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# HW00[1] - University of Illinois Fall 2009 ECE 313 Problem...

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University of Illinois Fall 2009 ECE 313: Problem Set 0: Solutions Calculus Review 1. [Geometric, MacLaurin, and Taylor series; L’Hˆ opital’s rule] (a) Let n denote a positive integer. Then, (1 - x ) 1 + x + x 2 + · · · + x n - 1 = 1 + x + x 2 + · · · + x n - 1 - x + x 2 + · · · + x n = 1 - x n . If x 6 = 1, divide both sides by (1 - x ) to get 1 + x + x 2 + · · · + x n - 1 = 1 - x n 1 - x . (b) At x = 1, the sum is 1 + 1 1 + 1 2 + · · · + 1 n - 1 = n . Now, since 1 - x n 1 - x is the indeterminate form 0 0 at x = 1, we apply L’Hˆ opital’s rule to get lim x 1 1 - x n 1 - x = lim x 1 - nx n - 1 - 1 = n . Therefore, the value of 1 + x + x 2 + · · · + x n - 1 at x = 1 does equal lim x 1 1 - x n 1 - x . (c) Since | x | < 1, 1 + x + x 2 + · · · = lim n →∞ 1 + x + x 2 + · · · + x n - 1 = lim n →∞ 1 - x n 1 - x = 1 1 - x . (d) Using the product rule, d dx exp( - x ) n X k =0 x k k ! = - exp( - x ) n X k =0 x k k ! + exp( - x ) n X k =1 kx k - 1 k ! = - exp( - x ) n X k =0 x k k ! - n - 1 X j =0 x j j ! = - exp( - x ) x n n ! . Next, since X k =0 x k k ! = exp( x ), we have that exp( - x ) X k =0 x k k ! = 1 and so its derivative is 0. 2. [The binomial theorem] (a) The 0-th derivative f (0) ( x ) is just f ( x ) = (1 + x ) n itself. The k -th derivative is f ( k ) ( x ) = n ( n - 1) · · · ( n - k + 1)(1 + x ) n - k for k n . Note that f ( n ) ( x ) = n ( n - 1) · · · 2 · 1 = n ! is a constant, and hence the higher-order derivatives are 0. Thus, the MacLaurin series for

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