/*! 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 13 Find the particular solution. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 13

Find the particular solution. \(x_{n+1}=x_{n}+2 x_{n-1} ; x_{0}=2, x_{1}=1\)

Short Answer

Expert verified
Particular solution includes terms: \(x_2 = 5\), \(x_3 = 7\), \(x_4 = 17\).

Step by step solution

01

- Understand the Recurrence Relation

The given recurrence relation is \(x_{n+1} = x_{n} + 2 x_{n-1}\). This means each term depends on the previous two terms. We also know the initial conditions are \(x_0 = 2\) and \(x_1 = 1\).
02

- Calculate \(x_2\)

Using the recurrence relation for \(n=1\), we get \(x_2 = x_1 + 2x_0\). Substituting the initial values: \[x_2 = 1 + 2(2) = 1 + 4 = 5\]
03

- Calculate \(x_3\)

Now calculate the next term using \(n=2\): \(x_3 = x_2 + 2x_1\). Use the value of \(x_2\) from the previous step: \[x_3 = 5 + 2(1) = 5 + 2 = 7\]
04

- Calculate \(x_4\)

Next term calculation for \(n=3\): \(x_4 = x_3 + 2x_2\). Using the known values: \[x_4 = 7 + 2(5) = 7 + 10 = 17\]

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 91影视!

Key Concepts

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

Initial Conditions
Initial conditions are crucial to any recurrence relation problem. They provide the starting values needed to calculate subsequent terms. For our given example, the initial conditions are x_0 = 2 and x_1 = 1. These values act like seeds. Without them, we can't grow the sequence. Understanding how to use these initial conditions forms the first step in solving any recurrence problem.
Recurrence Relation
The recurrence relation is the formula that describes the sequence. In this case, the relation is \(x_{n+1} = x_{n} + 2 x_{n-1}\). This relation tells us how we can derive the next term (x_{n+1}) based on the previous two terms (x_n and x_{n-1}). The recurrence relation acts like a recipe, guiding us on how to progress through the sequence. By repeatedly applying this relationship, we can generate all the terms of the sequence.
Term Calculation
To calculate each term, we need both the initial conditions and the recurrence relation. Let's walk through the calculations step-by-step:
  • First, to find x_2: Substitute the initial values into the recurrence relation: \[x_2 = x_1 + 2 x_0 = 1 + 2(2) = 1 + 4 = 5\]
  • Next, for x_3: Use the previously calculated value: \[x_3 = x_2 + 2 x_1 = 5 + 2(1) = 5 + 2 = 7\]
  • Finally, to find x_4: Incorporate the value of x_3: \[x_4 = x_3 + 2 x_2 = 7 + 2(5) = 7 + 10 = 17\]
This step-by-step process shows the importance of carrying forward calculated terms to derive the next. It ensures that each term is correctly built upon its predecessors, maintaining the relation and initial conditions as defined.

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

Show that if \(x_{n}\) and \(x_{n}^{\prime}\) are two solutions to the linear difference equation $$ x_{n+1}=a x_{n}+b x_{n-1} $$ then \(x_{n}+x_{n}^{\prime}\) and \(c x_{n}\) are also solutions, where \(c\) is a constant.

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 \(M_{0}=295 \mathrm{~g}\). a) Solve the equation $$ M=\frac{(1+\rho) M}{1+\theta M} $$ to determine a nonzero equilibrium value for the mass of the plant. b) Use the recursion relation to compute the total mass of the plant on days \(3,6,9,12\), and 15 \((n=1,2,3,4,5)\) c) Explain why your answers to part (b) are expected.

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 \(\lim _{n \rightarrow \infty} x_{n}=\infty\)

Let \(A=\left[\begin{array}{ll}4 & -1 \\ 7 & -9\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 2 & -2\end{array}\right], C=\left[\begin{array}{rr}8 & 3 \\ 0 & -3\end{array}\right]\) \(D=\left[\begin{array}{rrr}5 & -6 & 1 \\ 10 & 3 & -1\end{array}\right], E=\left[\begin{array}{rr}-7 & 4 \\ -3 & 2 \\ 2 & -1\end{array}\right]\) \(F=\left[\begin{array}{rrr}-4 & 2 & 3 \\ 0 & -1 & 2 \\ -7 & -2 & 5\end{array}\right]\), and \(v=\left[\begin{array}{r}2 \\\ -3\end{array}\right]\). Compute \(3 \mathrm{~A}\).

Multiply the matrix and the vector to determine if the vector is an eigenvector. If so, what is the eigenvalue? $$ \left[\begin{array}{rr} 5 & 0 \\ 3 & -2 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \end{array}\right] $$
See all solutions

Recommended explanations on Biology Textbooks

Bioenergetics

Read Explanation

Plant Biology

Read Explanation

Cells

Read Explanation

Biological Processes

Read Explanation

Microbiology

Read Explanation

Biology Experiments

Read Explanation
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.