anonymous
  • anonymous
Let \[f:\mathbb{R}^n \rightarrow \mathbb{R} \]be a differentiable function such that \[f(t\mathbf{x})=t^m f(\mathbf{x}) \text{ for all }\mathbf{x} \text{ in }\mathbb{R}^n \text{ and for all } t > 0\text{.}\]\[\text{Show that } x_1f_{x_1}(\mathbf{x})+...+x_nf_{x_n}(\mathbf{x})=mf(\mathbf{x}) \text{ where }\mathbf{x}=(x_1,...,x_n) \text{.}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
*
Loser66
  • Loser66
me too *
geerky42
  • geerky42
what is "*" ? Something like STAHP ? lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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\[x_1 f_{x_1}(\mathbf{x})+x_2 f_{x_2}(\mathbf{x})+...+x_n f_{x_n}(\mathbf{x})=m f(\mathbf{x})\]
anonymous
  • 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\[x. \nabla f(\mathbb{x})=k f(\mathbb{x})\]
Loser66
  • 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."

Looking for something else?

Not the answer you are looking for? Search for more explanations.