Is there an index-less notation (using gradiends, Jacobians, curls, hessians, anything) to describe a second-order term in the Taylor expansion of a vector field $\mathbf{f}(\mathbf{x}): \mathbb{R}^n \to \mathbb{R}^n$?
Up to the linear terms, it's nicely written using the Jacobian matrix:
$$
\mathbf{f}(\mathbf{x}) \approx \mathbf{f}(\mathbf{x_0}) + D_{\mathbf{x}}\mathbf{f}(\mathbf{x_0})(\mathbf{x} – \mathbf{x_0}) + O(\mathbf{x}^2)
$$
I don't know how is it better to express the quadratic part without ugly indices:
$$
O(\mathbf{x}^2)_j = \frac{1}{2}(\mathbf{x} – \mathbf{x_0})^TD^2f_j(\mathbf{x_0})(\mathbf{x} – \mathbf{x_0}) + O(\mathbf{x}^3)_j
$$
Best Answer
As one can see from $$df^i=\frac{\partial f^i}{\partial x^j}dx^j+\frac12\frac{\partial^2f^i}{\partial x^j\partial x^k}dx^jdx^k+\cdots$$ The second order term is actually a third-order tensor and we cannot express it using matrix notation. But forget matrix, just remember it is a tensor. Then by denoting $D^2f=\displaystyle \frac{\partial^2f^i}{\partial x^j\partial x^k}e_i\otimes e^j\otimes e^k$, we have $$f(x)\approx f(x_0)+Df(x_0)\cdot(x-x_0)+\frac12D^2f(x_0)\cdot(x-x_0)^2$$ where bullet $\cdot$ means inner product and $x^2=x\otimes x$.