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Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1. You may use the general solution given on P.342.

Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1. You may use the general solution given on P.342.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1. You may use the general solution given on P.342.

342 8 Sequences and Series is given in two parts: if a = 1, Sn = I+ nc for V n E N; if a + 1, Sn = a"A+ 1 for V n e N. a When a 1, any particular solution is obtained by determining a specific, numerical value for I. In fact, a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = 1+jc for I, I = J– jc. // since S; = I+jc // where So =I we get // One particular “particular solution" has I = 0. When a + 1, any particular solution is obtained by determining a specific, In fact, numerical value for A; if the starting value I is given, then A =I- 1 - a a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = Aa + for A, 1 - a we get 1 A = // But what if a = 0? // One particular “particular solution" has A = 0. Example 8.2.1: The Towers of Hanoi The recurrence equation for the number of moves in the Towers of Hanoi problem is a first-order linear recurrence equation: Tn = 2T,-1+1. %3D C 1 Here a = 2 and c = 1, so 1, and any sequence T that satisfies 1 - a 1 this RE is given by the formula Tn = 2" [I – (–1)]+(-1) = 2" [I+ 1] – 1. Assuming T has domain N and denoting To by I, we saw at the beginning of this chapter several particular solutions: Т 3 (0, 1,3, 7, 15, 31, ...); Т 3 (2,5, 11, 23, 47, 95, ...); if I = 0, then if I = 2, then if I = 4, then if I = -1, then T = (-1,–1,–1,–1, –1,...). // T, = 2" [–1+1] – 1 = // T, = 2" (0+1] – 1 = 2" - . // T, = 2" [2+1] – 1 = 3 × 2" – 1. T = (4,9, 19, 39, 79, 159, ...); // T, = 2" [4 + 1] – 1= 5 x 2" – 1. -1.
Transcribed Image Text:342 8 Sequences and Series is given in two parts: if a = 1, Sn = I+ nc for V n E N; if a + 1, Sn = a"A+ 1 for V n e N. a When a 1, any particular solution is obtained by determining a specific, numerical value for I. In fact, a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = 1+jc for I, I = J– jc. // since S; = I+jc // where So =I we get // One particular “particular solution" has I = 0. When a + 1, any particular solution is obtained by determining a specific, In fact, numerical value for A; if the starting value I is given, then A =I- 1 - a a particular solution is determined by a specific, numerical value J for any (particular) entry, S;. Solving the equation J = Aa + for A, 1 - a we get 1 A = // But what if a = 0? // One particular “particular solution" has A = 0. Example 8.2.1: The Towers of Hanoi The recurrence equation for the number of moves in the Towers of Hanoi problem is a first-order linear recurrence equation: Tn = 2T,-1+1. %3D C 1 Here a = 2 and c = 1, so 1, and any sequence T that satisfies 1 - a 1 this RE is given by the formula Tn = 2" [I – (–1)]+(-1) = 2" [I+ 1] – 1. Assuming T has domain N and denoting To by I, we saw at the beginning of this chapter several particular solutions: Т 3 (0, 1,3, 7, 15, 31, ...); Т 3 (2,5, 11, 23, 47, 95, ...); if I = 0, then if I = 2, then if I = 4, then if I = -1, then T = (-1,–1,–1,–1, –1,...). // T, = 2" [–1+1] – 1 = // T, = 2" (0+1] – 1 = 2" - . // T, = 2" [2+1] – 1 = 3 × 2" – 1. T = (4,9, 19, 39, 79, 159, ...); // T, = 2" [4 + 1] – 1= 5 x 2" – 1. -1.
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