/*! 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 1 Solve each of the following recu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 1

Solve each of the following recurrence relations. a) \(a_{n+1}-a_{n}=2 n+3, \quad n \geq 0, \quad a_{0}=1\) b) \(a_{n+1}-a_{n}=3 n^{2}-n, \quad n \geq 0, \quad a_{0}=3\) c) \(a_{n+1}-2 a_{n}=5, \quad n \geq 0, \quad a_{0}=1\) d) \(a_{n+1}-2 a_{n}=2^{n}, \quad n \geq 0, \quad a_{0}=1\)

Short Answer

Expert verified
The solutions for the recurrence relations are: a) \(a_{n+1} = n^2 + 4n + 4\), b) \(a_{n+1} = n^3 + n^2 + 3\), c) \(a_{n+1} = 5 脳 2^n + 4\), d) \(a_{n+1} = n 脳 2^n + 2^{n+1}\)

Step by step solution

01

Solve a)

We start solving the recurrence relation \(a_{n+1} - a_{n} = 2n + 3, a_{0} = 1\). Iterate the recurrence relation as follow: \(a_{n+1} = a_{n} + 2n + 3 = (a_{n-1} + 2(n-1) + 3) + 2n + 3 = ... = a_{0} + 3(n+1) + \sum_{i=0}^{n}2i\). Since \(a_{0} = 1\) and \(\sum_{i=0}^{n}2i = n(n+1)\), we have \(a_{n+1} = 1 + 3(n+1) + n(n+1) = n^2 + 4n + 4\).
02

Solve b)

We proceed similarly for the recurrence relation \(a_{n+1} - a_{n} = 3n^{2} - n, a_{0} = 3\). We iterate the recurrence: \(a_{n+1} = a_{n} + 3n^{2} - n = ... = a_{0} + \sum_{i=0}^{n}(3i^{2} - i)\). Since \(a_{0} = 3\) and \(\sum_{i=0}^{n}(3i^2 - i) = n(n+1)(2n + 1)/2 - n(n+1)/2 = n^2(n+1)\), we get \(a_{n+1} = 3 + n^2(n+1) = n^3 + n^2 + 3\).
03

Solve c)

Next, we solve for the recurrence \(a_{n+1} - 2a_{n} = 5, a_{0} = 1\). Iterating yields: \(a_{n+1} = 2a_{n} + 5 = ... = a_{0} 脳 2^{n} + 5 \sum_{i=0}^{n} 2^{i}\). Given that \(a_{0} = 1\) and \(\sum_{i=0}^{n}2^i = 2^{n+1} - 1\), we get \(a_{n+1} = 2^n + 5(2^{n+1} - 1) = 5 脳 2^n + 4\).
04

Solve d)

Lastly, we tackle the relation \(a_{n+1} - 2a_{n} = 2^{n}, a_{0} = 1\). Iterating this gives: \(a_{n+1} = 2a_{n} + 2^{n} = ... = 2^n 脳 a_{0} + \sum_{i=0}^{n}2^{n-i} 脳 2^i\). Given \(a_{0} = 1\) and \(\sum_{i=0}^{n}2^{n-i} 脳 2^i = (n + 1) 脳 2^n\), we find \(a_{n+1} = 2^n + (n + 1) 脳 2^n = (n + 2) 脳 2^n = n 脳 2^n + 2^{n+1}\).

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.

