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Chapter 7: Problem 15

Solve the given recurrence relation for the initial conditions given. \(a_{n}=6 a_{n-1}-8 a_{n-2} ; \quad a_{0}=1, \quad a_{1}=0\)

Short Answer

Expert verified
The solution to the given recurrence relation is \(a_{n} = (2)^n - (4)^n\).

Step by step solution

01

Derive the characteristic equation

The given recurrence relation is \(a_{n}=6 a_{n-1}-8 a_{n-2}\). From this, we can derive the characteristic equation as \(r^{2}-6 r+8=0\). Here r is the root of this equation.
02

Solve the characteristic equation

Let's solve for r in the characteristic equation, \(r^{2}-6 r+8=0\). If we factor, we get \((r-2)(r-4)=0\), which has solutions \(r=2\) and \(r=4\).
03

Get the general solution from the roots

The general solution of a recurrence relation is \(a_{n} = C*(r_{1})^n + D*(r_{2})^n\). Since \(r_{1}=2\) and \(r_{2}=4\), the general solution becomes \(a_{n} = C*(2)^n + D*(4)^n\).
04

Use the initial conditions to find the constants

Given that \(a_{0}=1\) and \(a_{1}=0\), we will use these conditions to find the values for C and D: - For \(n=0\), \(a_{0} = C*(2)^0 + D*(4)^0\), which becomes \(1 = C+D\).- For \(n=1\), \(a_{1} = C*(2)^1 + D*(4)^1\), which leads to \(0 = 2C + 4D\).Solving these equations, we obtain \(C=1\) and \(D=-1\).
05

Write the final solution

Now substitute the constants C and D into the general solution, therefore, the solution for the recurrence relation is \(a_{n} = (2)^n - (4)^n\).

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Most popular questions from this chapter

Solve the given recurrence relation for the initial conditions given. \(a_{n}=6 a_{n-1}-8 a_{n-2} ; \quad a_{0}=1, \quad a_{1}=0\)

Exercises \(22-24\) refer to the sequence \(S_{1}, S_{2}, \ldots,\) where \(S_{n}\) denotes the mumber of \(n\) -bit strings that do not contain the pattern \(010 .\) $$ \text { Compute } S_{1}, S_{2}, S_{3}, \text { and } S_{4} $$

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