Newton's method for a vector field

Question

Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be $C^2$ and let $f(x^*)=0$. Since

$$f(x^*) \approx f(x) + Df(x) (x^* - x)$$

we can have the iterative procedure

$$x_{k+1} = x_k - Df(x_k)^{-1} f(x_k)$$

Is $G(x): = x - Df(x)^{-1} f(x)$ invertible near $x=x_0$? Are there any results on the convergence of this procedure?

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I tried to use the inverse function theorem. However, I do not know how to prove that $$DG(x_0) = I - D(Df(x)^{-1} f(x)) \bigg |_{x_0}$$ is non-singular.

Answer 1

This even doesn't hold for 1-D function since $$DG(x_0)=1-\left(\dfrac{f(x)}{f'(x)}\right)'|_{x_0}=1-\dfrac{f'^2(x_0)-f(x_0)f''(x_0)}{f'^2(x_0)}=0$$which is singular. To show that for higher dimensions let's define $$Df^{-1}(x)=[a_{ij}(x)]\\Df^{-1}(x)f(x)=[c_{i}(x)]$$and $$D(Df^{-1}(x)f(x))=[b_{ij}(x)]$$therefore $$c_{i}(x)=\sum_{k=1}^{n}a_{ik}(x)f_{k}(x)$$where $$f(x)=\begin{bmatrix}f_1(x)\\f_2(x)\\.\\.\\.\\f_n(x)\end{bmatrix}$$also$$b_{ij}(x)=\dfrac{\partial c_i}{\partial x_j}=\sum_{k=1}^{n}\dfrac{\partial a_{ik}(x)}{\partial x_j}f_{k}(x)+\sum_{k=1}^{n}a_{ik}(x)\dfrac{\partial f_k(x)}{\partial x_j}$$when substituting $x=x_0$, $f_k(x_0)$ becomes zero since $f(x_0)=0$ therefore$$b_{ij}(x_0)=\sum_{k=1}^{n}a_{ik}(x_0)\dfrac{\partial f_k(x)}{\partial x_j}|_{x_0}$$but $\dfrac{\partial f_k(x)}{\partial x_j}|_{x_0}$ is the (k,j)th entry of $Df(x_0)$ which by substitution leads to $$D(Df^{-1}(x)f(x))|_{x_0}=Df^{-1}(x_0)Df(x_0)=I$$or $$\Large DG(x_0)=0$$

Answer 2

Hint.

The process

$$ x_k = x_{k-1}-Df^{-1}(x_{k-1})f(x_{k-1}) $$

can be written as

$$ x_k = \phi(x_{k-1})\\ x_{k-1} = \phi(x_{k-2}) $$

and then

$$ |x_k-x_{k-1}| = |\phi(x_{k-1})-\phi(x_{k-2})| = |\phi'(\xi)||x_{k-1}-x_{k-2}| $$

for a suitable $\xi\in (x_{k-1}-x_{k-2})$ hence is sufficient that $|\phi'(\xi)| < 1$ for convergence