सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

314 that moments I would reckon you as one who could be well compared with the goad that could be applied to the head of the wild elephants of puffed up astronomers. Verses 82, 83. Answer to the first question. Assume the H sine of the Unnatakāla to be Iṣṭa Hṛti in the first place. Multiply it by 12 s and divide by k² the square of the Viṣuvatkarṇa. Then you get an approximate value of H sin δ. Then compute with this, H sin δ, Charajyā etc. and thereby obtain a more correct value of the Iṣṭa Hṛti. Multiply this by the H sin δ got before and divide by the first Hṛti assumed. Then we have a nearer approximation of H sin δ. Repeat the process till a stationary value has been reached. That will be the correct H sin δ. Comm. We know that H sine of the Unnatakāla is nearly the Iṣṭāntyakā. It will be noted that Iṣṭāntyakā is the sum of two H sines namely (1) Carajyā (2) H sine of Unnatakāla minus Chara. The second H sine is called Sūtra or H cos h. (Vide page 278). Thus Sūtra + Carajya = Iṣṭāntyaka whereas H sine (Unnatakāla) = H sine of the Cāpas of Carajyā and Sūtra. In other words H sin (Unnatakāla) = H sin (H sin⁻¹ Carajyā + H sin⁻¹ (Sūtra)). Iṣṭa Hṛti is H cos δ Iṣṭāntyaka × ----------- . But we do not know H sin δ so R that Iṣṭa Hṛti could not be got. So we will not be far from truth in assuming the given Unnatakāla to be Iṣṭa Hṛti itself ie. Taddhṛti here, as the Sun is on the prime- vertical. The formula for Taddhṛti is R² H sin δ R². H sin δ K² H sin δ ----------------- = --------------------- = ------------ H cos ϕ H sin ϕ R. 12 s 12 s ----- × R × --- k k

315 ∴ (Taddhṛti × 12 s) / k² = sin δ. Thus assuming H sine of the given Unnatakāla to be Taddhṛti and multiplying it. by 12 s and dividing by k² we have the value of H sin δ. But this is approximate because the given Unnatakāla is not exactly Taddhṛti but only an approximate value. From this H sin δ, compute H cos δ, Charajyā, Kujyā and through the process indicated in verse 54 namely “Subtract the Characāpa from the Unnatakāla. The H sine of the result is called Sūtra. Multiply the Sūtra by H cos δ and divide by R; then we have Kalā. Add Kujyā to Kalā; we get Iṣṭa-Hṛti”, we obtain a more correct value of Taddhṛti. Then here we may cut short the process as follows namely ‘ If by the assumed Taddhṛti we had the previous H sin δ, what shall we have for this more approximate Taddhṛti ’? The result will be a more approximate value of H sin δ. Again form the Taddhṛti with this H sin δ and so repeating the process till we have a stationary value, we have the correct value of H sin δ. Note. This is a beautiful example of the method of successive approximations which is a modern technique but which was so much in vogue and favourite with the Hindu astronomers. (It will be noted how to cut short the method). Verses 84, 85. Answer to the second question. Obtain 12² R²/(R² − H sin² h) s² + 1 and divide R² by this and take the square root which gives H sin δ. Then (R × H sin δ) / (H sin ω) gives H sin λ whose Cāpa gives the longitude of the Sun. Comm. The H sine of the given Natakāla is H sin h and R² − H sin² h = H cos² h. Let H sin δ be x, which is required to be found. Then R² − x² = H cos² δ; H cos h = Sūtra and

316 (Sūtra × H cos δ) / R = Kalā = (H cos h . H cos δ) / R = ∴ (H cos h . √(R² — x²)) / R But Kalā is the Koti of the fifth latitudinal triangle of which H sin δ is Bhuja. Hence (Kalā × H sin ϕ) / (H cos ϕ) = H sin δ = x ie. (H cos h √(R² — x²)) / R × (H sin ϕ) / (H cos ϕ) = x; but (H sin ϕ) / (H cos ϕ) = s / 12 ∴ Squaring both sides [(R² — H sin² h) (R² — x²)] / R² × s² / 12² = x² ∴ 12² R² x² = s² R² (R² — H sin² h) — s² x² (R² — H sin² h) ie. x² {(12² R² + s² (R² — H sin² h)} = s² R² (R² — H sin² h) ∴ x² = [s² R² (R² — H sin² h)] / [12² R² + s² (R² — H sin² h)] = R² / [(12² R²) / (s² (R² — H sin² h)) + 1] ∴ x = R / √[(12² R²) / (s² (R² — H sin² h)) + 1] = H sin δ as given. From H sin δ, the method of obtaining λ is clear from the formula (H sin λ H sin ω) / R = H sin δ. In the given numerical example h = 5 nādīs = 360° / 12 = 30° since 60 nadis of time correspond to 360°. Thus H sin h = R / 2 ; the remaining work follows. Verse 86. Another question. When the Sun is on the prime-vertical the gnomonic shadow is noted to be 16 inches. The Unnatakāla is 8 nādis. If you could give the H sin δ and s, I shall con- sider you nothing short of one who is an adept in solving the totality of the diurnal problems. Verses 87 and 88. Answer to the question.

