lecture11 - Differential Equations & Linear...

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Unformatted text preview: Differential Equations & Linear Algebra Lecture 11 Jianli XIE Email: xjl@sjtu.edu.cn Department of Mathematics, SJTU Fall 2010 Contents Review of power series Radius of convergence Operations of convergent series Taylor series and Maclaurin series Shift of index of summation Contents Review of power series Radius of convergence Operations of convergent series Taylor series and Maclaurin series Shift of index of summation Series solutions near an ordinary point Method of series solutions near an ordinary point Linear equations with variable coefficients For a linear equation, only when the coefficients are constants, we have a systematic method to solve. If the coefficients are variable, it is necessary to extend our search for solutions beyond the elementary functions. The principal tool that we need is the representation of a given function by a power series. The basic idea is similar to the method of undetermined coefficients: we assume that the solutions of a given differential equation have power series expansions, and then we attempt to determine the coefficients so as to satisfy the differential equation. Convergence of a power series A power series ∞ X n =0 a n ( x- x ) n is said to converge at a point x if lim m →∞ m X n =0 a n ( x- x ) n exists for that x . The series certainly converges for x = x ; it may converge for all x , or it may converge for some values of x and not for others. The series ∞ X n =0 a n ( x- x ) n is said to converge absolutely at a point x if the series ∞ X n =0 | a n ( x- x ) n | = ∞ X n =0 | a n || x- x | n converges. Convergence of a power series It can be shown that if the series converges absolutely, then the series also converges; however the converse is not necessarily true. If the power series ∞ X n =0 a n ( x- x ) n converges at x = x 1 , it converges absolutely for | x- x | < | x 1- x | ; and if it diverges at x = x 1 , it diverges for | x- x | > | x 1- x | . Ratio test Ratio test for the absolute convergence of a power series: If a n 6 = 0 , and if, for a fixed value of x , lim n →∞ a n +1 ( x- x ) n +1 a n ( x- x ) n = | x- x | lim n →∞ a n +1 a n = | x- x | L, then the power series converges absolutely for all x such that | x- x | L < 1 and diverges for those x such that | x- x | L > 1 . Example Example For which values of x does the power series ∞ X n =1 (- 1) n +1 n ( x- 2) n converge? Solution Use the ratio test. Since lim n →∞ (- 1) n +2 ( n + 1)( x- 2) n +1 (- 1) n +1 n ( x- 2) n = | x- 2 | lim n →∞ n + 1 n = | x- 2 | , the series converges absolutely for | x- 2 | < 1 , or 1 < x < 3 , and diverges for | x- 2 | > 1 . When x = 1 or 3 , the series diverges since the general term does not approach zero as n → ∞ . Radius of convergence There is a nonnegative number ρ , called the radius of convergence , such that ∞ X n =0 a n ( x- x ) n converges absolutely for | x- x | < ρ and diverges for | x- x | > ρ ....
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lecture11 - Differential Equations & Linear...

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