Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider \(x=\sqrt{1+x}\). (a) Apply the Fixed-Point Algorithm starting with \(x_{1}=0\) to find \(x_{2}, x_{3}, x_{4}\), and \(x_{5} .\) (b) Algebraically solve for \(x\) in \(x=\sqrt{1+x}\). (c) Evaluate \(\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}\).

Step-by-Step Solution

Answer

(a) 1, \(\sqrt{2}\), \(\sqrt{2.414}\), \(\sqrt{2.553}\); (b) \(x = \frac{1+\sqrt{5}}{2}\); (c) \(\frac{1+\sqrt{5}}{2}\).

Step 1: Set Up the Fixed-Point Iteration

Given the equation \(x = \sqrt{1+x}\), we recognize this as a fixed-point problem where the next iteration \(x_{n+1}\) can be defined as \(x_{n+1} = \sqrt{1+x_n}\). We start with \(x_1 = 0\).

Step 2: Calculate \(x_2\)

Substitute \(x_1 = 0\) into the iteration formula: \(x_2 = \sqrt{1+0} = \sqrt{1} = 1\).

Step 3: Calculate \(x_3\)

Substitute \(x_2 = 1\) into the iteration formula: \(x_3 = \sqrt{1+1} = \sqrt{2}\). This gives a numerical value, \(x_3 \approx 1.414\).

Step 4: Calculate \(x_4\)

Substitute \(x_3 \approx 1.414\) into the iteration formula: \(x_4 = \sqrt{1+1.414} = \sqrt{2.414}\). This gives \(x_4 \approx 1.553\).

Step 5: Calculate \(x_5\)

Substitute \(x_4 \approx 1.553\) into the iteration formula: \(x_5 = \sqrt{1+1.553} = \sqrt{2.553}\). This results in \(x_5 \approx 1.597\).

Step 6: Algebraically Solve for \(x\)

Re-arrange the equation \(x = \sqrt{1+x}\) by squaring both sides: we obtain \(x^2 = 1 + x\). Re-arranging gives \(x^2 - x - 1 = 0\). The solutions to this quadratic equation are found using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -1, c = -1\). This results in \(x = \frac{1 \pm \sqrt{5}}{2}\). The acceptable solution (since \(x \geq 0\)) is \(x = \frac{1 + \sqrt{5}}{2}\).

Step 7: Evaluate Infinite Nested Radical

The expression \(x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}\) forms an infinite nested radical. Setting \(x = \sqrt{1+x}\) and solving as before, we confirm \(x = \frac{1 + \sqrt{5}}{2}\) is the value of this expression, which is the golden ratio, approximately \(1.618\).

Key Concepts

Algebraic Solutions

Algebraic solutions are analytical methods used to find precise answers to equations or problems. In the context of the equation \(x = \sqrt{1+x}\), to solve for \(x\) algebraically means teasing out the exact values by mathematical manipulation. Here, we start by both sides of the equation. By squaring, we eliminate the square root, thus forming \(x^2 = 1 + x\). This is a classic quadratic equation, which can be expressed as \(x^2 - x - 1 = 0\).
If you’re familiar with solving quadratic equations, this is where you use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For our equation:

• \(a = 1\)
• \(b = -1\)
• \(c = -1\)

Plugging into the formula gives us \(x = \frac{1 \pm \sqrt{5}}{2}\). Since we want the positive solution for \(x\), we choose \(x = \frac{1 + \sqrt{5}}{2}\). This value matches what is known as "the golden ratio," which is approximately 1.618. This algebraic approach is precise and aligns with the solution found through iterative methods.

Nested Radicals

Nested radicals appear in expressions like \(\sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}\), where there is an infinitely recurring radical term. These can be tricky to evaluate directly, but they can provide interesting insights into mathematical patterns and limits.
Here's the intuition: Suppose \(x\) is an infinite nested radical \(\sqrt{1 + \sqrt{1 + \cdots}}\). Then you can express this again as \(x = \sqrt{1+x}\). If we treat \(x\) this way, it becomes a fixed-point equation. By solving mathematically as done before, we rediscover \(x = \frac{1 + \sqrt{5}}{2}\), which emerges as a constant despite the complexity of the nesting.
This is one of the fascinating properties of nested radicals—they sometimes converge to simple and beautiful mathematical constants, like the golden ratio in this case. Understanding nested radicals not only helps you evaluate infinite series but also broadens your comprehension of limits and convergence in mathematics.

Iterative Methods

Iterative methods, like the Fixed-Point Algorithm, are useful for approaching solutions gradually through successive approximations. These mathematical tools are commonly employed when algebraic methods become cumbersome or exact solutions are unnecessary.
Starting with an initial guess—here, \(x_1 = 0\)—the computation proceeds by plugging the last result back into the function, calculate the next approximation. For example:

• Compute \(x_2 = \sqrt{1 + 0} = 1\).
• Calculate \(x_3 = \sqrt{1 + 1} \approx 1.414\).
• Find \(x_4 = \sqrt{1 + 1.414} \approx 1.553\).
• Determine \(x_5 = \sqrt{1 + 1.553} \approx 1.597\).

This process exhibits convergence, bringing us incrementally closer to an accurate solution. By \(x_5\), we're nearing the golden ratio \(\frac{1 + \sqrt{5}}{2}\). Iterative methods are incredibly practical for real-world applications, allowing for more flexible and adaptable solutions, especially when dealing with complex functions or systems that are resistant to straightforward solutions.