RecitationOneSols

# RecitationOneSols - Rensselaer Electrical Computer and...

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Rensselaer Electrical, Computer, and Systems Engineering Department ECSE 4500 Probability for Engineering Applications Recitation 1 Solutions Wednesday 27 January, 2010 Review of Relevant Mathematics ref. Appendix A in Stark and Woods (S&W) text 1. sum of Geometric Series and related (a) Consider the formula: X n =0 a n = 1 1 a Under what conditions on a is the formula correct? Ans: We must have | a | < 1 . Why? Ans: It ’sthesumo fthegeometr icser ies . (b) More generally, we have, for n 2 >n 1 : n 2 X n = n 1 a n = a n 1 a n 2 +1 1 a , which holds for any value of a ,except a =1 . Find this result by using the formula aS = S + a n 2 +1 a n 1 , where S , P n 2 n = n 1 a n . Ans: Bringing aS to the other side, we f nd 0= S (1 a )+ a n 2 +1 a n 1 , which implies S (1 a )= a n 1 a n 2 +1 , or S = a n 1 a n 2 +1 1 a for a 6 . Show why is this formula true for the geometric series. Ans: The formula is true because aS = a n 2 X n = n 1 a n = n 2 X n = n 1 a n +1 = n 2 X n = n 1 a n + a n 2 +1 a n 1 = S + a n 2 +1 a n 1 , as was to be shown. (c) Now de f ne a generating function as: G ( z ) , X n =0 a n z n . 1

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1. Show that G (1) = P n =0 a n = 1 1 a whenever | a | < 1 . Ans: Just plug z =1 into the de f nition of G ( z ) above. 2. Show that G 0 (1) = P n =0 na n = a (1 a ) 2 whenever | a | < 1 . Ans: We use the geometric series with a replaced with az, to get G ( z )= 1 1 az , then we take the derivative dG/dz to obtain a/ (1 az ) 2 . So G 0 (1) = a/ (1 a ) 2 . Meanwhile,
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RecitationOneSols - Rensselaer Electrical Computer and...

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