Week6 - S Tanny Non-Linear DFE Exercise 1 YnYn 2 = Y2 n 1 Y0 =1 Y1 = 2 MAT 344 Spring 1999 Follows that Yn 0 oen ^ Yn Yn 2 1 Yn 1 Yn Let Wn = Yn 1

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S. Tanny MAT 344 Spring 1999 83 Non-Linear DFE Exercise 1 Y n Y n+2 = Y 2 n + 1 Y 0 =1, Y 1 = 2 Follows that Y n 0 ± n. ² Y n % 2 Y n % 1 Y n % 1 Y n Let W n = . Then W n + 1 = W n , W 0 = 2 Y n % 1 Y n ² W n = 2 Y Y n + 1 = 2Y n Y Y n = 2 n (Y 0 = 1) N.B. Could also linearize with logs . Exercise 2 + - = 0 Y 2 n % 2 3Y 2 n % 1 4Y 2 n Set W n = Y n 2 W n + 2 + 3W n + 1 - 4W n = 0 W n = c 1 (-4) n + c 2 Y Y n = c 1 ( 4) n % c 2 Solving Recurrences Using Generating Fn *Basic Idea : G.F. is formal power series with the coeff of interest to us related by the recursion. Using formal manipulations this leads to a formal expression (hopefully a recognizable closed form) from which the coefficients of the FPS can be determined. Exercise: a n + 1 = 2a n + 1 a 0 = 0 , a 1 = 1 Define: A(x) = (“FPS”) j n $ 0 a n x n ² A(x) = a 0 + “Formal manipulation” j n $ 1 a n x n = a 0 + j n $ 0 a n % 1 x n % 1 = j n $ 0 (2a n % 1)x n % 1 = 2 j n $ 0 a n x n % 1 % j n $ 0 x n % 1
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MAT 344 Spring 1999 84 A(x) = 2x + = 2x A(x) + . j n $ 0 a n x n x 1 x x 1 x A(x) = “Closed Form” x (1 x)(1 2x) = x @ j n $ 0 x n @ j n $ 0 2 n x n = x j n $ 0 j n j 0 2 j x n j n $ 0 a n x n j n $ 0 2 n % 1 1 x n % 1 Comparing coefficients we see that a n = 2 n - 1 N.B . If 2 FPS are equal, they are the same coefficient by coefficient. Exercise: F n + 2 = F n + 1 + F n n $ 0 , F 0 = F 1 = 1 Let F(x) = “FPS” j n $ 0 F n x n = F 0 + F 1 x + j n $ 2 F n x n = 1 + x + j n $ 0 F n % 2 x n % 2 = 1 + x + x j n $ 0 F n % 1 x n % 1 % x 2 j n $ 0 F n x n = 1 + x + j n $ 0 F n % 1 % F n x n % 2 = 1 + x + x [F(x) - F 0 ] + x 2 F(x) F(x) = 1 + x F(x) + x 2 F(x) F(x) = 1 1 x x 2 How can we determine the coeff of the FPS for F(x) from this? Use Partial Fraction Expansion of RHS!
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This note was uploaded on 03/04/2012 for the course BIO 510 taught by Professor Miller during the Fall '06 term at Carnegie Mellon.

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Week6 - S Tanny Non-Linear DFE Exercise 1 YnYn 2 = Y2 n 1 Y0 =1 Y1 = 2 MAT 344 Spring 1999 Follows that Yn 0 oen ^ Yn Yn 2 1 Yn 1 Yn Let Wn = Yn 1

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