# Homogenous Functions

1. Jul 23, 2010

### psholtz

Is there such a thing as a homogenous function of degree n < 0?

Considering functions of two variables, the expression:

$$f(x,y) = \frac{y}{x}$$

is homogeneous in degree 0, since:

$$f(tx,ty) = \frac{ty}{tx} = \frac{y}{x} = f(x,y) = t^0 \cdot f(x,y)$$

and the expression:

$$f(x,y) = x$$

is homogenous in degree 1 since:

$$f(tx,ty) = tx = t^1 \cdot f(x,y)$$

and the expression:

$$f(x,y) = x^3y^2$$

is homogeneous in degree 5, since:

$$f(tx,ty) = t^5 \cdot f(x,y)$$

I suppose that in the same way I could construct a function something like:

$$f(x,y) = \frac{y}{x^2}$$

So that:

$$f(tx,ty) = t^{-1} \cdot f(x,y)$$

or in other words, "homogenous" in degree n = -1.

Does this ever really come up much?

2. Jul 24, 2010

### psholtz

I think the answer is "yes"..

For instance, the Euler theorem on homogeneous functions states, in relevant part, that if you have a function in two variables, x and y, which is homogeneous in degree n, then:

$$n f(x,y) = x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}$$

So taking the example I gave above, where you have:

$$f(x,y) = \frac{x}{y^2}$$

which, ostensibly, is homogeneous in degree -1. Plugging this equation in the Euler theorem above, you would have:

$$\frac{\partial f}{\partial x} = \frac{1}{y^2}$$

$$x\frac{\partial f}{\partial x} = \frac{x}{y^2}$$

and that:

$$\frac{\partial f}{\partial y} = -2\frac{x}{y^3}$$

$$y\frac{\partial f}{\partial y } = -2\frac{x}{y^2}$$

so that:

$$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = -\frac{x}{y^2} = -f(x,y)$$

which is what we would expect of a function which is homogenous in degree -1..

3. Jul 26, 2010

### l'Hôpital

Yes.

Consider the following:

$$U = \sum_{1 \leq i < j \leq n}^{n} \frac{m_i m_j}{||q_i - q_j||}$$

It usually appears in the n-body problem in the form of...

$$m_i \ddot{x_i} = \frac{dU}{dq_i}$$

The fact that U is homogeneous of degree -1 is key to prove that there are no equilibrium solutions for the n-body problem.