The recursion relation to solve x = e-x using Newton-Raphson method is

1
Xn + 1 = e-Xn
2
Xn+1 = Xn – e-Xn
3
\({X_{n + 1}} = \left( {1 + {X_n}} \right)\frac{{{e^{ - Xn}}\;}}{{1 + {e^{ - Xn}}}}\)
4
\({X_{n + 1\;}} = \frac{{X_n^2 - \;{e^{ - Xn}}\left( {1 + {X_n}} \right)\; - 1}}{{{X_n} - {e^{ - Xn}}}}\)

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