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Unformatted text preview: MATH 150 Series Cheung Man Wai Tutorial 9 The purpose of this note is to find two independent solutions around x of the following P ( x ) y 00 + Q ( x ) y + R ( x ) y = 0 , (1) where P ( x ), Q ( x ) and R ( x ) are polynomials with no common factors. We shall consider two cases respectively, say x is an ordinary point (if P ( x ) 6 = 0) or a singular point (otherwise) of (1). 1. Ordinary points We determine two power series solutions for y = y ( x ) around x . Let us see an example. Example 1. y 00 + k 2 x 2 y = 0 , x = 0 , k a constant . With y ( x ) = ∞ X n =0 a n x n , we find that y 00 ( x ) = ∞ X n =2 n ( n 1) a n x n 2 and the equation becomes ∞ X n =2 n ( n 1) a n x n 2 + k 2 ∞ X n =0 a n x n +2 = 0 . By shift of index one has ∞ X n =0 ( n + 2)( n + 1) a n +2 x n + k 2 ∞ X n =2 a n 2 x n = 0 . 1 The coefficient of each power of x vanishes, thus a 2 = a 3 = 0 , ( n + 2)( n + 1) a n +2 + k 2 a n 2 = 0 , n ≥ 2 . The recurrence relation can be equivalently written as a n +4 = k 2 ( n + 4)( n + 3) a n , n ≥ . Four sequences of coefficients, starting with a ,a 1 ,a 2 and a 3 decouple. In par ticular the sequences are a ,a 4 ,a 8 ,a 12 ,... ; a 1 ,a 5 ,a 9 ,a 13 ,... ; a 2 ,a 6 ,a 10 ,a 14 ,... ; a 3 ,a 7 ,a 11 ,a 15 ,.......
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 Spring '09
 T.Qian
 Math, Polynomials, Factors

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