/*! 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 21 . Let \(F_{n}\) denote the \(n\)... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 21

. Let \(F_{n}\) denote the \(n\)th Fibonacci number, for \(n \geq 0\), and let \(\alpha=(1+\sqrt{5}) / 2\). For \(n \geq 3\), prove that (a) \(F_{n}>\alpha^{n-2}\) and (b) \(F_{n}<\alpha^{n-1}\)

Short Answer

Expert verified
The results prove that for \(n \geq 3\), the \(n\)th Fibonacci number \(F_{n}\) is always larger than \(\alpha^{n-2}\) and smaller than \(\alpha^{n-1}\), where \(\alpha\) is the Golden Ratio, which is \((1 + \sqrt{5}) / 2\)

Step by step solution

01

Background Information

The Fibonacci sequence is defined by the recursive rule: \(F_{1}=1, F_{2}=1\) and \(F_{n}=F_{n-1}+F_{n-2}\) for \(n\geq 3\). The expression \(\alpha=(1+\sqrt{5}) / 2\) is actually the Golden Ratio. To solve this problem, we would use mathematical induction.
02

Base Case for Mathematical Induction

For \(n = 3\), we have \(F_{3} = 2\) and \(\alpha^{3-2} = \alpha\), so \(F_{3} > \alpha^{3-2}\). For part (b), we have \(F_{3} = 2\), \(\alpha^{3-1} = \alpha^2 > 2\), so \(F_{3} < \alpha^{3-1}\), which confirms that our base case is correct.
03

Inductive Step

For the inductive step, we need to prove the inequalities are true for \(n + 1\), assuming they are true for \(n\). Based on the recurrence, we have \(F_{n+1}=F_{n}+F_{n-1}\). Since \(F_{n} > \alpha^{n-2}\) and \(F_{n-1} > \alpha^{n-3}\), we can add these two inequalities to get \(F_{n+1} > \alpha^{n-2}+\alpha^{n-3} = \alpha^{n-3}(1+\alpha)\). Because \(\alpha^{n-3}(1+\alpha)>\alpha^{n-3}\cdot \alpha= \alpha^{n-2}\), we have confirmed that \(F_{n+1}>\alpha^{n-1}\), completing part (a) of the inductive step. Part (b) follows a similar method, and using the fact that \(F_{n} < \alpha^{n-1}\) and \(F_{n-1} < \alpha^{n-2}\), we have \(F_{n+1} < \alpha^{n-1} + \alpha^{n-2} = \alpha^{n-2}(1 + \alpha)\). But \(\alpha^{n-2}(1+\alpha)<\alpha^{n-2}\cdot \alpha^2= \alpha^n\), so we have confirmed that \(F_{n+1}<\alpha^{n}\), completing part (b) of the inductive step.

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Key Concepts

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

Mathematical Induction
Mathematical induction is a powerful method used to prove statements, especially those involving sequences. It has two main components:

1. **Base Case**: The statement holds for the initial value, usually for the smallest number in the set.
2. **Inductive Step**: If the statement holds for some arbitrary number, say \( n \), it should also be true for \( n+1 \).

In the context of proving properties about the Fibonacci sequence, induction helps ensure that inequalities involving the Fibonacci numbers and the powers of the Golden Ratio hold true for any integer \( n \).
For instance, in the given problem, the base case involves checking the truth of the statement for \( n = 3 \). After verifying this, the inductive step extends this truth to \( n+1 \).
  • This creates a domino effect where proving the base and inductive steps ensures that the statement is true for all subsequent numbers.
  • Induction is particularly useful in sequences because patterns or formulas from previous terms often directly affect future terms.
Recursive Formulas
Recursive formulas define sequences where each term is based on its preceding terms. They allow the computation of complex series through simple iterative processes.

In the Fibonacci sequence, each number is formed by adding up the two numbers before it. This can be expressed as:
\[ F_{n} = F_{n-1} + F_{n-2} \]
Where \( F_1 = 1 \) and \( F_2 = 1 \). This recursive relationship makes Fibonacci numbers inherently connected, meaning each term threads from those before.

The beauty of recursive formulas is their simplicity and power, allowing numbers such as the Fibonacci sequence to grow exponentially fast without missing any intermediate calculations.
  • They are a stepping stone between basic arithmetic patterns and complex mathematical concepts.
  • Recursive formulas also create a direct link to mathematical induction, as knowing the correctness of prior terms confirms future terms.
Golden Ratio
The Golden Ratio, denoted by \( \alpha \), is a special number approximately equal to 1.6180339887. Given by the formula \( \alpha = \frac{1 + \sqrt{5}}{2} \), it frequently appears in nature, art, and mathematics.

