Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a function $f(x)$, we can approximate $x_r$ where $f(x_r)=0$ , by using Newton's method:

$$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)} $$

The method only works when 'you choose an $x_0$ near enough to $x_r$'.

My questions are:

1.) When does this series converge?

2.) Given that the series converges, what is the order of convergence?

If it depends on $f(x)$, then let me define $f(x)=e^x-x-1.9\cos{x}$

share|cite|improve this question
up vote 5 down vote accepted

When $|f'(x)|<1 \ \forall x \in (a,b), \ f(x)$ will converges to a root in $(a,b)$ given a start value in $(a,b)$. The rate of convergence is quadratic for roots of multiplicity $m=1$. for $m>1$ convergence is linear but in that case $x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)}$ will have a quadratic convergence rate.

share|cite|improve this answer
Thanks a lot. What do you mean by roots of multiplicity of $m$ exactly? – Applied mathematician May 20 '13 at 20:29
@JoyeuseSaintValentin A root $x$ of multiplicity $m$ is a root for which $f^{(m-1)}(x)=0$ but $f^{(m)}(x)\neq 0$. The motivating example is polynomials; for instance, $f(x)=x^3-4x^2+5x-2=(x-1)^2(x-2)$ has a root of multiplicity $2$ at $x=1$, since $f(1)=0$ and $f'(1)=0$ but $f''(1)=-2\neq 0$. – Steven Stadnicki May 20 '13 at 20:39
tnx, i thought that is was something like that :) – Applied mathematician May 20 '13 at 20:41
@Angela Nicely explained. Could you explain me how $x_{n+1}=m\frac{f(x_n)}{f'(x_n)}$ for $m >1$? – srijan May 20 '13 at 23:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.