I am trying to find a solution for Newton's method for this function, $f(x)=1- e^{x}$, and figuring out if it converges for every initial starting point.
Now, from my understanding, Newton's method is defined as:
$x_{n+1} = x_{n}-\frac {f(x_{n})}{f^1(x_{n}}$
and to check if a function converges for every $x$, we use the error formula (where $P$ is the root):
$|x_{n+1} - P| =|x_{n}-P|*|x_{n}-P|*\frac {f^2(x_{n})}{2f^1(x_{n})}$
Convergence will happen when $|x_{n}-P|*\frac {f^2(x_{n})}{2f^1(x_{n})} <1$
Now, calculating the first and second derivatives of $f(x)=1- e^{x}$ we obtain:
$f^1(x_{n})=-e^{x_{n}}$
$f^2(x_{n})=-e^{x_{n}}$
With all this information, I am a little confused how to find a solution to the function and to check if for every initial starting point there is convergence. Any help would be great, thank you.
The function $f$ is concave, monotonic and it is easy to check that there is a solution, hence it is unique.
Note that $f'(x)<0$ for all $x$, hence the Newton iteration is well defined.
You can show that if you start at any $x_0$ then $x_1 \ge 0$.
You can show that if $x_n \ge 0$ then $0 \le x_{n+1} \le x_{n}$, so $x_n$ is a non increasing sequence bounded below, hence has a limit $x^* \ge 0$.
Since $x_{n+1} = x_n - {f(x_n) \over f'(x_n)}$ and $x_n \to x^*$, then this gives $x^* = x^* - {f(x^*) \over f'(x^*)}$, or $f(x^*) = 0$ (since $f'(x^*) \neq 0$).
Since the solution is unique (because of monotonicity) we must have $x_n \to 0$ for any starting point.