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Chapter 5: Problem 8

Solve the recurrence relation \(a_{n}=2 a_{n-1}-a_{n-2}, n \geq 2\) given \(a_{0}=40, a_{1}=37\)

Short Answer

Expert verified
The solution to the recurrence relation is \(a_n = 40 - 3n\).

Step by step solution

01

Identify the Nature of the Recurrence Relation

The given recurrence relation is a linear homogeneous relation with constant coefficients: \( a_n = 2a_{n-1} - a_{n-2} \). This suggests attention to characteristic polynomials to find the general solution.
02

Formulate the Characteristic Equation

For the relation \( a_n = 2a_{n-1} - a_{n-2} \), we set up the characteristic equation by substituting \(a_n = r^n\): \( r^n = 2r^{n-1} - r^{n-2} \). Dividing through by \(r^{n-2}\) gives \( r^2 = 2r - 1 \).
03

Solve the Characteristic Equation

Rewriting the characteristic equation \( r^2 - 2r + 1 = 0 \), notice it is a perfect square: \((r - 1)^2 = 0\). We have a double root at \(r=1\).
04

Write the General Solution

For a double root \(r=1\), the general solution of the recurrence relation is \( a_n = C_1 \cdot 1^n + C_2 \cdot n \cdot 1^n = C_1 + C_2n \).
05

Apply Initial Conditions

Substitute the initial conditions \(a_0 = 40\) and \(a_1 = 37\) into \(a_n = C_1 + C_2n\):- For \(n = 0\): \(a_0 = C_1 = 40\).- For \(n = 1\): \(a_1 = C_1 + C_2 = 37\).Substitute \(C_1 = 40\) into the second equation: \(40 + C_2 = 37\), so \(C_2 = -3\).
06

Construct the Particular Solution

With \(C_1 = 40\) and \(C_2 = -3\), substitute back into the general form: \( a_n = 40 - 3n \). This is the explicit formula for \(a_n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Homogeneous Recurrence
A linear homogeneous recurrence relation is a sequence where each term is a linear combination of previous terms. In our example, the relation is represented as \( a_n = 2a_{n-1} - a_{n-2} \). The coefficients of the previous terms (2 and -1) do not depend on \( n \); hence, it is called 'homogeneous'. Linear implies that each term is only multiplied by a constant and not raised to any power beyond the first.
  • "Linear" means no term is squared or cubed, just multiplied by coefficients.
  • "Homogeneous" means all terms are of the same type, depending only on previous terms.
  • No constants are added outside of the coefficients times sequence terms.
When dealing with a linear homogeneous recurrence, identifying it helps to apply consistent methods to find solutions, such as deriving characteristic equations.
Characteristic Equation
To solve a recurrence relation, we often use a characteristic equation. For our sequence, you start by assuming solutions of the form \( a_n = r^n \), leading us to the equation \( r^2 = 2r - 1 \). Solving the characteristic equation helps us find the values of \( r \) that satisfy the recurrence.
  • Convert the recurrence equation to characteristic form: \( r^n = 2r^{n-1} - r^{n-2} \).
  • Simplify by dividing all terms by \( r^{n-2} \).
  • The resulting equation \( r^2 - 2r + 1 = 0 \) is quadratic.
We then solve this quadratic equation as any standard quadratic, using factoring or the quadratic formula to find the roots of \( r \). These roots will guide us to the form of the general solution.
General Solution of Recurrence
The solution to a recurrence relation is generalized with constants that allow it to fit diverse initial conditions. For our root \( r = 1 \) (a double root), the general solution takes the form \( a_n = C_1 + C_2n \), which incorporates the initial values of the sequence through constants \( C_1 \) and \( C_2 \).
  • Double roots require modification: \( C_1 + C_2n \).
  • This accounts for sequences growth or decline due to multiplying terms by \( n \).
  • The general solution should cover possible sequences fitting the recurrence.
Each component of the solution functions accommodate different sequence behaviors. By plugging in initial conditions, we can solve for \( C_1 \) and \( C_2 \) to tailor the solution to specific cases.
Initial Conditions in Recurrence Relations
Initial conditions are specific values that define beginning terms of a sequence. In our case, they are \( a_0 = 40 \) and \( a_1 = 37 \). These values enable us to find constants \( C_1 \) and \( C_2 \) by substituting them into the general solution.
  • Starting values anchor the generalized formula to a specific sequence.
  • Substitute initial conditions back into the general solution to find constants.
  • This transforms a flexible form into a precise formula, like \( a_n = 40 - 3n \).
By using initial conditions, a recurrence relation shifts from a theory to a functioning sequence, valuable for predictions and calculations. Solving for these conditions is crucial to obtain exact terms in sequence construction.

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Most popular questions from this chapter

[BB] Use generating functions to find a formula for \(a_{n}\) given \(a_{0}=5\) and \(a_{n}=a_{n-1}+2^{n}\) for \(n \geq 1\)

(a) \([\mathrm{BB}]\) Solve the recurrence relation \(a_{n}=4 a_{n-1}, n \geq\) 1, given \(a_{0}=1\) (b) \([\mathrm{BB}]\) Solve the recurrence relation \(a_{n}=4 a_{n-1}+8^{n}\), \(n \geq 1\), given \(a_{0}=1\) (c) [BB] Verify that your answer to (b) is correct.

Let \(f_{1}=f_{2}=1, f_{k}=f_{k-1}+f_{k-2}\) for \(k>2\) be the Fibonacci sequence. Which terms of this sequence are even? Prove your answer.

Define \(g: \mathrm{N} \rightarrow \mathrm{N}\) by $$ g(m)=\left\\{\begin{array}{ll} \frac{m}{2} & \text { if } m \text { is even } \\ \frac{3 m+1}{2} & \text { if } m \text { is odd. } \end{array}\right. $$ This function is known as the \(" 3 m+1 "\) function. It is suspected that for any starting number \(m\), the sequence \(g(m), g^{2}(m), g^{3}(m), \ldots\) eventually terminates with \(1 .\) Verify this assertion for each of the five integers \(m=\) \(341,96,104,336\), and \(133 .^{3}\)

(a) Solve the recurrence relation \(a_{n}=5 a_{n-1}-6 a_{n-2}\), \(n \geq 2\), given \(a_{0}=2, a_{1}=11\) (b) Solve the recurrence relation \(a_{n}=5 a_{n-1}-6 a_{n-2}+\) \(3 n, n \geq 2\), given \(a_{0}=2, a_{1}=14\) (c) Verify that your answer to (b) is correct.
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