317 Here also assume H sin (Unnata) to be the Taddhṛti as formerly done. Then as the shadow is 16″, K̇ = √(16² + 12²) = 20″. Then H cos z = (12 R) / K = 12 / 20 × 3438 Unnatakāla = 8 nāḍīs = 48°. ∴ H sin (48°) = assumed Taddhṛti. Then from the fourth latitudinal triangle, (H sin 48) / (H cos z) = k / 12 ∴ Approximate value of k is (12 H sin 48) / (12/20 × 3438) = (20 H sin 48) / 3438 . This is a known quantity from which s could be computed since 12² + s² = k². Again from the fifth latitudinal triangle s / k = (H sin δ) / (S. S.) . Here s, k and S. S. (Samamandala-Sanku) are known ∴ H sin δ could be got approximately. Thus we have found approximately the required quantities s and H sin δ. From this H sin δ and s we have to compute again H cos δ, Carajyā, Kujyā, etc. and applying the procedure of verse 54 [H sin (Unnata-cāra) × H cos δ] / R + Kujyā = Iṣṭa Hṛti; this is nearer value of Iṣṭa Hṛti than the assumed Taddhṛti. From this again obtain as before s and H sin δ; we could not apply the proportion “ If by the assumed Taddhṛti we have the previous H sin δ, what shall we have for this computed Taddhṛti (Iṣṭa Hṛti) ” for the reason given below. So Repeat the entire process till an invariable quantity is got which will be the correct value of H sin δ. Here repeating the entire calculation is correct and not taking the proportion because Taddhṛti = (R sin δ) / (Sin ϕ cos ϕ) where sin δ and sin ϕ are both to be computed. Formerly we could take the proportion in verse 81 because the latitude of Ujjain being known, in the magni-

318 tude of Taddhṛti namely (R H sin δ) / (Sin ϕ cos ϕ) only H sin δ is variable and Taddhṛti is directly proportional to H sin δ. But in the present example both H sin δ and H sin ϕ are both variables so that, that kind of rule of three does not work. Verse 89. Oh! Mathematician! At a place where s = 5'', there 10 nādikas after Sun-rise the shadow S is observed to be 9''. Tell me what the longitude of the Sun would be, if you are an adept in computing as well as understanding the geometry of the sphere. Verses 90, 91. Answer to the question posed. Assume H sine (Unnatakāla) to be Iṣṭāntyakā. Then (K × H cos z × R) / (12 × I. A.) = H cos δ where I. A. = Iṣṭāntyakā. R² − H cos² δ = H sin² δ ; from this approximate H sin δ and the given s compute a more approximate I. A. Repeat the process till an invariable quantity is obtained for H sin δ, which will be its correct value. From this, using the formula H sin δ = (H sin λ H sin ω) / R , λ could be had. Comm. We know the formula for I. A. as (R³ H cos z) / (H cos φ H cos δ) Assuming Unnatakālajyā as I. A. H sin (Unnatakāla) = (R³ H cos z) / (H cos φ H cos δ) ∴ H cos δ = (R³ H cos z) / ((12 R. I.A.) / k) = (k × R × H cos z) / (12 × I. A.) ; from which obtaining H sin δ and proceeding as indicated we have λ. In the above proof we have used our formula. But Hindu Astronomers proceed from first principles. Let us hear Bhāskara. Since S = 9''; K = √(9² + 12²) = 15''

