## Homogeneous Polynomials and Euler.

### September 14, 2010

A small tangent, because I enjoyed this so much in class.

Let’s call a polynomial of $n$ variables a homogeneous polynomial if every term has the same degree.  Recall that we compute the degree of a term by summing the powers of each variable in the term.

For example $3x_{1}^{2}x_{2}^{4}x_{3}^{4}$ has degree 10.  For another example, $2x_{1}^{3}x_{2}^{3}$ has degree 6.  So, an example of a homogenous polynomial in three variables is

$f(x_{1}, x_{2}, x_{3}) = 2x_{1}^{3}x_{2}^{1} + 5x_{1}^{2}x_{3}^{2} + x_{3}^{4}$

and this polynomial is of degree 4.

We sometimes write a general homogeneous polynomial of degree $m$ the following way

$\displaystyle \sum_{|I| = m} a_{I}x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{n}^{i_{n}} = \sum_{|I| = m} a_{I}x^{I}$

where $I = (i_{1}, \dots, i_{n})$ and $|I| = \displaystyle \sum_{j = 1}^{n}i_{j}$.  This is a lot of notation to get used to, but basically we just want to say that all of these terms have powers that add up to $m$.  This sum is saying exactly that, and the right-most-side is shorthand.

Let’s just show one thing before we move onto Euler’s relation.  Suppose our field is the reals.  Then we can state

Property: Suppose $f(x_{1}, \dots, x_{n})$ is homogeneous of degree $m$.  Then, writing $f(x) = f(x_{1}, \dots, x_{n})$, we have $f(\alpha x) = \alpha^{m}f(x)$ for $\alpha\in {\mathbb R}$.

Proof. This is obvious by plugging in $\alpha x$ for $x$ and seeing how many we can pull out of the polynomial.

That’s pretty cool, right?  If we have an arbitrary field, this may not be so nice, but usually it’s pretty okay.

Unfortunately, if we take partial derivatives, we do not necessarily get homogeneous polynomials as our partials (why not?) BUT, if we do know all of our partials, we can figure out the original function by this really sweet relation.  Let’s just dive right into it.

Theorem (Euler’s Relation.):  If $f$ is a homogeneous polynomial in $n$ variables of degree $k$ and we have that $f(\alpha x) = \alpha^{n}f(x)$ then the following relation holds

$\displaystyle \sum_{j = 1}^{n} x_{j}\frac{\partial f}{\partial x_{j}} = kf(x_{1}, \dots, x_{n})$

Which, all things considered, is a really nice formula!  The proof is easy and elementary, but it takes a lot of writing down.  It’s really just taking partial derivatives, looking at the coefficients, and keeping track of indices.  I leave that to you, dear reader.