EXAMPLES.-1. Given the series 1, 4, 8, 13, 19, 26, &c. to find the several orders of differences. Thus 1, 4, 8, 13, 19, 26, &c. the given series. where the work evidently must terminate. 2. Given the series 1, 4, 8, 16, 32, 64, 128, &c. to find the several orders of differences. Here 1, 4, 8, 16, 32, 64, 128, &c. given series. And... 3. Find the several orders of differences in the series 1, 2, 3, 4, &c. Ans. First differences 1, 1, 1, 1, &c. Second diff. 0, 0, 0, &c. 4. To find the several orders of differences in the series 1, 4, 9, 16, 25, &c. Ans. First differences 3, 5, 7, 9, &c. Second. 2, 2, 2, &c. Third 0, 0, &c. 5. Required the orders of differences in the series 1, 8, 27, 64, 125, &c. 6. Given 1, 6, 20, 50, 105, &c. to find the several orders of differences. 7. Given the series 1, 3, 7, 13, 21, &c. to find the third and fourth orders of differences. 15. To find any term of a given series. RULE I. Let a, b, c, d, e, &c. be the given series; d1, d11, d111, di, &c. respectively, the first term of the first, second, third, fourth, &c. order of differences, as found by the preceding article; n=the number denoting the place of the term required. If the differences be very great, the logarithms of the quantities may be used, the differences of which will be much smaller than those of the quantities themselves; and at the close of the operation the natural number answering to the logarithmical result will be the answer. See Emerson's Differential Method, prop. 1. EXAMPLES.—1. To find the 10th term of the series 2, 5, 9, 14, 20, &c. Here (Art. 12.) 2, 5, 9, 14, 20, &c. series. Where d1=3, d11=1, d111=0, also a=2, n=10; wherefore 1=) 2+27+36=65=the 10th term required. 2. To find the 20th term of the series 2, 6, 12, 20, 30, &c. Here a=2, n=20; and Art. 12. 2, 6, 12, 20, 30, &c. series. 어떻 .d + 4, 6, 8, 10, &c. 1st diff. 2, 2, 2, &c, 2nd diff. or d'=4, d11=2; whence n-1n-2 1 2 19 19 18 · d2 = (2 + = x 4 + = = = = +342=420=the 20th term required. 3. Required the 5th term of the series 1, 3, 6, 10, &c. Ans. 15. 4. To find the 10th term of the series 1, 4, 8, 13, 19, &c. Ans. 64. 5. To find the 14th term of the series 3, 7, 12, 18, 25, &c. Ans. 133. 6. Required the 20th term of the series 1, 8, 27, 64, 125, . &c. Ans. 8000. 7. To find the 50th term of 1, 4, 8, 13, 19, &c. 8. To find the 10th term of 3, 7, 12, 18, 25, &c. 16. If the succeeding terms of a given series be at an unit's distance from each other, any intermediate term may be found by interpolation, as follows. • For the investigation of this rule, see Emerson's Differential Methods prop. 2. RULE I. Let y be the term to be interpolated, x its distance from the beginning of the series, d1, d11, d111, div, &c. the first terms of the several orders of differences. EXAMPLES.-1. Given the logarithms of 105, 106, 107, 108, and 109, to find the logarithm of 107.5. 5 Here x=(107.5-105=2.5) the distance of the term y, a=.0211893, d1=41166, d11=—387, d111= 0211893+102915-725-2.5+.078=.031407128, the logarithm required. 2. Given the logarithmic sines of 3o 41, 3° 51, 3o 61, 3o 71, and 3o 81, to find the sine of 3o 61 1511. 9 4 Here x=(3o 61 1511—3°41=2° 151=)—_—— the distance of the to be interpolated ; a=8.7283366, d'=23516, d11——126, This rule is investigated in Emerson's Differential Method, prop. 5. d1+=d11+ d111=)8.7283366+.0052911—.00001771875+ 45 15 32 128 .0000000117=8.73360999296, the log. sine required. 4. Given the logarithmic sines of 1° 0', 1o 1', 1o 2', and 1o 3', to find the logarithmic sine of 1o 11 4011, Ans. 8.2537533, 17. If the first differences of a series of equidifferent terms be small, any intermediate term may be found by interpolation, as follows. RULE I. Let a, b, c, d, e, &c. represent the given series, and n=the number of terms given. n-l II. Then will a−nb+n. C-n. n-in- -2 n-1n-2 -.d+n. n-3 4 2 3 -.e+, &c.=o, from whence, by transposition, &c. any re quired term may be obtained §. EXAMPLES.-1. Given the square root of 10, 11, 12, 13, and 15, to find the square root of 14. Here n=5, and e is the term required. a=(/10=)3.1622776 b=(/11=)3.3166248 c={√/12=)3.4641016 d=(✓✓13=)3.6055512 f=(15)3.8729833 And since n=5, the series must be continued to 6 terms. For the investigation of this rule, see Emerson's Differential Method, prop. 6. Whence, by transposition, in order to find e, we shall have ·2 n-3 n-4 2 3 4 -f; this in numbers becomes 5 c=-3.1622776 +5x3.3166248-10 x 3.4641016+10 × 3.6055512+3.8729833 =56.5116193-37.8032936=18.7083257, and e= 3.74166514 the root, nearly. 18.7083257 5 2. Given the square roots of 37, 38, 39, 41, and 42, to find the square root of 40. Ans. 6.32455532. 3. Given the cube roots of 45, 46, 47, 48, and 49, to find the cube root of 50. Ans. 3.684033. 4. Given the logarithms of 108, 109, 110, 111, 112, and 114, to find the logarithm of 113. Ans. 2.0530784. 18. To revert a given series. When the powers of an unknown quantity are contained in the terms of a series, the finding the value of the unknown quantity in another series, which involves the powers of the quantity to which the given series is equal, and known quantities only, is called reverting the series". RULE I. Assume a series for the value of the unknown quantity, of the same form with the series which is required to be reverted. II. Substitute this series and its powers, for the unknown quantity and its powers, in the given series. III. Make the resulting terms equal to the corresponding terms of the given series, whence the values of the assumed coefficients will be obtained. EXAMPLES.-1. Let ax + bx2+cx3 +dx++, &c.=z be given, to find the value of x in terms of z and known quantities, h Various methods of reversion may be seen, as given by Demoivre, in the Philosophical Transactions, No. 240. in Maclaurin's Algebra, p. 263, &c. Colson's Comment on Newton's Fluxions, p. 219; Horsley's Ed. of Newton's Works, vol. I. p. 291, &c. Stuart's Explanation of Newton's Analysis, p. 455. Simpson's Fluxions, &c. &c. |