For let x, p(x), be the co-ordinates of a point P on the curve y=4(x), and let x+h, p(x+h) be the co-ordinates of another point Q, also on the curve. And suppose the curve and the inclination of the tangent to the curve to the axis of x to be continuous and finite between P and Q; draw PM, QN perpendicular to OX and PL perpendicular to QN, then p(x+h)− p(x) ___ NQ — MP __ LQ h tan LPQ. Also, x+0h is the abscissa of some point R on the curve between P and Q, and p'(x+0h) is the tangent of the angle which the tangent line to the curve at R makes. with the axis of x. Hence the assertion that is equivalent to the obvious geometrical fact that there must be a point R somewhere between P and Q at which the tangent to the curve is parallel to the chord PQ. 126. The cases in which Taylor's Theorem is said to fail are those in which it happens (1) That p(x), or one of its differential coefficients, becomes infinite between the values of the variable considered; (2) Or that p(x), or one of its differential coefficients, becomes discontinuous between the same values; hn (3) Or that the remainder, "(x+0h), cannot be made Ex. If n! to vanish in the limit when n is taken sufficiently large, so that the series does not approach a finite limit. Hence Taylor's Theorem gives p(x+h)=√x+h= √x + 2 √√ x 1 .h+... 1 If, however, we put x= =0, becomes √h. becomes infinite, while √x+h 2√x Thus, as we might expect, we fail at the second term to expand √h in a series of integral powers of h. 127. In Art. 107 the proof of Taylor's Theorem is not general, the assumption being made that a convergent expansion in ascending positive integral powers of x is possible. The above article points out clearly when this assumption is legitimate. For any continuous function in which the (p+1)th differential coefficient is the first to become infinite or discontinuous for the value of the variable, the theorem hp p(x+h)=4(x)+hp'(x)+...+= p2(x+Oh), 20! which involves no differential coefficients of higher order than the pth, is rigorously true, although Taylor's Theorem, p(x+h)=4(x)+hp'(x)+...+ hp p hp+1 p! 13 (∞ ) + ( p + 1 ) ! 2 + 1 (x) + ... fails to furnish us with an intelligible result. But Equation 6 of Art. 120 gives the result 5 h2 (x+h− a) 3 = (x − a) 1⁄2 + 1⁄2 (x − a) 3⁄4 h+ 4 2! and this obeys the only limitation necessary, viz., that ✪ should be à proper fraction. 128. The remarks made with respect to the failure of Taylor's Theorem obviously also apply to the particular form of it, Maclaurin's Theorem, so that Maclaurin's Theorem is said to fail when any of the expressions ¿(0), p'(0), p′′(0), ... become infinite, or if there be a discontinuity in the function or any of its differential coefficients as a passes through the value zero, or if the remainder xn nip(x) does not become infinitely small when H n becomes infinitely large, for in this case the series is divergent and does not tend to any finite limit. 129. Examples of Expansions by Maclaurin's Theorem, with investigation of Remainder after n terms. Now x^ax(log.a)n n! 2! xn n! can be made smaller than any assignable quantity by sufficiently increasing n; hence the remainder, after n terms of Maclaurin's Theorem have been taken, ultimately vanishes when n is taken very large, and therefore Maclaurin's Theorem is applicable and gives x2 X3 a2 = 1+xlog.a+2, (log.a)2+3, (log.a)3+ ... to œ. Hence ..., .ƒ(0)=0, ƒ'(0) =1, ƒ"(0) = −1, ƒ'''(0)=2 ... And the Lagrange-formula for the remainder, after n terms of Mac laurin's Series have been subtracted from f(x), viz. xnƒn (0x) becomes n! is a proper frac tion, and therefore by making ʼn sufficiently large the above rẹmainder ultimately vanishes, and therefore Maclaurin's Theorem is applicable and gives log(1+x)=x x2 x3 x4 + + ... to ∞, 2 where x lies between 0 and 1 inclusive. It appears that if we consider f(x)=log(1 − x) the remainder is In this form it is not clear that the limit of the remainder is zero. But if we choose for this example Cauchy's form of remainder, Art. 123, it reduces to n unity, and therefore 1(-6x)" )" can be made as small as we like 1 by sufficiently increasing n. - Hence Maclaurin's series is applicable with a similar notation for higher differential coefficients. Then Maclaurin's Theorem gives Changing the sign of x we see that the left hand member of this equation remains unaltered; hence we have whence, by equating to zero the coefficients of the several powers of x, we infer that so that the expansion contains no odd powers of x. |