Explanation Theorem used: “Consider the second order linear differential equation with constant coefficients s n = a s n − 1 + b s n − 2 for n ≥ 2 with initial values s 0 and s 1 . Let r 1 and r 2 denotes the roots of x 2 = a x + b . Then (a) If r 1 ≠ r 2 , there exist constants c 1 and c 2 such that s n = c 1 r 1 n + c 2 r 2 n for n = 0 , 1 , 2 , … (b) If r 1 = r 2 = r , there exist constants c 1 and c 2 such that s n = ( c 1 + n c 2 ) r n for n = 0 , 1 , 2 , … .” Calculation: The given recurrence relation is, s n = s n − 1 + 2 s n − 2 which is a second order linear differential equation with the initial conditions s 0 = 9 and s 1 = 0 . Compare the recurrence relation s n = s n − 1 + 2 s n − 2 with that of s n = a s n − 1 + b s n − 2 and obtain the values of a and b as a = 1 and b = 2 . Therefore, the auxiliary equation x 2 = a x + b becomes x 2 = x + 2 . Now compute the roots of the equation x 2 = x + 2 as shown below. x 2 − x − 2 = 0 x 2 − 2 x + x − 2 = 0 x ( x − 2 ) + 1 ( x − 2 ) = 0 ( x − 2 ) ( x + 1 ) = 0 ( Factoring out the common terms ) x = 2 , x = − 1 Since the roots r 1 = 2 and r 2 = − 1 are not equal, by the above theorem there exists constants c 1 and c 2 such that s n = c 1 ( 2 ) n + c 2 ( − 1 ) n for n = 0 , 1 , 2 , … Now evaluate the constants c 1 and c 2 with the use of initial conditions as follows. Substitute n = 0 in s n = c 1 ( 2 ) n + c 2 ( − 1 ) n and obtain an equation relating c 1 and c 2 as shown below

5th Edition

ISBN: 9780134689562

Author: Dossey, John A.

Publisher: Pearson,

See similar textbooks

Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

Videos

Question

Chapter 9.3, Problem 13E

To determine

The formula that expresses sn as a function of n, provided the recurrence relation and initial condition.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.3, Problem 12E

Chapter 9.3, Problem 14E

Students have asked these similar questions

Prove that a relation X defined on a set A that is reflexive, symmetric and antisymmetric is an equivalence relation and determine the equivalence classes.

8:38 *** TEMU TEMU -3 -2 7 B 2 1 & 5G. 61% 1 2 -1 Based on the graph above, determine the amplitude, period, midline, and equation of the function. Use f(x) as the output. Amplitude: 2 Period: 2 Midline: 2 ☑ syntax error: this is not an equation. Function: f(x) = −2 cos(πx + 2.5π) +2× Question Help: Worked Example 1 ☑ Message instructor Submit Question ||| <

Let X be the relation defined on the power set of the set integers P(Z) by AXB whenever A U B is a finite set of integers. Prove whether or not X is reflexive, symmetric, antisymmetirc or transitive

