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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 $f(x)=e^x-x-1.9\cos{x}$

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up vote 4 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.

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Thanks a lot. What do you mean by roots of multiplicity of $m$ exactly? –  MSKfdaswplwq May 20 '13 at 20:29
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@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 :) –  MSKfdaswplwq 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
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