Newton's Method - Slow Convergence I'm using Newton's method to find the root of the equation $\frac{1}{2}x^2+x+1-e^x=0$ with $x_0=1$. Clearly the root is $x=0$, but it takes many iterations to reach this root. What is the reason for the slow convergence? Thanks for any help :)
 A: Newton's method has quadratic convergence near simple zeros, but the derivative of $\frac{1}{2}x^2+x+1-e^x$ at $x=0$ is zero and so $x=0$ is not a simple zero.
A: If
$f(x)
=\frac{1}{2}x^2+x+1-e^x
$,
$f'(x)
=x+1-e^x
$,
so
$f'(0) = 0$.
Since the iteration is
$x^*
= x-\dfrac{f(x)}{f'(x)}
$,
the next value will be
far away
if you start at $x=0$.
You were probably saved by rounding error.
It would have helped if
your question showed what your iterations
actually were.
A: Define ${\rm F}\left(x\right) \equiv 2\left({\rm e}^{x} - 1 - x\right)/x^{2}$. We  write the iteration as follows:
$$
x_{n + 1}
=
x_{n} - 1 + {1 \over {\rm F}\left(x_{n}\right)}
$$
The ${\rm F}\left(x\right)$ evaluation is performed 'carefully': For 'small $x$
$$
{\rm F}\left(x\right)
\approx
1 + {1 \over 3}\,x + {1 \over 12}\,x^{2} + {1 \over 60}\,x^{3}\,,
\qquad\qquad
\left\vert x\right\vert \gtrsim 0
$$
We neglect the cubic term whenever
$\left\vert x\right\vert^{3}/60 < \delta_{\rm mp}$ where $\delta_{\rm mp}$ is the ${\it\mbox{machine precision}}$. It means
$\left\vert x\right\vert <  60^{1/3}\delta_{\rm mp}^{1/3}$. The computer evaluation of ${\rm F}\left(x\right)$ is performed strictly according to the following definition:
$$
{\rm F}\left(x\right)
=
\left\{%
\begin{array}{ll}
1 + {x \over 3}\left(1 + {x \over 4}\right)\,,
&
\mbox{if}\qquad \left\vert x\right\vert < 60^{1/3}\,\delta_{\rm mp}^{1/3}
\\[1mm]
{2\left({\rm e}^{x} - 1 - x\right) \over x^{2}}\,,
&
\mbox{if}\qquad \left\vert x\right\vert \geq 60^{1/3}\,\delta_{\rm mp}^{1/3}
\end{array}\right.
$$
For example, in $\tt\mbox{C++}$, $\delta_{\rm mp}$ is defined in the $\tt\mbox{<cfloat>}$ library as $\tt\mbox{FLT_EPSILON}$, $\tt\mbox{DBL_EPSILON}$ and $\tt\mbox{LDBL_EPSILON}$ for $\tt\mbox{float}$, $\tt\mbox{double}$ and $\tt\mbox{long double}$ types, respectively.
