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Homework1 Solution - University of Wisconsin, Madison I SY E 645

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HW 1 Solution Problem 1 Denote T¯ as the average age. a) By denition, T¯ = X N i=1 1 N · i = 1 N (1 + N) · N/2 = (1 + N) /2 b) By denition, T¯ = X∞ i=0 α (1 − α) i ... (i + 1) = limn→∞ α Xn i=0 (1 − α) i (i + 1) Let x = Pn i=0 (1 − α) i (i + 1). αx = x − (1 − α)x = Xn i=0 (1 − α) i (i + 1) − Xn i=0 (1 − α) i+1 (i + 1) = Xn i=0 (1 − α) i − (1 − α) n+1 (n + 1) = (1 − α) 0 − (1 − α) n+1 1 − (1 − α) − (1 − α) n+1 (n + 1) = 1 − (1 − α) n+1 α − (1 − α) n+1 (n + 1) Since 0 < 1 − α < 1, limn→∞ (1 − α) n+1 = 0. By the L'Hôpital's rule, 1 limn→∞ (1 − α) n+1 (n + 1) = limn→∞ (n + 1) 1 (1−α) n+1 = limn→∞ 1 n+1 (1−α) n+2 =0. Therefore T¯ = limn→∞ αx = limn→∞ 1 − (1 − α) n+1 α − (1 − α) n+1 (n + 1) = 1 α Problem 2 a) By Problem 1 a), the fact that the average age is 2 indicates N = 3. Therefore F9 = 19+20+21 3 = 20, and F10 = 20+21+18 3 = 19 2 3 . MSE = 1 2 2 2 +  4 3 2 ! ≈ 2.89 b) By Problem 1 b), α = 1 2 . By iterating Ft+1 = αDt + (1 − α) Ft we obtain F9 ≈ 20.25 F [Show More]

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