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aptsol - UCSD ECE153 Handout#5 Prof Young-Han Kim Tuesday...

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Unformatted text preview: UCSD ECE153 Handout #5 Prof. Young-Han Kim Tuesday, March 29, 2011 Solutions to ECE153 Aptitude Test 1. ∞ summationdisplay n =0 r n = 1 1 − r . Solution: Let S = 1 + r + r 2 + ··· . (The series converges since 0 < r < 1.) Then rS = r + r 2 + r 3 + ··· . Taking the difference between two equations, we have S − rS = 1, or equivalently, S = 1 1 − r . 2. ∞ summationdisplay n =0 nr n = r (1 − r ) 2 . Solution: There are two ways to evaluate this. First, we can write S = r + 2 r 2 + 3 r 3 + ··· . Then rS = r 2 + 2 r 3 + 3 r 4 + ··· . By taking the difference between S and rS , we can easily see that (1 − r ) S = r + r 2 + r 3 + ··· = r 1 − r . Alternatively, we have ∞ summationdisplay n =0 nr n = r ∞ summationdisplay n =0 nr n − 1 = r ∞ summationdisplay n =0 d dr ( r n ) = r d dr parenleftBigg ∞ summationdisplay n =0 r n parenrightBigg = r d dr parenleftbigg 1 1 − r parenrightbigg = r (1 − r ) 2 . (Note: The interchange of the differentiation and the summation can be justified. We already learned a similar and more general method in ECE109 with moment generating functions; the given identity is nothing but the mean of a geometric random variable.) 1 3. ∞ summationdisplay n =0 r n n ! = e r . Solution: This is one of famous Maclaurin series (a special form of Taylor series eval- uated at 0). The Maclaurin series generated by a function f ( r ) is f ( r ) = ∞ summationdisplay n =0 f ( n ) (0) n ! r n where f ( n ) (0) denotes the n th derivative of f evaluated at 0....
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