Take the 2-minute tour ×
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.

I am having trouble understanding, why the $k$-th derivative of a map $F\colon\mathbb R^n \to\mathbb R^m$ is a symmetric multilinear map for each $x$ in $\mathbb R^n$. Can you please explain which vectors this map accepts as input where multilinearity comes from ? Also, why is symmetry mentioned ?

Thank you readingframe

share|improve this question

1 Answer 1

Well, in the general case where $F:\mathbb R^n\to V$ for some (normed) vector space $V$, the derivative $F'(x_0)$ at some point $x_0\in\mathbb R^n$ is a linear map $\mathbb R^n\to^{\mathrm{Lin}} V$ such that $$F(x_0+h)=F(x_0) + F'(x_0)(h) + o(h)$$ when $h\in\mathbb R^n$ goes toward $0$.

Thus, the function that takes every $x_0$ to the derivative at $x_0$ is of type $\mathbb R^n\to(\mathbb R^n\to^{\mathrm{Lin}} V)$. Because the space of $\mathbb R^n\to^{\mathrm{Lin}} V$ functions is itself a vector space, we can repeat the process to get higher-order derivatives. The $k$th derivative, if it exists, ends up having type $$\mathbb R^n \to (\underbrace{\mathbb R^n\to^{\mathrm{Lin}}(\mathbb R^n\to^{\mathrm{Lin}}(\cdots\to^{\mathrm{Lin}}(\mathbb R^n\to^{\mathrm{Lin}}}_{k\text{ times}} V)\cdots)))$$ which (if you know tensor products) is isomorphic to $$\mathbb R^n \to (\underbrace{\mathbb R^n\otimes\mathbb R^n\otimes\cdots\otimes\mathbb R^n}_{k\text{ times}}\to^{\mathrm{Lin}} V)$$ Thus, for each $x_0$, $F^{(k)}(x_0)$ is a multilinear map taking $k$ vectors in $\mathbb R^n$ to one vector in $v$.

That the map is symmetric means that $$F^{(k)}(x_0)(h_1)(h_2)\cdots(h_k)=F^{(k)}(x_0)(h_{\sigma(1)})(h_{\sigma(2)})\cdots(h_{\sigma(k)})$$ for any permutation $\sigma$ of the $h_i$'s. (This is true under appropriate smoothness conditions for the original $F$).

Generalizes ordinary higher derivatives. When $n=1$, this all reduces to ordinary higher derivatives because $\mathbb R\to^{\mathrm{Lin}} V$ is naturally isomorphic to $V$ (each $f:\mathbb R\to^{\mathrm{Lin}} V$ is fully determined by $f(1)$).

Connection to partial derivatives. A multilinear map is determined by its values on the standard basis vectors, and $$F^{(k)}(\vec x_0)(\mathbf e_i)(\mathbf e_j)\cdots(\mathbf e_l) = \frac{\partial^k F}{\partial x_i \partial x_j \cdots \partial x_l} (\vec x_0) $$

share|improve this answer

Your Answer

 
discard

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.