Homogeneous Functions

Homogeneous

To be Homogeneous a function must pass this test:

f(zx, zy) = zⁿ f(x, y)

In other words

Homogeneous is when we can take a function: f(x, y)

multiply each variable by z: f(zx, zy)

and then can rearrange it to get this: zⁿ f(x, y)

An example will help:

Example: x + 3y

Start with: f(x, y) = x + 3y

Multiply each variable by z: f(zx, zy) = zx + 3zy

Let's rearrange it by factoring out z: f(zx, zy) = z(x + 3y)

And x + 3y is f(x, y): f(zx, zy) = z f(x, y)

Which is what we wanted, with n=1: f(zx, zy) = z¹ f(x, y)

Yes, x + 3y is homogeneous!

The value of n is called the degree. So in that example the degree is 1.

Example: 4x² + y²

Start with: f(x, y) = 4x² + y²

Multiply each variable by z: f(zx, zy) = 4(zx)² + (zy)²

Which is: f(zx, zy) = 4z²x² + z²y²

Factoring out z²: f(zx, zy) = z²(4x² + y²)

And 4x² + y² is f(x, y): f(zx, zy) = z² f(x, y)

Yes, 4x² + y² is homogeneous.

And its degree is 2.

How about this one:

Example: x³ + y²

Start with: f(x, y) = x³ + y²

Multiply each variable by z: f(zx, zy) = (zx)³ + (zy)²

Which is: f(zx, zy) = z³x³ + z²y²

Factoring out z²: f(zx, zy) = z²(zx³ + y²)

But zx³ + y² is NOT f(x, y)!

So x³ + y² is NOT homogeneous.

And notice that x and y have different powers: x³ vs y². For polynomial functions that is often a good test.

Can it work for functions that are not polynomials? How about this one:

Example: the function x cos(y/x)

Start with: f(x, y) = x cos(y/x)

Multiply each variable by z: f(zx, zy) = zx cos(zy/zx)

Which is: f(zx, zy) = zx cos(y/x)

Factoring out z: f(zx, zy) = z(x cos(y/x))

And x cos(y/x) is f(x, y): f(zx, zy) = z¹f(x, y)

So x cos(y/x) is homogeneous, with degree of 1.

Notice that (y/x) is "safe" because (zy/zx) cancels back to (y/x)

Homogeneous, in English, means "of the same kind"

For example "Homogenized Milk" has the fatty parts spread evenly through the milk (rather than having milk with a fatty layer on top.)

Homogeneous applies to functions like f(x), f(x, y, z) etc. It is a general idea.

Homogeneous Differential Equations

A first order Differential Equation is homogeneous when it can be in this form:

homogeneous equation

In other words, when it can be like this:

M(x, y) dx + N(x, y) dy = 0

And both M(x, y) and N(x, y) are homogeneous functions of the same degree.

Find out more on Solving Homogeneous Differential Equations.