91Ó°ÊÓ

The given exercise deals with a second-order linear recurrence relation. A second-order linear recurrence relation is a mathematical expression where each term in a sequence is defined as a linear combination of the two preceding terms. It generally looks like this:

• \( x_{n+1} = a \, x_n + b \, x_{n-1} \)

In the given exercise, the coefficients are \(a = 0.1\) and \(b = 0.95\). To solve such relations and determine the asymptotic behavior, the characteristic equation needs to be analyzed. The characteristic equation is derived from assuming a solution of the form \( x_n = r^n \). This is common because geometric sequences simplify the behavior of recurrence relations.

The characteristic equation is a crucial step in solving second-order linear recurrence relations. It is formed by substituting the assumed solution \( x_n = r^n \) into the recurrence relation. For the given equation

• \( x_{n+1} = 0.1 \, x_n + 0.95 \, x_{n-1} \),

we substitute to get: \( r^{n+1} = 0.1 \, r^n + 0.95 \, r^{n-1} \).

Dividing through by \( r^{n-1} \) simplifies this to:

• \( r^2 = 0.1 \, r + 0.95 \)

This is our characteristic equation. The solutions to this equation (the roots) provide insight into the long-term behavior of the sequence. These roots are found using the quadratic formula.

The quadratic formula is used to solve the characteristic equation. The standard quadratic equation is of the form: \( ax^2 + bx + c = 0 \). Applying the quadratic formula:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \, \)

to the characteristic equation \( r^2 - 0.1 \, r - 0.95 = 0 \), we identify:\( a = 1, b = -0.1 \, \text{and} \, c = -0.95 \). Plugging these values in, we get:

• \( r = \frac{0.1 \pm \sqrt{(0.1)^2 + 4 \cdot 0.95}}{2} \)

which simplifies to:

• \( r = \frac{0.1 \pm \sqrt{3.81}}{2} \).

From here, we find the two roots:

• \( r_1 = 1.0243 \)
  \( r_2 = -0.9243 \)

Asymptotic analysis is the final step in understanding the long-term behavior of the recurrence relation. Asymptotic analysis examines the behavior of functions as inputs become large. In this case, as the term number \( n \) increases, the behavior of the solution is dominated by the largest root's magnitude. With roots \( r_1 \approx 1.0243 \) and \( r_2 \approx -0.9243 \), the root closest to one dominates. Because \( r_1 \approx 1.0243 \) is slightly greater than 1, the general solution will grow exponentially as \( (1.0243)^n \). Thus, we conclude that

• the sequence grows exponentially.

This means over a long period, the terms of the recurrence will become very large unless counteracted by initial conditions.