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Unformatted text preview: Lecture 22 Power series solutions for 2nd order linear ODEs (not necessarily with constant coecients) Recall a few facts about power series: n =0 a n z n This series in z is centered at z = 0. Here z can be real or complex. Denote b  z  its absolute value, which is a nonnegative real number. The following questions arise: (Q1) When does this converge absolutely, i.e., when does n a n  z  n con verge? Moreover, is there a real number R > 0 such that a n z n converge absolutely z with  z  < R , but diverges for  z  > R ? (Q2) When can we differentiate a n z n term by term, i.e., when do we have the equality ( a n z n ) = na n z n 1 ? (Q3) Given a function f ( z ) when can we express it as a power series a n bz n , for z in some region, say for  z  < R ? Given n =0 a n z n , we look at L = lim n (  a n +1   z  n +1  a n   z  n ) = ( lim n  a n +1   a n  )  z  If L makes sense, it will be a nonnegative real number. A basic result (cf. Chapter 2 of the Notes for Ma1a) is the following: Theorem 1 a n z n converges absolutely iff L < 1 . Definition The radius of convergence R is the largest nonnegative real number such that 1 a n z n converges absolutely for  z  < R , and a n z n diverges for  z  > R . We will allow R to be , in which case a n z n converges z . By this definition,  z  < R lim n  a n +1  a n   z  = 1 . So R = lim n  a n   a n +1  . This answers Q1. For Q2, here is another basic result (cf. Chap. 2 of the Notes for Ma1a): Theorem 2 For  z  < R , we have ( a n z n ) = a n nz n 1 . If  z  = R , anything can happen. To understand Q3 better, recall (from Ma1a) that if f is infinitely differ entiable at z = 0, then we have the associated Taylor series n a n z n , with a n = f ( n ) (0) n ! . Examples of infinitely differentiable functions at z = 0: e z , sin( z ) , cos( z ), polynomials, and rational functions P ( z ) /Q ( z ), with P, Q polynomials and Q (0) = 0. A Subtle Fact : The Taylor series of a function f (around z = 0 ) may not equal f ( z ) , for f infinitely differentiable. The simplest example of such a function is ( x ) which is e 1 /x if x > and 0 if x 0. You may check that is differentiable to any order at x = 0, but with ( n ) (0) = 0 for all n 0, making the Taylor expansion around 0 to be identically 0. But ( x ) is > 0 for x > 0, however small, and so cannot be represented by the Taylor expansion....
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 Fall '08
 Makarov,N
 Power Series

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