0
$\begingroup$

How do I show that

$$ f(\overrightarrow{x}) = \nabla f( \overrightarrow{0} ) \cdot \overrightarrow{x} $$ for every $x$, given that $f \space (t \cdot \overrightarrow{v})= t \space f( \overrightarrow{v})$?

I'm confused by the notion of the gradient of a single-variable function and the vector variable. I assume that since $f \space (t \cdot \overrightarrow{v})= t \space f( \overrightarrow{v})$, $\space$ $f({x})$ has the form $f({x})=ax$, but that's all I got so far. Any hints?

$\endgroup$
  • $\begingroup$ Why do you assume that? What if $f(x)$ was rotation by say $\pi/3$ in $\mathbb{R}^2$? $\endgroup$ – David Peterson Sep 28 '14 at 6:31
0
$\begingroup$

I'll leave the details to you, since this looks like homework.

Consider the function $g(t)=f(tx)$ then $g'(t)=\nabla f(tx)\cdot x$ (chain rule), but also $g'(t)= f(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.