$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}-\frac{1}{2}\frac{f(x_n)^2 f''(x_n)}{f'(x_n)^3}$
For obtaining the formula above, I did the following:
$f(x_n+ \Delta x_n) \approx f(x_n) + f'(x_n)\Delta x_n + \frac{f''(x_n)}{2} (\Delta x_n )^2 \iff -\frac{f(x_n)}{f'(x_n)} \approx \Delta x_n + \frac{f''(x_n)}{2 f'(x_n)} (\Delta x_n )^2 \iff -\frac{f(x_n)}{f'(x_n)} \approx \Delta x_n + \frac{f''(x_n)}{2 f'(x_n)} (- \frac{f(x_n)}{f'(x_n)} )^2 \iff -\frac{f(x_n)}{f'(x_n)} \approx \Delta x_n + \frac{f''(x_n)f'(x_n)^2}{2 f'(x_n)^3} \iff \Delta x_n \approx -\frac{f(x_n)}{f'(x_n)} -\frac{f''(x_n)f'(x_n)^2}{2 f'(x_n)^3} $
Now my goal is to determine its order/rate of convergence.
I wrote that:
$\Delta x_{n+1} = x - x_{n+1} = x - (x_n - \frac{f(x_n)}{f'(x_n)} - \frac{1}{2}\frac{f(x_n)^2 f''(x_n)}{f'(x_n)^3}$), but can't find a way to write it in other useful form.
Any suggestions?