319 ∴ Mahā-Sanku = H cos z = (R × 12) / K = (3438 × 12) / 15 = 2750 - 24. We know that Mahā-Sanku forms a latitudinal triangle with Iṣṭa Hṛti. So (H cos z) / (Iṣṭa Hṛti) = 12 / k ∴ Iṣṭa Hṛti = (H cos z × k) / 12 ∴ Iṣṭāntyā = (Iṣṭa Hṛti × R) / (H cos δ) ie. H cos δ = (Iṣṭa. Hṛti × R) / Iṣṭāntyā = (H cos z × k × R) / (12 × I. A.) substituting the above value of Iṣṭa Hṛti. Here H cos z is got above and I. A. has been assu- med above as H sin (Unnatakāla). Note. (1) Computing H cos δ = (12 R / 15) × (√(s² + 12²) × R) / (12 × H sin (60)) = (R² × 13) / (15 H sin 60) = (13 R × 2) / (15 × √3). Here H cos δ > R which is invalid. (2) This is the only place where Bhāskara gave a numerical example with a slight flaw. In other words, under the given circumstances the shadow must be greater than what is given. However, the procedure indicated is mathematically correct. (3) It is interesting to note that the flaw was noted by a commentator named Lakṣmīdāsa as reported by Munīśwara in his Marīchi Bhāṣya. Munīśwara also noted the flaw but argues away in an untenable way. Another commentator named Gaṇeśa who was the author of the commentary named Śiromaṇipracāśa, does not seem to have noticed the flaw, or even if he did notice, probably he fought shy of pronouncing that there was a flaw. In fact a simple flaw like this in numerical examples, is not in the least derogatory to the prestige of Bhāskara. So, the commentators who happened to notice the flaw need not have pointed the same.

320 (4) Or again in the given place, for the value of H cos δ to be valid the Unnatakāla x must be such that 15 H sin x > 13 R so that H cos δ might be less than R. This means sin x > 13/15 = .8667 so that x > 60°—4' ; so instead of 10 nādīkas, if the time were given to be just even one Vinādika greater, it would have been alright, or again if the latitude were given to be just a little less it would have been alright. Verse 92. Oh ! Mathematician ! please tell me the magnitudes of the equinoctial shadow and the longitude of the Sun if at a place on a particular day, Kujyā is 245 and Taddhṛti 3125. Verse 93. Answer to the questiou above. s = √[ 144 Kujyā / (Taddhṛti—Kujyā) ] and H sid δ = (12 Kujyā) / s and H sin λ = (R H sin δ) / (H sin ω) . Comm. From the fifth latitudinal triangle compared with third, Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12 (1) (2) (3) (4) Multiplying (1) by (2) Kujyā / (Taddhṛti — Kujyā) = s² / 12² ∴ s = √[ 144 Kujyā / (Taddhṛti — Kujyā) ] Also Equating (1) and (4) Krāntijyā = (12 / s) Kujyā. Verse 94. Given that H sin δ + S. S. + Taddhṛti—Kujyā = 6720, and Kujyā + Agrā + H sin δ = 1960. Then I shall consider him who finds s and the longitude of the Sun as the very Sun illuminating the lotuses of astronomers.

321 Verse 95. Answer to the question above. Divide 12 × Second sum by the first sum, that will be s. Again (12 × Second sum) / (12 + s + k) = H sin δ. From H sin δ, λ could be had as before. Comm. Comparing the third and fifth latitudinal triangles Kujyā / Krāntijyā = Krāntijyā / (Taddhṛti — Kujyā) = Agrā / S. S. = s / 12 (Kujyā + Krāntijyā + Agrā) / (Krāntijyā + Taddhṛti + S. S. — Kujyā) = 1960 / 6720 = 7 / 24 ∴ s = 7/24 × 12 = 7/2 = 3½″. Again comparing the third and the first latitudinal triangles s / Kujyā = 12 / Krāntijyā = k / Agrā = (s + 12 + k) / (Kujyā + Agrā + Krāntijyā) (1) (2) (3) (4) = (s + 12 + k) / 1960 (5) Equating (2) and (5) Krāntijyā = (12 × 1960) / (12 + s + k) = (12 × 1960) / (7/2 + 24/2 + 25/2) = (12 × 1960) / 28 = 840 since when s = 7/2 k = 25/2 which is the hypotenuse of the triangle formed by the equinoctial shadow with the gnomon. Equating (1) and (5) Kujyā = 245 ; equating (3) and (5) Agrā = 875. Now from the fourth latitudinal triangle compared with the first Agrā / s = S. S. / 12 = Taddhṛti / k (1) (2) (3) 41

