-x=h; and therefore x, xo+h, and h is the finite increment of xo; then

F(xo+h) F(x) = hr′(xo+Oh).

(3)

It is to be observed that, if h is infinitesimal, we must neglect Oh when added to the finite quantity o; and the result is, the derived-function as originally defined in Art. 18.

112.] THEOREM III.—If x and x, are two definite values of , being greater than ro and x-x being a finite quantity; and if r(x) and ƒ(x) are functions of x, which, as also their first-derived-functions, are finite and continuous for all values of a between x and x; and besides, if f'(x) does not change sign within these limits; then

where represents a fraction mean to 0 and 1.

Let the difference x-xo be divided into n parts, and let a1, 2,...,-1 be the values of a corresponding to the n-1 points of division; and let us moreover suppose n to be infinite, and let each of the divided elements, x-xo, X2 — X1,

...

Xn-Xn-1,

F′(x1)

=

ƒ'(x1) '

(4)

Now this being a series of fractions whose denominators are all of the same sign, according to Preliminary Theorem IV, the ratio of the sum of all the numerators of the left-hand members of the equations to the sum of all the denominators is equal to some mean value of the fractions; that is,

F(x) F(x) ƒ (xn) —ƒ (x0)

is equal to some mean value of the fractions which are the

right-hand members of the above equations; and such a value will be properly represented by

F'{xo+0(xn−xo)} ƒ'{xo+0(xn− xo)}'

where is a proper and positive fraction; therefore

F(x) F(xo) F'{xo+0(xnxo)}

=

f(xn)—f(xo) ƒ'{xo +0 (xn−xo)}

For the sake of convenience, let x-xo=h, so that xnxo+h; and therefore h is the finite increment of

Xo, and the functions under consideration include those for all values of a between ro and ro+h; then

--

F(xo+h) F(xo)
f(xo+h)-f(xo)

=

F'(xo + 0h) f'(xo+0h)*

(5)

The further condition to which f(x) is subject, viz. that ƒ'(x) does not change sign between x。 and xo+h, is necessary, in order that the sum of the denominators of the right-hand members of (4) may not vanish, for thereby the first member of (5) might be equal to an infinite quantity, and the equation might be useless.

To enable a student the better to grasp the full meaning of the conditions of this important theorem, the graphical illustration of fig. 11 is given.

Of the two curves therein delineated, let the upper one be that whose equation is y = F(x), and let the lower one represent y = f(x).

Let o be the origin, oм。 = x0, 0м,=,; therefore M, M。=h; Oм = x, x referring to any point intermediate to Mo and M2; and suppose that f'(x) is positive for all values of a between o and xn, which property is represented by the curve being such that the ordinate increases as the abscissa increases; now it is immaterial what forms, or branches, or points of discontinuity the curves may have outside the assigned limits; we consider them only within and between the limits, and restrict them to certain conditions within those limits, and which are expressed in the curves of the figure, viz. that they are finite and continuous; also

COR. I.-Suppose in the above equation (5), that F(x)=0 and f(x) = 0, which expresses the condition that both the curves in fig. 11 pass through the point Mo, then

F(xo+h) F'(xo + Oh)

=

f(xo+h) f'(xo + Oh)*

(6)

COR. II. Also suppose that xo, which is the lower limit of x, 0, which expresses the condition that in fig. 11 the origin of coordinates is at Mo, then

=

and writing x for h, as h is measured from the origin mo in this case, we have

F(x) - F(0) F'(0x)

=

f(x) —ƒ(0) f'(0x)

and if F(0) = 0, and ƒ(0) = 0, that is, if both the curves pass through the origin,

F(x) F'(0x)

=

f(x)

f'(0x)*

(8)

113.] By Corollary I. of the last Article and equation (6), it appears that if any two functions of a vanish when x = xo; then, subject to the necessary conditions,

replacing oh by h1 for the sake of convenience, and observing therefore that h1 is less than h.

Suppose now that r′(x) = 0, and f'(x) = 0, and that r"(x) and ƒ"(x), as well as r'(x) and f'(x), are finite and continuous for all values of x between co and xo+0h; and that f"(x) does not change sign within these limits; then, by virtue of (9), we have

F′(xo+h1) F" (xo + 0h1)

=

f'(xo+h1) f" (xo+0h)

;

and replacing th1 by h2, and observing that he is less than h1, we have

F" (xo+h2)

=

F′(xo+h1)
f'(xo+h1) f" (xo+ha)

(10)

Similarly again, if "(x) = 0 and ƒ"(x) = 0, and if the derived-functions F""(x), ƒ""(x) are finite and continuous for all values of a between x and x。+h2; and if, besides, ƒ"(x) always increases or decreases with x between these limits; then

PRICE, VOL. I.

C C

replacing the by hs; whence it appears, that hs, which is less than ha, which is also less than h1, and therefore less than h, is of the form Oh.

Suppose now that these several conditions as to the derivedfunctions, and others similar to them, hold good up to the (n−1)th derived-functions inclusively, then

But by (5) the left-hand member of this equation is equal to

so that we have the following proposition:

THEOREM IV.-If F(x) and f(x) are two functions of x, which, and also all the derived-functions up to the nth inclusively, are finite and continuous for all values of a between xo and xo+h; and if

COR. I.-Hence, if r(x) = 0, and f(x) = 0,

F(xo+h) F" (xo+ 0h)

(12)

(13)

COR. II. Also, if so, which is the inferior value of x, = 0; then, writing a for h, as h is measured from the origin, we have from (12),

F(x) — F (0) f(x) − f(0)

n

=

F" (0x) ƒ" (0x) *

(14)

COR. III. And if in addition F(0) = 0, and f(0) = 0, then

-

114.] In equation (12), which is the result including all others of the above Articles, let a specific form be assigned to f(x), which will satisfy the requisite conditions and render the equation applicable to future purposes; as, for instance, let ƒ (x) = (x−x。)”, wherein n is positive and integral; then

f(x) = 0, and f(xo+h) = h";

and as f(x) does not involve x, it has the same value whatever x is, therefore

f" (xo+0h) = n(n-1) ... 3.2.1;

and substituting these values in the general result, we have

F(xo+h) − F(x0)

=

hn
1.2.3... (n-1) n

F" (xo+0h);

(16)

the conditions to which this equation is subject being, that F(x), F′(x), F′(x),... F" (x) are finite and continuous for all values of a between x and xo+h, and that r′(x) = 0, r′′(x) = 0, up to p"-1(x)=0; and that F" (xo) does not vanish.

115.] COR. I.-Hence, if r (a) = 0, we have

which proposition may be enunciated as follows:

THEOREM V.-If r(x) is a function of x, which and also all its derived-functions up to the nth inclusively are finite and continuous for all values of a between 0 and x; and if F(x), F′(x), ... F-1(x) severally vanish when a = 0, then