Consider the fixed point iteration $$x_{n+1}=\phi(x_n), n\geq 0,\ \text{with}\ \phi(x)=3+(x-3)^3, \ x\in(2.5, 3.5)$$ and the initial approximation $x_0=3.25$. Then the order of convergence of the fixed point iteration is
Let $$f(x)=x^4+2x^3-11x^2-12x+36$$ for $x\in\mathbb{R}$. The order of convergence of the Newton-Raphson method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, n\geq 0$$ with $x_0=2.1$, for finding the root $\alpha=2$ of the equation $f(x)=0$ is
How to start with these problems? I have studied that Newton-Raphson method has quadratic convergence. Is that true in this case? Please help.
When you are using the fixed point method to approximate the solution $x^*$ of the equation $x = \phi(x)$, if the method is convergent and $$ \phi'(x^*) = \cdots =\phi^{(p-1)}(x^*) =0, \quad \phi^{(p)}(x^*)\ne 0, $$
the method will converge with order $p$. For instance, in your first example, since $x^*=3$, you have that $\phi'(x^*) = \phi''(x^*)=0$ and $\phi'''(x^*) \ne 0$, you conclude that the convergence is cubic (you still need to check convergence).