found with great facility by one of the more simple methods of limits given at pp. 471, 472, etc. Finally, since all the conditions required in the determination of a' = a+b1 -1 are embraced in the condition that u, as given by (d), shall be so small as to make the series (6) converge with sufficient rapidity, in order to find a to the required degree of exactness by using a few of the first terms of (b), it is clear that if we assume a and b, such that P and S T Qin (e) are small, and thence find that and are also R R very small, we shall manifestly have found suitable values for 01 and b1, or such values as will enable us to find a from (b) to the required degree of exactness. EXAMPLES. 1. To find all the roots of the equation 4y 8y-17=0. Putting y = a + x, the equation is easily reduced to 4a3 +12a'a2 + (12a-8)x = 17 + 8a' - 4a's, which is easily (1); and this, compared to (a), gives u= = 1 3a' 3a2-2' 3a2-2' b = c = 0, d = 0, etc. 178a4a's After a few trials, we find that by putting a' = 2, we shall 6 0.025, a = 0.6, b=0.1; consequently, we 10 = have au2 = 0.000375, (b2a)u3=- -0.0000096875, (c + 5a3 5ab)u= 0.00000030468, etc. Hence, using only four terms of (b), we shall get a = 0.025 — 0.000375 +0.0000096875 — 0.00000030468 = 0.024634 +, which is correct to six decimal places; consequently, we shall have y = a' + x = 2.024634+ for the real root of the proposed equation, which is correct to six decimal places. Again, if we put 'a' = a + b −1 = shall get 17+ Sa' 4a 1 and 12a" √ − 1), which give u = = 8(1+31-1) ; and we shall have a = 80 x= 4957-1 13-9-1 100 - 62591 1 12800000 Hence, using only three terms of (b), converting u, au2, and (b — 2a2)u3 into equivalent decimals, etc., we shall get ∞ = 0.0123173 +0.03647 -1; consequently, we shall have a x − 1.0123173 + 1.03647 √ −1 for one of the imaginary roots, and changing the sign of the imaginary - 1.0123173 — 1.03647 V part, we get 1 for the other imaginary root of the proposed equation; these roots being correct to about seven decimals in their real parts, and nearly to six places in their imaginary parts. 2. To find the imaginary roots of y-19y+1324-302y + 200 = : 0. By putting a' + x for y, the equation is easily reduced to + a'4 4a' - 19 203 4a3-57a+264a — 302 4a3 57a2+ 264a' 302 pared to (a), gives a = 302' 57a2+ 264a' d0, e0, etc. Also, if we put a' 132a2-302a + 200 132a,2 1326,2 — 302α1 + 200 + (4α ̧3b1 — 4α‚b‚3 — 57a11 + 6a,b12 + b1— 19a,3 + 57ab12+ = - 3 196 ̧3 + 264α ̧b1 — 30261) — 1, which will be found to give a1 = 8 and b, 4 for the proper values of a, and b1; which being put for a and b, the expression becomes 104 +8 1. Similarly, by putting a' = 8+ 4√-1, we shall. get 4a57a2+ 264a' - 302 = 414 + 224 √ — 1. the first approximation to the value of x; and adding this to a' = 8+41-1, we shall get a' 7.8+ 3.9 1-1 nearly, for a more correct value of a'. By putting a 7.8 +3.9 1, we shall get u = 5.97077.6674 V-1 369.158-171.756 V- 1 3521.053625+1804.9785 V 165777.7525 0.0212396028. + 0.0108879410. √ — 1, a = 38847.16500-46026.04500 V 165777.7525 0.277637043 — 1, b = − 1 = 0.010401480 0.047378420 369.158 +171.756 V 1 0.1301157 — 1, b—2a2 = 0.0339187 +0.2128530 √ — 1, and (b2a2)u30.0000029304 + 0.0000000247 — 1. = √ Hence, by using three terms of (b), we shall get x = u au2 - (b — 2a2)u3 = 0.02118619 +0.01108841 -1, and thence a' + x = 7.77881381 ± 3.91108841 v 1 are the imaginary roots, using for for one of them, and for the other; it may be noticed that these roots are correct to about seven decimal places, both in their real and imaginary parts. Remarks. Since the sum and product of