/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Assume that \(a>0\) and \(b&g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume that \(a>0\) and \(b>0\) and consider the characteristic equation \(r^{2}-a r-b=0\) with roots \(r_{1}\) and \(r_{2}\). Assume that \(a+b>1\). a) Show that \(r_{1}>1\). b) Show that if \(x_{n}=c_{1} r_{1}^{n}+c_{2} r_{2}^{n}\) is a solution of \(x_{n+1}=a x_{n}+b x_{n-1}\), then \(\lim _{n \rightarrow \infty} x_{n}=\infty\)

Short Answer

Expert verified
Given \( a+b>1 \) and \( a, b >0 \), we have \( r_1 > 1 \). Thus, \( \lim_{n \rightarrow \infty} x_{n} = \infty \).

Step by step solution

01

Write down the characteristic equation

The characteristic equation is given as: \[ r^2 - ar - b = 0 \]
02

Use the quadratic formula

The quadratic formula is used to find the roots of the equation \( ax^2 + bx + c = 0 \). The roots \( r_1 \) and \( r_2 \) are given by: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the characteristic equation \( r^2 - ar - b = 0 \), this simplifies to: \[ r = \frac{a \pm \sqrt{a^2 + 4b}}{2} \]
03

Calculate the roots \( r_1 \) and \( r_2 \)

\[ r_1 = \frac{a + \sqrt{a^2 + 4b}}{2} r_2 = \frac{a - \sqrt{a^2 + 4b}}{2} \]
04

(Part a): Show that \( r_1 > 1 \)

Since \( a > 0 \) and \( b > 0 \), we know that \( a^2 + 4b > a^2 \) which means \( \sqrt{a^2 + 4b} > a \). Then we have: \[ r_1 = \frac{a + \sqrt{a^2 + 4b}}{2} > \frac{a + a}{2} = a \] Given that \( a + b > 1 \) and both \( a \) and \( b \) are positive, \( a > 1 \) or close to 1 with positive \( b \), thus it follows from the inequality that \( r_1 > 1 \).
05

(Part b): Define the solution form

Given the solution of the recurrence relation \( x_{n+1} = a x_{n} + b x_{n-1} \) is: \[ x_n = c_1 r_1^n + c_2 r_2^n \] Where \( r_1 \) and \( r_2 \) are the roots from previous steps.
06

Analyze the solution behavior

As \( n \rightarrow \infty \), the term \( r_1^n \) dominates because \( r_1 > r_2 \). Since we have shown that \( r_1 > 1 \), raising \( r_1 \) to an increasingly large power will cause \( r_1^n \) to grow indefinitely. Therefore: \[ \lim_{n \to \infty} x_n = \infty \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Quadratic Formula
To solve quadratic equations like our characteristic equation, we use the quadratic formula. This formula is essential for finding the roots of any polynomial of the form \( ax^2 + bx + c = 0 \). The roots can be found using:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, the characteristic equation is \( r^2 - ar - b = 0 \). Here, \( a = 1 \), \( b = -a \), and \( c = -b \). Plugging these into the quadratic formula gives us:
\[ r = \frac{a \pm \sqrt{a^2 + 4b}}{2} \]
This formula helps us find the roots \( r_1 \) and \( r_2 \) efficiently. Remember, the roots are the values of \( r \) that satisfy the quadratic equation. Using these roots, we solve recurrence relations and analyze long-term behavior.
Recurrence Relation
A recurrence relation is a way of defining sequences based on previous terms. In this exercise, we are given the relation:
\[ x_{n+1} = a x_n + b x_{n-1} \]
To solve such a relation, we use the characteristic equation \( r^2 - ar - b = 0 \) to find the roots \( r_1 \) and \( r_2 \). The general solution of the recurrence relation is then:
\[ x_n = c_1 r_1^n + c_2 r_2^n \]
where \( c_1 \) and \( c_2 \) are constants determined by initial conditions. This form leverages the roots from our characteristic equation to build a formula for future terms of the sequence.
Asymptotic Behavior
Asymptotic behavior examines how a sequence behaves as \( n \) approaches infinity. For the sequence \( x_n = c_1 r_1^n + c_2 r_2^n \), the most important factor is the value of the roots \( r_1 \) and \( r_2 \). If \( |r_1| > 1 \), the term \( r_1^n \) grows exponentially, dominating the behavior of the sequence.
Given that \( r_1 > r_2 \) and \( r_1 > 1 \), \( r_1^n \) will grow larger very quickly, causing \( x_n \) to approach infinity. Thus, we conclude:
\[ \lim_{n \to \infty} x_n = \infty \]
Understanding the asymptotic behavior is critical for predicting how sequences evolve over time, especially in fields like computer science and economics, where we often deal with iterative processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]\). The matrix \(A\) is the identity matrix with the first two rows exchanged. a) Lel \(B=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right]\). Compute AB. b) Let \(C=\left[\begin{array}{rrr}4 & -2 & 7 \\ 1 & 1 & -3 \\ -4 & 3 & 6\end{array}\right]\). Compute AC. c) Based on parts (a) and (b), what is the effect of multiplying \(A\) on the left with another \(3 \times 3\) matrix? Explain why.

Solve using Gaussian elimination. $$ \begin{array}{r} x-y+3 z=2 \\ 2 x+3 y-z=5 \\ -x-9 y+11 z=1 \end{array} $$

The San Diego fairy shrimp live in ponds that fill and dry many times during a year. While the ponds are dry, the fairy shrimp survive as cysts. The population is divided into three groups. Cysts that survive one dry period are in group 1, cysts that survive two dry periods are in group 2 , and cysts that survive three or more dry periods are in group 3. The Leslie matrix representing the survivability and fecundity is given below. \({ }^{18}\) In the third dry spell, there are 8365,1095, and 310 individuals in groups 1,2, and 3, respectively. $$G=\left[\begin{array}{ccc} 3.6 & 0.98 & 0.65 \\ 0.5 & 0 & 0 \\ 0 & 0.5 & 0.49 \end{array}\right]$$ Compute the long-term percentage growth rate between dry periods.

Multiply the matrix and the vector to determine if the vector is an eigenvector. If so, what is the eigenvalue? $$ \left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right] \cdot\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$

In the absence of competitors and herbivores, plant growth can be modeled by the recursion relation $$ M_{n+1}=\frac{(1+\rho) M_{n}}{1+\theta M_{n}} $$ where \(M_{n}\) is the total plant mass after \(3 n\) days, \(\rho\) is the maximum growth rate, and \(\theta\) is a constant. For a plant, \(\rho=0.3, \theta=0.001\), and the starting mass is \(M_{0}=1 \mathrm{~g}\) a) Plot the points \(\left(n, x_{n}\right)\) for \(n=0,1,2,3, \ldots, 40\). b) Does this graph look familiar?

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.