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Chapter 6: Problem 233

The sequence of Fibonacci numbers $$ 0,1,1,2,3,5,8,13, \ldots $$ is defined by its first two terms \(F_{0}=0, F_{1}=1,\) and by the recurrence relation: $$ F_{n+2}=F_{n+1}+F_{n} \text { when } n \geqslant 0 $$ Prove by induction that, for all \(n \geqslant 0\) $$ F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}}, \text { where } \alpha=\frac{1+\sqrt{5}}{2} \text { and } \beta=\frac{1-\sqrt{5}}{2} $$

Short Answer

Expert verified
The formula is proved by mathematical induction.

Step by step solution

01

Base Case Check

First, let's verify that the formula holds for the initial terms. For \(n=0\), we have \(F_0=0\). According to the given formula, the value should be \(\frac{\alpha^0-\beta^0}{\sqrt{5}} = \frac{1-1}{\sqrt{5}} = 0\). For \(n=1\), we have \(F_1=1\). The formula gives: \(\frac{\alpha^1-\beta^1}{\sqrt{5}} = \frac{\alpha-\beta}{\sqrt{5}}\). Upon simplifying, \(\alpha-\beta=\sqrt{5}\), so the result is \(\frac{\sqrt{5}}{\sqrt{5}} = 1\). The base cases hold true.
02

Inductive Hypothesis

Assume that the formula holds true for some arbitrary positive integer \(k\), i.e., \(F_k = \frac{\alpha^k - \beta^k}{\sqrt{5}}\) and \(F_{k+1} = \frac{\alpha^{k+1} - \beta^{k+1}}{\sqrt{5}}\). We need to prove it holds for \(k+2\).
03

Inductive Step

Using the recurrence relation, we express \(F_{k+2}\) as \(F_{k+2} = F_{k+1} + F_k\). Substituting the expressions from the inductive hypothesis: \[ F_{k+2} = \frac{\alpha^{k+1} - \beta^{k+1}}{\sqrt{5}} + \frac{\alpha^k - \beta^k}{\sqrt{5}} \] \[ = \frac{\alpha^{k+1} + \alpha^k - \beta^{k+1} - \beta^k}{\sqrt{5}} \] Using the property \(\alpha^2 = \alpha+1\) and \(\beta^2 = \beta+1\), we can factor out:\[ = \frac{\alpha^k(\alpha + 1) - \beta^k (\beta + 1)}{\sqrt{5}} \] \[ = \frac{\alpha^{k+2} - \beta^{k+2}}{\sqrt{5}} \].
04

Conclusion

We have shown that if the formula holds for \(k\) and \(k+1\), it must hold for \(k+2\) as well. Since it holds for \(n=0\) and \(n=1\), by the principle of mathematical induction, the formula \(F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}\) is true for all \(n \geq 0\).

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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 or formulas that are asserted to be true for all natural numbers. It consists of two main parts:
  • Base Case: Verify if the statement is true for the initial value, often starting with 0 or 1.
  • Inductive Step: Assume that the statement holds for an arbitrary integer, say \( k \), and then prove that it also holds for \( k+1 \).
The idea is similar to dominoes falling over: if the first domino falls (the base case), and each domino causes the next to fall (the inductive step), then all dominoes will fall.
In the Fibonacci problem, the base case verifies the formula for \( n = 0 \) and \( n = 1 \). For the inductive step, we assume the formula holds for some \( k \) and \( k+1 \), then show it holds for \( k+2 \). Thus, by induction, it's true for all \( n \).
Recurrence Relation
A recurrence relation is an equation that expresses a term in a sequence using preceding terms. Understanding these relations is crucial in defining sequences analytically. In our case, the Fibonacci sequence uses a recurrence relation:
  • \( F_{n+2} = F_{n+1} + F_n \)
This relation tells us that every term in the Fibonacci sequence is the sum of the two preceding terms. It gives us a straightforward way to compute Fibonacci numbers step-by-step, starting from known initial conditions \( F_0 = 0 \) and \( F_1 = 1 \).
Recurrence relations like this are pivotal when using induction, as they help in expressing a sequence in terms of previous values. They serve as a backbone in not only calculating the elements of a sequence but also in proving things about the sequence, such as providing a closed-form solution.
Binet's Formula
Binet's Formula is a fascinating formula that directly calculates the nth Fibonacci number without the iterative process of adding prior terms in the sequence. The formula is given as:\[F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}\]where \( \alpha = \frac{1+\sqrt{5}}{2} \), known as the golden ratio, and \( \beta = \frac{1-\sqrt{5}}{2} \).
Why is it useful? It provides a quick way to find any Fibonacci number, whether it is the 5th or the 100th, without needing to compute all previous numbers. The expression involves powers of \( \alpha \) and \( \beta \), which relate back to the roots of the characteristic equation derived from the Fibonacci recurrence relation.
Binet's Formula elegantly combines algebraic concepts with the recursive nature of Fibonacci numbers, showcasing the beauty and depth of mathematics by linking seemingly simple sequences to golden ratio dynamics.