322 From (1) and (2) S. S. = 12 / (7/2) × Agrā = 24/7 × 875 = 3000 From (1) and (3) Taddhṛti = k / s × Agrā = 25/2 × 2/7 × 875 = 3125. Note. This is a beautiful example exhibiting Bhās- kara's dexterity in algebra. Verse 96. Given that the sum of H sin δ, S. S. and Taddhṛti – Kujyā = 1440, and the sum of Agrā, S. S. and Taddhṛti = 800, I shall deem him whoever finds s and the longitude of the Sun, as the very Sun illuminating the lotuses of astronomers. Verse 97. Answer to the problem above. The second sum divided by the first and multiplied by 12 gives k from which s could be got. Then the first sum divided by s + 12 + k̄ gives H sin δ from which the longitude of the Sun could be got. Comm. Comparing the third and the fifth latitudinal triangles, we have Agrā / Krāntijyā = S. S. / (Taddhṛti – Kujyā) = Taddhṛti / S. S. (1) (2) (3) = k / 12 = (Agtā + S. S. + Taddhṛti) / (Krāntijyā + Taddhṛti – Kujyā + S. S.) = 1800 / 1440 = 5 / 4 I (4) (5) (6) Equating (4) and (6) k = (12 × 5) / 4 = 15 ∴ k² = 225 = 12² + s² ∴ s = 9. Again comparing the fourth latitudinal triangle, with the fundamental,

328 Agrā / s = S. S. / 12 = Taddhṛti / k = (Agrā + S. S. + Taddhṛti) / (s + 12 + k) (1) (2) (3) = 1800 / (9 + 12 + 15) = 1800 / 36 = 50 II (4) Equating (1) and (4) Agrā = 9 × 50 = 450 Equating (2) and (4) S. S. = 12 × 50 = 600 Thirdly Taddhṛti = 15 × 50 = 750 Again Equating (1) and (6) of I Agrā / Krāntijyā = 5 / 4 = 450 / Krāntijyā ∴ H sin δ = (450 × 4) / 5 = 360 from which λ could be computed. Verse 98. The chara at a place where s = 9, is equal to 3 nādis. If you could compute the longitude of the Sun, then certainly you are a leader among astronomers, Oh ! Scholar ! Verse 99. Answer to the problem above. 12 Carajyā / √((12 × Carajyā / R)² + s²) = H sin δ where from λ the longitude of the Sun could be computed. Comm. Let H sin δ = x; then from the third lati- tudinal triangle Kujyā / Krāntijyā = s / 12 = 9 / 12 = 3 / 4 ∴ Kujyā = 3 x / 4 since Krāntijyā means H sin δ ∴ Carajyā = 3 x / 4 × R / (H cos δ) = 3 R x / (4 √(R² - x²)) = H sin (3 × 6) = H sin 18° ∴ Squaring 9 R² x² = 16 (R² - x²) H sin² 18 = 16 Carajyā² (R² - x²)

324 ∴ x² (9 R² + 16 Carajyā²) = 16 R² Carajyā² ∴ x² = (16 R² Carajyā²) / (9 R² + 16 Carajyā²) ∴ x = (4 R Carajyā) / √(9 R² + 16 Carajyā²) = (12 Carajyā) / √(81 + (12² Carajyā² / R²)) = (12 Carajyā) / √(9² + (12 Carajyā / R)²) Here Carajyā being known, H sin δ could be computed. Verse 100. If you studied what is known as Madh- yamāharaṇa, then compute λ the longitude of the Sun given that H sin δ + H cos δ + H sin λ = 5000. Verse 101. Answer to the problem above. Let the given sum multiplied by 4 and divided by 15 be Ādya ; then H sin δ = Ādya − √(910678 − (2 square of the given sum / 337)) . Comm. Let H sin δ = x ; then H cos δ = √(R² − x²) and since H sin δ = (H sin ω H sin λ) / R ∴ H sin λ = (x R) / (H sin ω) = (x R) / 1397 ∴ The given sum = x + √(R² − x²) + (x R / 1397) = 5000 ∴ √(R² − x²) = 5000 − x (1 + R / 1397) = 5000 − (4835 / 1397) x ∴ R² − x² = 5000² + x² (4835 / 1397)² − (2 × 5000 × 4835 / 1397) x ∴ x² {1 + 4835² / 1397²} − (2 × 5000 × 4835 / 1397) = R² − 5000²

