Could someone possibly help me understand how I can derive the Hessian matrix of a twice-differentiable function $f$ defined on $\mathbb{R}^n$ using the limit definition of the second derivative. Namely, how does: $\lim_{h -> 0}\frac{\nabla f(x+h) – \nabla f(x)}{h}$ result in the Hessian $\nabla^2 f(x)$. If I happen to be wrong about this, could you please point out what I am misunderstanding?
Thank you very much!
Best Answer
To extend user251257's answer, we have that, for any vector $v\in\mathbb{R}^n$, $$\lim_{h\rightarrow 0} \frac{\nabla f (x + hv) - \nabla f(x)}{h} = \nabla^2 f(x)v$$ We can deduce this directly from his/her answer and subsequent comments since, as he/she suggested, $$\lim_{h\rightarrow 0} \frac{\nabla f (x + he_i) - \nabla f(x)}{h} = \nabla^2 f(x)e_i = \begin{bmatrix} \frac{\partial^2 f(x)}{\partial x_i \partial x_j}\end{bmatrix}_j \in \mathbb{R}^n$$
For more on this topic, I recommend reading about directional derivatives in a multivariate analysis text such as that of Loomis & Sternberg.