Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider \(x=1+\frac{1}{x}\). (a) Apply the Fixed-Point Algorithm starting with \(x_{1}=1\) to find \(x_{2}, x_{3}, x_{4}\), and \(x_{5}\). (b) Algebraically solve for \(x\) in \(x=1+\frac{1}{x}\). (c) Evaluate the following expression. (An expression like this is called a continued fraction.) \(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}\)

Step-by-Step Solution

Answer

The solution to (a) gives x5 ≈ 1.6; in (b), x = (1 + √5)/2; and, for (c), the expression is (1 + √5)/2.

Step 1: Initialize the Fixed-Point Algorithm

Start with the initial value given: \( x_1 = 1 \). We will use this initial value to calculate the following values in the sequence.

Step 2: Calculate x2 using the Fixed-Point Relation

Use the formula \( x = 1 + \frac{1}{x} \) to find \( x_2 \). Substitute \( x_1 = 1 \) in:\[ x_2 = 1 + \frac{1}{x_1} = 1 + \frac{1}{1} = 2. \]

Step 3: Calculate x3 using the Fixed-Point Relation

Now find \( x_3 \) by substituting \( x_2 = 2 \) into the equation:\[ x_3 = 1 + \frac{1}{x_2} = 1 + \frac{1}{2} = 1.5. \]

Step 4: Calculate x4 using the Fixed-Point Relation

Substitute \( x_3 = 1.5 \) to find \( x_4 \):\[ x_4 = 1 + \frac{1}{x_3} = 1 + \frac{1}{1.5} = 1 + 0.6667 = 1.6667. \]

Step 5: Calculate x5 using the Fixed-Point Relation

Substitute \( x_4 = 1.6667 \) to find \( x_5 \):\[ x_5 = 1 + \frac{1}{x_4} = 1 + \frac{1}{1.6667} \approx 1.6. \] We observe that the sequence continues to oscillate towards the real solution.

Step 6: Algebraically Solve for x

Consider the equation \( x = 1 + \frac{1}{x} \). Multiply both sides by \( x \) to eliminate the fraction:\[ x^2 = x + 1. \]Rearrange to form a quadratic equation:\[ x^2 - x - 1 = 0. \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -1 \):\[ x = \frac{1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{1 \pm \sqrt{5}}{2}. \]The positive root is \( x = \frac{1 + \sqrt{5}}{2} \) (since \( x \) must be positive).

Step 7: Evaluate the Continued Fraction Expression

The expression is essentially the infinite continued fraction representation of the solution to \( x = 1 + \frac{1}{x} \).The solution is the golden ratio which is \( \frac{1 + \sqrt{5}}{2} \). Thus, \[ 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}} = \frac{1 + \sqrt{5}}{2}. \]

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents the variable. These equations are crucial because they can represent a wide variety of real-world phenomena, from projectile motion in physics to the calculation of areas in geometry. Solving a quadratic equation involves finding the value(s) of \( x \) that make the expression equal to zero.

Quadratics can be solved using several methods, including factoring, completing the square, and the quadratic formula. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is derived from the process of completing the square and provides an efficient way to find solutions when factoring is not straightforward. It shows that every quadratic equation has two roots, which may be real or complex.

For instance, in the problem above, the equation \( x = 1 + \frac{1}{x} \) simplifies to the quadratic equation \( x^2 - x - 1 = 0 \). By using the quadratic formula, we discover that the solutions are \( \frac{1 \pm \sqrt{5}}{2} \), with the positive solution corresponding to the famous golden ratio.

Continued Fractions

Continued fractions are a way to express numbers through a sequence of integer quotients. They are a bit like a supercharged version of fractions and provide deeper insights, especially into irrational numbers. A continued fraction looks like this:

• Numerator: Always 1.
• Denominator: Consists of a whole number plus another fraction.
• The pattern continues indefinitely.

For example, in the expression \( 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \cdots}}} \), the continued pattern continues infinitely.

This particular continued fraction actually converges to a well-known number: the golden ratio. This is because as you continue expanding the pattern, you get closer and closer to \( \frac{1 + \sqrt{5}}{2} \), which is the same solution you would find solving the earlier quadratic equation derived from \( x = 1 + \frac{1}{x} \). Continued fractions can provide approximations for irrational numbers, making them valuable tools in mathematical analysis and number theory.

Golden Ratio

The golden ratio, often denoted by the Greek letter \( \phi \), is a special number approximately equal to 1.618033988749895. It has been studied and admired because of its unique mathematical properties and its frequent appearance in nature, art, architecture, and other fields.

The golden ratio is derived from the relationship \( x = 1 + \frac{1}{x} \), which resolves to the quadratic equation \( x^2 - x - 1 = 0 \). Solving this quadratic using the quadratic formula gives the positive root \( x = \frac{1 + \sqrt{5}}{2} \), which is the definition of the golden ratio.

This ratio is particularly fascinating because it appears in various growth patterns, such as the arrangement of leaves on a stem, the pattern of seeds in a sunflower, and the spiral shells of certain mollusks. Artists and architects have also used the golden ratio to create aesthetically pleasing works, as it is believed to produce a sense of harmony and balance in compositions. Understanding the golden ratio not only deepens appreciation for mathematical beauty but also uncovers patterns that connect mathematics to the world around us.