2 k 1 k 2 ak2 2ak1 k 1 ak1 2ak k

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Unformatted text preview: ⇒ ∞ n (n − 1) an z n−2 ∞ −3 n=2 ∞ nan z n=1 k k=0 +2 k an z n = 0 ⇐⇒ ∞ (k + 1) ak+1 z + 2 k=0 ∞ ∞ n=0 ∞ (k + 2) (k + 1) ak+2 z − 3 n−1 ak z k = 0 ⇐⇒ k=0 [(k + 2) (k + 1) ak+2 − 3 (k + 1) ak+1 + 2ak ] z k = 0 ⇐⇒ k=0 (k + 2) (k + 1) ak+2 − 3 (k + 1) ak+1 + 2ak = 0, k ≥ 0. ⇐⇒ 2 (k + 1) [(k + 2) ak+2 − 2ak+1 ] = (k + 1) ak+1 − 2ak , k ≥ 0. Let bk = (k + 1) ak+1 − 2ak . Then the above relation becomes (k + 1) bk+1 = bk , k ≥ 0 ⇐⇒ bk+1 = bk+1 = = So 1 bk , k ≥ 0 ⇒ (k + 1) 1 1 1 1 bk = bk − 1 = bk − 2 = . . . = b0 (k + 1) (k + 1) k (k + 1) k (k − 1) (k + 1)! 1 1 (a1 − 2a0 ) = (2 − 2) = 0. (k + 1...
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This note was uploaded on 12/05/2013 for the course MATH 334 taught by Professor Thind during the Summer '13 term at University of Toronto- Toronto.

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