of the powers of x, (3) becomes x=Aax+ Ab x2+2Babx+ Ac Because (4) is an identical equation, we must have the equations Aa=1, Ba2 + Ab= 0, Ca3 + 2Bab + Ac= 0, and which agrees with the formula given by Cagnoli, at p. 46 of his Trigonometry. If the coefficients of the even powers of x in (1) equal 0, then it will be reduced to the form y= ax + bx3 + cx3 + da + etc., (1); and instead of (5) we shall have x= y b a 362 ac 12b38abc + a2d y' + etc., (5'). Remarks.-1. If we have y' = m + ax + bx + cx2 + etc., then, by transposition, and putting y=y'-m, we shall have y = ax + bx2 + etc.; consequently, the series can be reversed by (5) or (5'). 2. If we have a'z + b'z2 + c'z3 + etc. = ax + bx2 + cx3 + etc.; then, by putting ya'z + b'z2 + c'z3 + etc., we shall have y ax + bx2 + ca3 + etc., and thence x = y by2 α as + etc.; consequently, putting a'z+b'2+ etc., for y, we shall To illustrate what has been done, take the following com mon etc.; then, by (5'), we shall have = y + 1 c = 2.3 6' etc., a = 1, b=1, c=1, etc., 2100 + etc. 207 2.3.4.5.6.7 1 1 = 2.3.4.5 120' 3. We shall suppose that we have the equation an+ axr+m+ bxn+2m + can +3 1+3m + dan+1m + etc. = ±y", (1), and that it is hence proposed to develop the value of x. It is easy to perceive that we may assume = y [A + Bym + Cy3m + Dy3m + etc.], (2), in which A, B, C, etc., are independent of y; and it is clear that we may regard Bym as being an increment of A, Cy" as an increment of B, Dym as an increment of C, and so on. x From the substitution of the value of a from (2) in (1), we get y"[A+By" + etc.]" + ayn+m[A + ByTM + etc.]n+m+ byn + 2m [A + By + etc.]"+2m+ etc.y", or omitting the factor y", which is common to the two members of the equation, we have [A + ByTM + etc.]” + ayTM[A + By” + etc.]”+” +by[A+By+ etc.]"+m+ etc. 1, (3), which must be an identical equation. It is easy to perceive that the developments required in the first member of (3) can be obtained from A” + αA”+”yTM +bAn+2mm + cAn+smysm + etc., by changing A into A+ By", B into B-+Cy", C into C+Dy", and so on, successively, and developing according to the rule given in Involution. a An+m Thus, we get A+ A-1B | 3" etc. 1, (4). = + 2m (n + m)a An+m-1B 1 2 nAn-1C Because (4) is an identical equation, we must have the equations A±1, (5), (noticing that must be used in this for, when the right member of the given equation is +, and vice versa), and aAn+m + nAn−1 B = 0, bAn+2m + n(n−1) An-2B2. (n + m)a An+m−1B+ B2+nA”-1C=0, and so on. 1 2 a From these equations we get B - Am+1, C= (n + 2m + 1)a2 — 2nb 2n2 = x A2m+1, and so on. n If y represents the arithmetical root of y", and A the algebraic root of equation (5), then, if we put Ay = y', (6), and substitute the values of B, C, D, etc., in (2), we shall [(n + 4m + 1). (2n + 4m + 1). (3n + 4m2 + 1)a^ 2 4n1 d 3 n + 4m + 1 (b2+2ac) —; 2n2 n a2b+ + (n+5m+1). (2n+5m+1).(3n+5m+1). (4n+5m+1), (n + 5m + 1). (2n + 5m + 1). (3n + 5m + 1) 1 2 as 3n (n + 5m + 1). (2n + 5m + 1)(a2c + ab3) + 1 n manifest. 2n3 15m+1 + etc., (a), whose law of continuation is If n is a positive whole number, and m = −1, then (1) becomes a axn−1 + bxn-can-s+ etc. y", (6'); and (a) is reduced to x = y' - a n 3 + a2b+ (1a + 2ac) — 24] 2n2 (n—4)(n—2)(3n—4)(n−1) a3 + 3 2) 5n5 n n (n−4)(n−2)(3n—4) 3n1 'a3b (a2c + ab2) + (ad + be) — 213 n3 *] + + (3ab + 20°c) (2ae + 2bd + c2) — n a If in (6') we put instead of x, the resulting equation n will be of the same form as (6'), with the exception that the term which corresponds to ax-1 will be wanting, because its coefficient will equal 0; consequently, (6') may represent the resulting equation, if we put a = 0. Hence, putting a = 0, (b) is reduced to xy' + [ n n 7 4n1 + -6 h1 n 2n2 -(ď2 + 2bf + 2ce) (n − 4) (n 3n+ (de + bg + of) a much simpler form than (b), while it has the same gener ality. Supposing n to be a positive whole number, and a, b, c, d, etc., together with ±y", to be real numbers or quantities, it is clear that (6') may represent any algebraic equation of the nth degree, observing that one or more of the letters a, b, c, etc., may be naught or negative, if required. = Because±y"=±1× y" = (Ay)" = y'", it results that the right member of (6') equals y'". To get y' Ay, we must (according to what has been shown) multiply the positive root of y" by the value of A, as found from the solution of the binomial equation (5). Since it is proved in the Treatises on Analytical Trigonometry, that (5) has n different roots, it follows that y' must have n different values. Also, because the imaginary roots of (5) occur in pairs of the forms p+q√ −1 and p−q √ — 1 (in which p may equal 0, if required), it follows that there must be similar corresponding values of y'. Hence, it is clear, from (b) or (c), that the equation of the nth degree represented by (6) has n roots, whose developments are given by (b) or (c), by putting the different values of y' for y' in them; it is also clear, if (6') has imaginary roots, that they must occur in pairs of the forms P+Q-1 and PQ1, observing that P may equal 0, if required. It may be added, if a, b, c, etc., and y'" are one or more of them imaginary, then it may be shown, in much the same way as before, that (6') has n roots. To show the use of what has been done, take the following |