Understanding multi-variable Taylor formula I am reading Joseph Taylor's Foundations of analysis, in which (Brook) Taylor's formula is given as the following (for a function $f:\mathbb{R}^n \to \mathbb{R}^m$ approximated at a point $a$ in its domain) $$f(x) = f(a) + df(a)(x-a) + \frac{1}{2}d^2f(a)(x-a)^2 + ... + \frac{1}{n!}d^nf(a)(x-a)^n + R_n$$ where $R_n$ is the remainder term (not really relevant here).
First question: What is the meaning of the term $(x-a)^n$? Is it the $n$-th power of the vector $(x-a)$? That wouldn't make sense because $(x-a)$ has size $1 \times n$, but I can't imagine what else it means.

I'm trying to understand an example in J. Taylor's book in which he applies the above formula to the function $f(x, y) = \ln(x+y)$ at the point $a = (0, 1)$. This is what he writes:
\begin{align}
\ln(x+y) & = (1, 1) \begin{pmatrix} x \\ y-1 \end{pmatrix} - \frac{1}{2} (x, y-1) \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} \begin{pmatrix} x \\ y-1 \end{pmatrix} + R_2 \\
& = x+y-1-\frac{1}{2}(x+y-1)^2+R_2 \\\\
\end{align} 
Second question: How is the quadratic term in B. Taylor's formula equal to what J. Taylor writes on the first line of his example; i.e. how is it that $$\frac{1}{2}d^2f(a)(x-a)^2 = - \frac{1}{2} (x, y-1) \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} \begin{pmatrix} x \\ y-1 \end{pmatrix}$$
 A: Denote $V = \Bbb{R}^n$ and $W = \Bbb{R}^m$. If $f:V \to W$ is $k$ times differentiable at a point $a$, then the $k^{th}$ differential at $a$ is a multilinear map $d^kf_a: \underbrace{V \times \cdots \times V}_{k \text{ times}} \to W$. Usually, for convenience, if $\xi \in V$, then by $(\xi)^k$, we mean the $k$-tuple $(\xi, \dots, \xi) \in \underbrace{V \times \cdots \times V}_{k \text{ times}}$. 
Another thing to note is that $d^kf_a$ is symmetric with respect to all its arguments. More precisely, if $\sigma:\{1, \dots k\} \to \{1, \dots, k\}$ is any bijection (i.e a permutation of $S_k$, the set of $k$ elements), then for any $\xi_1, \dots \xi_k \in V$ we have
\begin{align}
d^kf_a(\xi_{\sigma(1)}, \dots, \xi_{\sigma(k)}) = d^kf_a(\xi_1, \dots, \xi_k)
\end{align}
(this can be seen as the reason for why the mixed partial derivatives of a sufficiently differentiable are equal). Because of this, we say that $d^kf_a$ is a symmetric, $k$-linear map from $V^k$ into $W$.

For your second question, you need to recall some linear algebra. In general, if $V$ and $W$ are real vector spaces (in your particular example, $V = \Bbb{R}^2, W = \Bbb{R}$), and $g: V \times V \to W$ is a bilinear map, and $\xi,\eta \in V$, then to compute the quantity
\begin{align}
g(\xi,\eta) \in W
\end{align}
we can do the following: choose a basis $\{e_1, \dots, e_n\}$ for $V$. Then, in terms of this basis, we can "expand" the vectors $\xi$ and $\eta$
\begin{align}
\xi = \sum_{i=1}^n \xi_i e_i \qquad \text{and} \qquad \eta = \sum_{i=1}^n \eta_i e_i
\end{align}
for some $\xi_i, \eta_i \in \Bbb{R}$. So, now using bilinearity of $g$ we can compute things easily:
\begin{align}
g(\xi,\eta) &= g \left(\sum_{i=1}^n \xi_i e_i, \sum_{j=1}^n \eta_j e_j \right) \\
&= \sum_{i=1}^n \sum_{j=1}^n \xi_i \eta_j \cdot g(e_i, e_j) \\
&= \begin{pmatrix}
\xi_1 & \dots & \xi_n
\end{pmatrix}
\cdot [g] \cdot
\begin{pmatrix}
\eta_1 \\
\vdots \\
\eta_n
\end{pmatrix}
\end{align}
where $[g]$ is the $n \times n$ matrix whose $ij$ entry is $g(e_i,e_j)$. Hence what this says is that to compute the value of a bilinear map on two vectors, $g(\xi,\eta)$, we can think of it as matrix multiplication:
\begin{equation}
g(\xi,\eta) = \xi^t \cdot [g] \cdot \eta.
\end{equation}
Here we think of the vectors $\xi,\eta \in V= \Bbb{R}^n$ as column vectors.
Now we have sufficient theory to apply it to your question. Here our bilinear map is $d^2f_a: \Bbb{R}^2 \times \Bbb{R}^2 \to \Bbb{R}$, and $\xi = \begin{pmatrix} x\\y \end{pmatrix} \in \Bbb{R}^2$, and $a = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \in \Bbb{R}^2$, so that $\xi-a = \begin{pmatrix} x\\y-1 \end{pmatrix}$. The second term in the Taylor expansion is (ignoring the $1/2$)
\begin{align}
d^2f_a(\xi-a, \xi-a) &= (\xi-a)^t \cdot [d^2f_a] \cdot (\xi-a) \\\\
&= (x,y-1) \cdot
\begin{pmatrix}
\partial_{1,1}f(a) & \partial_{1,2} f(a) \\
\partial_{2,1} f(a) & \partial_{2,2} f(a)
\end{pmatrix}
\cdot
\begin{pmatrix}
x \\
y-1
\end{pmatrix} \\\\
&= (x,y-1) \cdot
\begin{pmatrix}
-1 & -1 \\
-1 & -1
\end{pmatrix}
\cdot
\begin{pmatrix}
x \\
y-1
\end{pmatrix}
\end{align} 
The first equal sign is because of everything I've said above. For the second equal sign, you need to know that $d^2f_a(e_i,e_j) = \partial_{i,j}f(a)$ (the order of $i,j$ doesn't matter because it is a symmetric bilinear map). In words this says, the second differential evaluated on the standard basis vectors gives the corresponding second partial derivatives.
