What do you call third order derivative matrix and what does it geometrically signify?

Question

The first-order derivatives matrix is known as Jacobian, gives the gradient of the graph. Similarly, the second-order derivatives matrix is Hessian, which gives the curvature of the plot. What next? i.e., is there a third-order derivative matrix? If yes, what does it denote?

Answer 1

For a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, the Taylor series expansion is \begin{align*} f(\mathbb{x}) = f(\mathbb{z}) + \color{blue}{\sum_{i=1}^{n}(x_i - z_i)f_{i}(\mathbb{z})} + \frac{1}{2}\color{orange}{\sum_{i,j} (x_i - z_i)(x_j-z_j)f_{ij}(\mathbb{z})} + \frac{1}{6}\color{greeN}{\sum_{ijk}(x_i-z_i)(x_j-z_j)(x_k-z_k)f_{ijk}(\mathbb{z})} + \cdots \end{align*} where $f_i(\mathbb{z}) = \frac{\partial}{\partial z_i}f(\mathbb{z})$, $f_{ij}(\mathbb{z}) = \frac{\partial^2}{\partial z_i\partial z_j}f(\mathbb{z})$, and so on. We normally represent the blue expression as \begin{align*} (\mathbb{x} - \mathbb{z})^\intercal [\nabla f(\mathbb{z})] \end{align*} and the orange expression as \begin{align*} (\mathbb{x} - \mathbb{z})^\intercal [\nabla^2f(\mathbb{z})](\mathbb{x} - \mathbb{z}) \end{align*} here, $\nabla f(\mathbb{z})$ and $\nabla^2 f(\mathbb{z})$ denote the gradients and Hessians, respectively. If we could, we wish to represent the green portion in a similar manner, perhaps \begin{align*} (\mathbb{x} - \mathbb{z})^\intercal \underset{(\mathbb{x} - \mathbb{z})^\dagger}{[\nabla^3f(\mathbb{z})]}(\mathbb{x} - \mathbb{z}) \end{align*} in which case, we see the matrix of third derivatives appear. Note that we haven't defined how to conduct this 3D matrix multiplication, and I'm simply using the above notation to give you the idea that $(\mathbb{x} - \mathbb{z})$ must be utilized thrice (i.e. 3rd-power products). The formal definition leads to the n-mode product.