has been put as the function f, and the values of ƒ (a+no) have been computed for a—3 w, a—2 ∞, a, a+w, a + 2 w, &c......a +233 w, a +234 w, in which a = January 0, 1810, += January 21, 1810. It will be observed that the constants C, C'-, have taken the place in the table of 'f (a) and "f(a — 1); in order to obtain It is not necessary to go farther back than ƒ"", because the higher differences are insensible, and the terms are wanting in the beginning of the series to obtain them, except when two computations made from different elements meet in this place. By means of the above values, C_ = +0.02493 0.00004 = + 0.02489 and = a + (8 + 1 ∞), whence :+ 0.00007. 1 4M=+137.93477 is the amount of M from January 0, 1810, to January 13, 1811. Again, the argument for July 3, 1836, is (a +230 ∞), and "ƒ (a+230) 4821.8356 We may employ the same functions to furnish an example of the single integral, remembering, as before mentioned, that when a single integral is derived from a table of functions comprising the factor o2, it must be divided by @. 424μ25.34094 4μ 0.603356 from January 0, 1810, to Jan Δμ = And for the argument (a + 232 ∞) we have f(a+232 ∞): 232)= f(a+232 ∞) 23.7706 ) = — 0.0131.2 in which is a whole number, has been introduced, it may be well to notice that the numerical coefficients of the different series are changed from +1, + 24, − złšo, to + 1, − 12, +2. Should we desire to take immediately from the table the first and second integrals for any other time, as, for example, for February 10, 1811, or for the argument a +(9+), we must interpolate for 42 μ as follows: ; 'ƒ (a + 9+ })='ƒ (a+9+1 − ƒ' (a + 9 + 1) = ƒ' (a + 9 + 1 − 1) ) = + 26.09809 }) = − 0.29740 ƒ''"' (a + 9 + }) = ƒ'""' (a + 9 + 1 − 1) = — 0.00175; and for 4 M, "ƒ (a + 9 + 1) = + 154.99961 f(a+9+b)= 1.01754 ƒ" (a + 9 + }) = 0.00636, from which we should obtain and 424μ26.09809 0.01239.2 0.00000 . 5 4μ0.621088 from Jan. 0, 1810, to Feb. 10, 1811; = 0.00002.6 +155.08438 from Jan. 0, 1810, to Feb. 10, 1811, agreeing with the interpolation from the previous values. When the amount of the remaining perturbations is computed, the actual value of μ for January 0, 1810, is lastly to be added to 4, and so also the value which M would each time is to be added to 4 M. Mo, and the form of the last will be only be an unnecessary addition to place of the constant C, the value of C the value C++ Mo, for the purpose of obtaining the whole result immediately from the table. have without the perturbations at Моя Finally, the connection of the two parts of the above table may be shown by computing the value of the formula M=L―π+ μ (t − T) according to the precept. And by the elements answering to the date June 12, 1836. M 319 13 π 250 7 56.4 L= M+π= 209 20 58.2 ON THE DEPENDENCE OF NAPIER'S RULES. By Rev. ANTHONY VALLAS, Phil. Dr., New Orleans, La. THE five parts of a right-angled spherical triangle being placed on the circumference of a circle, NAPIER'S Rules are as follows: RULE I. The sine of the middle part equals the product of the cosines of the opposite parts. RULE II. The sine of the middle part is equal to the product of the tangents of the adjacent parts. It must be remembered that, instead of the hypothenuse and the two acute angles, their complements are used, the right angle not being counted among the parts. That the second of these rules may be deduced from the first has been shown by Mr. SAFFORD, in No. I. Vol. I. We are going to show the same in the following way. Apply the first rule to the three parts a, a, a, and then to a, a', w. Eliminate one of the common parts, for instance a, and the result, properly transformed, will give the second rule. From ', In order to find the second rule from the first, take the three adjacent parts a, μ, a', and μ, a', a', and eliminate u from the two equations Replacing the tangents by their equivalents in cosines, we have |