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SalasSV_11_04_ex_ans

# 5 1 x 2 k 1 2k 1 53 k 0 ak k x k 1 1

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Unformatted text preview: 0 ≤ k ≤ n, P (k ) (0) = k !ak ; for k > n, P (k ) (0) = 0. Thus P (x) = k =0 P (k ) (0) xk . k! 51. d 2k ( sinh x) dx2k x =0 = sinh (0) = 0 : d 2k +1 ( sinh x) dx2k +1 ∞ k =0 x =0 = cosh (0) = 1 Therefore sinh x = x + ∞ x3 x5 + + ··· = 3! 5! ∞ 1 x 2 k +1 (2k + 1)! ∞ 53. k =0 ak k x, k! (−∞, ∞) 1+ 1− x −n e1/x2 1 3 1 3 55. k =0 ( − 1)ka2k 2k x, (2k )! 1 3 3 (−∞, ∞) 5 57. ln a + k =1 ( − 1)k −1 k x, kak (−a, a] 59. In 2 = In ∼2 1 + 1 = 3 3 + 1 5 1 3 = 842 1215 842 ∼ = 0. 693 1215 61. routine; use u = (x − t )k and dv = f (k +1) (t ) dt cxk e1/x2 63. (b) f (x) = and lim f (x) has the form ∞ x→0 ∞. Successive applications of L opital’s rule will ﬁnally produce a quotient of the form ’H ˆ x→0 , where k is a nonnegative integer and c is a constant. It follows that lim f (x) = 0. (c) f (0) = lim e −1/x2 −0 x→0 x x→0 = 0 by part (b). Assume that f (k ) (0) = 0. Then f (k +1) (0) = lim f (k ) ( x ) − 0 f (k ) ( x ) = lim . x→0 x x...
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