Chapter 9 Solutions

Discrete Mathematics

Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

Ch. 9.1 - Prob. 11E Ch. 9.1 - Prob. 12E Ch. 9.1 - Prob. 13E Ch. 9.1 - Prob. 14E Ch. 9.1 - Prob. 15E Ch. 9.1 - Prob. 16E Ch. 9.1 - Prob. 17E Ch. 9.1 - Prob. 18E Ch. 9.1 - Prob. 19E Ch. 9.1 - Prob. 20E Ch. 9.1 - Prob. 21E Ch. 9.1 - Prob. 22E Ch. 9.1 - Prob. 23E Ch. 9.1 - Prob. 24E Ch. 9.1 - Prob. 25E Ch. 9.1 - Prob. 26E Ch. 9.1 - Prob. 27E Ch. 9.1 - Prob. 28E Ch. 9.1 - Prob. 29E Ch. 9.1 - Prob. 30E Ch. 9.1 - Prob. 31E Ch. 9.1 - Suppose that you have an unlimited supply of red,...Ch. 9.1 - Prob. 33E Ch. 9.1 - Prob. 34E Ch. 9.1 - Prob. 35E Ch. 9.1 - Prob. 36E Ch. 9.1 - Prob. 37E Ch. 9.1 - Prob. 38E Ch. 9.2 - Prove by mathematical induction that n2 − n + 2 is...Ch. 9.2 - Prob. 2E Ch. 9.2 - Prob. 3E Ch. 9.2 - Prob. 4E Ch. 9.2 - Prob. 5E Ch. 9.2 - Prove by mathematical induction that is a...Ch. 9.2 - Prob. 7E Ch. 9.2 - Prob. 8E Ch. 9.2 - Prob. 9E Ch. 9.2 - Prob. 10E Ch. 9.2 - In Exercises 11–24, use the method of iteration to...Ch. 9.2 - Prob. 12E Ch. 9.2 - In Exercises 11–24, use the method of iteration to...Ch. 9.2 - In Exercises 11–24, use the method of iteration to...Ch. 9.2 - Prob. 15E Ch. 9.2 - In Exercises 11–24, use the method of iteration to...Ch. 9.2 - In Exercises 11–24, use the method of iteration to...Ch. 9.2 - Prob. 18E Ch. 9.2 - Prob. 19E Ch. 9.2 - Prob. 20E Ch. 9.2 - Prob. 21E Ch. 9.2 - Prob. 22E Ch. 9.2 - Prob. 23E Ch. 9.2 - Prob. 24E Ch. 9.2 - Prob. 25E Ch. 9.2 - Prob. 26E Ch. 9.2 - Prob. 27E Ch. 9.2 - Prob. 28E Ch. 9.2 - Prob. 29E Ch. 9.2 - Consider the sequences of n terms in which each...Ch. 9.2 - Prob. 31E Ch. 9.2 - Prob. 32E Ch. 9.2 - Prob. 33E Ch. 9.2 - Prob. 34E Ch. 9.3 - In Exercises 1–24, find an explicit formula for sn...Ch. 9.3 - Prob. 2E Ch. 9.3 - Prob. 3E Ch. 9.3 - Prob. 4E Ch. 9.3 - Prob. 5E Ch. 9.3 - Prob. 6E Ch. 9.3 - Prob. 7E Ch. 9.3 - Prob. 8E Ch. 9.3 - Prob. 9E Ch. 9.3 - Prob. 10E Ch. 9.3 - Prob. 11E Ch. 9.3 - Prob. 12E Ch. 9.3 - Prob. 13E Ch. 9.3 - Prob. 14E Ch. 9.3 - Prob. 15E Ch. 9.3 - Prob. 16E Ch. 9.3 - Prob. 17E Ch. 9.3 - Prob. 18E Ch. 9.3 - Prob. 19E Ch. 9.3 - Prob. 20E Ch. 9.3 - Prob. 21E Ch. 9.3 - Prob. 22E Ch. 9.3 - Prob. 23E Ch. 9.3 - Prob. 24E Ch. 9.3 - Prob. 25E Ch. 9.3 - Prob. 26E Ch. 9.3 - Prob. 28E Ch. 9.3 - Prob. 30E Ch. 9.3 - Prob. 31E Ch. 9.3 - Prob. 32E Ch. 9.3 - Prob. 33E Ch. 9.3 - Prob. 34E Ch. 9.3 - Prob. 35E Ch. 9.3 - Prob. 36E Ch. 9.3 - Prob. 37E Ch. 9.3 - Prob. 38E Ch. 9.4 - Prob. 1E Ch. 9.4 - Prob. 2E Ch. 9.4 - Prob. 3E Ch. 9.4 - Prob. 4E Ch. 9.4 - Prob. 5E Ch. 9.4 - Prob. 6E Ch. 9.4 - Prob. 7E Ch. 9.4 - Prob. 8E Ch. 9.4 - Prob. 9E Ch. 9.4 - Prob. 10E Ch. 9.4 - Prob. 11E Ch. 9.4 - Prob. 12E Ch. 9.4 - Prob. 13E Ch. 9.4 - Prob. 