Newton's Method - Why is there slow convergence with a high multiplicity I'm using a calculator to observe the sluggishness with which Newton's method converges with $f(x) = (x-1)^8$. I let $x_0 = 1.1$. Clearly it's taking forever to get to the root $x=0$. I'm not completely sure why this is. I know it has to deal with the multiplicity of 8. 
I'm looking to reconcile this slow convergence with the theory. Thanks for any help :)
 A: Newton's method converges quadratically to the root for any initial approximation provided the root is a simple zero.
For roots that are not simple (higher multiplicity), we do not get quadratic convergence.
We can modify Newton's Method to work with those to:
$$g(x) = x - \dfrac{f(x)f'(x)}{[f'(x)]^2 - [f(x)][f''(x)]}$$
The obvious drawback is the need for the second derivative, but it will converge quadratically if $g$ has the required continuity  conditions. This second order term can also cause serious roundoff errors when we have small differences.
A: Why not use $f(x) = x^8$ instead?
Let's look at one iteration of Newton's method: $$x_1 = x_0 - \frac{x_0^8}{8x_0^7} = x_0 - \frac{1}{8}x_0 = \frac{7}{8}x_0.$$
So each iteration just multiplies the previous one by 7/8; repeated application gives a geometric series converging to the root 0.
If you replace the exponent 8 by $n$, then the geometric series has ratio $1-(1/n)$; so the larger $n$ is, the closer the ratio is to 1, and the slower the convergence.
A: $\newcommand{\pars}[1]{\left( #1 \right)}$
If we know the multiplicity $m$, we solve ${\rm f}^{1/m}\pars{x} = 0$ instead of ${\rm f}\pars{x} = 0$. Newton-Rapson yields
$$
x_{n + 1}
=
x_{n}
-
{{\rm f}^{1/m}\pars{x_{n}}
 \over
 \pars{1/m}\,{\rm f}^{\pars{1/m} - 1}\pars{x_{n}}{\rm f}'\pars{x_{n}}}
\quad\Longrightarrow\quad
\color{#ff0000}{\large x_{n + 1}}
\color{#000000}{\large\ =\ }
\color{#ff0000}{\large x_{n}- m\,{{\rm f}\pars{x_{n}} \over {\rm f}'\pars{x_{n}}}}
$$
Notice the multiplicative factor $m$. It is faster than the original one
$
x_{n + 1}
=
x_{n}
-
{\rm f}\pars{x_{n}}/{\rm f}'\pars{x_{n}}
$.
$$
\begin{array}{rrclcrcl}
\mbox{Modified Newton-Rapson}:&
x_{n + 1} & = & x_{n} - 8\,{\pars{x_{n} - 1}^{8} \over 8\pars{x_{n} - 1}^{7}}
& \Longrightarrow &
\color{#ff0000}{\Large x_{n + 1}} & = & \color{#ff0000}{\Large 1}
\\
\mbox{Original Newton-Rapson}:&
x_{n + 1} & = &x_{n} - {\pars{x_{n} - 1}^{8} \over 8\pars{x_{n} - 1}^{7}}
& \Longrightarrow &
x_{n + 1} & = & {7x_{n} + 1 \over 8}
\end{array}
$$
