# Finite differences of polynomials

It is interesting watching my kids go through the school math curriculum. Since I’m a math professor, one would think that I would know all of the school-aged math. While that is mostly true, sometimes the teachers and textbooks use unfamiliar terminology for familiar mathematical ideas. (“Oh, ____ is just ___,” I’ve said multiple times.)

My son is now in Algebra 2, and for the first time, he showed me something that I’ve never seen before—the relationship between polynomials and finite differences.

Take any polynomial, such as $f(x)=7x^3+4x^2-2x+15,$ and any arithmetic sequence, such as 0, 2, 4, 6,… Plug these values into the polynomial. Take the neighboring pairwise differences. So, for instance, $f(2)-f(0)=83-15=68.$ Then take the neighboring pairwise differences of those values, and so on. It turns out that the nth level will consist entirely of the same nonzero value if, and only if, the polynomial has degree n. Wow! That’s so cool!

Here’s my worked-out degree-3 example. Notice that after three levels, we yield the constant value 336:

My son’s textbook (Algebra 2, by Larson, Boswell, Kanold, and Stiff) shared this fact but provided no explanation for why it is true (which, as a mathematician, I find very disappointing). So, I had to work it out myself.

It turns out that more is true—if the polynomial has degree n with leading coefficient c, and is the difference between terms in the arithmetic sequence, then the final value is $a^n c\cdot n!.$ In particular, if a=1 and c=1, then the pairwise difference process will terminate with n!. Notice that in my example, the final value is $2^3\cdot 7\cdot 3!=336.$

Why is this true?

Here’s a proof by induction on the degree of the polynomial. As the base case, consider a degree-1 polynomial: p(x)=cx+b. Then,

$p(x+a)-p(x)=(c(x+a)+b)-(cx+b)=ac=a^1c\cdot 1!,$

so the base case holds.

Now, assume that the result is true for any polynomial of degree n-1, for some n≥2. We will prove that it is true for a polynomial of degree n. Let $p(x)=cx^n+dx^{n-1}+\text{l.o.t.},$ where c≠0 and “l.o.t.” means “lower order terms.” We see that

$p(x+a)-p(x)=(c(x+a)^n+d(x+a)^{n-1}+\text{l.o.t.})-(cx^n+dx^{n-1}+\text{l.o.t.})=cx^n+acnx^{n-1}+dx^{n-1}+\text{l.o.t.}-cx^n-dx^{n-1}+\text{l.o.t.}=acnx^{n-1}+\text{l.o.t.}.$

Let us call this polynomial q(x).

Notice that because the leading coefficient acn is nonzero,  q(x) has degree n-1. By our inductive hypothesis, after n-1 pairwise differences, the polynomial q(x) will yield a constant value $acn\cdot a^{n-1}(n-1)!=a^ncn!.$ Thus, for p, the process terminates after steps with the constant value $a^n c n!.$ This proves the theorem.

After playing around with this, I googled it, and—no surprise—the mathematics of finite differences has a long history. Also, it is not difficult to see the resemblance of these calculations to the calculation of the derivative (using the definition of the derivative).