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Let $m$ be a positive integer. Assume $f: R^n \to R$ has continuous first partials and satisfies the following property:

$f(t\vec{x})=t^mf(\vec{x})$ for all $\vec{x}$ and $t$

Show that $\vec{x} \cdot \nabla{f(\vec{x})}=mf(\vec{x})$

So far I have

$\vec{x} \cdot \nabla{f(\vec{x})}= \frac{1}{t^m}\vec{x} \cdot \nabla{f(t\vec{x})}$ but I'm not sure how to proceed from here.

  • 0
    That's [Euler's homogeneous function theorem](http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html).2011-08-12

1 Answers 1

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hint

Let $g(t;x,y,z) = f(tx,ty,tz)$, then

$ \frac{d}{dt} g(t;x,y,z) = x(\partial_xf)(tx,ty,tz) + y(\partial_yf)(tx,ty,tz) + z(\partial_z f)(tx,ty,tz) $

by the chain rule. So $\vec{x}\cdot \nabla f(\vec{x}) = \frac{d}{dt}g(t;\vec{x}) |_{t=1}$. On the other hand $g(t;x,y,z) = t^m f(x,y,z)$ by hypothesis, which allows you a different way to compute the $t$ derivative of $g$.