Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1. You may use the general solution given on P.342. **Sequences and Series****General Formulation:**The sequence is given in two parts:- If $ a = 1 $, \[ S_n = I + nc \quad \text{for } \forall n \in \mathbb{N}; \]- If $ a \neq 1 $, \[ S_n = a^n A + \frac{c}{1-a} \quad \text{for } \forall n \in \mathbb{N}. \]**Particular Solutions:**When $ a = 1 $:- Any *particular solution* is found by determining a specific numerical value for $ I $. If a specific numerical value $ J $ for any entry $ S_j $ is given, solving $ J = I + jc $ for $ I $ gives: \[ I = J - jc. \] One specific solution is $ I = 0 $.When $ a \neq 1 $:- Any *particular solution* is found by determining a specific numerical value for $ A $. If the starting value $ I $ is given, then $ A = I - \frac{c}{1-a} $. Solving for $ A $ using the equation $ J = A a^j + \frac{c}{1-a} $ gives: \[ A = \frac{1}{a^j} \left[ J - \frac{c}{1-a} \right]. \] One specific solution is $ 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:\[ T_n = 2T_{n-1} + 1. \]Here $ a = 2 $ and $ c = 1 $, resulting in:\[ \frac{c}{1-a} = \frac{1}{1-2} = -1. \]Any sequence $ T $ satisfying this recurrence equation (RE) is given by:\[ T_n = 2^n [I - (-1)] + (-1) \]\[ = 2^n [I + 1] - 1. \]Assuming $ T $ has domain \( \mathbb{N} \

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Solve the first-order linear recurrence relation: Sn+1 = Sn + 2, with S0=1.

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

Problem 1PE

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Concept explainers

Recurrence relation

A recurrence relation or recursive relation is an equation that represents a function in terms of the values of its smaller inputs. Every recurrence relation T(n) is a recursive function of integer n and consists of a base case and a recursive case.

Question

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

**Sequences and Series**

**General Formulation:**

The sequence is given in two parts:

- If $ a = 1 $,
\[ S_n = I + nc \quad \text{for } \forall n \in \mathbb{N}; \]

- If $ a \neq 1 $,
\[ S_n = a^n A + \frac{c}{1-a} \quad \text{for } \forall n \in \mathbb{N}. \]

**Particular Solutions:**

When $ a = 1 $:

- Any *particular solution* is found by determining a specific numerical value for $ I $. If a specific numerical value $ J $ for any entry $ S_j $ is given, solving $ J = I + jc $ for $ I $ gives:
\[ I = J - jc. \]

One specific solution is $ I = 0 $.

When $ a \neq 1 $:

- Any *particular solution* is found by determining a specific numerical value for $ A $. If the starting value $ I $ is given, then $ A = I - \frac{c}{1-a} $. Solving for $ A $ using the equation $ J = A a^j + \frac{c}{1-a} $ gives:
\[ A = \frac{1}{a^j} \left[ J - \frac{c}{1-a} \right]. \]

One specific solution is $ 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:

\[ T_n = 2T_{n-1} + 1. \]

Here $ a = 2 $ and $ c = 1 $, resulting in:
\[ \frac{c}{1-a} = \frac{1}{1-2} = -1. \]

Any sequence $ T $ satisfying this recurrence equation (RE) is given by:
\[ T_n = 2^n [I - (-1)] + (-1) \]
\[ = 2^n [I + 1] - 1. \]

Assuming $ T $ has domain \( \mathbb{N} \

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