. Differential and integral calculus. f{x) +f(x)y +f\x)f +f(x)£ + etc., Li L3 118 Differential Calculus and making y — — x and transposing, we have, f{x) = /(o) + xf\x) - ff(x) + Li 3. 14. By aid of this formula we are enabled to expand a function of asingle variable into a series. Thus, ,4 e~x = i jc2 __ xz _— ^ z — ?—e ? — ^[4 Dividing through by e~x and transposing, we have,-L = ^=i+* + ^ + -- + -+--- ll Li [4See Ex. 6, p. in. 98. We have referred in a previous article (90) to the factthat the term development was inapplicable to a divergent series,and the student was cautioned not to accep
. Differential and integral calculus. f{x) +f(x)y +f\x)f +f(x)£ + etc., Li L3 118 Differential Calculus and making y — — x and transposing, we have, f{x) = /(o) + xf\x) - ff(x) + Li 3. 14. By aid of this formula we are enabled to expand a function of asingle variable into a series. Thus, ,4 e~x = i jc2 __ xz _— ^ z — ?—e ? — ^[4 Dividing through by e~x and transposing, we have,-L = ^=i+* + ^ + -- + -+--- ll Li [4See Ex. 6, p. in. 98. We have referred in a previous article (90) to the factthat the term development was inapplicable to a divergent series,and the student was cautioned not to accept an infinite seriesobtained by any of the foregoing methods as the development ofthe function, unless the question of its convergency or divergencyhad been previously settled. It remains to show how series areexamined for divergency and convergency. 99. Lemma. 7//(x)]a = o andf(x)\ = o andf(x) is continu-ous between these fruiting values of x, then f(x) = o for somevalue of x intermediate betiveen x = a and x — Fig. 14. Let APB be the locus of y =f(x), and let OA — a, andOB = b; then /(*)]. =*o, and/(*)], Series 119 Since, by hypothesis, f{x) is continuous between the pointsA(a,6) and Bib, o), there must be some intermediate point(OC, CP) where the tangent to the curve is II to the ^ at such a point | =/(*) = o. Hence the proposition. 100. Lagranges Theorem on the limits of Taylors Taylors formula may be written,— ]2 « — I + ^ ••••(>) in which P is some function of x and y, and P— = Remainder after ?i terms. It is desired to find the value of P and thence the value ofPf^. Let j = X — x. Substituting in (i) in every term, including P, and transposingwe have (x-xy f(X)-f(x)-f\x)(X-x) -/(*) 1 -f*-\x)±-. - P^—, ^- = o (2) K y — 1 v in which P is now a function of x and X. Replacing x by z in every term in the preceding expression, 1 This discovery of Lagrange placed Taylors theorem on a satisfactory basis for t
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