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anonymous

  • one year ago

Let \[f:\mathbb{R}^n \rightarrow \mathbb{R} \]be a differentiable function such that \[f(t\mathbf{x})=t^m f(\mathbf{x}) \text{ for all }\mathbf{x} \text{ in }\mathbb{R}^n \text{ and for all } t > 0\text{.}\]\[\text{Show that } x_1f_{x_1}(\mathbf{x})+...+x_nf_{x_n}(\mathbf{x})=mf(\mathbf{x}) \text{ where }\mathbf{x}=(x_1,...,x_n) \text{.}\]

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  1. anonymous
    • one year ago
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    *

  2. Loser66
    • one year ago
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    me too *

  3. geerky42
    • one year ago
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    what is "*" ? Something like STAHP ? lol

  4. anonymous
    • one year ago
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    once differentiate both sides of \(f(t\mathbf{x})=t^m f(\mathbf{x})\) with respect to \(x_i\) to get\[t f_{x_1} (t\mathbf{x})= t^m f_{x_1}(\mathbf{x}) \\ f_{x_1} (t\mathbf{x})= t^{m-1} f_{x_1}(\mathbf{x}) \ \ \ \star\]Now differentiate both sides of \(f(t\mathbf{x})=t^m f(\mathbf{x})\) with respect to \(t\)\[x_1f_{x_1} (t\mathbf{x})+x_2f_{x_2} (t\mathbf{x})+...+x_nf_{x_n} (t\mathbf{x})=mt^{m-1} f(\mathbf{x})\]Use \(\star\) to rewrite the LHS of last equation\[x_1t^{m-1} f_{x_1}(\mathbf{x})+x_2t^{m-1} f_{x_2}(\mathbf{x})+...+x_nt^{m-1} f_{x_n}(\mathbf{x})=mt^{m-1} f(\mathbf{x})\]\(t>0\) so you can divide both sides of last equation by \(t^{m-1}\) to get\[x_1 f_{x_1}(\mathbf{x})+x_2 f_{x_2}(\mathbf{x})+...+x_n f_{x_n}(\mathbf{x})=m f(\mathbf{x})\]

  5. anonymous
    • one year ago
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    For more information, It's Euler's theorem and says The differentiable function \(f\) of \(n\) variables is homogeneous of degree \(k\) if and only if\[x. \nabla f(\mathbb{x})=k f(\mathbb{x})\]

  6. Loser66
    • one year ago
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    @geerky42 It means "the problem is interesting but I don't know how to solve yet. I leave my signal here so that if someone solves it, I can be back to learn."

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