# $f^{(2)}$ is not defined in an analysis book, where $f$ is a mapping from $\mathbb{R}^n$ to $\mathbb{R}^m$. Why?

Let $$f$$ be a function from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$.

In general, mathematicians like to generalize a mathematical concept very very much.

I think they like generalization more than food.

They define $$f'$$ but they don't define $$f^{(2)}, f^{(3)}, \cdots$$ in their analysis books.

Why?

• It is defined in all the analysis books that I know. Are you referring to a specific book? If not, I don't really see the point of this question. Commented Feb 18, 2020 at 12:33
• I cannot find $f^{(2)}$ in Baby Rudin, for example. Commented Feb 18, 2020 at 12:41
• Google works, which led me (2nd highest result) to this Wikipedia page that includes, in the section In higher dimensions the following sentence: "By repeatedly taking the total derivative, one obtains higher versions of the Fréchet derivative, specialized to $R^p.$" If your question is about some specific book or books (as @Thorgott asks), then you need tell us what they are. Commented Feb 18, 2020 at 12:41
• @Thorgott Rudin defined $f^{(2)}$, where $f$ is a real valued function. But I think he didn't define $f^{(2)}$, where $f$ is a function from $R^{n}$ to $R^{m}$. Commented Feb 18, 2020 at 12:45
• I doubt you'll find the notation $f^{(2)}$ to denote the second derivative of a function in the context of multivariable analysis anywhere.
– user239203
Commented Feb 18, 2020 at 12:47

Note that if $$f:X\rightarrow Y$$ is a differentiable map between normed spaces, then its derivative is a mapping $$f':X\rightarrow \mathcal{L}(X,Y),x\mapsto f'(x)$$. Indeed for $$x_0\in X$$ the derivative is the linear map $$f'(x_0)\in\mathcal{L}(X,Y)$$ such that $$\frac{f(x)-f(x_0)-f'(x_0)(x-x_0)}{||x-x_0||_X}\rightarrow 0,x\rightarrow x_0$$ where the convergence of the images under $$f$$ is understood to be that with respect to $$||.||_Y$$ and that of $$x\rightarrow x_0$$ that with respect to $$||.||_X$$. Now $$\mathcal{L}(X,Y)$$ is of course again a normed space. So exactly the $$\textbf{same}$$ definition yields a second derivative: $$f^{(2)}:X\rightarrow \mathcal{L}(X,\mathcal{L}(X,Y))$$ and a third $$f^{(3)}:X\rightarrow \mathcal{L}(X, \mathcal{L}(X,\mathcal{L}(X,Y)))$$ ...and so on, if they exist. Actually this looks more complicated than it is, since instead of $$f^{(2)}(x_0)\in \mathcal{L}(X,\mathcal{L}(X,Y))$$ You can consider the bilinear operator $$\phi:X\times X\rightarrow Y$$ defined by $$\phi(x_1,x_2)=(f^{(2)}(x_0)x_1)x_2$$ that for example in the case of finite dimensional spaces can be represented by a matrix again.
For a sufficiently smooth function $$f$$, Taylor's theorem in one dimension may be written $$f(x+u)=\sum_{k=0}^l\frac1{k!}f^{(k)}(x)u^k +\varepsilon(x,u,l),$$where the “error” term $$\varepsilon(x,u,l)$$ is treated, for any given $$x\in\Bbb R$$, $$l\in\Bbb N$$, and for any sufficiently small $$u$$, as negligible for the current purpose—and we ignore it henceforth. Now taking $$f$$ to be a $$\Bbb R^m$$-valued function on $$\Bbb R^n$$, Taylor's theorem becomes $$\pmb f(\pmb x+\pmb u)=\sum_{k=0}^l\frac1{k!}\pmb f^{(k)}(\pmb x)\pmb u^k +\pmb\varepsilon(\pmb x,\pmb u,l),$$under suitable interpretation of the “powers” $$\pmb f^{(k)}$$ and $$\pmb u^k$$ and their multiplication. Since we are working in $$\Bbb R^n$$ and $$\Bbb R^m$$ with their natural column-vector representation, we can think of $$\pmb f^{(1)}$$ as having values in $$\Bbb R^{m\times n}$$, the space of $$m\times n$$ matrices, with $$\pmb u^{(1)}:=\pmb u$$. Generally, for $$k=2,3,.. .$$ , we may interpret $$\pmb f^{(k)}$$ as living in a space of $$m\times(n\times(\cdots\times n)\cdots)$$ tensors, which can be thought of as highly row-structured $$m\times n^k$$ matrices, where there are $$k$$ $$n$$s and $$k$$ applications of $$\times$$, while $$\pmb u^{(k)}$$ lives in a tensor space with (columnar) dimensional structure $$(\cdots(n\otimes n)\otimes \cdots \otimes n)\times 1$$, where there are $$k$$ $$n$$s and $$k-1$$ applications of the column-building operation $$\otimes$$. The multiplication between these scary beasts is relatively simple, as most of the structure collapses through addition, leaving simply an $$m\times1$$ matrix—namely an $$m$$-vector. The components of this vector are sums of terms of the form $$a_{\pmb i}(\pmb x) \pmb u^{\pmb i},$$where $$a_{\pmb i}(\pmb x)\in\Bbb R$$, $$\pmb i =(i(1),...,i(m))\in \Bbb N^m$$ with $$i(1)+\cdots+i(m)=k$$, and $$\pmb u^{\pmb i}:=u_1^{i(1)}\cdots u_k^{i(k)}$$. Here the $$a_{\pmb i}(\pmb x)$$ are individual differential coefficients of the form$$\frac1{h(1)!\cdots h(n)!}\frac{\partial^k f_j(\pmb x)}{\partial x_1^{h(1)}\cdots\partial x_n^{h(n)}},$$where $$h(1),...,h(n)\in \Bbb N$$ with $$h(1)+\cdots+h(n)=k$$. The initial factor accounts for permutations under which the differential coefficient is invariant.