# Math Help - Multivariable Differential Mapping

1. ## Multivariable Differential Mapping

If $p:\mathbb{R}^n\to\mathbb{R}^m$ is a linear map plus some constant and $f:A\subset\mathbb{R}^m\to\mathbb{R}^s$ is $k$ times differentiable, prove that:

$\textbf{D}^k(f\circ p)(x_0)(x_1,...,x_k)=\textbf{D}^k f(p(x_0))(\textbf{D}p(x_0)(x_1),...,\textbf{D}p(x_ 0)(x_k))$

So I started by induction. Using the chain rule,

$\textbf{D}(f\circ p)(x_0)(x_1)=(\textbf{D}f(p(x_0))\textbf{D}p(x_0)) (x_1)$

Terrific. So next I assumed that the $\textbf{D}^k$ statement was true. Proceeding for $\textbf{D}^{k+1}$, I looked at:

$\textbf{D}\bigg(\textbf{D}^k(f\circ p)(x_0)(x_1,...,x_{k+1})\bigg)$

Here's where I run into trouble though. I would like to use the induction hypothesis on the stuff inside the big parenthesis, but the fact that there are now $k+1$ components is throwing me off. Another thing that's not helping is the fact that I'm not really comfortable with this notation. I understand that $\textbf{D}^k(f\circ p)(x_0)$ is the $k$th derivative of $f\circ p$ evaluated at the point $x_0$, and I'm applying it to $(x_1,...,x_k)$ but is $(x_1,...,x_k)$ a vector, or are $x_1,...,x_k$ each vectors on their own?

So that's where I stand on this problem. Any help is appreciated.

2. Is there any relation between n, m and s? It just seems weird, seeing as you're talking about derivatives and composition.

3. It's a typo. Should be $p:\mathbb{R}^n\to\mathbb{R}^m$. But there's no relation other than that.

4. Have you seen that $\textbf{D} ^kp : \mathbb{R} ^n \rightarrow \mathcal{M}( \left( \mathbb{R} ^n \right) ^k , \mathbb{R} ^m )$ $:= \{ g: \left( \mathbb{R} ^n \right) ^k \rightarrow \mathbb{R} ^m : g \ is \ a \ multilinear \ continous \ function \}$ ?

Knowing this I think it's easier to see what you're trying to do, since from this point of view it's clear that $(x_1,...,x_k) \in \left( \mathbb{R} ^n \right) ^k$ and you're basically trying to prove a chain rule for $\textbf{D} ^k f \circ p$.

On another note, isn't $\textbf{D}^kp =0$ for all $k\geq 2$ since $\textbf{D} p(x_0)=L$ for all $x_0$ where $p(x)=Lx +b$?

5. I need to quit being so copy-paste happy (or just get some sleep). I just checked the problem in the book; there aren't supposed to be $^k$ on the $\textbf{D}$.

So can I split up the $\textbf{D}\bigg(\textbf{D}^k(f\circ p)(x_0)(x_1,...,x_{k+1})\bigg)$ into $\textbf{D}\bigg(\textbf{D}^k(f\circ p)(x_0)(x_1,...,x_k)\bigg)(x_{k+1})$?