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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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This note was uploaded on 09/16/2011 for the course ECE ECE153 taught by Professor Younghankim during the Spring '11 term at UCSD.
- Spring '11