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$F(x,y)$ 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?

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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 $g''$.

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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

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