It is fascinating how this ratio relates closely with the Fibonacci sequence. As the Fibonacci numbers grow, the ratio of consecutive Fibonacci numbers (i.e., \( \frac{F_{n}}{F_{n-1}} \)) approaches the Golden Ratio.
  • This relationship highlights the Golden Ratio as a limit of the Fibonacci sequence's growth rate.
  • The sequence's recursive nature ensures that large Fibonacci numbers, when divided, give values that converge to \( \alpha \).
The properties of \( \alpha \) serve as a crucial element in mathematical proofs involving Fibonacci numbers. In the given problem, the inequalities demonstrate the minimal and maximal bounds of Fibonacci numbers around powers of \( \alpha \).
Its pervasive quality makes the Golden Ratio a vital concept in understanding not only Fibonacci's properties but also broader mathematical and natural phenomena.

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Most popular questions from this chapter

Find and solve a recurrence relation for the number of ways to park motorcycles and compact cars in a row of \(n\) spaces if each cycle requires one space and each compact needs two. (All cycles are identical in appearance, as are the cars, and we want to use up all the \(n\) spaces.)

In Corollary \(10.2\) we were concerned with finding the appropriate "big-Oh" form for a function \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}^{+} \cup\\{0\\}\) where \(\begin{aligned} f(1) \leq & c, \quad \text { for } c \in \mathbf{Z}^{+} \\\ f(n) \leq & a f(n / b)+c, \\ & \text { for } a, b \in \mathbf{Z}^{+} \text {with } b \geq 2 \text {, and } n=b^{k}, k \in \mathbf{Z}^{+} \end{aligned}\) Here the constant \(c\) in the second inequality is interpreted as he amount of time needed to break down the given problem of size \(n\) into \(a\) smaller (similar) problems of size \(n / b\) and to combine the \(a\) solutions of these smaller problems in order to set a solution for the original problem of size \(n\). Now we shall examine a situation wherein this amount of time is no longer a) Let \(a, b, c \in \mathbf{Z}^{+}\), with \(b \geq 2\). Let \(f: \mathbf{Z}^{+} \rightarrow \mathbf{R}^{+} \cup\\{0\\}\) be a monotone increasing function, where $$ \begin{aligned} &f(1) \leq c \\ &f(n) \leq a f(n / b)+c n, \quad \text { for } n=b^{k}, \quad k \in \mathbf{Z}^{+} \end{aligned} $$ Use an argument similar to the one given (for equalities) in Theorem \(10.1\) to show that for all \(n=1, b, b^{2}, b^{3}, \ldots\), $$ f(n) \leq c n \sum_{t=0}^{k}(a / b)^{\prime} $$ b) Use the result of part (a) to show that \(f \in O(n \log n)\), where \(a=b\). (The base for the log function here is any real number greater than 1.) c) When \(a \neq b\), show that part (a) implies that $$ f(n) \leq\left(\frac{c}{a-b}\right)\left(a^{k+1}-b^{k+1}\right) $$ d) From part (c), prove that (i) \(f \in O(n)\), when \(ab\). [Note: The "big-Oh" form for \(f\) here and in part (b) is for \(f\) on \(\mathbf{Z}^{+}\), not just \(\left\\{b^{k} \mid k \in \mathbf{N}\right\\} .\) ]

a) Let \(n\) lines be drawn in the plane such that each line intersects every other line but no three lines are ever coincident. For \(n \geq 0\), let \(a_{n}\) count the number of regions into which the plane is separated by the \(n\) lines. Find and solve a recurrence relation for \(a_{n}\). b) For the situation in part (a), let \(b_{n}\) count the number of infinite regions that result. Find and solve a recurrence relation for \(b_{n}\).

For \(n \in \mathbf{N}\), let \(s_{n}\) count the number of ways one can travel from \((0,0)\) to \((n, n)\) using the moves \(\mathrm{R}:(x, y) \rightarrow(x+1, y)\), \(\mathrm{U}:(x, y) \rightarrow(x, y+1), \mathrm{D}:(x, y) \rightarrow(x+1, y+1)\), where the path can never rise above the line \(y=x\). (a) Determine \(s_{2}\). (b) How is \(s_{2}\) related to the Catalan numbers \(b_{0}, b_{1}, b_{2} ?\) (c) How is \(s_{3}\) related to \(b_{0}, b_{1}, b_{2}, b_{3} ?\) What is \(s_{3} ?\) (d) For \(n \in \mathbf{N}\), how is \(s_{n}\) related to \(b_{0}, b_{1}, b_{2}, \ldots, b_{n}\) ? (The numbers \(s_{0}, s_{1}, s_{2}, \ldots\) are known as the Schr枚der numbers.)

Solve the recurrence relation \(a_{n+2}^{2}-5 a_{n+1}^{2}+4 a_{n}^{2}=0\), where \(n \geq 0\) and \(a_{0}=4, a_{1}=13\).
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