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Chapter 8: Problem 11

The Lucas numbers satisfy the recurrence relation $$ L_{n}=L_{n-1}+L_{n-2} $$ and the initial conditions \(L_{0}=2\) and \(L_{1}=1\) a) Show that \(L_{n}=f_{n-1}+f_{n+1}\) for \(n=2,3, \ldots,\) where \(f_{n}\) is the \(n\) th Fibonacci number. b) Find an explicit formula for the Lucas numbers.

Short Answer

Expert verified
a) Use induction to show \( L_n = f_{n-1} + f_{n+1} \). b) The formula is \( L_n = \left( \frac{2 + \sqrt{5}}{2} \right)\alpha^n + \left( \frac{2 - \sqrt{5}}{2} \right)\beta^n \).

Step by step solution

01

Understand the recurrence relation and initial conditions

The Lucas numbers satisfy the recurrence relation \[ L_n = L_{n-1} + L_{n-2} \] and have initial conditions \( L_0 = 2 \) and \( L_1 = 1 \). The Fibonacci numbers satisfy the recurrence relation \[ f_n = f_{n-1} + f_{n-2} \] with initial conditions \( f_0 = 0 \) and \( f_1 = 1 \).
02

Define the relation to prove for part a

We need to show that \( L_n = f_{n-1} + f_{n+1} \) for \( n \geq 2 \).
03

Prove by induction

Prove the statement by mathematical induction.**Base Case**: For \( n = 2 \), \( L_2 = L_1 + L_0 = 1 + 2 = 3 \) and \( f_1 + f_3 = 1 + 2 = 3 \), so \( L_2 = f_1 + f_3 \).**Inductive Step**: Assume \( L_k = f_{k-1} + f_{k+1} \) and \( L_{k+1} = f_k + f_{k+2} \). Then for \( L_{k+2} \): \[ L_{k+2} = L_{k+1} + L_k \] Using the induction hypothesis, \[ L_{k+2} = (f_k + f_{k+2}) + (f_{k-1} + f_{k+1}) \] Simplify it: \[ L_{k+2} = f_k + f_{k+1} + f_{k+2} + f_{k-1} \] Using the Fibonacci recurrence relation, \( f_{k+2} = f_{k+1} + f_k \), So, \[ L_{k+2} = f_{k+3} + f_{k+1} \]. Thus, the statement holds for \( n=k+2 \). Hence, by induction, the statement is true for all \( n \geq 2 \).
04

Recognize the form of the characteristic equation

To find an explicit formula, recognize that the Lucas sequence is a second-order linear homogeneous recurrence relation with characteristic equation: \[ x^2 - x - 1 = 0 \].
05

Solve the characteristic equation

Solve the characteristic equation for its roots: \[ x = \frac{1 \pm \sqrt{5}}{2} \]. These are the same as the Fibonacci sequence roots often denoted as \( \alpha = \frac{1 + \sqrt{5}}{2} \) and \( \beta = \frac{1 - \sqrt{5}}{2} \).
06

Form the general solution

The general solution for the Lucas numbers is a combination of the roots: \[ L_n = A\alpha^n + B\beta^n \].
07

Apply initial conditions

Use the initial conditions to solve for constants \( A \) and \( B \):\( L_0 = 2 = A + B \) and \( L_1 = 1 = A\alpha + B\beta \).
08

Solve the system of equations

Solve the system of equations for \( A \) and \( B \):\[ A + B = 2 \] and \[ A\alpha + B\beta = 1 \]. Solving these gives: \[ A = \frac{2 + \sqrt{5}}{2} \] and \[ B = \frac{2 - \sqrt{5}}{2} \].
09

Write the explicit formula

Thus, the explicit formula for the Lucas numbers is: \[ L_n = \left( \frac{2 + \sqrt{5}}{2} \right)\alpha^n + \left( \frac{2 - \sqrt{5}}{2} \right)\beta^n \].

