Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Consider \(x=\sqrt{5+x}\). (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 \(x=\sqrt{5+x}\). (c) Evaluate \(\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}\).

Step-by-Step Solution

Answer

(a) Iterates are \(x_2 = \sqrt{5}\), \(x_3 = \sqrt{5 + \sqrt{5}}\), \(x_4 = \sqrt{5 + \sqrt{5 + \sqrt{5}}} \), and \(x_5 = \sqrt{5 + \sqrt{5 + \sqrt{5 + \sqrt{5}}}}\). (b) \(x = \frac{1 + \sqrt{21}}{2}\). (c) \(\frac{1 + \sqrt{21}}{2}\).

Step 1: Define the Fixed-Point Iteration Function

Given the equation \(x = \sqrt{5+x}\), rewrite it as a function: \(g(x) = \sqrt{5+x}\). This will be used for the fixed-point iteration.

Step 2: Start with Initial Value for Iteration

Use the initial value \(x_1 = 0\) as suggested.

Step 3: Calculate the Next Iteration Value

Plug \(x_1 = 0\) into the function:\[x_2 = g(0) = \sqrt{5+0} = \sqrt{5}\]

Step 4: Calculate the Second Iteration Value

Use \(x_2\) to compute \(x_3\):\[x_3 = g(x_2) = \sqrt{5 + \sqrt{5}}\]

Step 5: Calculate the Third Iteration Value

Next, use \(x_3\) to find \(x_4\):\[x_4 = g(x_3) = \sqrt{5 + \sqrt{5 + \sqrt{5}}}\]

Step 6: Calculate the Fourth Iteration Value

Finally, use \(x_4\) to compute \(x_5\):\[x_5 = g(x_4) = \sqrt{5 + \sqrt{5 + \sqrt{5 + \sqrt{5}}}}\]

Step 7: Algebraically Solve \(x = \sqrt{5+x}\)

Square both sides of the equation:\[x^2 = 5 + x\]Rearrange and solve the quadratic equation:\[x^2 - x - 5 = 0\]Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -1\), and \(c = -5\).This gives:\[x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}\]Since \(x\) must be positive, choose \(x = \frac{1 + \sqrt{21}}{2}\).

Step 8: Evaluate the Infinite Nested Radical

From the previous solutions, we know \(x\) satisfies \(x = \sqrt{5+x}\). Therefore, the expression \(\sqrt{5 + \sqrt{5 + \sqrt{5 + \cdots}}}\) converges to \(x = \frac{1 + \sqrt{21}}{2}\).

Key Concepts

Iterative Methods

Iterative methods are a powerful tool for solving equations, especially when an exact solution is difficult to find analytically. In these methods, we generate a sequence of approximations that converge to the desired solution.
For instance, consider the problem of finding a number that satisfies the equation \(x = \sqrt{5+x}\). Instead of directly solving this equation, we define a function \(g(x) = \sqrt{5+x}\) and iterate:

• Start with an initial guess, \(x_1 = 0\).
• Use this value to find the next approximation, \(x_2 = g(0) = \sqrt{5}\).
• Continue the process to calculate \(x_3, x_4, x_5\), etc.

Each step brings us closer to the actual solution, narrowing down the possible values through iteration.
Iterative methods are particularly useful when dealing with non-linear equations where direct solutions are more complex.
This approach allows us to see how the solution evolves, refining our approximation with each iteration. These methods are prevalent in both numerical analysis and real-world problem-solving.

Quadratic Equation

Quadratic equations are ones that take the form \(ax^2 + bx + c = 0\). These equations have a wide range of applications, and they are solved using the quadratic formula
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula provides solutions for \(x\) by considering the coefficients \(a\), \(b\), and \(c\).
In the exercise of solving \(x = \sqrt{5+x}\), the problem is converted into a quadratic equation:

• Square both sides to eliminate the square root: \(x^2 = 5 + x\).
• Rearrange to form \(x^2 - x - 5 = 0\).
• Apply the quadratic formula to find \(x\).

The discriminant \(b^2 - 4ac\) determines the nature of the roots. If it's positive, as in our case, we get two real solutions. However, since the context is a positive number, only \(x = \frac{1 + \sqrt{21}}{2}\) is considered because a length (or root) must be positive.
This solution highlights the utility of quadratic equations in analytically finding elegant solutions to seemingly complex problems.

Nested Radicals

Nested radicals are expressions involving a radical (square root, cube root, etc.) within another radical. This can create infinitely recursing expressions, which at first glance seem difficult to evaluate. However, they often converge to a specific number.
The problem \(\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}\) is a classic example of a nested radical. Despite their complex appearance, nested radicals can simplify elegantly.
In these scenarios:

• We assume that the expression converges to a value \(x\).
• From our algebraic solution, we derived that \(x = \frac{1 + \sqrt{21}}{2}\).

This approach helps us tackle expressions that might seem intractable at first. By recognizing and applying forms that are known to converge, we simplify calculation and find meaningful solutions.
Nested radicals, although intimidating, follow patterns and mathematical logic that make them solvable with the right techniques.