Linear Recurrence Relation
A linear recurrence relation is a sequence of numbers where each term is a linear combination of previous terms. It usually comes with an initial condition to kick off the sequence. The given exercise presents several linear recurrence relations with different coefficients and right-hand side expressions. For example, in part a) the relation is first-order (as it involves only one previous term, a_n) with a non-homogeneous part (2n + 3) that depends on the index n. These relations are crucial for understanding how sequences evolve over time, and they appear in various fields like computer science, economics, and physics.
Iterative Method
The iterative method is a process for computing values of a sequence according to a recursive formula. In the recursive sequences given, we start from the initial term and use the recurrence relation to find subsequent terms. This process is akin to climbing stairs: we reach the next step using the previous one. For instance, part a) of our problem shows how to iteratively build the sequence a_{n+1} from a_n, a_{n-1}, down to a_0. Iteration is a foundational concept for algorithm design, where you often need to repeat steps until you reach a desired outcome.
Summation Formulas
Summation formulas are mathematical expressions that give the sum of a series of terms based on a certain pattern. They are essential in simplifying the expressions we get from iterating recurrence relations. In the provided solutions, formulas such as \(\sum_{i=0}^{n} 2i = n(n+1)\) for part a) are applied to express the sum of consecutive integers as a simple quadratic function. These formulas save time and effort, allowing for a concise representation of the solution. They also serve as powerful tools in mathematical proofs and solving complex problems.
Sequence Iteration
Sequence iteration refers to the step-by-step calculation of each term in a sequence based on a recurrence relation. This concept is exemplified in each part of the exercise, where the next term of the sequence is computed by applying the formula to the current term and known sums. Iterating a sequence enables the understanding of the pattern and growth of sequences, particularly when studying their long-term behavior. For students, getting hands-on experience with sequence iteration develops problem-solving and analytical skills, valuable in both academic studies and real-world applications.

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

Determine the number of \(n\)-digit quaternary \((0,1,2,3)\) sequences in which there is never a 3 anywhere to the right of a 0 .

In each of the following, \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}\). Solve for \(f(n)\) relative to the given set \(S\), and determine the appropriate "big- Oh" form for \(f\) on \(S\). a) \(f(1)=5 } \\\\{f(n)=4 f(n / 3)+5, \quad n=3,9,27, \ldots} \\\\{S=\left\\{3^{t} \mid i \in \mathbf{N}\right\\}} \end{array}\) b) \(f(1)=7\) \(f(n)=f(n / 5)+7, \quad n=5,25,125, \ldots\) \(S=\left\\{5^{\prime} \mid i \in \mathbf{N}\right\\}\)

Let \(a, b, c \in \mathbf{Z}^{+}\)with \(b \geq 2\), and let \(d \in \mathbf{N}\). Prove that the solution for the recurrence relation $$ \begin{aligned} &f(1)=d \\ &f(n)=a f(n / b)+c, \quad n=b^{k}, \quad k \geq 1 \end{aligned} $$ satisfies a) \(f(n)=d+c \log _{b} n\), for \(n=b^{k}, k \in \mathbf{N}\), when \(a=1\). b) \(f(n)=d n^{\log _{b} a}+(c /(a-1))\left[n^{\log _{h} a}-1\right]\), for \(n=b^{k}\), \(k \in \mathbf{N}\), when \(a \geq 2\)

Paul invested the stock profits he received 15 years ago in an account that paid \(8 \%\) interest compounded quarterly. If his account now has \(\$ 7218.27\) in it, what was his initial investment?

In Exercise 12 of Section \(4.2\) we leamed that \(F_{0}+F_{1}+\) \(F_{2}+\cdots+F_{n}=\sum_{i=0}^{n} F_{i}=F_{n+2}-1\). This is one of many such properties of the Fibonacci numbers that were discovered by the French mathematician Fran莽ois Lucas (1842-1891). Although we established the result by the Principle of Mathematical Induction, we see that it is easy to develop this formula by adding the system of \(n+1\) equations $$ \begin{aligned} F_{0} &=F_{2}-F_{1} \\ F_{1} &=F_{3}-F_{2} \\ \cdots & \ldots, \quad \cdots \\ F_{n-1} &=F_{n+1}-F_{n} \\ F_{n} &=F_{n+2}-F_{n+1} \end{aligned} $$Develop formulas for each of the following sums, and then check the general result by the Principle of Mathematical Induction. a) \(F_{1}+F_{3}+F_{3}+\cdots+F_{2 n-1}\), where \(n \in \mathbf{Z}^{+}\) b) \(F_{0}+F_{2}+F_{4}+\cdots+F_{2_{x}}\), where \(n \in \mathbf{Z}^{+}\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Calculus

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.