of the series (2) continued to an infinite number of terms. For the (v + 1)th term in the series (1), supposing v to be a number, between r and n, much greater than r, and, in fact, comparable with n, will not converge to the expression 1 . fv (x), 3 n (-) (1-2) (1-2) (1-1) will not converge to unity, but, in fact, for values of v very nearly in a ratio of equality to n, will be reduced to zero. We may however always suppose that If we determine R, being an unknown function of x and h. the conditions that R, may lie between limits, which, as r increases indefinitely, become less than any assignable quantities, we shall have thereby ascertained the conditions under which the function f(x + h) may be considered to be equivalent to the series (2) continued to infinity. 83. With this object in view, we remark that, if a function (h) is equal to zero when h = 0, and if its derivative p'(h) retains always the same sign between the limits 0, h, without passing through infinity, the function (h) will have always for this interval the same sign as '(h). For and therefore 84 (h), when dh is indefinitely small, must have always the same sign as '(h) Sh, that is, (h) must, with the continuous increase of h, be continuously increasing or continuously diminishing accordingly as '(h) is positive or negative, but, by hypothesis, (h) = 0 when h= 0; it follows therefore that (h) must always have the same sign as p′(h). Suppose now that we assume, as we are evidently at liberty being a symbol of unknown functionality. It is manifest then that we shall have determined inferior and superior limits to the value of R,, if we ascertain two quantities P, Q, such that, for all values of h, Let ø,(h), p,(h), represent the former members of these two inequalities: the functions p,(h), p,(h), both vanish when h = 0; hence, from what has been said above, we know that the inequalities will be satisfied if we have, for all values of h, But p''(h), p,'(h), are also functions of h which vanish when h = 0: hence the inequalities (4), and consequently the inequalities (3), will be satisfied if p."(h)>0, p," (h) < 0, Continuing this reasoning, it is evident that after r+ 1 successive differentiations, we shall arrive at the inequalities Now these final inequalities, and consequently the primitive inequalities (3), will be satisfied, if we assign to P and Q respectively the least and the greatest value assumed by the derived function fr+1 (2), for the values of z which lie between x and x + h. Hence ℗ denoting an unknown numerical quantity lying between 0 and 1. Now, if the numerical values of the function fr1(z) never exceed a certain finite magnitude λ, for all values of z from 2 = x to z = xh, and for all values of r the index of differentiation, it is plain that R. < hr+1.λ and then, whatever be the value of h, we may always take r so large that R, may become less than any assignable magnitude. Hence, in this case, the function f(x + h) will be equal to the series (2) continued to infinity, or This formula is commonly called Taylor's theorem, after the name of its discoverer, Brook Taylor; it was published in the year 1715, in his Methodus Incrementorum. Another Demonstration of Taylor's Theorem. 84. By the application of the elementary principles of integration, a demonstration of Taylor's formula may be given, which has the advantage of determining not only the limits of R,, but also its actual value expressed by a definite integral. The student may pass over this demonstration until he has become acquainted with the Integral Calculus. Whatever be the function f, provided that it remains continuous between the values of the variable designated by x and xh, we shall have f(x + h) − f(x) = ['* ̈*ƒ'(x) . dx. - In order to express more clearly the distinction between the general value of x under the sign of integration, and the value of x which enters into the expression of the limits, we may write hence h2 1 ƒ(x + h) = f(x) + ' ƒ'(x) + 1,2 ƒ ̈(x) + 1, ["^ 2° ƒTM" (x + h − 2) dz. 1.2 0 1.2.3 S 2ƒ" (x + h − 2) dz, 0 and generally, by a series of successive steps, we see that Now, supposing P and Q to represent the least and the greatest value of the function fr+1 (x + h − z), for the series of values of z comprised between 0 and h, it is evident that we shall have It appears therefore that the value of R, falls between 85. Putting z for x, and then replacing h by x - z in the formula (7), we get |