Question

Let \( f \) be a twice continuously differentiable function on \( [a, b] \) such that \( f'(x)<0 \) and \( f''(x)<0 \) for all \( x \in (a, b) \). Let \( f(\zeta) = 0 \) for some \( \zeta \in (a, b) \). The Newton-Raphson method to compute \( \zeta \) is given by \[ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}, \quad k = 0, 1, 2, ...... \] for an initial guess \( x_0 \). If \( x_k \in (\zeta, b) \) for some \( k \geq 0 \), then which of the following statements is/are correct?

• A. \( x_{k+1}>\zeta \)
• B. \( x_{k+1}<\zeta \)
• C. \( x_{k+1}<x_k \)
• D. For every \( \eta \in (\zeta, x_k) \), \( \frac{f''(\eta)}{f'(x_k)}>1 \)

Hint:

For Newton-Raphson problems, remember the geometric interpretation. The next iterate, \( x_{k+1} \), is the x-intercept of the tangent line to the curve at \( x_k \). Visualizing the graph of a decreasing, concave down function can quickly help you determine the relative positions of \( \zeta \), \( x_k \), and \( x_{k+1} \).

Answer 1

The Correct Option is A, C

Step 1: Understanding the Concept:
The problem analyzes the behavior of the Newton-Raphson method under specific conditions for the function \( f(x) \). We are given that \( f(x) \) is decreasing (\( f'(x)<0 \)) and concave down (\( f''(x)<0 \)). The root is \( \zeta \), and the current iterate \( x_k \) is to the right of the root (\( x_k>\zeta \)).
Step 2: Key Formula or Approach:
The Newton-Raphson iteration formula is \( x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \).
The error in the Newton-Raphson method can be analyzed using Taylor's theorem. The error at step \( k+1 \) is given by \( e_{k+1} = x_{k+1} - \zeta \). Taylor's expansion of \( f(\zeta) \) around \( x_k \) is: \[ f(\zeta) = f(x_k) + f'(x_k)(\zeta - x_k) + \frac{f''(\eta)}{2}(\zeta - x_k)^2 \quad \text{for some } \eta \in (\zeta, x_k) \] Step 3: Detailed Explanation or Calculation:
Analysis of Statement (C): \( x_{k+1}<x_k \)
We are given \( x_k \in (\zeta, b) \), which means \( x_k>\zeta \).
Since \( f'(x)<0 \), the function \( f \) is strictly decreasing.
Because \( x_k>\zeta \) and \( f \) is decreasing, we have \( f(x_k)<f(\zeta) \).
We know \( f(\zeta) = 0 \), so \( f(x_k)<0 \).
We are also given that \( f'(x_k)<0 \).
Now consider the term \( \frac{f(x_k)}{f'(x_k)} \). The numerator is negative and the denominator is negative, so the fraction is positive. \[ \frac{f(x_k)}{f'(x_k)}>0 \] From the Newton-Raphson formula: \[ x_{k+1} = x_k - \left( \frac{f(x_k)}{f'(x_k)} \right) = x_k - (\text{a positive value}) \] This directly implies that \( x_{k+1}<x_k \). Therefore, statement (C) is correct.
Analysis of Statement (A): \( x_{k+1}>\zeta \)
We use the Taylor expansion. Since \( f(\zeta) = 0 \): \[ 0 = f(x_k) + f'(x_k)(\zeta - x_k) + \frac{f''(\eta)}{2}(\zeta - x_k)^2 \] Rearranging for \( f(x_k) \): \[ f(x_k) = -f'(x_k)(\zeta - x_k) - \frac{f''(\eta)}{2}(\zeta - x_k)^2 = f'(x_k)(x_k - \zeta) - \frac{f''(\eta)}{2}(x_k - \zeta)^2 \] Substitute this into the Newton-Raphson formula: \[ x_{k+1} = x_k - \frac{f'(x_k)(x_k - \zeta) - \frac{f''(\eta)}{2}(x_k - \zeta)^2}{f'(x_k)} \] \[ x_{k+1} = x_k - (x_k - \zeta) + \frac{f''(\eta)}{2f'(x_k)}(x_k - \zeta)^2 \] \[ x_{k+1} - \zeta = \frac{f''(\eta)}{2f'(x_k)}(x_k - \zeta)^2 \] We are given \( f''(x)<0 \) and \( f'(x)<0 \). Thus, for \( \eta \in (\zeta, x_k) \), \( f''(\eta)<0 \) and \( f'(x_k)<0 \). This makes the fraction \( \frac{f''(\eta)}{f'(x_k)}>0 \).
Also, since \( x_k \neq \zeta \), the term \( (x_k - \zeta)^2 \) is positive.
Therefore, the entire right-hand side is positive: \[ x_{k+1} - \zeta>0 \implies x_{k+1}>\zeta \] Therefore, statement (A) is correct. Statement (B) is consequently incorrect.
Analysis of Statement (D):
We found that \( \frac{f''(\eta)}{f'(x_k)}>0 \). However, there is no information given in the problem to conclude that this ratio must be greater than 1. For example, if \( f(x) = -x - x^2/4 \) and \( \zeta=0 \), then \( f'(x)=-1-x/2 \) and \( f''(x)=-1/2 \). If \( x_k=1 \), \( f'(x_k) = -1.5 \). For any \( \eta \in (0,1) \), \( f''(\eta) = -0.5 \). The ratio is \( -0.5 / -1.5 = 1/3 \), which is not greater than 1. So, statement (D) is not always correct.
Step 4: Final Answer:
Based on the analysis, statements (A) and (C) are correct.
Step 5: Why This is Correct:
The conditions \( f'(x)<0 \) and \( f''(x)<0 \) define a function that is decreasing and concave down. For an initial guess \( x_k>\zeta \), the tangent line at \( (x_k, f(x_k)) \) will intersect the x-axis at a point \( x_{k+1} \) that is to the left of \( x_k \) but still to the right of the root \( \zeta \). This geometric interpretation matches the algebraic derivation.