  13′′  17′′′700^{\prime}\,\,13^{\prime\prime}\,\,17^{\prime\prime\prime}
(b)22\sqrt{\frac{(b^{\ast})^{2}}{2}} = 699  46′′  43′′′699^{\prime}\,\,46^{\prime\prime}\,\,43^{\prime\prime\prime}
Since a>ba^{\ast}>b^{\ast}, we consider C4C4\frac{C^{\ast}}{4}-\frac{C}{4}.
Δ=jyā(C4C4)=(a)2(b)22\Delta=\text{jy\={a}}(\tfrac{C^{\ast}}{4}-\tfrac{C}{4})=\sqrt{\frac{(a^{\ast})}{2}}-\sqrt{\frac{(b^{\ast})^{2}}{2}} = 0  26′′  24′′′0^{\prime}\,\,26^{\prime\prime}\,\,24^{\prime\prime\prime}
δ=Δ+1D2Δ36\delta=\Delta+\frac{1}{D^{2}}\frac{\Delta^{3}}{6} = 0  26′′  24′′′0^{\prime}\,\,26^{\prime\prime}\,\,24^{\prime\prime\prime}
4δ4\delta = 1  45′′  36′′′1^{\prime}\,\,45^{\prime\prime}\,\,36^{\prime\prime\prime}
C=D4δC=D-4\delta = 4398  14′′  24′′′4398^{\prime}\,\,14^{\prime\prime}\,\,24^{\prime\prime\prime}
Figure 4: Computations

8.2 Illustrative example

Rāmavarma (Maru) Tampurā has illustrated the procedure with the a numerical example. He has used the procedure to get the circumference CC of a circle of diameter D=1400D=1400^{\prime}. As an approximation to CC, he has used the value of the circumference computed using the value 227\frac{22}{7} for π\pi. For details of the computations, see Figure 4.

The value of the circumference CC obtained here is still only an approximate value. The correct value of the circumference is 3.14159265×1400=4398  13′′  47′′′.3.14159265\times 1400^{\prime}=4398^{\prime}\,\,13^{\prime\prime}\,\,47^{\prime\prime\prime}.

References

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