325 ie. x² (1397² + 4835²) - 2 × 4835 × 1397 × 5000 x = 1397² (R² - 5000²) ie. x² (25328834) - 2 × 4835 × 1397 × 5000 x = 1397² (R² - 5000²) ∴ x² - (2 × 4835 × 1397 × 5000 x) / 25328834 = (1397² (R² - 5000²)) / 25328834 ∴ x² - 2 × 5000 x × 6754495 / 25328834 = ,, Converting 675 / 2533 into a continued fraction we have 1/(3 +) 1/(1 +) 1/3 = 4 / 15 so that the equation could be written as x² - 2 × 5000 (x 4) / 15 = (1397² (R² - 5000²)) / 25328834 Here (5000 × 4) / 15 is symbolized as Ādya so that we have x² - 2 Ādya x = (1397² (3438² - 5000²)) / 2532883 ∴ (x - Ādya)² = Ādya² + (1397² (3438² - 5000²)) / 2532883 = 5000² × 16 / 225 - (5000² × 1397²) / 2532883 + (1397² × 3438²) / 2532883 = 5000² (16 / 225 - 1397² / 2532883) + (1397² × 3438²) / 2532883 Here 16 / 225 - 1397² / 2532883 is approximated to -2 / 337 and (1397² × 3438²) / 2532883 is approximated to 910678 so that we have x = Ādya ± √(910678 - (2 s²) / 337) where s is the given sum. Since the positive sign of the radical is invalid because H sin δ ≯ R, so the negative sign is taken.

326 Verse 102. In a place where s = 5″, the sum of H sin δ, S. S., Taddhṛti, Kujyā and Agrā is 6500 ; find them individually oh, mathematician, if thou art adept in understanding the sphere and dealing with the latitudinal triangles. Verse 103. Answer to the problem above. Assuming H sin δ to be equal to 12 s and computing the various quantities cited ; take their sum. Then by rule of three “ If for this sum got, the individual magni- tudes are such and such what will they be for the given sum ” each can be had. Comm. The cited magnitudes are respectively H sin δ, (R H sin δ) / (H sin ϕ), (R² H sin δ) / (H sin ϕ H sin ϕ), (H sin δ H sin ϕ) / (H sin φ) and (R H sin δ) / (H cos ϕ) which are all proportional to H sin δ, ϕ being given through ‘ s ’. With this idea of proportionality at the back of his mind, Bhāskara sets this ingenious ques- tion, and gives an easy way of solving it by assuming H sin δ to be 5 × 12 = 60, so that the others can be got rationally. With this H sin δ, S. S. (3438 × 60) / (3438 × 5/13) = 156, Taddhṛti = (3438² × 60) / (3438 × 5/13 × 3438 × 12/13) = 169 ; Kujyā = (60 × 5) / (13 × 12/13) = 25 Agrā = (3438 × 60) / (3438 × 12/13) = 65 The sum of these is 475. So, by the rule of three men- tioned above, H sin δ = 1200, S. S. = 3120, Taddhṛti = 3380, Kujyā = 500 and Agrā = 1300. Or alternatively given s = 5, k = 13 so that H sin ϕ = (3438 × 5) / 13, H cos ϕ = (3438 × 12) / 13. Hence the values of

327 the various magnitudes are H sin δ, (H sin δ × 13) / 5 , (H sin δ × 13²) / 60 , (H sin δ × 5) / 12 and (H sin δ × 13) / 12 The sum of these is H sin δ (1 + 13/5 + 169/60 + 5/12 + 13/12) = H sin δ ((60 + 156 + 169 + 25 + 65) / 60) = 475/60 H sin δ = 95/12 H sin δ = 9500 ∴ H sin δ = 1200 from which by substi- tution the remaining magnitudes could be obtained. Verse 104. If the sum of Agrā, H sin δ and Kujyā be 2000 find them individually, oh ! mathematician if thou be an adept in the geometry of the sphere and compu- tation. Comm. Here the quantities are respectively (R H sin δ) / (H cos φ), H sin δ and (H sin δ H sin φ) / (H cos φ) so that their sum is H sin δ (1 + R / (H cos φ) + (H sin φ) / (H cos φ)) = (H sin δ (H sin φ + H cos φ + R)) / (H cos φ) Here also we are to presume s = 5 so that the above sum is (H sin δ (s + 12 + k)) / 12 (by proportion of the first and second latitudinal triangles) = (H sin δ (5 + 12 + 13)) / 12 = 5/2 H sin δ = 2000 ∴ H sin δ = 800. Substituting this value in the above formula, Agrā = (RH sin δ) / (H cos φ) = (3438 × 800 × 13) / (3438 × 12) = 10400 / 12 = 866-40 ; Kujyā = (H sin δ H sin φ) / (H cos φ) = (800 × 5) / 12 = 4000 / 12 = 333-20

