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 21. The fluents of fluxions of the forms &c, or &c, where c and n are whole numbers, will be found in finite terms, by dividing the numerator by the denominator, using the variable letter r as the first term in the divisor, continuing the division till the powers of a are exhausted; after which, the last remainder will be the fluxion of a logarithm, or of a circular arc, &c. Exam. 1. The find the fluent of a + x or xx x + a evidently a x the fluxion of the hyperbolic logarithm of a + x: therefore the whole fluent of the proposed fluxion is t a x hyp. log. of (a + x). In like manner it will be found that, Ex. 2. The fluent. of Ex. 3. The fluent of (ax). Ex. 4. The fluent of is x2-ax + a2 × hyp.log. of (a + x). a + x Ex. 10. The fluent of, is - }ax3 + ‡a2x2 x2x = Ex. 14. The fluent of (by division) ✯ — cir. arc of radius a and tang x, or t a x cir. arc of rad. 1 and cosine manner, Ex. 15. The fluent of a x + a × h. 1. 2+, by form 10. And a-x 22. In like manner for the fluents of Ex. 17. The fluent of is by form, r3 - a'r + a - or ƒx3 — a2x+ža3 × cir. arc a + x − 4x3 — a3x + ža33 x hyp. log. by form 10. Also ±x3 + a2x + ža3 x hyp. log., by form 10. 23. And in general for the fluent of ati is is where n is any even positive number, by dividing till the powers of r in the numerator are exhausted, the fluents will be found as before. And first for the denominator 2 + a2, as in 2”—2 ✯ — a2x”—ˆ¿ + a2x2- - &c ± a"- the ε F number of terms in the quotient being in, and the remainder or according as that number of terms is odd or even. Hence, as before, the fluent is 2 + &c... ± a2 -2 x Ħ a2 -2 x arc to rad. arc to rad. 1 and cos. Ex. 21. In like manner, the fluent of is – x hyp. log. N- 3 24. In a similar manner we are to proceed for the fluents of when n is any odd number, by dividing by the denominator inverted, till the first power of x only be found in the remainder, and when of course there will be one term less in the quotient than in the foregoing case, when n was an even number; but in the present case the log. fluent of the remainder will be found by the 8th form in the table of fluents in the 2d volume. Ex. 22. Thus, for the fluent of where n is an odd number, the quotient by division as before, is "-2 je a2x2-4 * + a1.x”—- &c ±a"-3rx, the number of terms being + Ex. 23. The fluent of is obtained in the same 12. a2 manner, and has the same terms, but the signs are all positive, and the remainder is +a"-x hyp. log. x2 - a2. I a2-x2 Ex. 24. Also the fluent of is still the same, but the signs are all negative, and the remainder is a"-1× hyp. |
