Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wonder what is the meaning of the second derivative or what kind of object it is when we have a function $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$.

The first derivative is the Jacobian matrix, but then, what is the second derivative? How can I treat them when I write $f''$ or $D^2 f$?

Thanks a lot for your help!

share|cite|improve this question
The second derivative is something very big and complicated! :-) – Giuseppe Negro Dec 12 '12 at 13:24
More seriously, the second derivative of a mapping like that is an object with 3 indices, a kind of "cubic matrix". Tensor algebra could help in giving it a more precise name, but still, it is a complicated object. – Giuseppe Negro Dec 12 '12 at 13:25
Roughly speaking, the second derivative is a tensor. – Siminore Dec 12 '12 at 13:26

$f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ is differentiable iff its increment has the form $$f(x+h)-f(x)=Df(x)h+\alpha(x,\,h),$$ where $Df(x)$ is linear mapping $Df(x)\colon \; \mathbb{R}^n \to \mathbb{R}^m$ and $\alpha$ satisfies $$\| \alpha(x,\,h) \|_{\mathbb{R}^m}=o(\|h\|_{\mathbb{R}^n}).$$ Analogously, the second order derivative is bilinear mapping $D^{2}f(x)\colon \; \mathbb{R}^n \to \mathbb{R}^m,$ which acts on a pair of vectors $(h_1,\,h_2), \quad h_1, \,h_2\in \mathbb{R}^n.$ Derivative of $k^{th}$ order is polylinear (more precisely, $k$-linear) mapping from $ \mathbb{R}^n $ to $\mathbb{R}^m$.

share|cite|improve this answer

The first thing that comes to mind is the Hessian matrix which is the special case for $m=1$:

$$f: \mathbb{R}^n \rightarrow \mathbb{R}$$

The case you asked about occurs in differential geometry, where it is involved in defining the covariant derivative (see this) and related to the connection coefficients.

In a rough sense it is a curvature. Much like the second derivative of a single variabled function tells you how quickly the curve is changing.

As for notation, I usually write this tensor as

$$\left[\frac{\partial^2 f^k(\xi)}{\partial x^i \partial x^j}\right] \;\;\;\ \text{or more simply} \;\;\; \left[\partial_i \partial_j f^k(x)\right]$$

If writing the entries individually, would probably write a Hessian for each $k$.

share|cite|improve this answer

The Jacobian matrix is the best linear approximation to $f$ at a particular point. However, if you change the point, you get a different Jacobian. The second derivative quantifies how the Jacobian changes as the point of approximation changes - the "change of the change".

To that end, we can think of the derivative $D$ as a mapping from the domain of the original function to the space of linear maps $\mathcal{L}(X,Y)$ with domain $X$ and range $Y$ the same as the original function: \begin{align} f:& X \rightarrow Y \\ Df:& X \rightarrow \mathcal{L}(X,Y) \end{align} The derivative of $f$, $Df$, is a function where you put in a point and it gives you a linear function, $$Df(x_0) = \text{best linear function approximating $f$ near }x_0.$$

In matrix form, $Df(x_0)$ is the Jacobian matrix $J$ at $x_0$: $Df(x_0)(y) = J|_{x_0} y$.

Since $Df$ is itself a function, we can take it's derivative, and so on, getting a tower of higher and higher derivatives as follows:

\begin{align} f:&\mathbb{R}^n \rightarrow \mathbb{R}^m \\ Df:&\mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \\ D(Df) = D^2f:&\mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n, \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)) \\ D(D^2f) = D^3f:&\mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n, \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m))) \\ \dots \end{align}

Now this gets confusing fast (spaces of linear maps mapping to spaces of linear maps mapping to... ack!!). Luckily there is an isometric isomorphism theorem saying that everything just boils down to multilinear maps: $$\mathcal{L}^n(X,\mathcal{L}^m(X,Y)) \cong \mathcal{L}^{n+m}(X,Y),$$ where $\mathcal{L}^k(X,Y)$ is the space of $k$-linear maps from $X$ to $Y$, and $\cong$ denotes an isometric isomorphism of function spaces. In more detail, what it means for $g$ to be in $\mathcal{L}^k(X,Y)$ is that $g : X \times \dots \times X \rightarrow Y$, and $g$ is independently linear in each of it's entries: $$g(x_a + x_b,z,w) = g(x_a,z,w) + g(x_b,z,w),$$ $$g(x,y_a+y_b,w) = g(x,y_a,w) + g(x,y_b,w),$$ and so on.

So, now we can simplify our tower of derivatives using spaces of multilinear functions: \begin{align} f:&\mathbb{R}^n \rightarrow \mathbb{R}^m \\ Df:&\mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n, \mathbb{R}^m) \\ D^2f:&\mathbb{R}^n \rightarrow \mathcal{L}^2(\mathbb{R}^n, \mathbb{R}^m) \\ D^3f:&\mathbb{R}^n \rightarrow \mathcal{L}^3(\mathbb{R}^n, \mathbb{R}^m) \\ \dots \end{align}

So, from this picture it is pretty clear what the second derivative of your function $f:X \rightarrow Y$ is at a point. It is a bilinear map from $X \times X$ to $Y$. You put in two vectors from $X$, and it gives out a vector in $Y$, and does so in a way that is linear in each input independently.

If you have a basis $\{ b_i\}$ of $n$ vectors for $X$ and basis $\{e_i\}$ of $m$ vectors for $Y$, you could completely characterize the second derivative by a 3D $n$-by-$n$-by-$m$ array of numbers $T_{ijk}$ where the $(i,j,k)$'th entry is found by applying the bilinear function with $b_i$ in the first argument and $b_j$ in the second argument, and then taking the component of the vector you get out in the $e_k$ direction: $$T_{ijk} = e_k^T D^2f(x_0)(b_i,b_j).$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.