Question

To solve the equation \( x \log x = 1 \), using Newton-Raphson method, the iterative formula and the first approximate \( x_1 \), when \( x_0 = 1 \) is (Assuming \(\log\) is natural logarithm \(\ln\))

• A. \( x_{n+1} = \frac{x_{n}-1}{1+\log x_n} ; x_1 = 0 \)
• B. \( x_{n+1} = \frac{x_n + 2x_n \log x_n + 1}{1+\log x_n} ; x_1 = 2 \)
• C. \( x_{n+1} = \frac{x_n - 2x_n \log x_n + 1}{1+\log x_n} ; x_1 = 2 \)
• D. \( x_{n+1} = \frac{x_n+1}{1+\log x_n} ; x_1 = 2 \)

Hint:

Newton-Raphson: \(x_{n+1} = x_n - f(x_n)/f'(x_n)\).
Remember derivative rules (e.g., product rule, \(\frac{d}{dx}\ln x = 1/x\)).
\(\ln 1 = 0\).

Answer 1

The Correct Option is D

Let \(f(x) = x \ln x - 1\). We want to solve \(f(x)=0\). The Newton-Raphson formula is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). First, find \(f'(x)\). Using product rule for \(x \ln x\): \(\frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\). So, \(f'(x) = \ln x + 1\). The iterative formula is: \(x_{n+1} = x_n - \frac{x_n \ln x_n - 1}{\ln x_n + 1} = \frac{x_n(\ln x_n + 1) - (x_n \ln x_n - 1)}{\ln x_n + 1}\) \(x_{n+1} = \frac{x_n \ln x_n + x_n - x_n \ln x_n + 1}{\ln x_n + 1} = \frac{x_n + 1}{1 + \ln x_n}\). This matches the formula part of option (d) (with \(\log = \ln\)). Now, find \(x_1\) given \(x_0 = 1\): \(x_1 = \frac{x_0 + 1}{1 + \ln x_0} = \frac{1 + 1}{1 + \ln 1} = \frac{2}{1 + 0} = \frac{2}{1} = 2\). So, \(x_1 = 2\). This matches option (d). \[ \boxed{x_{n+1} = \frac{x_n+1}{1+\log x_n} ; x_1 = 2} \]