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Thread: Multivariable Differential Mapping

  1. #1
    Super Member redsoxfan325's Avatar
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    Multivariable Differential Mapping

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

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

    So I started by induction. Using the chain rule,

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

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

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

    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 $\displaystyle k+1$ 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 $\displaystyle \textbf{D}^k(f\circ p)(x_0)$ is the $\displaystyle k$th derivative of $\displaystyle f\circ p$ evaluated at the point $\displaystyle x_0$, and I'm applying it to $\displaystyle (x_1,...,x_k)$ but is $\displaystyle (x_1,...,x_k)$ a vector, or are $\displaystyle x_1,...,x_k$ each vectors on their own?

    So that's where I stand on this problem. Any help is appreciated.
    Last edited by redsoxfan325; Nov 19th 2009 at 10:19 PM.
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  2. #2
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    Is there any relation between n, m and s? It just seems weird, seeing as you're talking about derivatives and composition.
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  3. #3
    Super Member redsoxfan325's Avatar
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    It's a typo. Should be $\displaystyle p:\mathbb{R}^n\to\mathbb{R}^m$. But there's no relation other than that.
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  4. #4
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    Have you seen that $\displaystyle \textbf{D} ^kp : \mathbb{R} ^n \rightarrow \mathcal{M}( \left( \mathbb{R} ^n \right) ^k , \mathbb{R} ^m )$$\displaystyle := \{ 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 $\displaystyle (x_1,...,x_k) \in \left( \mathbb{R} ^n \right) ^k$ and you're basically trying to prove a chain rule for $\displaystyle \textbf{D} ^k f \circ p$.

    On another note, isn't $\displaystyle \textbf{D}^kp =0$ for all $\displaystyle k\geq 2$ since $\displaystyle \textbf{D} p(x_0)=L $ for all $\displaystyle x_0$ where $\displaystyle p(x)=Lx +b$?
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  5. #5
    Super Member redsoxfan325's Avatar
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    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 $\displaystyle ^k$ on the $\displaystyle \textbf{D}$.

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