the roots found are nearly equal to 15.55762762 and 75.80655683, and that 19 is the negative of the sum of all the roots of the proposed equation, while 200 equals their product; it follows that if we subtract 15.557, etc., from 19, we shall get 3.44237238 for the sum of the remaining two roots, and that if we divide 200 by 75.80655683, we shall get 2.63829552 for the product of the same roots. Hence, since the sum and product of the remaining roots are known, we easily find them to be 2.2905598 and 1.1518125, which are correct to about six decimal places. Thus, we perceive how we may sometimes find the real roots of an equation from its imaginary roots; and at the same time we have a verification of the correctness of the rules which have been given for the treatment of imaginary expressions, when they are to be added, subtracted, multiplied, etc. It may be added, that if we put 2.29 and 1.15 separately, for a' in the values of u, a, b, etc., we shall, from (b), easily get the remaining roots of the equation to any degree of exactness that may be required. 3. To find the roots of y3- 5y2 + 8y · 1 = : 0. Also, putting a' = a + b1 = 1, we get a's - 5a2+ 8a' — 1 = a,3 — 5a2 — 3a,b,2 + 5b — 1 + (3a,b,- 10a,b1 - · 3 8b1) √ — 1, and 3a”—10a′+8=3(a ̧2—b12)—10α1+8+ (6α1b1— 106) -1. From the forms of these equations, it is clear that a and b can not differ much from 2 and 1; consequently, putting these values for a and b, in the equations, we soon find that 2.4 and 1.2 are more correct values of a and band putting these values of a and b, in the equations, they give a3 - 5a2 + 8a' - 10.056 0.192 √ — 1, Hence, we 0.288 - 1 = 37.12 0.142241 -1; u2 = 0.00095719 0.000494945 √ — 1, and au2 = 1; u3 = 0.000026706 = 0.00001686 + 0.00074601 √ -0.00002321 - 1, b2a2 = 0.600326 +0.949092 √ — 1, and (b2a)u3 = 0.00003806 +0.00001141 -1. Substituting the preceding values in (b), and taking only three terms of the series, we get x = u — au2 — (b — 2a3)u3 0.03187535 -0.0085159 — 1, which is correct to four decimal places; consequently, ax gives 2.43187535 ± 1.1914841 - 1 for the expression of the imaginary roots, which are correct to four decimal places, both in their real and imaginary parts. Since 5 is the negative of the sum of all the roots of the equation, and that 4.8637507 is that of the roots found, we have 54.8637507 0.1362493 the remaining root, correct to four decimal places. = 4. To find the roots of "y — 5y3 + 13y2 — 17y + 12 = 0.” Here, by putting a' + x for y, and a' = a1 + b1 √ − 1, and proceeding as in the last question, we find that it will answer our purpose to put a = 1, and b, 1.4, or a'=1+1.41-1; then, from the substitution of 1 + 1.4 -1 for a', etc., we soon find a' + x = 0.9999, etc. ± 1.4142, etc. V 1 for two of the imaginary roots, which indicate that 1-2 are two of the imaginary roots. = Dividing the equation by (y-1+ √ − 2) × (y−1 — √ — 2) = (y − 1)2 + 2 = y3 — 2y + 3, we get y2 - 3y+4 for the quotient; consequently, the given equation is equivalent to (y2 — 2y + 3) × (y2 — 3y + 4) = 0. Hence, if we put the factor y 3y+ 4 equal to 0, and solve the resulting quad3±17 ratic, we shall get for the two remaining roots of 2 the equation. 5. To find the integral roots of a3 - x2-41x + 105 = 0. If a stands for an integral root, we know (from principles heretofore given) that 105 must be exactly divisible by a, and that the first member of the equation must be exactly divisible by x a. Hence, if we put 1, 0, and 1, successively, for x in the given equation, the results must be severally divisible by 1-a, -a, and 1-a. a, and 1-a. If a is positive, |