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Chapter 5: Problem 39

Show that, for \(n \geq 2\), the \(n\) th term of the Fibonacci sequence is less than \((7 / 4)^{n-1}\). [Use the definition of the Fibonacci sequence, not the approximation to \(f_{n}\) given in equation (9).]

Short Answer

Expert verified
The inequality holds true by mathematical induction: \( f_n < (\frac{7}{4})^{n-1} \) for all \( n \geq 2 \).

Step by step solution

01

Fibonacci Sequence Definition

Start by recalling that the Fibonacci sequence is defined as \( f_1 = 1 \), \( f_2 = 1 \), and \( f_n = f_{n-1} + f_{n-2} \) for \( n \geq 3 \).
02

Set Up the Inductive Proof Base Case

For an inductive proof, initialize with the base case. Check for \( n = 2 \):- We have \( f_2 = 1 \).- Calculate the inequality: \( (\frac{7}{4})^{2-1} = \frac{7}{4} \).- Clearly, \( f_2 = 1 < \frac{7}{4} \), so the base case holds.
03

Inductive Hypothesis

Assume that the inequality \( f_k < (\frac{7}{4})^{k-1} \) holds for some arbitrary integer \( k \geq 2 \).
04

Inductive Step

Prove that if the assumption holds for \( k \), then it must also hold for \( k + 1 \):- We know \( f_{k+1} = f_k + f_{k-1} \).- By the inductive hypothesis, \( f_k < (\frac{7}{4})^{k-1} \) and \( f_{k-1} < (\frac{7}{4})^{k-2} \).- Thus, \( f_{k+1} < (\frac{7}{4})^{k-1} + (\frac{7}{4})^{k-2} = (\frac{7}{4})^{k-2} \left( \frac{7}{4} + 1 \right) \).
05

Simplify the Inductive Step

Simplify the expression:- \( \frac{7}{4} + 1 = \frac{11}{4} \).- Therefore, \( (\frac{7}{4})^{k-2} \left( \frac{11}{4} \right) = (\frac{7}{4})^{k-2} \cdot \frac{11}{4} \).- Show that \( \frac{11}{4} < \frac{7}{4} \times \frac{7}{4} = \frac{49}{16} \) for \( k \geq 2 \), which rearranges to \( \frac{11}{4} < \frac{49}{16} \), verifying the inequality for \( f_{k+1} \).
06

Conclude the Induction

Since the base case holds and the inductive step is proven, by the principle of mathematical induction, the inequality \( f_n < (\frac{7}{4})^{n-1} \) is true for all \( n \geq 2 \).

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Key Concepts

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

Fibonacci Sequence
The Fibonacci sequence is a famous series of numbers where each term is the sum of the two preceding ones. It starts with two terms: both 1. This makes it easy to remember the first few terms: 1, 1, 2, 3, 5, 8, and so on. The formal way to express the sequence is with a formula:
  • \( f_1 = 1 \)
  • \( f_2 = 1 \)
  • For \( n \geq 3 \), \( f_n = f_{n-1} + f_{n-2} \)
These numbers appear frequently in nature and mathematics. Their simplicity makes them a great way to learn about sequences. Understanding them is key when proving mathematical properties like inequalities.
Inequality Proof
An inequality proof shows that one mathematical expression is less than or greater than another. In this exercise, we're proving the Fibonacci sequence grows slower than a specific exponential function for values starting from 2. ### Understanding the Steps- **Base Comparison**: Here, we start by comparing specific values, like checking if \( f_2 < (\frac{7}{4})^{1} \).- **Logical Deduction**: The main challenge is reasoning through the steps. We assess whether the inequality stands up to logical scrutiny when we use different numbers. These proofs are both elegant and practical. They help demonstrate relationships between complex mathematical ideas using simpler approximations.
Inductive Hypothesis
The inductive hypothesis is a central part of mathematical induction. It's similar to making a leap-of-faith assumption about the truth of a proposition for some arbitrary integer. ### How it Works:- We assume the statement is true for a particular \( k \). For example, in this exercise, we propose \( f_k < (\frac{7}{4})^{k-1} \) is valid.- This assumption allows us to try and prove the statement for \( k + 1 \).In doing so, we bridge the specific case (the base) to an arbitrary instance. This helps build confidence in logical reasoning by extending a particular truth to the general case.
Base Case
The base case is where any inductive proof begins. It's the foundation that enables logical progression through larger sets of numbers or terms. Without it, we can't affirm that the more complex statements we derive hold true for initial conditions. ### Explanation:- **Purpose**: It verifies the proposition for an initial value, ensuring the logic can start correctly.- **Example in Exercise**: Checking for \( n = 2 \), we showed \( f_2 = 1 < (\frac{7}{4}) \). This comparison provides a simple starting point.By establishing the base, every step following it hinges on its correctness, allowing us to extend proof to many other numbers. This simple verification is crucial for the strength of an inductive argument.

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Most popular questions from this chapter

(a) Prove that for any integer \(n \geq 1\), any set of \(n\) positive real numbers has a smallest element. (b) Prove that the result of (a) is not true for infinite sets of positive real numbers in general but that it is true for some infinite sets. (c) What is the name of the principle which asserts that any nonempty set of natural numbers has a smallest element?

For any integer \(n \geq 0\), recall that \(n Z=\\{k n \mid k \in Z\\}\) denotes the set of multiples of \(n\). (a) Prove that \(n Z\) is an ideal of the integers. (b) Let \(A\) be any ideal of \(Z\). Prove that \(A=n Z\) for some \(n \geq 0\) by establishing each of the following statements. i. If \(A\) contains only one element, then \(A\) is of the desired form. Now assume that \(A\) contains more than one element. ii. Show that \(A\) contains a positive number. iii. Show that \(A\) contains a smallest positive number \(n\). iv. \(n Z \subseteq A\), where \(n\) is the integer found in iii. v. \(A \subseteq n\) Z. [Hint : The division algorithm, 4.1.5.]

Prove that \(A \cap\left(\bigcup_{i=1}^{n} B_{i}\right)=\bigcup_{i=1}^{n}\left(A \cap B_{i}\right)\) for any sets \(A, B_{1}, B_{2}, \ldots, B_{n}\)

[BB] Use the method of generating functions to solve the recurrence relation \(a_{n}=5 a_{n-1}-6 a_{n-2}\), given \(a_{0}=1\), \(a_{1}=4\). Verify your answer by comparing with Problem 18 .

What is wrong with the following argument, which purports to prove that all the Fibonacci numbers after the first two are even? Let \(f_{n}\) denote the \(n\) th term of the Fibonacci sequence. We prove that \(f_{n}\) is even for all \(n \geq 3\) using the strong form of the Principle of Mathematical Induction. The Fibonacci sequence begins \(1,1,2 .\) Certainly \(f_{3}=2\) is even and so the assertion is true for \(n_{0}=3\). Now let \(k>3\) be an integer and assume that the assertion is true for all \(n, 3 \leq n
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