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Chapter 10: Problem 29

A particle moves horizontally to the right. For \(n \in \mathbf{Z}^{+}\), the distance the particle travels in the \((n+1)\) st second is equal to twice the distance it travels during the \(n\)th second. If \(x_{n}, n \geq 0\), denotes the position of the particle at the start of the \((n+1)\) st second, find and solve a recurrence relation for \(x_{n}\), where \(x_{0}=1\) and \(x_{1}=5\).

Short Answer

Expert verified
The recurrence relation for the position of the particle at the start of the \((n+1)\) st second is given by \(x_{n}= 2x_{n-1}+x_{n-2}\). This can be used to calculate \(x_{2}, x_{3},...\).

Step by step solution

01

Determine the Recurrence Relation

Given the behaviour of the particle, we can derive a recurrence relation. The position of the particle at the start of the \((n+1)\) st second, denoted as \(x_{n}\), would be twice the position it was at the \((n)\)th second, so we can write it as \(x_{n}=2x_{n-1}+x_{n-2}\) for \(n \geq 2\).
02

Use the Initial Conditions

Now, plug in the values of the initial conditions given to us in the problem, i.e., \(x_{0}=1\) and \(x_{1}=5\).
03

Solve the Recurrence Relation

Solving this recurrence relation will give us the value for \(x_{n}\) which describes the distance the particle travels at the start of the \((n+1)\) st second. Since the recurrence relation is of second order, we can take the first initial values \(x_{0}=1\) and \(x_{1}=5\) and then use the recurrence relation multiple times to find other terms.
04

Validation

To ensure the solution is correct, the recurrence relation should consistently work for all values of \(n \geq 2\) and initial conditions.

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Most popular questions from this chapter

a) For \(n \in \mathbf{Z}^{+}\), let \(a_{n}\) count the number of palindromes of \(2 n\). Then \(a_{n+1}=2 a_{n}, n \geq 1, a_{1}=2\). Solve this first-order recurrence relation for \(a_{n}\). b) For \(n \in \mathbf{Z}^{+}\), let \(b_{n}\) count the number of palindromes of \(2 n-1\). Set up and solve a first-order recurrence relation for \(b_{n}\).

Renu wants to sell her laptop for \(\$ 4000\). Narmada offers to buy it for \(\$ 3000\). Renu then splits the difference and asks for \(\$ 3500\). Narmada likewise splits the difference and makes a new offer of \(\$ 3250\). (a) If the women continue this process (of asking prices and counteroffers), what will Narmada be willing to pay on her 5 th offer? 10th offer? \(k\) th offer, \(k \geq 1 ?\) (b) If the women continue this process (providing many, many new asking prices and counteroffers), what price will they approach? (c) Suppose that Narmada was willing to buy the laptop for \$3200. What should she have offered to pay Renu the first time?

d) For \(n \geq 2\), show that g) Prove that for \(n \geq 2\) \(E_{n}=\sum_{i=1}^{\lfloor n / 2\rfloor}\left(\begin{array}{c}n-1 \\ 2 i-1\end{array}\right) E_{2 t-1} E_{n-2 k}, \quad E_{0}=E_{1}=1 . \quad E_{n}=\left(\frac{1}{2}\right) \sum_{i=0}^{n-1}\left(\begin{array}{c}n-1 \\\ i\end{array}\right) E_{1} E_{n-1-1}, \quad E_{0}=E_{1}=1\) e) Where do we find 1 in a rise/fall permutation of h) Use the result in part (g) to find \(E_{6}\) and \(E_{7}\). \(1,2,3, \ldots, n ?\) f) For \(n \geq 1\), show that i) Find the Maclaurin series expansion for \(f(x)=\sec x+\) \(\tan x\). Conjecture (no proof required) the sequence for \(E_{n}=\sum_{i=0}^{\lfloor(n-1) / 2]}\left(\begin{array}{c}n-1 \\ 2 i\end{array}\right) E_{2} E_{n-2 i-1}, \quad E_{0}=1$$\mathrm{A}\) one-to-one function \(f:\\{1,2,3, \ldots, n\\} \rightarrow\\{1,2,3\) \(\ldots, n\\}\) is often called a permutation. Such a permutation is termed a rise/fall permutation when \(f(1)f(3)\) \(f(3)1\)

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 \(n \in \mathbf{Z}^{+}\)

Solve the following systems of recurrence relations. $$ \text { a) } \begin{aligned} a_{n+1}=&-2 a_{n}-4 b_{n} \\ b_{n+1}=& 4 a_{n}+6 b_{n} \\ n \geq 0, \quad a_{0}=1, \quad b_{0}=0 \end{aligned} $$ $$ \text { b) } \begin{aligned} &a_{n+1}=2 a_{n}-b_{n}+2 \\ &b_{n+1}=-a_{n}+2 b_{n}-1 \\ &n \geq 0, \quad a_{0}=0, \quad b_{0}=1 \end{aligned} $$
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