EXAMPLES. 1. To develop the roots of +2x=932 x 1. A2 Comparing the equation to (6'), we have n = 2, a = 2, b = 0, c = 0, etc., y'" = y'2 = 32 × 1, or, since y'2= y2 × A2, we put y 32 and A2 = 1; consequently, the positive root of y is 3, and the algebraic roots of A= 1 give A = ± 1, and we shall have y± 3. 1 = Substituting the preceding values in (b), we get x=y'−1+ etc., and putting ± 3 for y' in this, we 1 + 1 2y' 8y's 16y's have 1 1 1 + etc.3-1± 0.16666.0.004629, 0.000257 etc. Hence, using the upper signs in the ambiguous signs (±, F), in the preceding series, we get x = 2.1622; and using the lower signs, we have x = - 4.1622. These roots, which are correct to four decimal places, have been obtained by using only five terms of the preceding series. 2. To find z, from the equation 22 - 700259829. To simplify the equation, we put 2100, and thence get a-7x=5.9829, or, multiplying by a-1, and transposing, we have x 5.9829x-17. In this equation we have n = 1, a = 0, b = 5.9829, c= 0, d= 0, etc., and y' = y = 7. Hence, by substituting the preceding values in (c), we get x = 7+ 2(5.9829) 5(5.9829)* 75 77 5.9829 7 2 + etc. = 7 +0.8547-0.10435887 + 0.0254844 0.0074875 + etc. = = 7.768308 +, when we use only five terms of the series; consequently, we have 2 = 100x776.8308+. The preceding value of z is correct to two places of figures, and nearly to three places, for the correct value is z=777. (See Young's Theory of Equations, p. 103.) It may be added, that the remaining root of the equation is easily found to be 77. 3. To find a from the equation a3-3a2 5. Multiplying the terms by - and transposing, the equation becomes -5x-23; which gives a = 0, b = 0, c = — 5, d = 0, e = 0, etc., y' = y = 3 and n = 1. Substituting the preceding values in (c), we get x = 3 + 5 50 875 9 etc. + - etc. 30.5555.0.2058 +0.1333. — 3.42., correct to two decimal places nearly. x 4. To find a from the equation — 10x3 +6x+1=0. Multiplying by x-3 and transposing, the equation becomes x2 + 6x2 + x¬3 = 10; which gives n = 2, a = 0, b = 0, c = 0, d = 6, e = 1, ƒ = 0, g=0, etc., y = 10 and A2= 1, or y= √10 and A = ± 1 and y' = ± √10. 3 Hence, √10 10 1/10 200 8 103/10 2 3.162277. F 0.094868. - 0.005 0.0071151. Taking the upper signs in the ambiguous signs, we get x 3.05, correct to two decimal places; and using the lower signs we have x = — 3.06, which is also correct to two decimal places. 5. To find a from the equation x-3x2 + 75x = 10000. Here we have n = 4, a = 0, b = 3, c=75, d = 0, e = 0, etc., also, y' 10000 10 x 1 = y × A, or y = 10* and A 1, so that y= 10, A = ± 1, and A = ± √−1; consequently, y' has the four values expressed by y' = ± 10 and y'10 -1. 9 By the substitution of the preceding values in (c), we get 22473 675 270405 + 128y' 32y' + 10 for y', we get x=10±0.075-0.1875 0.00175570 + 0.00002109 0.00001320 Hence, we get x - 10.26099 for the ap 9.8860 and x = proximate values of two of the roots; the first being correct to four, and the second to three decimal places. Again, by putting ± 10-1 for y', we get x = ± 10 √−1 0.00028125_0.00175570, 0.00002109 F. 1)2 (-1)3 (-1) (N-1) etc. 10-1 0.075 √1 +0.1875 = 0.0000132 -1 ± etc. = 0.18747 ± 9.9270 √ — 1, which is correct to four decimal places, both in the real and imaginary parts; consequently, we have a 0.1874 +9.9270 √1 and 0.1874 9.9270 -1 for the approximate values of the imaginary roots of the given equation. Remarks. We have taken this question from p. 106 of Young's Theory of Equations, where one value of x is found to be 9.8860027. The question appears to have been originally proposed by Dr. Halley, who found 9.8862603936495 for the approximate value of one of the roots. (See a small tract on the Solution of Equations, at the end of Newton's Universal Arithmetic, Example 1.) 