Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$F(x,y)$ is homogenous of degree n if $f(tx,ty)=t^nf(x,y)$. Verify that

  • $xf_x(x,y)+yf_y(x,y)=nf(x,y)$
  • $x^2f_{xx}(x,y)+2xyf_{xy}+y^2f_{yy}(x,y)=n(n-1)f(x,y)$

Looks like I need enlightenment again... Hopefully, its not another embarrassingly simple thing I missed out in another question

What I tried:



$xt^nf_x(x,y) + yt^nf_y(x,y) = t^n(xf_x(x,y) + yf_y(x,y))$

Doesn't look like I am doing the right thing?

share|cite|improve this question
You can prove it by showing it is true for every monomial $x^k y^l$ (a direct calculation) and then argue by linearity that it folds for every polynomial $f(x,y)$. – Michael Joyce Apr 12 '12 at 1:11
up vote 4 down vote accepted

Consider $g(t)=f(tx,ty)$. By the chain rule, $g'(t)=x f_x(tx,ty)+y f_y(tx,ty)$.

On the other hand, since $f$ is homogenous of degree $n$, we have $g(t)=t^n f(x,y)$ and so $g'(t)=nt^{n-1}f(x,y)$.

Now take $t=1$ and conclude that $x f_x(x,y)+y f_y(x,y) = g'(1) = n f(x,y)$.

For the second result you mention, consider $g''$.

share|cite|improve this answer
This proof appears also in – lhf Apr 11 '12 at 14:29
I think the part I don't get is why $g'(t)=x f_x(tx,ty)+y f_y(tx,ty)$? Or in the link, why is $x'=xt$, similar for $y'$ – Jiew Meng Apr 12 '12 at 0:43
Also, how did you get $g(t)=f(tx,ty)=t^nf(x,y)$? – Jiew Meng Apr 12 '12 at 0:56
You get $g'$ using the chain rule. I've edited my answer to reflect this. – lhf Apr 12 '12 at 1:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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