[[Newton's Method]]:
$\huge \begin{align}
x_{n+1} &= x_{n} - \frac{f(x)}{f'(x)}
\end{align}$
Halleys Method:
$\huge \begin{align}
x_{n+1} &= x_{n} - \frac{f(x_{n})}{f'(x_{n}) - \frac{f''(x_{n})f(x_{n})}{2f'(x_{n})}}
\end{align}$
To avoid calculating $f'$ and $f''$, we will use divided differences.
$\huge \begin{align}
f[x_{1},\dots,x_{n}] &= \frac{1}{n!}\deriv{^{n}f}{x^{n}}(0) \\
\end{align}$
For small intervals, $[x_{1},x_{n}]$
$\huge
f[x_{1},\dots,x_{n}] \approx \frac{1}{n!}\deriv{^{n}f}{x^{n}}(x_{1})
$
Therefore:
$\large \begin{align}
f[x_{i},x_{i+1}] &= \frac{1}{1!}f'(x_{i}) =
\frac{f(x_{i})-f(x_{i+1})}{x_{i}-x_{i+1}} \\
f[x_{i}, x_{i+1}, x_{i+2}] &= \frac{1}{2!}f''(x_{i}) =
\frac{f[x_{i},x_{i+1}]-f[x_{i+1,i+2}]}{
x_{i} - x_{i+2}
}
\end{align}$
This gives us the Oiler's Method:
$\huge \begin{align}
x_{1} &= f(x_{0}) \\
x_{2} &= x_{1} - \frac{f(x_{1})}{f[x_{0},x_{1}]}
\end{align}$
$\huge
x_{i+1} = x_{i} - \frac{
f(x_{i})f[x_{i-1},x_{i}]
}{
(f[x_{i-1},x_{i}]^{2})
- \frac{1}{2}f(x_{i})f[x_{i-2},x_{i-1},x_{i}]
}
$