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Let $f:\mathbb{R}^n\to\mathbb{R}^m$ be a function at least $C^k$ and fix some point $x_0\in\mathbb{R}^n$. I want an explicit formula for its derivatives of higher order.

I already read this threads:

What are higher derivatives?

Using higher order derivatives

One thing is clear, $D^kf(x_0)$ is a $k$-linear map from $\underbrace{\mathbb{R}^n\times\ldots\times\mathbb{R}^n}_{k \text{ times}}$ to $\mathbb{R}^m$. Yet, the answers given do not address my question, they only give the abstract description but I want a concrete description.

For $k=1$ we have that $Df(x_0)v = [Df(x_0)]\cdot v$, where $ [Df(x_0)] = \left(\frac{\partial f_i(x_0)}{\partial x_j}\right)$ stands for the Jacobian matrix of $Df(x_0)$. So we have an explicit formula.

For $k\geq 2$, it's not clear how to get an explicit formula for $D^kf(x_0)(v_1,\ldots,v_k)$ from the definition and the case $k=1$. There must be some way to do this.

I hope you can help me. Thank you very much.

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Write $${\bf v}_1 =\sum_{i_1=1}^n v_1^{i_1}{\bf e}_{i_1}, \cdots , {\bf v}_k =\sum_{i_k=1}^n v_k^{i_k}{\bf e}_{i_k},$$where $\{{\bf e}_1,\ldots, {\bf e}_n\}$ is the standard basis. Also, let $\{{\bf e}^\ast_1,\ldots, {\bf e}^\ast_n\}$ be the dual basis. Then:

$$\begin{align} D^kf({\bf x}_0)({\bf v}_1,\cdots,{\bf v}_k) &= D^kf({\bf x}_0)\left(\sum_{i_1=1}^n v_1^{i_1}{\bf e}_{i_1},\cdots, \sum_{i_k=1}^n v_k^{i_k}{\bf e}_{i_k}\right) \\ &= \sum_{i_1,\ldots,i_k=1}^n v_1^{i_1}\ldots v_k^{i_k} D^kf({\bf x}_0)({\bf e}_{i_1},\cdots,{\bf e}_{i_k}) \\ &=\sum_{i_1,\ldots,i_k=1}^n v_1^{i_1}\ldots v_k^{i_k} \frac{\partial^k f}{\partial x_{i_1}\cdots \partial x_{i_k}}({\bf x}_0) \\ &= \sum_{i_1,\ldots,i_k=1}^n \frac{\partial^k f}{\partial x_{i_1}\cdots \partial x_{i_k}}({\bf x}_0) {\bf e}^\ast_{i_1}\otimes \cdots \otimes {\bf e}^\ast_{i_k}({\bf v}_1,\cdots,{\bf v}_k) \end{align}$$Hence: $$D^kf({\bf x}_0) = \sum_{i_1,\cdots,i_k=1}^n \frac{\partial^k f}{\partial x_{i_1}\cdots \partial x_{i_k}}({\bf x}_0) {\bf e}^\ast_{i_1}\otimes \cdots \otimes {\bf e}^\ast_{i_k}. $$

We could have used any basis instead of the standard one - the only difference is that we'd have iterated directional derivatives as the coefficients of that tensor in the new basis. I guess that's as concrete we can get.

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  • $\begingroup$ maybe you know some book where this topic is shown in the level of detail of your answer $\endgroup$
    – Masacroso
    Sep 13, 2017 at 2:27
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    $\begingroup$ Well, this book explains quite well derivatives of higher orders (some brazillian books happen to be good), but for the relation of that with tensor products I just applied what I knew about multilinear maps (see Roman's Advanced Linear Algebra, for example) to the particular map $D^kf({\bf x}_0)$... $\endgroup$
    – Ivo Terek
    Sep 13, 2017 at 2:35
  • $\begingroup$ $\sum_{i_1,\ldots,i_k=1}^n v_1^{i_1}\ldots v_k^{i_k} D^kf({\bf x}_0)({\bf e}_{i_1},\cdots,{\bf e}_{i_k}) =\sum_{i_1,\ldots,i_k=1}^n v_1^{i_1}\ldots v_k^{i_k} \frac{\partial^k f}{\partial x_{i_1}\cdots \partial x_{i_k}}({\bf x}_0)$. I know how to prove this but could you please tell how do you know this fact? $\endgroup$
    – Hamilton
    Jul 2, 2022 at 7:36

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