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Solve the recurrence relation with the given initial conditions. $$a_{0}=4, a_{1}=1, a_{n}=a_{n-1}-a_{n-2} / 4 \text { for } n \geq 2$$

Short Answer

Expert verified
The sequence decreases and becomes negative after a few terms.

Step by step solution

01

Understanding the Recurrence Relation

The given recurrence relation is \( a_n = a_{n-1} - \frac{1}{4}a_{n-2} \) for \( n \geq 2 \). We have the initial conditions \( a_0 = 4 \) and \( a_1 = 1 \). We will use these to find the next few terms in the sequence to understand its behavior.
02

Compute \( a_2 \)

Substitute the initial values into the recurrence relation to find \( a_2 \).\[a_2 = a_1 - \frac{1}{4}a_0 = 1 - \frac{1}{4} \times 4 = 1 - 1 = 0\]
03

Compute \( a_3 \)

Substitute \( a_2 = 0 \), \( a_1 = 1 \) into the recurrence relation to find \( a_3 \).\[a_3 = a_2 - \frac{1}{4}a_1 = 0 - \frac{1}{4} \times 1 = -\frac{1}{4}\]
04

Compute \( a_4 \)

Substitute \( a_3 = -\frac{1}{4} \), \( a_2 = 0 \) into the recurrence relation to find \( a_4 \).\[a_4 = a_3 - \frac{1}{4}a_2 = -\frac{1}{4} - \frac{1}{4} \times 0 = -\frac{1}{4}\]
05

Compute \( a_5 \)

Substitute \( a_4 = -\frac{1}{4} \), \( a_3 = -\frac{1}{4} \) into the recurrence relation to find \( a_5 \).\[a_5 = a_4 - \frac{1}{4}a_3 = -\frac{1}{4} - \frac{1}{4} \times (-\frac{1}{4})\]\[a_5 = -\frac{1}{4} + \frac{1}{16} = -\frac{1}{4} + \frac{1}{16} = -\frac{4}{16} + \frac{1}{16} = -\frac{3}{16}\]
06

Analyze the Calculated Values

The calculated values are \( a_0 = 4 \), \( a_1 = 1 \), \( a_2 = 0 \), \( a_3 = -\frac{1}{4} \), \( a_4 = -\frac{1}{4} \), and \( a_5 = -\frac{3}{16} \). The sequence shows that the values become negative and fractions after the initial integers, suggesting a decreasing trend.

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Key Concepts

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

Initial Conditions
When working with recurrence relations, initial conditions are crucial as they serve as the starting point or seed of your sequence. For the provided exercise, the initial conditions are \( a_0 = 4 \) and \( a_1 = 1 \). These values are essential because they allow us to compute all subsequent terms in the sequence using the recurrence formula.
Initial conditions determine the specific trajectory or path the sequence will take as it evolves. In essence, they define the sequence's uniqueness, as different initial conditions can result in entirely different sequences, even if they share the same recursive formula.
By starting with \( a_0 = 4 \) and \( a_1 = 1 \), we initiate the sequence's unique path, which guides our calculations for \( a_2 \), \( a_3 \), and so forth. It's analogous to setting a course for a journey; your initial direction and location are paramount.
Sequence Behavior
Understanding sequence behavior entails observing how the sequence evolves with increasing values of \( n \). In this case, after calculating the first few terms \( a_2, a_3, \) and \( a_4 \), you get a sense of the pattern the sequence follows.
In the provided example, the calculations show that:
  • \( a_2 = 0 \)
  • \( a_3 = -\frac{1}{4} \)
  • \( a_4 = -\frac{1}{4} \)
  • \( a_5 = -\frac{3}{16} \)
These terms suggest that the sequence begins with positive integers but swiftly transitions into negative numbers and fractional values. This indicates a change in behavior from initial stable positive terms to decreasing negative values.
Analyzing the trend and behavior of a sequence can reveal its overarching pattern, stability, or instability, whether it stabilizes to a certain value or perpetually decreases or increases. This can be crucial for predicting future terms or determining the sequence's long-term behavior.
Linear Recursive Sequences
The recurrence relation in this exercise is a linear recursive sequence. This means each term is expressed as a linear combination of previous terms. Our specific relation \( a_n = a_{n-1} - \frac{1}{4}a_{n-2} \) fits this definition.
Linear recursive sequences, especially those with constant coefficients, exhibit predictable structural characteristics which make them more manageable to analyze. The multiplier for \( a_{n-1} \) is 1, and for \( a_{n-2} \) is \(-\frac{1}{4} \). These coefficients guide how each term is derived from its predecessors.
Importantly, the properties of linear recursion allow for the potential discovery of a closed-form formula, although it's not always simple to find. Recognizing and working with such sequences helps in different areas of mathematics, including computer algorithms, where predicting next steps based on the current state is common.
Being familiar with linear recursive sequences can be remarkably practical, as it aids you in solving various real-world problems efficiently by understanding how past information can forecast future outcomes.

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Most popular questions from this chapter

Two species, \(X\) and \(Y\), live in a symbiotic relationship. That is, neither species can survive on its own and each depends on the other for its survival. Initially, there are 15 of \(X\) and 10 of \(Y\). If \(x=x(t)\) and \(y=y(t)\) are the sizes of the populations at time \(t\) months, the growth rates of the two populations are given by the system \\[ \begin{array}{l} x^{\prime}=-0.8 x+0.4 y \\ y^{\prime}=0.4 x-0.2 y \end{array} \\] Determine what happens to these two populations.

The matrices either are not diagonalizable or do not have a dominant eigenvalue (or both). Apply the power method anyway with the given initial vector \(\mathbf{x}_{0}\) performing eight iterations in each case. Compute the exact eigenvalues and eigenvectors and explain what is happening. $$A=\left[\begin{array}{lll} 4 & 0 & 1 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$

Write out the first six terms of the sequence defined by the recurrence relation with the given initial conditions. $$y_{0}=0, y_{1}=1, y_{n}=y_{n-1}-y_{n-2} \text { for } n \geq 2$$

Use the power method to approximate the dominant eigenvalue and eigervector of A. Use the given initial vector \(\mathbf{x}_{0},\) the specified number of iterations \(k,\) and three-decimal-place accuracy. $$A=\left[\begin{array}{ll} 7 & 2 \\ 2 & 3 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right], k=6$$

It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>0 .\) Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form \\[ \left[\begin{array}{ll} B & C \\ O & D \end{array}\right] \\] $$A=\left[\begin{array}{llll} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$$

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