Solve $\displaystyle e^{-2x}-2x-2 = 0$ using Newton's method, giving your answer correct two decimal places.

$\displaystyle f(x) = e^{-2x}-2x-2$

$\displaystyle f'(x) = -2e^{-2x}-2$

$\displaystyle x_{n+1} = x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}$

I can choose an initial value and what-not, no problem. My problem is, when do we stop? When we have reached a stage where $\displaystyle x_{n+k}$ has no effect on $\displaystyle x_{n+k-1}$, where the latter is correct to two decimal places? (Lipssealed)