Jacobian Matrix and dot product question

Question

I have a question on how the Jacobian matrix is used in a specific example from Wiki. I present definitions first and then a question.

The Jacobian is defined as:

$J = \begin{bmatrix} \frac {\partial f}{\partial x_1} ... \frac {\partial f}{\partial x_n} \end{bmatrix} = \begin{bmatrix} \Delta^T f_1 \\ ... \\ \Delta^T f_m \end{bmatrix} = \begin{bmatrix} \frac {\partial f_1}{\partial x_1} ... \frac {\partial f_1}{\partial x_n} \\ ... \\ \frac {\partial f_m}{\partial x_1} ... \frac {\partial f_m}{\partial x_n} \end{bmatrix}$

Wiki then says: $f(y) = f(x) + J(x) \cdot (y - x)$ is the best linear approximation for $f(y)$ for points close to $x$.

Questions

1. Does $J(x)$ represent scalar multiplication of the initial point $x$ with the $J$? I.e. to get the PDs at a point.

2. If (1) above, how does one distinguish this operation from matrix multiplication?

3. For $J(x) \cdot (y - x)$ can the dot product, $\cdot$, be omitted as matrix multiplication will result in the same value? The third form of $J$ is a matrix and dot product $\cdot$ is a vector operation, so this is confusing.

Answer 1

The Jacobian matrix $J$ contain partial derivatives $\frac{\partial f_i}{\partial x_j}$ which are themselves functions. The entries of $J(x)$ are these partial derivatives evaluated at $x$, i.e. $\frac{\partial f_i}{\partial x_j}(x)$.

Answer 2

As has been said before, $J(x)$ is the Jacobian matrix at the point $x$, as all the functions depend on $x$. The dot does not below there, as we do not take the dot product of a matrix and a vector. Just delete the dot and multiply the Jacobian matrix by the (column) vector $y-x$.

Answer 3

You don't need to worry about defining Jacobian for this. This is simple Taylor's theorem:

$$ f_i\left(y\right)=f_i\left(x+(y-x)\right)=f_i\left(x\right)+\sum_j \partial_j f_i\left(x\right)\,\left(y_j-x_j\right)+\mathcal{O}\left(\left|y-x\right|^2\right) $$

for all $i$