Question

Newton-Raphson method is used to compute the inverse of the number 1.6. Among the following options, the initial guess of the solution that results in non-convergence of the iterative process is:

• A. 0.55
• B. 0.75
• C. 1.15
• D. 1.25

Hint:

In the Newton-Raphson method, choosing an initial guess too far from the correct value, especially one where the derivative is zero (as in \( x_0 = 1.25 \)), can cause the method to fail to converge.

Answer 1

The Correct Option is D

Step 1: Understanding the Newton-Raphson method. 
The Newton-Raphson method for finding the inverse of a number \( x \) involves the following iterative formula: \[ x_{n+1} = x_n \left( 2 - a x_n \right) \] where \( a \) is the number whose inverse we want to find (in this case, \( a = 1.6 \)). 
Step 2: Set up the iteration. 
We want to find \( \frac{1}{1.6} \), and we start with an initial guess \( x_0 \). The iterative formula becomes: \[ x_{n+1} = x_n \left( 2 - 1.6 \cdot x_n \right) \] Step 3: Analyze each initial guess. 
Let's check the convergence of the method for each guess: For \( x_0 = 0.55 \): 
The first iteration:
\[ x_1 = 0.55 \left( 2 - 1.6 \times 0.55 \right) = 0.55 \left( 2 - 0.88 \right) = 0.55 \times 1.12 = 0.616 \] The iteration produces a value that moves closer to the desired value of \( \frac{1}{1.6} \approx 0.625 \), resulting in convergence. For \( x_0 = 0.75 \):
The first iteration: \[ x_1 = 0.75 \left( 2 - 1.6 \times 0.75 \right) = 0.75 \left( 2 - 1.2 \right) = 0.75 \times 0.8 = 0.6 \] This is a reasonable approximation, and further iterations will converge to the correct value. For \( x_0 = 1.15 \): 
The first iteration: \[ x_1 = 1.15 \left( 2 - 1.6 \times 1.15 \right) = 1.15 \left( 2 - 1.84 \right) = 1.15 \times 0.16 = 0.184 \] This initial guess also leads to convergence after a few iterations. For \( x_0 = 1.25 \):
The first iteration: \[ x_1 = 1.25 \left( 2 - 1.6 \times 1.25 \right) = 1.25 \left( 2 - 2.0 \right) = 1.25 \times 0 = 0 \] This produces zero, leading to further non-convergence. 
Step 4: Conclusion. 
The initial guess \( x_0 = 1.25 \) results in non-convergence in the Newton-Raphson method. Therefore, the correct answer is: \[ \boxed{{(D) } 1.25} \]