Does Newton's Method converge for f(x) Does Newton's Method converge for: $$f(x)=(x-5)^2e^{x-5}$$
So Newton's Method is:
$$x_{n+1}=x_n−\frac{f(x_n)}{f′(x_n)}$$
$$x_{n}=x_{n-1}-\frac{x_{n-1}-5}{x_{n-1}-3}$$
Error:
$$e_n = x_{n} - 5=x_{n-1}-\frac{x_{n-1}-5}{x_{n-1}-3}-5=\frac{(x_{n-1}-5)(x_{n-1}-4)}{x_{n-1}-3}$$
So:
$$ e_n = \frac{e_{n-1}(e_{n-1}+1)}{e_{n-1}+2}$$
But how to show that this converges to 0?
 A: For any sucession, if you have $a_{n+1}=f(a_n)$ the $f$ has a fixed point at $x$ if $f(x)=x$, which means that $a_{n+1}=a_{n}$.
If you have a fixed point $L$, solution of $L=f(L)$, the sucession will converge to such fixed point if there exists a neighbourhood in which $|a_{n+1}-L|<|a_n-L|$   (or $|f(x)-L|<|x-L|$) and you start from within a point of such neighbourhood.
In this case we know that a fixed point for the error is $e_n=0$, but will the sucession converge to $e_n=0$? well in this case $L=0$ so for what values of $e_n$ (or x) will $|f(x)|<|x|$?
Below is a graphical representation of it.

So the answer is that Newton method will converge if your initial error is $e_0>-3/2$
In all the above reasoning you should actually replace 'converge' with 'converge uniformly' for $e_n$ will converge to $0$ for any value of $e_0>-2$ but if you start from $e_0<-3/2$ the absolute value of the error will increase after the first iteration. If you start from $e_0<-2$ the error will become more negative each time. You can get that information from the graphic below

A: Note that $$|\frac{e_{n-1}+1}{e_{n-1}+2}| < 1$$
So if we start with any error $e_0$, it will at each step be multiplied by a factor whos aboslute value is less than one, and thus will converge closer to zero for each iteration.
