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By Herbert S. Wilf

ISBN-10: 1568811780

ISBN-13: 9781568811789

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1. 4. Recurrence Relations 33 Then K is finite, and clearly |xj | ≤ Kαj for j ≤ p. We claim that |xn | ≤ Kαn for all n, which will complete the proof. Indeed, if the claim is true for 0, 1, 2, . . , n, then: |xn+1 | ≤ b0 |x0 | + · · · + bp |xn−p | + G(n) ≤ b0 Kαn + · · · + bp Kαn−p + G(n) = Kαn−p {b0 αp + · · · + bp } + G(n) = Kαn−p {αp+1 − t} + G(n) = Kαn+1 − {tKαn−p − G(n)} ≤ Kαn+1 . 1 - Exercises 1. Solve the following recurrence relations (a) xn+1 = xn + 3 (n ≥ 0; x0 = 1) (b) xn+1 = xn /3 + 2 (c) xn+1 = 2(n + 1)xn (n ≥ 0; x0 = 0) (n ≥ 0; x0 = 1) (d) xn+1 = ((n + 1)/n)xn + n + 1 (e) xn+1 = xn + xn−1 (n ≥ 1; x1 = 5) (n ≥ 1; x0 = 0; x1 = 3) (f) xn+1 = 3xn − 2xn−1 (n ≥ 1; x0 = 1; x1 = 3) (g) xn+1 = 4xn − 4xn−1 (n ≥ 1; x0 = 1; x1 = ξ) 2.

Hence the road to the solution of such a differential equation begins by trying a solution of that form and seeing what the constant or constants α turn out to be. 34) calls for a trial solution of the form xn = αn . 34) and cancel a common factor of αn−1 , we obtain a quadratic equation for α, namely: α2 = aα + b. 34) will look like xn = c1 αn+ + c2 αn− (n = 0, 1, 2, . ). 36) The constants c1 and c2 will be determined so that x0 , x1 have their assigned values. 2. The recurrence for the Fibonacci numbers is: Fn+1 = Fn + Fn−1 (n ≥ 1; F0 = 0; F1 = 1).

Now look 54 2. Recursive Algorithms at the two ‘recursive calls,’ which really aren’t quite recursive. The first one of them sorts the array to the left of xi . That is indeed a recursive call, because we can just change the ‘n’ to ‘i − 1’ and call Quicksort. The second recursive call is the problem. It wants to sort a portion of the array that doesn’t begin at the beginning of the array. The routine Quicksort as written so far doesn’t have enough flexibility to do that. So we will have to give it some more parameters.

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Algorithms and Complexity (Second edition) by Herbert S. Wilf


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