In multivariate Taylor expansion formula, what is meant by $(x-a)^k$?

Question

I am studying the taylor expansion for multivariate functions but I am confused about the notation of the formula. According to my book, it's $f(x)= f(a)+df(a)(x-a)+(1/2) {d^2 f(a)(x-a)^2}+ \dots$. But I am confused about what is meant by $(x-a)^2$ here since $x-a$ is a vector. How can you take powers of a vector?

Answer 1

Let's say that the map is $f:V \to W$, where $V,W$ are (say finite-dimensional) normed vector spaces. In this case $d^kf(a) : \underbrace{V \times \dots \times V}_{k \text{ times}} \to W$ is a $k$-times multilinear map, so it needs to eat $k$ vectors and gives you another vector. So, in general, $(x-a)^k$ doesn't mean any sort of multiplication, it simply means the "k-tuple" of vectors $\underbrace{(x-a, \dots, x-a)}_{k \text{ times}} \in V^k$.

So, explicitly, the first few terms are: \begin{align} f(x) &= f(a) + df_a(x-a) + \dfrac{1}{2!}d^2f_a(x-a, x-a) + \dfrac{1}{3!} d^3f_a(x-a,x-a,x-a) +\dots \end{align}