1 1 vote The recursion relation to solve $x=e^{-x}$ using Newton-Raphson method is $x_{n+1}=e^{-x_{n}}$ $x_{n+1}=x_{n}-e^{-x_{n}}$ $x_{n+1}=\left(1+x_{n}\right) \frac{e^{-x_{n}}}{1+e^{-x_{n}}}$ $x_{n+1}=\frac{x_{n}^{2}-e^{-x_{n}}\left(1+x_{n}\right)-1}{x_{n}-e^{-x_{n}}}$ Numerical Methods gate2008-ec numerical-methods calculus + – admin 297 views answer comment Share Follow Add Sync Questions Print 0 reply Please log in or register to add a comment.