$f(x)=x^5-x^3+3$. Find an interval such that $\forall x_0$ in this interval , the newton raphson method is convergent. EDIT:
Now I see that my attempt isn't get $f(x)=0$,
then I think my solution is correct , I will be grateful for some help.
$f(x)=x^5-x^3+3$.
Find an interval such that $\forall x_0$ in this interval , the newton raphson method is convergent.
I know this thoerem:
$f\in C^2(\mathbb{R}), \forall x\in \mathbb{R},f'(x)>0,f''(x)>0 \implies \forall x_0$ the the newton raphson method is convergent.
In my problem:
The graph of $f(x)=x^5-x^3+3$ is :

The interval $(\sqrt{\frac{3}{5}},\infty)$ exists $ f'(x)>0,f''(x)>0 ,   \forall x\in (\sqrt{\frac{3}{5}},\infty)$.
Is it correct to say that the newton raphson method is convergent $\forall x\in (\sqrt{\frac{3}{5}},\infty)$.
In case I am wrong how to approach this problem ?
Thanks !
 A: For a polynomial, the Newton method is always convergent, either to a root, or to a stable cycle around some root.
The fact you cite can be extended to all intervals with a constant sign in the second derivative by reflection about the axes. One has to be careful to select the correct side of the interval for the initial value.

*

*In the given situation of the theorem, if $f(x_0), f'(x_0),f''(x_0)$ are all positive and there is actually a root in the interval, the iteration falls monotonously to the root. Think of the right side of the graph and the equation $f(x)=4$ for such a situation.

*After reflection about the $x$ axis one has all of these values negative.

*After reflection about the $y$ axis $f(x_0)>0$, $f'(x_0)<0$, $f''(x_0)>0$.

The pattern that appears is that $f(x_0)f''(x_0)>0$ for a good initial value on such an interval. For the given polynomial the relevant interval is $(-\infty,-\sqrt{0.3})$ with negative curvature there. Thus good initial points have negative value.
A: If $f$ has enough regularity, $f' \ne 0$ and $f''$ does not change sign, a sufficient condition for the convergence of Newton's method for all initial approximation $x_0 \in [a,b]$ is that
$$
\frac{|f(a)|}{|f'(a)|} < b - a\quad \textrm{and}\quad \frac{|f(b)|}{|f'(b)|} < b - a
$$
One such interval can be $I =[-2,-1.2]$.
