the classical Newton's method for non-linear systems of equations is $x_{k+1} =x_k-J_F(x_n)^{-1} F(x_n)$. In pratice, rather than compute the inverse of the Jacobian matrix, one solves the systems $J_F(x_k) (x_{k+1}-x_k)=-F(x_k)$, for the unknown $x_{k+1}-x_k$.
In my notes (about ODE) I found:
Newton's method requires the computation of the Jacobian matrix and its "inversion" at every step $k$. This could be too expensive ($\mathcal{O}(N^3)$), where $N$ is the dimension of the matrix.
My doubt is how to get that computational complexity. Is it talking about the way to invert a matrix using LU decomposition, which I know to be $\mathcal{O}(N^3)$ ?
Then it states:
A standard way to reduce computational complexity is to use always the same Jacobian matrix, compute its LU decomposition and use it to solve the linear systems. This is $\mathcal{O}(N^2)$
Here I have still a question: the complexity of the computation of the LU decomposition of $J_F$ should be $\mathcal{O}(\frac{N^3}{3})$. While the computational complexity of the resolution of a triangular system is $\mathcal{O}(\frac{N^2}{2})$. Since there are two triangular systems, it amounts to $2 \mathcal{O}(\frac{N^2}{2})$.
Shouldn't it be, totally, $\mathcal{O}(\frac{N^3}{3})$ instead of $\mathcal{O}(N^2)$?