71

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

I

(a+x).

√(a+x)

or 6x

٦٠

This example being compared with the general form xen-x(a + x)", in the several corresponding parts of the first series, gives these following equalities: viz, a = a, n=1, cn − 1 = 1, or c - 1 = 1, or c = 2; m=-1; y = a+x,

d = m+

y

1

+ c = 2- =,=

La

n

here the series ends, as all the terms after this become equal to nothing, because the following terms contain the factor c-2 = 0. These values then being substi

which multiplied by 6, the given coefficient in the proposed

example, there results (4x-8a). (a+x), for the fluent

required.

The several parts of this quantity being compared with the

corresponding ones of the general form, give a = a2, n = 2,

and d = m + c =-=-=-2, which being a negative integer, the fluent will be obtained by the 3d or last form of series; which, on substituting these values of the letters,

m=

5x37-12 √(b+an)

= 5(6+)-3-1.

Here, by proceeding as before, we have a = b, n = n, - c = 3, and d = c + m = 2; where c being a positive integer, this case belongs to the 2d series; into

1

which therefore the above values being substituted, it becomes 5(b+x)x2n

in

X

26

+

1)=2√(6+2)x

2.1.
즐

3x2-4bxn+82

3η

I

19. Exam. 4. Let the proposed fluxion be 5(+-z2)z-. Here, proceeding as above, we have a = =, n = 2, m

en m=

1 or 2c

1=

m=, - 8, and c = - 1, x = -2, d = c + - 3, which being a negative integer, the case belongs to the 3d or last series; which therefore, by substituting these values, becomes

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.

T

Ex. 5. Το find the fluent of - 3x3x(a2-x2).
Ex. 6. To find the fluent of - 6.xx. (a2-x),
Ex. 7. To find the flu.of√(a-") or (4-2)-2n+1%.

-1

7

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) x N(2 +2y2), or (4xx + 8yj) x (x2 + 2y2), the root being x2 + 2y2, 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 (x2 + 2y2), for the required fluent of the proposed fluxion.

:

Exam. 2. In like manner, the fluent of

I

T

(x2 + y + z) x (6xx + 12y3j + 18zz) is

(x2+ y + 26)3+1 x (bai+12y3j+18zz)

(2x+4y3j+6z5z) x

= (x2 + + + z).

4

Exam. 3. In like manner, the fluent of

2x2(xy2 + xyj + x2x)/(x2 + 2y2), is (x2+2x22).

21. The fluents of fluxions of the forms

&c, or, &c, where cand 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 x 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

xx
xx
or
a + x
x + a

where the remainder

evidently = a x the fluxion of the hyperbolic logarithm of a + x: therefore the whole fluent of the proposed fluxion is r - a x hyp. log. of (a + r). In like manner it will be found that,

Ex. 9. The fluent of a3 x hyp. log. of (a - x). Ex. 10. The fluent of a3x + a x hyp. log. (a + x). Ex. 11. The fluent of , is + &c ± a x h. 1. (a + x).

24

n

03

zax2 - ax +

ax3 + a2x2

is x3 - ax2 + a2x - a3 x

x2
x2 + a2

Ex. 14. The fluent of = (by division) *

cir. arc of radius a and tang x,

is, (by form 11 voi 2) x –
or x - a × cir. arc of rad. 1 and cosine

24
a2 + x2

*+

ax a2

is x + a x

Thus,

a'z Q2+xas

a222 = x2x - a2 + × cir. arc to rad. a and tang. x, to rad. 1 and cosine And

a2+x2

Ex. 17. The fluent of is by form, r3- a2x + a2 or 4x3- a2x+a3 x cir. arc Ex. 18. The fluent of - x2x - a2 + 3a'x + a3 x hyp. log. by form 10. Also Ex. 19. The fluent of

*x3+ a'x + a3 × hyp. log. , by form 10.

x+a

23. And in general for the fluent of

where n is

any even positive number, by dividing till the powers of x in the th numerator are exhausted, the fluents will be found as

before. And first for the denominator x2 + a2, as in

απά

x2 + a2

Ex. 20. For the fluent of = (by actual division)

number of terms in the quotient being in, and the remainder viz,

or + according as that number of terms is

odd or even. Hence, as before, the fluent

is

2

-2

+ &c... + a"-2xd-2 x arc to rad.

n-1

n-3

Ex. 21. In like manner, the fluent of ,is

a+x

&c + a"- x hyp. log..

24. In a similar manner we are to proceed for the fluents of

xi when n is any odd number, by dividing by the de

212

nominator 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.

xnx

Ex. 22. Thus, for the fluent of where

is an odd - αχ

number, the quotient by division as before, is "* + a*x&c ± a"-3xx, the number of terms being m-1 2, and the remainder 干

an-xx
x2 + a2
απ-3x2
2
ax
22-a2

is obtained in the same

Ex. 23. The fluent of manner, and has the same terms, but the signs are all positive, and the remainder is + a"- x hyp. log. x2 - a2.

Ex. 24. Also the fluent of is still the same, but the

signs are all negative, and the remainder is - a"- x hyp. log. a2 - x2. Hence also,