## 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{.}$

1. anonymous

*

2. Loser66

me too *

3. geerky42

what is "*" ? Something like STAHP ? lol

4. anonymous

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

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

@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."