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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?

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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Commented Feb 18, 2020 at 12:33
  • $\begingroup$ I cannot find $f^{(2)}$ in Baby Rudin, for example. $\endgroup$
    – tchappy ha
    Commented Feb 18, 2020 at 12:41
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    $\begingroup$ 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. $\endgroup$ Commented Feb 18, 2020 at 12:41
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    $\begingroup$ @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 $R^{m}$. $\endgroup$
    – tchappy ha
    Commented Feb 18, 2020 at 12:45
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    $\begingroup$ I doubt you'll find the notation $f^{(2)}$ to denote the second derivative of a function in the context of multivariable analysis anywhere. $\endgroup$
    – user239203
    Commented Feb 18, 2020 at 12:47

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much for your answer, Peter Melech. $\endgroup$
    – tchappy ha
    Commented Feb 20, 2020 at 9:57
  • $\begingroup$ You're welcome! $\endgroup$ Commented Feb 20, 2020 at 14:53
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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.

A precise description of the above would require a foray into the generalities of tensor algebra.

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  • $\begingroup$ Thank you very much for your answer, John Bentin. $\endgroup$
    – tchappy ha
    Commented Feb 20, 2020 at 9:57

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