How to expand the Taylor series of functions of several vectors? We know that the Taylor series expansion of the function of several scalars around zero is
$$
f(x,y)=f(0,0)+f_x(0,0)\cdot x+f_y(0,0)\cdot y+\frac{1}{2!}f_{xx}(0,0)\cdot x^2+\dots
$$
Then, how about the function of several vectors $f({\bf x,y}):\mathbb R^n\times\mathbb R^n\to\mathbb R$? What is the best way to express its Taylor series that it maintains the same structure of the scalar case, i.e. similar to the above expression?
Thanks in advance.
 A: Hints
If $f:\mathbb R^n \to \mathbb R$ is a function denoted by $f({\bf{x}})$then we have by Taylor series
$$f({\bf{x}})=f({\bf{x}}_0)+\nabla f({\bf{x}}_0) \cdot ({\bf{x}}-{\bf{x}}_0) + \frac{1}{2} ({\bf{x}}-{\bf{x}}_0) \cdot \nabla \nabla f ({\bf{x}}_0) \cdot ({\bf{x}}-{\bf{x}}_0) + \cdot \cdot \cdot \tag{1}$$
Now, if $f:\mathbb R^n \times \mathbb R^n \to \mathbb R$ denoted by $f({\bf{x}},{\bf{y}})$ we can use the above formula to get
$$f({\bf{x}},{\bf{y}})=f({\bf{x}}_0,{\bf{y}})+\nabla_{\bf{x}} f({\bf{x}}_0,{\bf{y}}) \cdot ({\bf{x}}-{\bf{x}}_0) + \frac{1}{2} ({\bf{x}}-{\bf{x}}_0) \cdot \nabla_{\bf{x}} \nabla_{\bf{x}} f ({\bf{x}}_0,{\bf{y}}) \cdot ({\bf{x}}-{\bf{x}}_0) + \cdot \cdot \cdot \tag{2}$$
then you should repeat the expansion for ${\bf{y}}$ in each term of the RHS of $(2)$. Specifically, you may write the expansions of $f({\bf{x}}_0,{\bf{y}})$, $\nabla_{\bf{x}} f({\bf{x}}_0,{\bf{y}})$ and $\nabla_{\bf{x}} \nabla_{\bf{x}} f ({\bf{x}}_0,{\bf{y}})$ at ${\bf{y}}={\bf{y}}_0$. Notice that in the expansions in ${\bf{y}}$ you have $\nabla_{\bf{y}}$ instead of $\nabla_{\bf{x}}$. I just leave the arithmetic to you.
