Question

Consider the fixed-point iteration $x_{k+1} = g(x_k)$ with $g(x) = \frac{x}{3} + \frac{4}{3x}$. Which root-finding problem is this equivalent to?

• A. $x - \frac{1}{3} + \frac{4}{3x^2} = 0$
• B. $\frac{x}{3} + \frac{4}{3x} = 0$
• C. $\frac{1}{3} - \frac{4}{3x^2} = 0$
• D. $x^2 - 2 = 0$

Hint:

To verify fixed-point convergence, you can check the derivative of $g(x)$ near the root:
\[ g'(x) = \frac{1}{3} - \frac{4}{3x^2} \]
For $x = \sqrt{2}$, we have $g'(\sqrt{2}) = \frac{1}{3} - \frac{4}{6} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$.
Since $|g'(\sqrt{2})| = \frac{1}{3} < 1$, the iteration is guaranteed to converge to the root $\sqrt{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question asks us to find the root-finding equation $f(x) = 0$ that is mathematically equivalent to the given fixed-point iteration scheme.

Step 2: Key Formula or Approach:
In a fixed-point iteration scheme $x_{k+1} = g(x_k)$, the sequence converges to a fixed point $x$ that satisfies the relation:
\[ x = g(x) \]
By rearranging this algebraic equation into the form $f(x) = 0$, we find the equivalent root-finding problem.

Step 3: Detailed Explanation:

• The given fixed-point iteration function is:
\[ g(x) = \frac{x}{3} + \frac{4}{3x} \]

• To find the fixed point, we set $x = g(x)$:
\[ x = \frac{x}{3} + \frac{4}{3x} \]

• We can solve this algebraic equation by clearing the denominators. Multiply every term in the equation by the common denominator $3x$ (assuming $x \neq 0$):
\[ 3x(x) = 3x \left( \frac{x}{3} + \frac{4}{3x} \right) \]
\[ 3x^2 = 3x \left( \frac{x}{3} \right) + 3x \left( \frac{4}{3x} \right) \]

• Simplifying each term on the right-hand side:
\[ 3x^2 = x^2 + 4 \]

• Subtract $x^2$ from both sides of the equation:
\[ 3x^2 - x^2 = 4 \]
\[ 2x^2 = 4 \]

• Divide the entire equation by 2:
\[ x^2 = 2 \]

• Rearranging the equation to the standard root-finding form $f(x) = 0$:
\[ x^2 - 2 = 0 \]

• This shows that the fixed-point iteration scheme is designed to find the roots of the equation $x^2 - 2 = 0$ (which are $x = \pm \sqrt{2}$).

Step 4: Final Answer
The root-finding problem is equivalent to $x^2 - 2 = 0$, which corresponds to option (D).