328 OG = Gnomon ; OP = Shadow of the gnomon ; PM = the Bhuja drawn from the extremity of the shadow P perpendicular on the East-west line ; OM = Koti of the shadow extending along the East-west line. AB is the Nalaka placed along the Chayakarṇa PG. The eye is placed at A and the planet ☉ is visible through the tube of the Nalaka AB. Fig. 62 Verses 105, 106 and 107. The method of observing through the instrument called Nalaka, the planetary position. On a horizontal plane mark a point and through it draw the East-west line and also the North-south ; if the planet is in the East mark off the computed Koti of the shadow towards on the East-west line; if the planet is in the Western hemisphere, mark this Koṭi towards the East. From the extremity of the Koti mark the computed Bhuja perpendicular to the East-west line and draw the computed shadow from the point so as to form a right-angled

329 triangle with the Bhuja and Koti. Extend a thread from the point of intersection of the bhuja and shadow to meet the gnomon's top so as to form the Chāyākarṇa or the hypotenuse of the right-angled triangle of which the other sides are the gnomon and the shadow. Along this thread place the Nalaka such that the lower extremity of the Nalaka coincides with the eye. Seeing through the Nalaka, the planet is to be seen. I shall tell how the planet could be seen in water as well. Comm. The Nalaka is a simple tube formed generally of bamboo. The purpose of this is to verify the correct- ness of the computation of the shadow and its bhuja. If the computation is wrong the planet will not be seen in that direction. It might be asked how the shadow and bhuja are pertinent with respect to a planet, whose shadow cannot be observed as that of the Sun. True, but the computation of the shadow and bhuja are done as will be done with respect to the Sun, knowing the declination etc. as in the case of the Sun. Computation does not depend on the observation of the actual shadow. Computing the magnitudes of the Bhuja and Koti, the direction of the Chāyākarṇa points to the planet in the sky. Verse 108. Observing the planet through the Nalaka in water. Fig. 63 42

330 Place the Śaṅku at the point of intersection of the Bhuja and shadow and holding the Nalaka along the join of the top of Śaṅku and the point, the planet could be seen in a basin of water placed at the point. Comm. Let P be the planet casting the shadow AC of the gnomon AB. C the extremity of the shadow is the point of intersection of the shadow and the Bhuja. Though we have shown the gnomon in the position AB, it need not have been placed there in as much as we have the computed magnitudes of the shadow, the Bhuja and the Koti. Now we are directed to place the Śaṅku act- ually at C the point of intersection of the shadow and the Bhuja. Thus CD is the Śaṅku. Since CD=AB and and both are vertical evidently Δs DBA and DCB are congruent. Hence DĈB = DÂB. But DĈB = zenith- distance of the planet and as such is equal to BÂD (also the zenith-distance of the planet) ∴ DÂB = BÂP. Hence if a tray of water is placed at A, the planet will be visible as seen through DE, the Nalaka since the angle DAB is the angle of incidence and BÂP the angle of reflection are equal. Verse 109. The planet is to be shown to the king, who has an eye of appreciation for the same, either direct- ly (as shown in fig. 62) in the sky or through water as shown in the fig. 63, having finished the preliminaries indicated. Comm. Clear. End of the Tripraśnādhyāya.

