# Fixed point iteration to find root of Arctan

I am trying to find the root of $f(x)=\arctan(x)$ by using successive iteration. There are some conditions to apply this in successive iteration .

1) The function has to be continuous.

2) $\sup|f'(x)|<1$

3) $f(a,b)\subset (a,b)$, where $f(x)$ is continuous in the interval $(a,b)$.

When I applied it in Matlab, it went to $0$ but the problem is that $0$ is a root but $f'(0)=1$ and it is not smaller than $1$. According to fixed point theorem, it must be $|f(x)|<1$.

So can we apply $f(x)$ in fixed point theorem? Or am I wrong? Can you help me please? Thank you for your helping.

• This is because you confuse necessary conditions and sufficient condition. The conditions you listed are sufficient to guarantee convergence but, as you observed, not necessary. – Surb Mar 19 '15 at 8:35
• Also note that $f'(x)<1$ as long as $x\ne0$, i.e. while you iterate (actually, convergence is slower and slower and you never reach $0$). – Yves Daoust Mar 19 '15 at 8:43
• Do you want to find a root or a fixed point? The listed conditions are for fixed points. And of course, the only root of $\arctan$ is $0$. – LutzL Mar 21 '15 at 9:14
• Yes i do .I wondered whether the function is suitable for fixed point theorem to find root. And now i know 0 is root beetwen $\pi/2$ and $-\pi/2$. Function $tan(x)$ has period $\pi$. Is it possible to find another root because of the period of tangent ? Thank you for your helping. – izaag Mar 22 '15 at 16:05
• what about f''(x) and you can up to Newten-Fourier method ,look conditions that you want [here] [here]:en.wikipedia.org/wiki/Newton%27s_method#Newton-Fourier_method – zeraoulia rafik Jun 10 '15 at 1:53