Multivariate Taylor with 2 parts: $$\vec F(\vec x)=\vec F(\vec a)+\nabla\vec F(\vec a)\cdot(\vec x-\vec a)$$ Inverse Taylor which is the multivariate Newton-Raphson method: $$\vec x =\vec a + (\nabla \vec F(\vec a))^{-1}\cdot(\vec F(\vec x)-\vec F(\vec a))$$
Multivariate Taylor with 3 parts: $$\vec F(\vec x)=\vec F(\vec a)+\nabla\vec F(\vec a)\cdot(\vec x-\vec a)+\frac 1 2 \{\nabla[\nabla\vec F(\vec a)\cdot (\vec x-\vec a)]\cdot (\vec x-\vec a)\}$$ which become using $\nabla(a\cdot b)=\nabla a\cdot b+a\cdot\nabla b$: $$\vec F(\vec x)=\vec F(\vec a)+\frac 3 2\nabla\vec F(\vec a)\cdot(\vec x-\vec a)+\frac 1 2 [\nabla(\nabla\vec F(\vec a))\cdot (\vec x-\vec a)]\cdot (\vec x-\vec a)$$ which gives a tensor of rank 3.
Now, I want an expansion to Newton-Raphson, which use all of Taylor parts. How can I solve for $(\vec x-\vec a)$ ?
If this was single-dimensional will be something like: $$x = \frac{-f'(a)+a f''(a)\pm\sqrt{f'(a)^2+2(f(x)-f(a))f''(a)}}{f''(a)}$$ but with such tensor algebra, I cannot solve.
Can you help please?