taylor expansion of scale space function

Question

I see the following expression from http://en.wikipedia.org/wiki/Scale-invariant_feature_transform

The quadratic Taylor expansion of the Difference-of-Gaussian scale-space function, with the candidate keypoint as the origin is

$$D(\textbf{x}) = D + \frac{\partial D^T}{\partial \textbf{x}}\textbf{x} + \frac{1}{2}\textbf{x}^T \frac{\partial^2 D}{\partial \textbf{x}^2} \textbf{x}$$

where D and its derivatives are evaluated at the candidate keypoint and $\textbf x = (x,y,\sigma)$ is the offset from this point.

I'm very confused about this expression. I don't know why $D^T$ appears and don't understand everything. Is there someone to help me?

Answer 1

The quantity $\frac{\partial D}{\partial \textbf{x}}$ is a vector, since it is the derivative of the scalar function $D(\textbf{x})$ w.r.t. all the elements of $\textbf{x}$. In the formula it is assumed that all vectors are column vectors, so in order to compute the dot product of the derivative $\frac{\partial D}{\partial \textbf{x}}$ and the vector $\textbf{x}$, you need to transpose one of them, which gives you a matrix product row times column (= scalar). The expression $\frac{\partial D^T}{\partial \textbf{x}}$ is simply the transpose of the vector of derivatives. Maybe it would have been clearer to write it as $\left(\frac{\partial D}{\partial \textbf{x}}\right)^T$.

Similarly, for the last term you have a row vector times a matrix (second derivative of $D$ w.r.t. $\textbf{x}$) times a column vector, which again results in a scalar value.