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Unformatted text preview: 1 Notes for Expansions/Series and Differential Equations In the last discussion, we considered perturbation methods for constructing solutions/roots of algebraic equations. Three types of problems were illustrated starting from the simplest: regular (straightforward) expansions, nonuniform expansions requiring modification to the process through inner and outer expansions, and singular perturbations. Before proceeding further, we first more clearly define the various types of expansions of functions of variables. 1. Convergent and Divergent Expansions/Series Consider a series, which is the sum of the terms of a sequence of numbers. Given a sequence { } 1 2 3 4 5 , , , , ,...., .. n a a a a a a , the n th partial sum S n is the sum of the first n terms of the sequence, that is, 1 . n n k k S a = = & (1) A series is convergent if the sequence of its partial sums { } 1 2 3 , , ,...., .. n S S S S converges. In a more formal language, a series converges if there exists a limit such that for any arbitrarily small positive number >0, there is a large integer N such that for all n N , n S  & . (2) A sequence that is not convergent is said to be divergent. Examples of convergent and divergent series: The reciprocals of powers of 2 produce a convergent series: 1 1 1 1 1 1 ......... 2 1 2 4 8 16 32 + + + + + + = . (3) The reciprocals of positive integers produce a divergent series: 1 1 1 1 1 1 ......... 1 2 3 4 5 6 + + + + + + (4) Alternating the signs of the reciprocals of positive integers produces a convergent series: 1 1 1 1 1 1 ......... ln 2 1 2 3 4 5 6 + + + = . (5) The reciprocals of prime numbers produce a divergent series: 1 1 1 1 1 1 ......... 2 3 5 7 11 13 + + + + + (6) 2 Convergence tests: There are a number of methods for determining whether a series is convergent or divergent . Comparison test : The terms of the sequence { } 1 2 3 4 5 , , , , ,...., .. n a a a a a a are compared to those of another sequence { } 1 2 3 4 5 , , , , ,...., .. n b b b b b b . If, for all n , 0 n n a b , and 1 n n b = & converges, then so does 1 n n a = & . However, if, for all n , 0 n n b a , and 1 n n b = & diverges, then so does 1 n n a = & . Ratio test : Assume that for all n , a n > 0. Suppose that there exists an r > 0 such that 1 lim n n n a r a + = . (7) If r <1, the series converges. If r >1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge....
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This note was uploaded on 12/28/2011 for the course ME 580 taught by Professor Na during the Fall '10 term at Purdue UniversityWest Lafayette.
 Fall '10
 NA

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