11. By the above example it appears, that such form of Auxion admits of a fluent in finite terms, when the index (n-1) of the variable quantity (x) without the vinculum, is less by 1 than n, the index of the same quantity under the vinculum. But it will also be found, by a like process, that the same thing takes place in such forms as (a+x")TMxn−1x, where the exponent (cn 1) without the vinculum, is 1 less than any multiple (c) of that (n) under the vinculum. And further, that the fluent, in each case, will consist of as many terms as are denoted by the integer number c; viz, of one term when c = 1, of two terms when c = 2, of three terms when c3, and so on. 12. Thus, in the general form, ż = {a + x»)” x‹−1ƒ‚ putting as before, a + x" = y; then is "y-a, and its fluxion nx-1=j, or r"-1; = ў and ren- or xn-n n 1 n x"−1 ; = — ( y − a)c−1ÿ;'; also (a + .x")" = ym: these values being now substituted in the general form proposed, give x = · (y — aɣc—1ymj. Now, if the compound quantity (ya)- be expanded by the binomial theorem, and each term multiplied by yy, that fluxion becomes &c); then the fluent of every term, being taken by art. 36, it is ym+c c-1 aym+c-1 c-1 c-2 4zym %= = 2 m+c-2 c-1.c-2 (12 + m+c-1 d-2 2y2 2 - &c), c-1.c-2.c-3 Q3TM &c), putting d = m+c, for the general form of the fluent; where, c being a whole number, the multipliers c— 1, c−2, c-3, &c, will become equal to nothing, after the first c terms, and therefore the series will then terminate, and exhibit the fluent in that number of terms; viz, there will be only the first term when c=1, but the first two terms when c = 2, and the first three terms when c3, and so on.Except however the cases in which m is some negative number equal to or less than c; in which cases the divisors, m +€, m + c − 1, m+c 2, &c, becoming equal to nothing, before the multipliers c-1, c-2, &c, the corresponding terms of the series, being divided by 0, will be infinite: and then the fluent is said to fail, as in such case nothing can be determined from it. 13. Besides this form of the fluent, there are other me thods of proceeding, by which other forms of fluents are VOL. III. derived, of the given fluxion &= (a + x")TM xc# −1, which are of use when the foregoing form fails, or runs into an infinite series; some results of which are given both by Mr. Simpson and Mr. Landen. The two following processes are after the manner of the former author. 14. The given fluxion being (a + x”)”xn−1; its fluent may be assumed equal to (a + ")" + multiplied by a general series, in terms of the powers of r combined with assumed unknown coefficients, which series may be either ascending or descending, that is, having the indices either increasing or decreasing; I 35 viz, (a + ")+1× (A.x" + Ex2-s+ C.x2-25 + Dt'—35 + &c), or (a + x")" +1× (Ax" + Bx' +'3 + cx2 + 25 + Dx” + 3o + &c). And first, for the former of these, take its fluxion in the usual way, which put equal to the given fluxion (a + xn)m -x, then divide the whole equation by the factors that may be common to all the terms; after which, by comparing the like indices and the coefficients of the like terms, the values of the assumed indices and coefficients will be determined, and consequently the whole fluent. Thus, the former assumed series in fluxions is, n(m + 1)x"1x(a + x")" × (Ax2 + Bx2- ' + C2-25 &c)+ (a+x")+1× (r'Ax ̃−1+(r−s) Bx ̃—♪~1 + (r−2s) c.r” — 25 — &c); this being put equal to the given fluxion (a + xo1)mxcn=1ƒ, and the whole equation divided by (a+"), there results n(m+1).x′′ × (Ax2 +B.x2- + cx2-25+ D.x2-35+ &c) ܨܐ 1511 -S +(a+x")× (rax” + (r− s)вxTM-s+(r−2s)cx2-2&c) ( Hence, by actually multiplying, and collecting the coefficients of the like powers of x, there results +r n(m + 1)} cn -I +r-25 +(r—s)aBx*—3 &c Here, by comparing the greatest indices of r, in the first and second terms, it gives r + n = cn, and r + n − s = r; which give r = (c 1)n, and n = s. Then these values being substituted in the last series, it becomes - (c+m)naxTM+(c+m−1)nâxTM — n +(c+m- 2) ncx -x+(c−1)n@AxTM”—”+ (c−2)naBxTM — &c Now, comparing the coefficients of the like terms, and ting c+m=d, there result these equalities: == &c; which values of A, B, C, &c, with those of r and s, being now substituted in the first assumed fluent, it becomes &c, the true fluent of (a + ")-1, exactly agreeing with the first value of the 19th form in the table of fluents in my Dictionary. Which fluent therefore, when c is a whole positive number, will always terminate in that number of terms; subject to the same exception as in the former case. Thus, if c = 2, or the given fluxion be (a + x")” x2−1 x ; then, c+m or d being = m + 2, the fluent becomes 312 And if c = 3, or the given fluxion be (a + x")" x3n-1 x ; then m + c or d being = m + 3, the fluent becomes ;). 