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

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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 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...
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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.

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

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