PARVASAMBHAVĀDHIKĀRA Investigation into the occurence of an eclipse Verses 1–2. Multiply the number of years that have elapsed from the beginning of the Kaliyuga by twelve and add the number of months elapsed from the beginning of the luni-solar year. Let the result be x. Then add [2 x (1 - 1/898)] / 65 to x. Let the result by y. Then the longi- tude of what is called Sapāta-Sūrya or the longitude of the Sun with respect to a node will be x Rasis + [(2 y + 503) (1 + 1/169)] / (3 × 30) Rasis. If this longitude be less than 14°, then a lunar eclipse is likely to occur. Comm. The first operation indicated above in direct- ing x to be added to [2 x (1 - 1/898)] / 65 is intended to obtain the lunations that have elapsed from the beginning of the Kaliyuga. In this behalf we are asked to multiply the elapsed years by twelve to get the number of solar months. Here there is one subtlety to be noticed. The years that have elapsed are not entirely solar. In fact the years reckoned according to the luni-solar system were all originally luni-solar; but according to the convention of intercalary months, they were rendered solar upto the point of the latest intercalation, for, solar months plus intercalary months are equal to the elapsed lunations. From the moment of the end of the latest intercalary month, the subsequent years or year or fraction thereof would be luni-solar only. Nonetheless, no difference will be there in the computed Adhikamāsas in adding a few lunar months to the solar and taking them all to be solar. The maximum error committed in so doing will be of the order of (no. of days in a solar month minus no.

of days in a lunar month) multiplied by 36 × ²/₆₅ × ¹/₃₀ of an adhikamāsa, assuming that an adhikamāsa would occur at the latest in 36 solar months. (In fact, an adhikamāsa would occur on the average in 32½ solar months, but we have taken 36 roughly as the maximum figure in as much as the occurence of the Adhikamāsa might be belated on account of the convention stipulated). Thus the error would be 36 × 2 × ²/₆₅ × ¹/₃₀ = ¹/₁₃th of an adhikamāsa at the maximum. Hence, we are directed not only to construe that all the years elapsed to be solar but also the subsequent lunations of the current luni-solar year also to be solar months. Thus getting the number of elapsed months from the beginning of the Kaliyuga, the computation of the Adhikamāsas is formulated as follows. If in the course of 51840000 solar months of the Yuga there be 1593300 Adhikamāsas then during the elapsed solar months x, what is the number of elapsed Adhika- māsas? The result is x × 1593300 x × 1593300 ----------- = --------------- 51840000 796650

51840000

796650 = 2 × x --------- . Since Bhāskara knows that there will be two 65-4-21 Adhikamāsas roughly in 65 solar months, he performed the above operation. This shows that for every 65 solar months roughly there occur two Adhikamāsas or more accurately a little less than two Adhikamās. So, taking, in the first instance 2/65 as the ratio of Adhikamāsas to the number of solar months, Bhāskara tries to find as to what quantity is to be subtracted from 2. That is found as follows. If there be A adhikamāsas in s solar months what will be the number of Adhikamāsas in x solar months? The result is A x --- . Again if there be two Adhikamāsas roughly in 65 s solar months, how many will be there in x solar months? 2 x The answer is --- . But we have seen about that the 65

accurate number should be (2 x / 65) - λ ie. a little less than (2 x / 65) . The question is now to find the value of λ. So, equating (2 x / 65) - λ to (A x / s), λ = (2 x / 65) - (A x / s) = x ((2 / 65) - (A / s)) = x ((2 s - 65 A) / (65 s)). Substituting for 2 s - 65 A namely 2 × 51840000 - 65 × 1593300 = 115500 λ = (x × 115500) / (65 × 51840000) = (x × 2 × 57750) / (65 × 51840000) = (2 x / 65) × 1 / (51840000 / 57750) = (2 x) / (65 × 898) ∴ (A x) / s = (2 x / 65) - λ = (2 x / 65) - (2 x) / (65 × 898) = (2 x / 65) (1 - 1 / 898) as given. The procedure, adopted as above, is in a way a short cut in Hindu Astronomy to obtaining a convenient con- vergent to a continued fraction. Let us use the method of continued fractions; the number of Adhikamāsas in x solar months is (A x / s) ie. x × A/s = (x × 1593300) / 51840000 = (x × 5311) / 172800 . Converting 172800 / 5311 into a continued fraction we have 32 + 1/(1+) 1/(1+) 1/(6+) 1/2 + 1/(1+) 1/(1+) 1/(18+) 1/4 to which 65/2 is a convergent but a good convergent is 245 / 69 . As this good convergent is unwieldy, Bhāskara used 2/65 and made amends for the roughness introduced by adopting it. Wherever a con- venient convergent is not available, an easy and rough convergent is used and amends will be made for the rough-