2a2 + And so on, when c is other m+3.m+2 m+3.m+2.m + 1 I 15. Again, for the latter or ascending form, (a+x")" + 1× (A.x2 + BX'+s+ cx2+25 + Dxr+ 35 + &c), by making its fluxion equal to the proposed one, and dividing, &c, as before, equating the two least indices, &c, the fluent will be obtained in a different form, which will be useful in many cases, when the foregoing one fails, or runs into an infinite series. Thus, if r+s, r+ 2s, &c, be written instead of r ́— ́s, r — 2s, &c, respectively, in the general equation in the last case, and taking the first term of the 2d line into the first line, there results 25 +raAx+(r+s)aв.x*+3+ (r+2s)ac.x*+ &c }· = 0. Here, comparing the two least pairs of exponents, and the coefficients, we have r=cn, and s=n; then A= 1 1 = (c + 1)cną3 3 C= Therefore, denoting e+m by d, as before, the fluent of the same fluxion (a + x)mxcn. *, will also be truly expressed by agreeing with the 2d value of the fluent of the 19th form in my Dictionary. Which series will terminate when d or c + m is a negative integer; except when c is also a negative integer less than d; for then the fluent fails, or will be infinite, the divisor in that case first becoming equal to nothing. To show now the use of the foregoing series, in some example of finding fluents, take first, 16. Example 1. To find the fluent of (a + x) 3. 61x √(a+x) or 6x* This example being compared with the general form xn−1 x(a + x”)”, in the several corresponding parts of the first series, gives these following equalities: viz, a=a,n=1, cn-11, or c − 1 = 1, or c = 2; m = ; y = a+x, d = m + c = 21, yd = = y (a + x)3, 1_2 c-1 q=33 -1° = ; here the series ends, as all the terms after this 2a a+z become equal to nothing, because the following terms contain the factor c 2 = 0. These values then being substi(1.), it becomes (a + x)2 × stituted in yd 12 2a 2x - 2a) (a+x)2 × (a + x) * = 2x=4a √ (a+x); which multiplied by 6, the given coefficient in the proposed example, there results (4x-8a).√(a+x), for the Auent required. 17. Exam. 2. To find the fluent of The several parts of this quantity being compared with the corresponding ones of the general form, give a = a2, n = 2, 16, whence c = 2 d=m+c=-=-=-2, which being a nega tive integer, the fluent will be obtained by the 3d or last form of series; which, on substituting these values of the letters, 3(a2 + x2) 3⁄4,x−5 18. Exam. 3. Let the fluxion proposed be Here, by proceeding as before, we have ab, n = n, 3, and d = c + m = ; where c being a positive integer, this case belongs to the 2d series; into which therefore the above values being substituted, it becomes I 19. Exam. 4. Let the proposed fluxion be 5(÷— z2)1z−3%. Here, proceeding as above, we have a = }, n = 2, m = 1, en 1 or 2c 1= - 8, and c = - z, x= -z, d = c + m3, which being a negative integer, the case belongs to the 3d or last series; which therefore, by substituting these values, becomes 15(}—z2)} -7x7 2x2 + G = × (1 + 12x2 + 94z1) = −3(-—z1)3×(5+12x2+2481), 1229 the true fluent of the proposed fluxion. And thus may many other similar fluents be exhibited in finite terms, as in these following examples for practice. Ex. 5. To find the fluent of 3x3x√(a2 — x2). Ex. 6. To find the fluent of - 6x5%. (a2— x2)−‡. . · Ex. 7. To find the flu. of√(a") or (u-2)x−kn+1%. 2-1 20. The case mentioned in art. 37, vol. 2, viz, of compound quantities under the vinculum, the fluxion of which is in a given ratio to the fluxion without the vinculum, with only one variable letter, will equally apply when the compound quantities consist of several variables. Thus, Example 1. The given fluxion being (4x + 8yj) * √(x2 +2y3), or (4xx + 8yÿ) × (x2 + 2y2), the root being x2+2y, the fluxion of which is 2xx+4yj. Dividing the former fluxional part by this fluxion, gives the quotient 2: next, the exponent increased by 1, gives: lastly, dividing by this, there then results 4(x2 + 2y2), for the required fluent of the proposed fluxion. Exam. 2. In like manner, the fluent of (x2 + y2+zo)3 × (6x* + (x2+y++ £6)š+1 × (6zi+ 12y3ý + 18z3z) Exam. 3. In like manner, the fluent of 2x2(ży2 + xyj + x2*)√(x2 + 2y2), is ‡(x2 + 2x2y°)3⁄4., 21. The |