MATH 152 Midterm: MATH 152 TAMU 2009a Exam 3a Solutions
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Solutions-form a: a, d: t3(x) = f (i)(1) i! Pi=0 (x 1)i f (1) f (1) (x 1)2 + (x 1)3. 3: a: taylor"s inequality says an upper bound on the absolute value of the remainder in using tn(x) to approximate f (x) in an interval containing a is. M = max|f (n+1)(x)| for x in an interval containing: here n = 3 and a = (n + 1)!|x a|n+1 where. Thus |r3(x)| x where m = max|f (4)(x)| for 0 x f (4)(x) = sin x, hence the maximum of |f (4)(x)| oc- cures when x = 1: c: recall an = lim n sn, where sn is the sequence of partial sums. We are given that sn = 4 + ln(2n) ln(n + 1), thus. Pn=1 n (cid:18)4 + ln (4 + ln(2n) ln(n + 1)) n + 1(cid:19) = 4 + ln 2 an = lim n .