6. To find a from the equation a3-52+8x-1=0. =-1, the equation is easily changed to 1 By putting = = 8 1 x 1 -3 = 8 y'-1; which gives y' = 8. Com paring this equation to (6'), we have n = 1 ; consequently, from (b), we have x' = 8— 65 etc.7.3379; correct to three decimal places. 16384 Hence, we have x = mate value of x, which is correct in all its figures. METHOD OF APPROXIMATING TO THE RATIO OF TWO NUMBERS OR QUANTITIES. 1. Let A and A' represent the numbers or quantities; then if we divide A by A', and divide A' by the remainder, and so on, as in finding the greatest common divisor, by using a, b, c, etc., for the successive quotients, we shall have the following process: (1 + be)A − (a + c + abc)A'|(1+ab)A'—bA[d and so on, to any extent that may be required. It is clear, from (1), that any dividend and its corresponding divisor (neglecting their signs) may be expressed by the forms PA'QA and QA-PA', (2); consequently, if m stands for the quotient resulting from the division, the remainder will evidently be of the form P1A' - Q1A: (P。 + mP)A' — (Qo + mQ)A; which gives P1 = mP + P。 and Q1 = mQ + Qo, (3). = 2. Supposing A and A' to be positive, and a, b, c, etc., the greatest positive integers in the successive quotients, it is clear that the remainders in (1) will all be positive. Because A the remainder AaA' is positive, it is plain that is A' greater than a =; and in like manner from the remainder, a 1 1 + ab a + c + abc A , etc., or they are limits to it; being b 1+ be greater than the first of the preceding ratios, less than the second, greater than the third, and so on. It is hence clear a 1 + ab a + c + abc 1' b that we may call 1 + be A etc., the first, second, third, etc., approximate ratios of ; observing that A' A A" is which expresses the exact geometrical ratio of A to A', sometimes called the exact or complete quotient, while the approximate ratios are called inexact or incomplete quotients. 3. If we represent any successive three approximate ratios. Po P and then from (3) these will become by and to Qo' Q' mP + Po mQ + Qo' P. (4). Po P Qo Q Since m represents the quotient in (1), which corresponds P1 Q1 (the last of the three ratios), it follows, from (4), that if we multiply the numerator of the second by m, and increase the product by the numerator of the first, the result will give the numerator of the third; and in the very same way the denominator of the third can be found from the denominators of the second and first fractions or ratios. Thus, if we take the first two approximate ratios or fractions, a 1 + ab and represent the third by ; then, since c is the Q1 b third quotient, we must put c for m, and we shall have P1 (1 + ab)c + a = a + c + abc as it ought to be. 4. If we multiply the first of (3) by Q and the second by P, we shall have P.Q+ mPQ = QP, and QoP + mPQ = PQ, which, by subtraction, give PQ-QP=-(PQ1—QP1); and if P2 P2 Hence, if for have PQ1-QP1 =-(P1Q2-Q1P2), and so on. PoQQP 0 1 — 1; con etc., ratios, we Qo' Q Qi Q2" represent the first, second, third, etc., approximate shall have the equations P,Q QoP1, PQ, QP, = 1, P1Q2 — Q1P2 = − 1, etc., (5); observing that the right members of (5) are negative when their first members are obtained from the last two of an even number of approximating ratios, and that the right members are positive when their first members are obtained from the last two of an odd number of approximating ratios. 5. If the first of (5) is divided by QQ, the second by QQ1, and so on, they will be reduced to Р 1 P Po Qo Because a, b, c, etc., are positive integers, and that Q。 = 1, Q=b, Q1 = 1+ be, etc., it is clear that QQ is less than QQ1, and that QQ, is less than QQ2, and so on; consequently, it Po P P follows, from (6), that the ratios 0 Q' Q' Qu' etc., will continu |