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anonymous
 one year ago
Let \[f:\mathbb{R}^n \rightarrow \mathbb{R} \]be a differentiable function such that \[f(t\mathbf{x})=t^m f(\mathbf{x}) \text{ for all }\mathbf{x} \text{ in }\mathbb{R}^n \text{ and for all } t > 0\text{.}\]\[\text{Show that } x_1f_{x_1}(\mathbf{x})+...+x_nf_{x_n}(\mathbf{x})=mf(\mathbf{x}) \text{ where }\mathbf{x}=(x_1,...,x_n) \text{.}\]
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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geerky42
 one year ago
Best ResponseYou've already chosen the best response.0what is "*" ? Something like STAHP ? lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0once differentiate both sides of \(f(t\mathbf{x})=t^m f(\mathbf{x})\) with respect to \(x_i\) to get\[t f_{x_1} (t\mathbf{x})= t^m f_{x_1}(\mathbf{x}) \\ f_{x_1} (t\mathbf{x})= t^{m1} f_{x_1}(\mathbf{x}) \ \ \ \star\]Now differentiate both sides of \(f(t\mathbf{x})=t^m f(\mathbf{x})\) with respect to \(t\)\[x_1f_{x_1} (t\mathbf{x})+x_2f_{x_2} (t\mathbf{x})+...+x_nf_{x_n} (t\mathbf{x})=mt^{m1} f(\mathbf{x})\]Use \(\star\) to rewrite the LHS of last equation\[x_1t^{m1} f_{x_1}(\mathbf{x})+x_2t^{m1} f_{x_2}(\mathbf{x})+...+x_nt^{m1} f_{x_n}(\mathbf{x})=mt^{m1} f(\mathbf{x})\]\(t>0\) so you can divide both sides of last equation by \(t^{m1}\) to get\[x_1 f_{x_1}(\mathbf{x})+x_2 f_{x_2}(\mathbf{x})+...+x_n f_{x_n}(\mathbf{x})=m f(\mathbf{x})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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\[x. \nabla f(\mathbb{x})=k f(\mathbb{x})\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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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