Multivariable Calculus: Applications of Grad (and the Chain Rule?)

gadje
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Homework Statement


We say that a differentiable function f : \mathbb{R}^n \rightarrow \mathbb{R} is homogenous of degree p if, for every \mathbf{x} \in \mathbb{R}^n and every a>0,
f(a\mathbf{x}) = a^pf(\mathbf{x}).

Show that, if f is homogenous, then \mathbf{x} \cdot \nabla f(\mathbf{x}) = p f(\mathbf{x}) .

Homework Equations


The chain rule (not sure if I need it): \displaystyle \frac{d}{dt} f(\mathbf{x}(t)) = \Sigma_{i = 1}^{n} f_{x_i}\dot{x_i} = \dot{\mathbf{x}} \cdot \nabla f

The Attempt at a Solution



Well, I see the resemblance between the rightmost hand side of the chain rule I wrote down, but I don't really understand how the chain rule is applied in this situation, seeing as there isn't anything about x being a function of something else here.

Any ideas?
Cheers.
 
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Try differentiating

f(a\mathbf{x}) = a^pf(\mathbf{x})

with respect to a.
 
\frac{\partial}{\partial a} f(a \mathbf{x}) = pa^{p-1}f(\mathbf{x})

Okay. I'm still clueless.

EDIT:

Hang on. \frac{\partial}{\partial a} f(a \mathbf{x}) = \frac{\partial}{\partial a} (a \mathbf{x}) \frac{\partial}{\partial \mathbf{x}} f(a\mathbf{x}) = \mathbf{x} \cdot \nabla f (a \mathbf{x}) (if you'll forgive the abuse of notation), right?
 
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Good. Now let a=1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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