14E Ch. 9.4 - Prob. 15E Ch. 9.4 - Prob. 16E Ch. 9.4 - Prob. 17E Ch. 9.4 - Prob. 18E Ch. 9.4 - Prob. 19E Ch. 9.4 - Prob. 20E Ch. 9.4 - Prob. 22E Ch. 9.4 - Prob. 24E Ch. 9.4 - Prob. 26E Ch. 9.4 - Prob. 30E Ch. 9.4 - Prob. 32E Ch. 9.4 - Prob. 34E Ch. 9.4 - Prob. 36E Ch. 9.4 - Prob. 38E Ch. 9.5 - Prob. 1E Ch. 9.5 - Prob. 2E Ch. 9.5 - Prob. 3E Ch. 9.5 - Prob. 4E Ch. 9.5 - Prob. 5E Ch. 9.5 - Prob. 6E Ch. 9.5 - Prob. 7E Ch. 9.5 - Prob. 8E Ch. 9.5 - Prob. 9E Ch. 9.5 - Prob. 10E Ch. 9.5 - Prob. 11E Ch. 9.5 - Prob. 12E Ch. 9.5 - Prob. 13E Ch. 9.5 - Prob. 14E Ch. 9.5 - Prob. 15E Ch. 9.5 - Prob. 16E Ch. 9.5 - Prob. 17E Ch. 9.5 - Prob. 18E Ch. 9.5 - Prob. 19E Ch. 9.5 - Prob. 20E Ch. 9.5 - Prob. 21E Ch. 9.5 - Prob. 22E Ch. 9.5 - Prob. 23E Ch. 9.5 - Prob. 24E Ch. 9.5 - Prob. 25E Ch. 9.5 - Prob. 26E Ch. 9.5 - Prob. 27E Ch. 9.5 - Prob. 28E Ch. 9.5 - Prob. 29E Ch. 9.5 - Prob. 30E Ch. 9.5 - Prob. 31E Ch. 9.5 - Prob. 32E Ch. 9.5 - Prob. 33E Ch. 9.5 - Prob. 34E Ch. 9.5 - Prob. 35E Ch. 9.6 - Prob. 1E Ch. 9.6 - Prob. 2E Ch. 9.6 - Prob. 3E Ch. 9.6 - Prob. 4E Ch. 9.6 - Prob. 5E Ch. 9.6 - Prob. 6E Ch. 9.6 - Prob. 7E Ch. 9.6 - Prob. 8E Ch. 9.6 - Prob. 9E Ch. 9.6 - Prob. 10E Ch. 9.6 - Prob. 11E Ch. 9.6 - Prob. 12E Ch. 9.6 - Prob. 13E Ch. 9.6 - Prob. 14E Ch. 9.6 - Prob. 15E Ch. 9.6 - Prob. 16E Ch. 9.6 - Prob. 17E Ch. 9.6 - Prob. 18E Ch. 9.6 - Prob. 19E Ch. 9.6 - Prob. 20E Ch. 9.6 - Prob. 21E Ch. 9.6 - In Exercises 21–26, find constants a and b such...Ch. 9.6 - Prob. 23E Ch. 9.6 - Prob. 24E Ch. 9.6 - Prob. 25E Ch. 9.6 - Prob. 26E Ch. 9.6 - Prob. 27E Ch. 9.6 - Prob. 28E Ch. 9.6 - Prob. 29E Ch. 9.6 - Prob. 30E Ch. 9.6 - Prob. 31E Ch. 9.6 - Prob. 32E Ch. 9.6 - Prob. 34E Ch. 9.6 - Prob. 35E Ch. 9.6 - Prob. 36E Ch. 9.6 - Prob. 37E Ch. 9.6 - Prob. 38E Ch. 9.6 - Prob. 39E Ch. 9.6 - Prob. 40E Ch. 9.6 - Prob. 41E Ch. 9.6 - Prob. 42E Ch. 9 - Prob. 1SE Ch. 9 - Prob. 2SE Ch. 9 - Prob. 3SE Ch. 9 - Prob. 4SE Ch. 9 - Prob. 6SE Ch. 9 - A data processing position pays a starting salary...Ch. 9 - Prob. 8SE Ch. 9 - Prob. 9SE Ch. 9 - Prob. 10SE Ch. 9 - Prob. 11SE Ch. 9 - Prob. 12SE Ch. 9 - Prob. 13SE Ch. 9 - Prob. 14SE Ch. 9 - Prob. 17SE Ch. 9 - Prob. 18SE Ch. 9 - Prob. 19SE Ch. 9 - Prob. 21SE Ch. 9 - Prob. 22SE Ch. 9 - Prob. 23SE Ch. 9 - Prob. 41SE Ch. 9 - Prob. 42SE Ch. 9 - Prob. 43SE Ch. 9 - Prob. 44SE Ch. 9 - Prob. 45SE Ch. 9 - Prob. 46SE Ch. 9 - Prob. 47SE Ch. 9 - Prob. 48SE

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The formula that expresses s n as a function of n , provided the recurrence relation and initial condition.

Concept explainers

Videos

The formula that expresses sn as a function of n, provided the recurrence relation and initial condition.

Want to see the full answer?

Chapter 9 Solutions