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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 0: Solutions Calculus Review 1. [Geometric, MacLaurin, and Taylor series; LHopitals 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 at x = 1, we apply LH opitals 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 0th derivative f (0) ( x ) is just f ( x ) = (1 + x ) n itself. The kth derivative is f ( k ) ( x ) = n ( n 1) ( n k + 1)(1 + x ) n k...
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This note was uploaded on 10/25/2010 for the course ECE 313 taught by Professor Milenkovic,o during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Milenkovic,O

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