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UDE, Fakultät für Mathematik: Discrete Mathematics (C1) [Recursions] 1 Solve recursions by the generating function method Define generating function for a sequence, defined by a recursion Determine the sequence (an ) defined by an+2 = 8 · 3n+1 + 9an − 8an+1 , Define the generating function: A(x) := n ∈ N0 ∞ X a0 := 2, a1 := −6 ak x k k=0 Separate the terms involving initial conditions A(x) = 2 − 6x + ∞ X ak+2 xk+2 k=0 Substitute the recursion equation A(x) = 2 − 6x + ∞ X 8 · 3k+1 + 9ak − 8ak+1 xk+2 k=0 Rewrite using the generating function A(x) = 2 − 6x + ∞ X 8 · 3k+1 xk+2 + k=0 = 2 − 6x + 8x ∞ X k=0 ∞ X 9ak xk+2 − 3 x + 9x 2 8ak+1 xk+2 k=0 k=0 k+1 k+1 ∞ X ∞ X k ak x − 8x k=0 ∞ X ak+1 xk+1 k=0 1 = 2 − 6x + 8x( − 1) + 9x2 A(x) − 8x(A(x) − 2) 1 − 3x Solve for generating function 1 2 − 6x + 8x( 1−3x − 1) + 16x 1 + 8x − 9x2 (2 + 10x)(1 − 3x) + 24x2 = (1 + 8x − 9x2 )(1 − 3x) −6x2 + 4x + 2 = (1 + 8x − 9x2 )(1 − 3x) A(x) = UDE, Fakultät für Mathematik: Discrete Mathematics (C1) [Recursions] 2 (1 − x)(6x + 2) (1 − x)(9x + 1)(1 − 3x) 6x + 2 = (9x + 1)(1 − 3x) = Note: To obtain the factorization above it may be helpful to rewrite the poylnomial, e.g.: 8 1 1 + 8x − 9x2 = (−9)(x2 − x − ) 9 9 Then determine the zeros: 4 y= ± 9 r 16 1 4 5 + = ± 81 9 9 9 Hence: 1 1 + 8x − 9x2 = (−9)(x − 1)(x + ) = (1 − x)(1 + 9x) 9 Expand into a power series For this step one may use partial fraction decomposition, Taylor-series-expansion or operations with known power series. Here we use partial fraction decomposition: A(x) = a b + ⇔ 6x + 2 = a(1 + 9x) + b(1 − 3x) = (9a − 3b)x + (a + b) 1 − 3x 1 + 9x The coefficients may be determined by solving the system 9a − 3b = 6, a+b=2 ⇐⇒ a = 1, b = 1 Alternatively one may insert suitable values for x, e.g. x = −1/9 and x = 1/3. Now A(x) can be expanded with the help of the geometric series: ∞ X 1 1 3k + (−9)k xk A(x) = + = 1 − 3x 1 + 9x k=0 Read off the sequence (and check if desired) So the solution is: an = 3n + (−9)n The initial conditions a0 = 2 and a1 = −6 are satisfied. So is the recursion equation: an+2 − 9an + 8an+1 = 3n+2 + (−9)n+2 − 9 (3n + (−9)n ) + 8 3n+1 + (−9)n+1 = 3n (9 − 9 + 24) + (−9)n (81 − 9 − 72) = 8 · 3n+1 D01