Discussion
The relation to solve \(x = e^{-x}\) using Newton-Raphson method is :
Answer: A
The iteration is \(x_{n+1} = \dfrac{(1 + x_{n})\, e^{-x_{n}}}{1 + e^{-x_{n}}}\).
Write the equation as \(f(x) = x - e^{-x} = 0\), so \(f'(x) = 1 + e^{-x}\). Newton-Raphson gives
\[ \begin{aligned} x_{n+1} &= x_{n} - \frac{x_{n} - e^{-x_{n}}}{1 + e^{-x_{n}}} \\ &= \frac{x_{n} + x_{n}e^{-x_{n}} - x_{n} + e^{-x_{n}}}{1 + e^{-x_{n}}} \\ &= \frac{(1 + x_{n})\, e^{-x_{n}}}{1 + e^{-x_{n}}} \end{aligned} \]
The options with \(1 - e^{-x_{n}}\) in the denominator use the wrong sign for \(f'\) (as if \(f = x + e^{-x}\)), and the last option is not a rearrangement of Newton's step. Starting from \(x_{0} = 0.5\) the iteration converges quickly to the root 0.5671.
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