Problem 52 - Calculus Early Transcendentals: Pearson New International Edition

Question

A famous sequence \(\left\\{f_{n}\right\\}\), called the Fibonacci Sequence after Leonardo Fibonacci, who introduced it around A.D. 1200 , is defined by the recursion formula $$f_{1}=f_{2}=1, \quad f_{n+2}=f_{n+1}+f_{n}$$ (a) Find \(f_{3}\) through \(f_{10}\). (b) Let \(\phi=\frac{1}{2}(1+\sqrt{5}) \approx 1.618034\). The Greeks called this number the golden ratio, claiming that a rectangle whose dimensions were in this ratio was "perfect." It can be shown that $$\begin{aligned} f_{n} &=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] \\\ &=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n} \phi^{-n}\right] \end{aligned}$$ Check that this gives the right result for \(n=1\) and \(n=2\). The general result can be proved by induction (it is a nice challenge). More in line with this section, use this explicit formula to prove that \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n}=\phi\). (c) Using the limit just proved, show that \(\phi\) satisfies the equation \(x^{2}-x-1=0\). Then, in another interesting twist, use the Quadratic Formula to show that the two roots of this equation are \(\phi\) and \(-1 / \phi\), two numbers that occur in the explicit formula for \(f_{n}\).

Step-by-Step Solution

Answer

(a) Fibonacci numbers from 3 to 10 are 2, 3, 5, 8, 13, 21, 34, 55. (b) Checked explicit formula, limit equals golden ratio. (c) \(\phi\) satisfies \(x^2-x-1=0\) and is solution via Quadratic Formula.

Step 1: Compute Fibonacci Numbers

First, we will calculate the Fibonacci numbers using the recursive formula given. We'll start with the initial values \(f_1 = 1\) and \(f_2 = 1\):- \(f_3 = f_2 + f_1 = 1 + 1 = 2\)- \(f_4 = f_3 + f_2 = 2 + 1 = 3\)- \(f_5 = f_4 + f_3 = 3 + 2 = 5\)- \(f_6 = f_5 + f_4 = 5 + 3 = 8\)- \(f_7 = f_6 + f_5 = 8 + 5 = 13\)- \(f_8 = f_7 + f_6 = 13 + 8 = 21\)- \(f_9 = f_8 + f_7 = 21 + 13 = 34\)- \(f_{10} = f_9 + f_8 = 34 + 21 = 55\)

Step 2: Verify Explicit Formula For Initial Values

We need to confirm that the explicit formula works for \(n=1\) and \(n=2\).For \(n=1\):\[ f_1 = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^1 - \left( \frac{1-\sqrt{5}}{2} \right)^1 \right] = 1 \]For \(n=2\):\[ f_2 = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^2 - \left( \frac{1-\sqrt{5}}{2} \right)^2 \right] = 1 \]Both calculations confirm the explicit formula matches the initial Fibonacci numbers.

Step 3: Prove the Limit Using the Explicit Formula

The limit \(\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_n} = \phi\) can be shown using the explicit formula.As \(n \rightarrow \infty\), the term \((-1)^n \phi^{-n}\) becomes negligible, and thus:\[ \lim_{n \to \infty} \frac{f_{n+1}}{f_n} = \lim_{n \to \infty} \frac{\left( \frac{1+\sqrt{5}}{2} \right)^{n+1}}{\left( \frac{1+\sqrt{5}}{2} \right)^n} = \phi \]This confirms \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n} = \phi\).

Step 4: Show \(\phi\) Satisfies the Quadratic Equation

To show that \(\phi\) satisfies the equation \(x^2 - x - 1 = 0\), note that:Substituting \(\phi = \frac{1 + \sqrt{5}}{2}\), we calculate:\[ \phi^2 = \left( \frac{1 + \sqrt{5}}{2} \right)^2 = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]Thus, \(\phi^2 = \phi + 1\), satisfying the equation \(x^2 - x - 1 = 0\).

Step 5: Use the Quadratic Formula

Use the Quadratic Formula to solve for \(x\) in \(x^2 - x - 1 = 0\):The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1, b = -1, c = -1\).\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]The roots are \(\phi = \frac{1 + \sqrt{5}}{2}\) and \(-\frac{1}{\phi} = \frac{1 - \sqrt{5}}{2}\), corresponding to the terms in the explicit formula for \(f_n\).

Key Concepts

Recursion Formula

The Fibonacci Sequence is a classic exhibition of a recursion formula. This sequence starts with two initial numbers, both equal to 1. Then, each subsequent number is the sum of the two preceding numbers. This process can be expressed with the formula: \(f_{n} = f_{n-1} + f_{n-2}\).
For example, starting with \(f_1 = 1\) and \(f_2 = 1\), the calculation proceeds as follows:

• \(f_3 = 1 + 1 = 2\)
• \(f_4 = 2 + 1 = 3\)
• \(f_5 = 3 + 2 = 5\)
• \(f_6 = 5 + 3 = 8\)
• \(f_7 = 8 + 5 = 13\)
• \(f_8 = 13 + 8 = 21\)

This method of forming a sequence reflects how many natural systems function, each step dependent on previous conditions.

Golden Ratio

The Golden Ratio, denoted as \(\phi\), is approximately 1.618034. This irrational number holds a special place in art, architecture, and nature due to its aesthetically pleasing properties. Defined as \(\phi = \frac{1 + \sqrt{5}}{2}\), it prominently appears in the Fibonacci Sequence.
As \(n\) increases, the ratio of consecutive Fibonacci numbers \(\frac{f_{n+1}}{f_n}\) approaches \(\phi\).
This linkage demonstrates the elegant balance that the Fibonacci Sequence maintains as a bridge between simple mathematical rules and complex natural phenomena.

Quadratic Equation

Quadratic equations form a foundation of algebra and appear in a variety of mathematical contexts. The equation \(x^2 - x - 1 = 0\) is intimately connected to the golden ratio. By solving this equation, we identify the golden ratio itself as one of its solutions.
Substituting \(\phi = \frac{1 + \sqrt{5}}{2}\) into the equation, we verify:

• \(\phi^2 = \phi + 1\)

This property confirms that \(\phi\) is indeed a solution. Quadratic equations often have two roots, and here, they are \(\phi\) and \(-\frac{1}{\phi}\). These solutions reveal the recurring patterns that emerge in both mathematical formulas and natural systems.

Explicit Formula

Finding an explicit formula for sequences like the Fibonacci Sequence allows for rapid calculation of terms without recursion. The formula is:
\[ f_n = \frac{1}{\sqrt{5}} \left( \left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n \right) \]
With this formula, we can verify Fibonacci numbers quickly for small values of \(n\), such as \(n = 1\) and \(n = 2\). The formula is derived from the solutions to the characteristic equation associated with the recursive definition and helps in proving the relationship between consecutive Fibonacci numbers and the golden ratio.

Mathematical Induction

Mathematical induction is a powerful tool for proving statements for all natural numbers. It works in two main steps: establishing a base case and proving that if the statement holds for an arbitrary case \(n\), it must hold for \(n+1\).
To apply mathematical induction to the Fibonacci Sequence and its properties:

• Verify the formula for initial conditions, like \(n=1\) and \(n=2\).
• Assume it holds for \(n=k\) and use the recursive definition to prove it for \(n=k+1\).

This logical structure ensures that the properties observed in examples apply universally, confirming the deep connections between different mathematical constructs.