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Most popular questions from this chapter

Prove by induction that $$ 5^{2 n+1} \cdot 2^{n+2}+3^{n+2} \cdot 2^{2 n+1} $$ is divisible by \(19,\) for all \(n \geqslant 0\)

Problem 232 The sequence $$ 2,5,13,35, \ldots $$ is defined by its first two terms \(u_{0}=2, u_{1}=5,\) and by the recurrence relation: $$ u_{n+2}=5 u_{n+1}-6 u_{n} $$ (a) Guess a closed formula for the \(n^{\text {th }}\) term \(u_{n}\). (b) Prove by induction that your guess in (a) is correct for all \(n \geqslant 0\).

Let \(f_{1}=2, f_{k+1}=f_{k}\left(f_{k}+1\right)\). Prove by induction that \(f_{k}\) has at least \(k\) distinct prime factors.

Prove the statement: \(" 5^{2 n+2}-24 n-25\) is divisible by \(576,\) for all \(n \geqslant 1 "\) When trying to construct proofs in private, one is free to write anything that helps as 'rough work'. But the intended thrust of Problem 229 is two-fold: \- to introduce the habit of distinguishing clearly between (i) the statement \(\mathbf{P}(n)\) for a particular \(n,\) and (ii) the statement to be proved \(-\) namely \(" \mathbf{P}(n)\) is true, for all \(n \geqslant 1 " ;\) and to draw attention to the "induction step" (i.e. the third bullet point above), where (i) we assume that some unspecified \(\mathbf{P}(k)\) is known to be true, and (ii) seek to prove that the next statement \(\mathbf{P}(k+1)\) must then be true. The central lesson in completing the "induction step" is to recognize that: to prove that \(\mathbf{P}(k+1)\) is true, one has to start by looking at what \(\mathbf{P}(k+1)\) says.

Problem 259 A tree is a connected graph, or network, consisting of vertices and edges, but with no cycles (or circuits). Prove that a tree with \(n\) vertices has exactly \(n-1\) edges. The next problem concerns spherical polyhedra. A spherical polyhedron is a polyhedral surface in 3-dimensions, which can be inflated to form a sphere (where we assume that the faces and edges can stretch as required). For example, a cube is a spherical polyhedron; but the surface of a picture frame is not. A spherical polyhedron has \- faces (flat 2-dimensional polygons, which can be stretched to take the form of a disc), \- edges (1-dimensional line segments, where exactly two faces meet), and \- vertices (0-dimensional points, where several faces and edges meet, in such a way that they form a single cycle around the vertex). Each face must clearly have at least 3 edges; and there must be at least three edges and three faces meeting at each vertex. If a spherical polyhedron has \(V\) vertices, \(E\) edges, and \(F\) faces, then the numbers \(V, E, F\) satisfy Euler's formula $$ V-E+F=2 $$ For example, a cube has \(V=8\) vertices, \(E=12\) edges, and \(F=6\) faces, and \(8-12+6=2\)
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