Euler's theorem and Cobb-Douglas When I apply Euler's Theorem onto the Cobb-Douglas equation I receive the Cobb-Douglas equation back. That is, there is no change after applying the theorem. Can someone please explain what the intent and point of this is, seeing no difference?
You are probably talking about Euler's homogeneous function theorem http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html , which shows that you can decompose a function into its partial derivatives based on its homogeneity. 

So a typical Cobb Douglass $U(x,y)=x^ay^{1-a}$

This function is homogeneous of degree 1 because when we multply each argument by a scalar:

$U(cx,cy)=(cx)^a(cy)^{1-a}=cx^ay^{1-a}=cU(x,y)$

So if we apply the theorem to this function we get:

$U(x,y)=xU_x+yU_y=xax^{a-1}y^{1-a}+x^ay(1-a)y^{-a}=x^ay^{1-a}$

So the result that we get back the original utility is expected.