then these two being substituted in the value of z, give z or case, therefore the arc in the latter case must be double the arc in the former. But, by trigonometry, the cosine of the double arc to secant s and radius 1, is 2 1; and, by the $2 2 α Or, the same fluent will be X arc to cosine√ be ηνά xn cause the cosine of an arc, is the reciprocal of its secant. n mn ут ; hence by art. 36, ź = n mn 59. By the above example it appears that such form of fluxion 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+xn)men-1, where the exponent (en-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 1, of two terms when c=2, of three terms when c 3, and so on.. 60. Thus, in the general form, z= (a+xn)men, putting as before, a+x=y; then is nya, and its fluxion or xen-n x2-1 nxn and these values being now sub n n' (y-a)-1y; also (a+x)y (a+x^)m—ym stituted in the general form proposed, give z Now, if the compound quantity (y-a)1 be expanded by the binomial theorem, and each term multiplyed by yy, that fluxion becomes &c.); then the fluent of every term, being taken by art. 36, it is 1 ym+c n mfc 1 m+c-1 1 c−2 a2ym+c=2 yd 1.c-2 a2 2m+c-2 N d d-1y d-2 242 &c.), putting d me, 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 c 3, and so on.-Except however the cases in which m is some negative number equal to or less than c; in which case the divisors, m+c, 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. z จ 61. Besides this form of the fluent, there are other methods of proceeding, by which other forms of fluents are derived, of the given fluxion = (a + n)men-x, 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. 62. The given fluxion being (a+x)men; its fluent may bé be assumed equal to (a + x)+1 multiplied by a general series, in terms of the powers of x combined with assumed unknown coefficients, which series may be either ascending or descending, that is, having the indices either increasing or decreasing : (a+x2) α And first, for the former of these, take its fluxion in the usual way, which put equal to the given fluxion (a+x)men-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) x2-1 (a + xn)m × (Ax2 + вx + cx2-25 &c.) + (a + xn)m• 1xx (raxTM-1 + (r—s) BX1+ (r-28)cx-28-1 &c.); this being put equal to the given fluxion (a+xn) mxcn−1x, and the whole equation divided by (a+x)mx-1 there results n (m + 1) x2 × (Ax2+Bx2¬3 + cx2¬2s + Dx-3s +&c.) + (α + x2) × (rAx+(r—s)Bx+(r-2s) cx-2$ &c.) Hence, by actually multiplying, and collecting the coefficients of the like powers of x, there results -28 3s z(m+1) } ax+n+n(m+1) { Bxṛ+n—s+n(m+1) 》 cx2+n−2s&c. +rs +―rs + } 0. +r-2s ( raАxr rɑax” ・・・ + (r−s) αвx2-s &C. Here, by comparing the greatest indices of x, in the first and second terms, it gives r+ n = cn, 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)naxTM2+(c+m—1)nBxn−n+(c+m− 2)nczen−2n&c. } -xon+(c-1)nasxonn(c2)naвxon-2n &c. =0. Now, comparing the coefficients of the like terms, and putting c+m=d, there results these equalities: 1 dn' dn B= c-1.αA c-1.α d-1 d— 1.dn; c= + + C-2.αB c-1.c-2.α2 d-2 d-1.d-2.dn' &c.; which values of A, B, C, &c. with those of r and s, being now substituted in the first assumed fluent, it becomes (a + x2) m + 1 x cnn 1 c-1.a c-1.c-2.a2 c-1.c-2.c-3.α3 d-1.xnd-1.d-2.xan d-1.d-2.d-3.x3" +&c. the true fluent of (a + x)men-x, 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. VOL. II. Thus, 44 Thus, if c≈2, or the given fluxion be (a+xn)mq2n−1; (a+xn) m + 1 xn axn m+1 n x(1. (m+2)n m+ 1.m+2 And if = 3, or the given fluxion be (a+xn) з n − 1 x ; then m + c or d being m+3, the fluent becomes m+cord c (a+xn) m + 1x2n (m+3)n ×(1 2αx-n 2α2x— în (a+xn)m+1x2n 2αx 2a3 + And so on, when c is other m+3.m+2m+3.m+2.m+1 whole numbers: but, when c denotes either a fraction or a negative number, the series will then be an infinite one, as none of the multipliers c➡ 1, c−2, c-3, can then be equal to nothing. 63. Again, for the latter or ascending form, (a+x2)m+1× (Ax? + вxrts + cxr+es+ Dxr+ss + &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 rts, r + 2s, &c. be written instead of r Ꭶ, ↑ 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 −xcn+n(m+1} +n)m+1) 7 Axr‡n trts Bærtnts &c. =0. +rasx+(r+s)αвxr+s+(r+-2s)αcxr+2s &c. Here, comparing the two least pairs of exponents, and the coefficients, we have r = cn, and s=n; then a= ~~______ r+n(m+1); ^== c+m+1 A ing cm by d, as before, the fluent of the same fluxion (a+xn)men-x, 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 To show now the use of the foregoing series, in some examples of finding fluents, take first, 1 64. Example 1. To find the fluent of (a+x)*. 6xx or 6xx √(a+x) This example being compared with the general form xen -1x (a+xn)m, in the several corresponding parts of the first series, gives these following equalities; viz. a = a, n=1, 1, or c= 2; m = a + x2 1=1, or c 1 d—m+c—2—1—3, −yd—(a + x) n 2 C 1 α 2a d 3'd-1 ya+x here the series ends, as all the terms after this become equal yd 1 C 1 ind 1 2a+2x _2a)x(a+x) *=* 3 2x-4a 3 (a+x); which multiplied by 6, the given co efficient in the proposed example, there results (4x—8α) · √ (a+x), for the fluent required. 17. Exam. 2. To find the fluent of 3x √ (a2+x2 ) = (a2 +x3) * × 3 x − x The several parts of this quantity being compared with the corresponding ones of the general form, give a = a2, n= m=1, cn-1 or 2c-1=-6, whence c 1-6 2, § and d = m+c=1}=-=-2, which being a negative integer, the fluent will be obtained by the 3d or last form of series; which for the required fluent of the proposed fluxion. 66. Exam. 3. Let the fluxion proposed be Here, by proceeding as before, we have a=b, n=n, m=−}, c=3, and d=c+m; where c being a positive integer, this case י די 1 |
