पञ्चसिद्धान्तिका (आचार्य वराहमिहिर - सूर्य, रोमक, पौलिश, वासिष्ठ एवं पितामह सिद्धान्त)

14 PAÑCASIDDHĀNTIKĀ I.13 should be included for greater accuracy and it can be done by an appropriate addition in the S-days, by the proportion: If 10/9761 intercalary month is got for one S-day, by how many S-days is (1 + 1/550)/9761 intercalary month got? Thus we get S-days equal to, (1 + 1/550)/9761 ÷ 10/9761 = (1 + 1/550)/10 = 1/10 + 1/10 × 1/550 . This is for every 107 years, and so, for every 107 years, 1/10 S-day has to be added for greater accuracy in getting the intercalary months and for every 550 such addi- tions one more tenth is to be added, which is the instruction given. (This is the reason for our giving as the correct reading, 'tithidaśamāṁśam where tithi according to the context means S-day). Now we proceed to explain the part of the formula relating to the elided days. We got before that there are 11,40,37,61,190 elided days in a period of 7,28,80,32,70,590 lunar tithis or simply tithis. Cancelling out a factor 30, we have 38,01,25,373 elided days for 24,29,34,42,353 tithis. So, to obtain the elided days for tithis gone we have the proportion, 24,29,34,42,353: 38,01,25,373 :: tithis gone: elided days during the period, i.e. elided days = tithis gone × 38,01,25,373 ÷ 24,29,34,42,353. The multiplying fraction 38,01,25,373/24,29,34,42,353 can be expressed as a continued fraction thus: 1 38,01,25,373 24,29,34,42,353 63 1 3,45,81,519 34,55,43,854 9 2,71,336 3,43,10,183 126 .... .... 38,01,25,373/24,29,34,42,353 = 1/(63+) 1/(1+) 1/(9+) 1/(1+) 1/(126+) ......... The successive convergents are 1/63, 1/64, 10/639, 11/703, 1396/89217 etc. Of these, our author has taken 11/703 (note that this is the same as that of the Romaka) as being enough for a first approx- imation. By taking this, 38,01,25,373/24,29,34,42,353 − 11/703 = 2,71,336/(24,29,34,42,353 × 703) elided day is left out for every tithi. In the period of 245 years, given in the rule, there are, from the constants given before, 7,28,80,32,70,590 × 245 ÷ 1,96,40,88,000 tithis. So in this period the left out elided day is {2,71,336/24,29,34,42,353 × 703} × {72,88,03,70,590 × 245 ÷ 1,96,40,88,000} = 16,61,933/(16,36,740 × 703). This can be included in the formula by making a proportionate change in the tithi thus: To get 11 elided days we have to take 703 tithis, to get the elided days left out in 245 years, we must take tithis equal to 703 × 16,61,933 ÷ (16,36,740 × 703 × 11) = 16,61,933 ÷ (16,36,740 × 11) = (1 + 25,193/16,36,740)/11 = 1/11 + 25,193/(16,36,740 × 11). In this the first term 1/11 is given by the instruction to add an eleventh of a tithi every 245 years. The second term does not agree with the instruction to omit adding one eleventh for every addition of 2,03,279 elevenths. This may be due to several reasons. It may be that the mean motion for 3031 days is given to the nearest minute, and small as this is, it can affect the value of the correction which itself is very very small. Or the Paulīśa Moon is slightly different from the Vāsiṣṭha Moon, which we have assumed for the Paulīśa. Or there is some error in the text here. We must be satisfied with the other and more important items of agreement. It must be remembered here that TS have omitted even the translation of these two verses, as a hopeless task.