Question
don't need to respond I just saw the error of my reasoning!
that theorem isn't applicable the function is homogeneuos on x and y not on y/x and even if it was it would be Phi ( y/x * x) not x^k right? i didn't pay attention to that ;p

As I noted in my second response, you cannot treat x like a constant. Even if you wanted to treat x like a constant, you could not pass it into φ because φ has degree zero . What that means is that φ((cy)/(cx)) = φ(y/x), which is trivial.

If you have cφ(y/x), or x^k φ(y/x), or such a thing there is no mechanism for removing or manipulating the c or x^k based on the homogeneity of φ. You can basically delete the c from the expression φ(cy/cx), but that should be clear. You cannot delete or move the c in the expression cφ(y/x).

And even if you could treat x^k like a constant, you cannot move or manipulate the x^k in the expression x^k φ(y/x). The only thing you might be able to say is:

x^k φ(y/x) = x^k (x^k)^0 φ( (y x^k) / (x x^k) ) = x^k φ(y/x)

which tells you nothing.

Refer back to the example from before, but I'll change it a little bit:

F(x,y) = ( x^3 y^2 + x^2 y ) / ( 3x^2 + 2xy )

Then you have:

F(x,y) = x ( (y/x)^2 + y/x ) / ( 3(y/x)^2 + 2(y/x) )

If you let φ(u) = (u^2+u)/(3u^2+2u), i.e. u=y/x, then this gives you:

F(x,y) = x φ(y/x)

The original F is homogeneous with degree 1. The function φ is homogeneous with degree 0, because φ( (cy)/(cx) ) = φ(y/x). That is not a surprising fact.

But when you have F(x,y) = x^k φ(y/x), the x^k cannot be removed. If it could, you'd have:

F(x,y) = φ(y/x * x) = φ(y)

That is impossible, since you've just eliminated the x from F(x,y) entirely.

Likewise, you might do something like:

F(x,y) = φ(y/x * x^k) or F(x,y) = φ(y/x * x^(k-1) ) or whatever

And that seems to imply that F must have degree 0 (since φ does) or that F is not homogeneous at all, and that can't be right. We know F is homogeneous with degree k.
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#### Clyde Oliver

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