I have been tasked with attempting to find properties with the function f(x) that would make It such that using newton's method would converge to a particular root at least cubically.
I don't exactly understand what this is asking me because I'm not sure where to even start with this to find a the mystery properties of this function. Can anyone point me in the right direction?
Given enough regularity, a convergent fixed point iteration of the form $x_{n+1} = g(x_n)$ will converge with order $p$ if
$$ g'(x^*) = \cdots = g^{(p-1)}(x^*)=0, g^{(p)}(x^*) \ne 0 $$
For Newton's method, where $g(x)=x -\frac{f(x)}{f'(x)}$, we already know that $g'(x^*)= 0$. The convergence would be at least cubic if $g''(x^*)=0$. If you write down $g''$ and compute it at $x^*$, you will see that the convergence is at least 3 when $f'(x^*) \ne 0$ and $f''(x^*)=0$.
You might want to look at Halley's method.
In general, apply Newton's method to some product $f(x)g(x)$. Then $$ N(x)=x-\frac{f(x)g(x)}{f'(x)g(x)+f(x)g'(x)} $$ has derivative \begin{align} N'(x)=\frac{f(x)g(x)(f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x))}{(f'(x)g(x)+f(x)g'(x))^2}. \end{align} For this to have a double root whenever $f$ has a simple root, you would need $f''(x)g(x)+2f'(x)g'(x)=0$ at the root. Demanding that this identity holds on some interval containing the root leads to $f'(x)g(x)^2=C$.
