TAYLOR'S THEOREM AND MACLAURIN'S THEOREM. Let y be a function of x, which it is possible to develop in a series of positive ascending powers of that variable, and let us suppose that y= A + Bx + Cx2 + Dx3 + Ex1 + and when a becomes x + h, let y becoms y'; then we have =A+ B + C + D + +Bh+2Cxh + 3Dx2h + + Ch2+3Dxh2 + + Dh3 + (1) (2) But by eq. (1) we have the first, second, &c. differential co-efficients, as follows: h2 hence by multiplying these differential co-efficients respectively by h, 1.2 h3 (3) 1.2.3' &c. and substituting the results in equation (2) we have finally dy day h2 d3y h3 d'y h4 y' = y + h+ dx + Again, since the co-efficients A, B, C, D, &c. in (1) do not involve x, they will remain unchanged whatever value be assigned to x. Let then the particular values of y, and its successive differential co-efficients, be expressed by means of brackets, and when x =0, we shall have by (1) and (3) (du) x + 1.2 (B) dx The former of these equations (A) is Taylor's theorem, and the latter (B) is Maclaurin's theorem; and the demonstrations we have given of these most important theorems will be readily comprehended by the student. We regret that room will not permit us to exemplify the latter of these theorems. 1.2.3 Cases in which Taylor's Theorem fails. In the preceding demonstration of Taylor's theorem, we have supposed with regard to f (x) that x remains indeterminate, and .. that ƒ (x) has as many values as f (x + h). But when we assign particular values to x, the above reasonings will not always hold good, and we shall not in all cases obtain the true expansion of ƒ (x + h) . .... 1. Let ƒ (x) = x when x = a, then the expansion of ƒ (x + h) must contain negative powers of h. For a will be determined from the equation n being any positive whole number whatever, and Q (x) some function of r which does not become 0 or a when xa 2. Let f (x) contain a radical which disappears when x = a. In this case, either the radical itself must vanish when a = a, or its co-efficient must vanish. m m If the radical itself vanish in ƒ (x) when x = a, it must be of the form (2 — α)ñ' mand n being whole numbers; hence f(x+h) will contain the corresponding radical (x- a + h)"' which, on making x = a becomes h' so that the developement of ƒ (a + h) according to powers of h may contain the radical hand its powers. m If the co-efficients of the radical vanish when xa, then this co-efficient must be of the form (x -— a)", n being a whole number, in this case the radical will disappear in the differential co-efficients ƒ' (a), ƒ" (a) . . . . ƒ ¤— 1 (@), but will be found in those of higher orders. In general the following proposition will hold good: .... When we assign a particular value to x in the developement of f (x + h), if a term appear containing a fractional power of h which lies between h" and +1, then Taylor's theorem will hold good for the first n terms only. Let f(a + h) = A + B + Ch2 + Dh3 + + M/" + N/1⁄2"++++. Take the differential co-efficients regarding has the variable, and let us denote the successive co-efficients by f'(a + h), ƒ"(a + k), &c. The co-efficients A, B. C, are the values which ƒ (2) and its differential co-efficients assume when x = a, precisely as in the series of Taylor. But at each differentiation the first term disappears because it is constant. When we arrive at the nth differential co-efficient, on the supposition that h = 0. and all the succeeding differential co-efficients will, in like manner, be infinite. It only now remains for us to show how we can obtain the developement of f(x + h) when Taylor's theorem fails. If, then, we wish to obtain the developement of f (x + h) when x = a, we must calculate the terms of the series but if, in effecting this calculation, we find that one of the differential coefficients becomes infinite upon the supposition that x = a, we must employ the following process. Substitute (x + h) for x in f (x); then the term which contains x — a in the denominator, will now contain x — a + h, and will no longer become infinite when xa, but will become a term involving a fractional power of h. For example, let d'y d3y Substitute this value and the values of das, &c. in Taylor's theorem, we shall then find a2 ах ƒ (x + h) = 2ax − x2 + a √ Œ = x + { 2 (a− x) + √2}+ But, when xa the term multiplied by h becomes infinite, hence Taylor's theorem fails, and the developement is no longer possible. But in the above case, since ƒ (x) = 2ax − x2 + a √x* — a2 According to the rule just given, substitute x + h for x, then ƒ (x + h) = 2ax+2ah x2 2xh h2 + a.√ x2 + 2xh + h3 - supposition that x = a becomes a2 which, upon the Expanding √2+h by the binomial, and representing the co-efficients by A, B, C. √2a + h = (2a + h) = A + Bh + Ch2 + Dh3 + . . . Substituting ƒ (a + h) = a* — ha + aAh3 + aBhi+aChi+ a series which gives the true developement of ƒ (a + h) but which does not proceed by integral powers of h. CHAPTER VII. APPLICATIONS OF THE DIFFERENTIAL CALCULUS. ON THE THEORY OF VANISHING FRACTIONS. when a a) is a P WHEN a fraction both of whose terms are functions of a becomes Q particular value is assigned to the variable as x = a, it shows that (x common factor both of numerator and denominator, and in order to real value of the fraction, we must make this factor disappear from one or both terms. find the Now, when Taylor's theorem can be applied to expand ƒ (a + h) and (ah), we have Hence the rule to find the value of a vanishing fraction. Differentiate both terms of the fraction the same number of times, until one or other ceases to become 0, on the supposition that x = a. Then substitute a for x in both terms of the fraction, and the result will be the value required. It is necessary to take the second differential co-efficient, because the common factor of the two terms of the original fraction is (≈ — c)3. therefore the true value of the fraction is 0, the factor of the numerator is (x-a), that of the denominator is (x a). 2a3x+2ax3 α- 2x 2a3 + 6ax |