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Key Concepts

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

Recurrence Relation
A recurrence relation is a way of defining sequences using previous terms. The Lucas numbers have the recurrence relation \[ L_n = L_{n-1} + L_{n-2} \]. This means each term is the sum of the two preceding terms. The initial conditions for the Lucas numbers are \(L_0 = 2\) and \(L_1 = 1\). Similarly, the Fibonacci numbers are defined by \[ f_n = f_{n-1} + f_{n-2} \] with initial conditions \(f_0 = 0\) and \(f_1 = 1\). Knowing these helps us understand how recurrence relations build sequences over time.
Fibonacci Numbers
Fibonacci numbers are a sequence where each number is the sum of the two preceding ones. This sequence starts with 0 and 1. The Fibonacci sequence is: 0, 1, 1, 2, 3, 5, 8, 13, ... . Understanding Fibonacci numbers is crucial for working with Lucas numbers, as there is a strong relationship between the two. Specifically, there's a formula that links them: \( L_n = f_{n-1} + f_{n+1} \). This connection shows the integral relationship between these sequences in number theory and combinatorics.
Mathematical Induction
Mathematical induction is a powerful method for proving statements about integers. It involves two steps: the base case and the inductive step. For the Lucas numbers, we used induction to show that for all \( n \ge \ 2 \), the relation \( L_n = f_{n-1} + f_{n+1} \) holds.
  • Base Case: Show the statement is true for the initial value (e.g., \( n = 2 \)).
  • Inductive Step: Assume it's true for \( n = k \) and then prove it holds for \( n = k+1 \).
By completing these steps, you can confidently say that the formula applies for all subsequent values.
Characteristic Equation
The characteristic equation is a polynomial equation derived from a linear recurrence relation that helps find an explicit formula for the sequence. For the Lucas numbers, the recurrence relation \( L_n = L_{n-1} + L_{n-2} \) translates into the characteristic equation: \[ x^2 - x - 1 = 0 \]. Solving this quadratic equation gives the roots \( \alpha = \frac{1 \pm \sqrt{5}}{2} \), which are essential for creating the explicit formula.
Explicit Formula
An explicit formula provides a direct way to compute terms in a sequence without recursion. For Lucas numbers, the general form is: \[ L_n = A\alpha^n + B\beta^n \], where \( \alpha \) and \( \beta \) are the roots of the characteristic equation. Using the initial conditions \( L_0 = 2 \) and \( L_1 = 1 \), you can solve for constants A and B:
  • \( A = \frac{2 + \sqrt{5}}{2} \)
  • \( B = \frac{2 - \sqrt{5}}{2} \).
Therefore, the explicit formula for Lucas numbers is: \[ L_n = \left( \frac{2 + \sqrt{5}}{2} \right)\alpha^n + \left( \frac{2 - \sqrt{5}}{2} \right)\beta^n \]. This formula allows you to compute any term in the Lucas sequence directly.

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Most popular questions from this chapter

Use generating functions to determine the number of different ways 15 identical stuffed animals can be given to six children so that each child receives at least one but no more than three stuffed animals.

This exercise deals with the problem of finding the largest sum of consecutive terms of a sequence of n real numbers. When all terms are positive, the sum of all terms provides the answer, but the situation is more complicated when some terms are negative. For example, the maximum sum of consecutive terms of the sequence \(-2,3,-1,6,-7,4\) is \(3+(-1)+6=8 .\) (This exercise is based on \([\mathrm{Be} 86] .\) Recall that in Exercise 56 in Section 8.1 we developed a dynamic programming algorithm for solving this problem. Here, we first look at the brute-force algorithm for solving this problem; then we develop a divide- and-conquer algorithm for solving it. a) Use pseudocode to describe an algorithm that solves this problem by finding the sums of consecutive terms starting with the first term, the sums of consecutive terms starting with the second term, and so on, keeping track of the maximum sum found so far as the algorithm proceeds. b) Determine the computational complexity of the algorithm in part (a) in terms of the number of sums computed and the number of comparisons made. c) Devise a divide-and-conquer algorithm to solve this problem. [Hint: Assume that there are an even number of terms in the sequence and split the sequence into two halves. Explain how to handle the case when the maximum sum of consecutive terms includes terms in both halves.] d) Use the algorithm from part (c) to find the maximum sum of consecutive terms of each of the sequences: \(-2,4,-1,3,5,-6,1,2 ; 4,1,-3,7,-1,-5, \quad 3,-2 ;\) and \(-1,6,3,-4,-5,8,-1,7\) e) Find a recurrence relation for the number of sums and comparisons used by the divide-and-conquer algorithm from part (c). f ) Use the master theorem to estimate the computational complexity of the divide-and-conquer algorithm. How does it compare in terms of computational complexity with the algorithm from part (a)?

a) What is the generating function for \(\left\\{a_{k}\right\\},\) where \(a_{k}\) is the number of solutions of \(x_{1}+x_{2}+x_{3}+x_{4}=k\) when \(x_{1}, x_{2}, x_{3},\) and \(x_{4}\) are integers with \(x_{1} \geq 3,1 \leq\) \(x_{2} \leq 5,0 \leq x_{3} \leq 4,\) and \(x_{4} \geq 1 ?\) b) Use your answer to part (a) to find \(a_{7}\)

Find a closed form for the exponential generating function for the sequence \(\left\\{a_{n}\right\\},\) where \(\begin{array}{ll}{\text { a) } a_{n}=(-2)^{n} .} & {\text { b) } a_{n}=-1} \\\ {\text { c) } a_{n}=n .} & {\text { d) } a_{n}=n(n-1)} \\ {\text { e) } a_{n}=1 /((n+1)(n+2))}\end{array}\)

Use generating functions to solve the recurrence relation \(a_{k}=3 a_{k-1}+2\) with the initial condition \(a_{0}=1\)
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