Questions on a self-made theorem about polynomials

I recently came up with this theorem:

For any complex polynomial $P$ degree $n$:

$$ \sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) = 0\quad \forall a,b \in\mathbb{C}$$

Basically, if $P$ is quadratic, $P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$ (inputs of $P$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.

• Has this been discovered? If yes, what's the formal name for this phenomenon?
• Is it significant/Are there important consequences of this being true?
• Can this be generalized to non-polynomials?

Solution 1:

In brief: this is well-known, but definitely important.

It's easiest to write this in terms of the finite difference operator $\Delta$: $\Delta P(x)=P(x+1)-P(x)$. You use $P(x+b)$ instead of $P(x+1)$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $\Delta_b$ for your operator.

The most important feature of the $\Delta_b$ operator is how it affects the degree of a polynomial:

Theorem: for any nonconstant polynomial $P(x)$, the degree of $\Delta_b P(x)$ is one less than the degree of $P(x)$.

Proof outline: Note that the degree of $\Delta_b P(x)$ is no greater than the degree of $P(x)$. Now, write $P(x) = a_dx^d+Q(x)$, where $Q(x)$ is a polynomial of degree $d-1$ or less. Then $P(x+b) =a_d(x+b)^d+Q(x+b)$, so $\Delta_b P(x) = a_d\left((x+b)^d-x^d\right)+\Delta_b Q(x)$; by the binomial theorem $(x+b)^d=x^d+{d\choose 1}bx^{d-1}+\ldots$, so $(x+b)^d-x^d={d\choose 1}bx^{d-1}+\ldots$ is a polynomial of degree at most $d-1$, and thus $\Delta_bP(x)$ is the sum of two polynomials of degree at most $d-1$ (namely, $a_d\left((x+b)^d-x^d\right)$ and $\Delta_b Q(x)$), so it's of degree at most $d-1$ itself.

(It's slightly more challenging to prove that the degree of $\Delta_bP(x)$ is exactly $d-1$ when $b \neq 0$, but this can also be shown.)

Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $\Delta_b$ operator $d+1$ times, where $d$ is the degree of the polynomial; since each application of $\Delta_b$ reduces the degree by one, then $(\Delta_b)^dP(x)$ is a polynomial of degree zero — a constant — and thus $(\Delta_b)^{d+1}P(x)$ will be identically zero. This is exactly your identity.

Now, you may know that the derivative of a polynomial of degree $d$ is also a polynomial of degree $d-1$. It turns out that this isn't a coincidence; $\Delta$ is very similar to a derivative in many ways, with the Newton polynomials ${x\choose d}=\frac1{d!}x(x-1)(x-2)\cdots(x-d)$ playing the role of the monomial $x^d$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.

In fact, we can also prove the converse (and this answers the question about generalizing to non-polynomials in the negative). I'll work in terms of $\Delta$, rather than $\Delta_b$, but again all the results generalize readily.

Note that $\Delta^n P(x)$ only depends on the values of $P(x+i)$ for $i$ an integer between $0$ and $n$; thus, a function can take arbitrary values for $0\lt x\lt1$ and still satisfy the identity; we can't say much about general points. However, it does constrain the values at integers:

Theorem: suppose that $\Delta^{d+1}f(x)\equiv 0$ identically. Then there exists a polynomial $P(x)$ of degree $d$ such that $f(n)=P(n)$ for all integers $n$.

The proof works by induction. For simplicity's sake, I'll consider all functions as being on $\mathbb{Z}$ now, and not consider non-integer values at all. Note first of all that if $\Delta f(x)=g(x)$, then $f(n)=f(0)+\sum_{i=0}^{n-1}g(i)$. (Proof by induction: the case $n=1$ is true by definition, since $g(0)=\Delta f(0)=f(1)-f(0)$ implies that $f(1)=f(0)+g(0)$. Now, assuming it's true for $n=k$, at $n=k+1$ we have $f(k+1)=f(k)+g(k)$ $=f(0)+\sum_{i=0}^{k-1}g(i)+g(k)$ $=f(0)+\sum_{i=0}^kg(i)$.) In particular, if $\Delta f(x)\equiv 0$ identically, then $f(n)=f(0)$ for all integers $n$; $f()$ is constant on $\mathbb{Z}$.

This gives us the base case for our induction; to induct we just need to show that if $\Delta f(x)$ is a polynomial of degree $d$, then $f(x)$ is polynomial of degree $d+1$. But suppose for concreteness that $\Delta f(x)=P(x)=\sum_{i=0}^da_ix^i$. Then $f(n)=f(0)+\sum_{k=0}^{n-1}P(k)$ $=f(0)+\sum_{k=0}^{n-1}\left(\sum_{i=0}^da_ik^i\right)$ $=f(0)+\sum_{i=0}^da_i\left(\sum_{k=0}^{n-1}k^i\right)$. Now, for each $i$ the sum $\sum_{k=0}^{n-1}k^i$ in parentheses in this last expression is known to be a polynomial of degree $i+1$ (see e.g. https://en.wikipedia.org/wiki/Faulhaber%27s_formula ), so the whole expression is a polynomial of degree $d+1$, as was to be proved.

Solution 2:

What you wrote is the finite difference operator of order $n+1$, acting on $P$, $$ \sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) =\Delta^{n+1} P(a+kb).$$

Notice that by linearity it suffices for the property to hold for all monomials $k^m, m\le n$ and it is easily explained by the fact that the first order difference of a polynomial is a polynomial of degree one less.

$$(k+1)^m-k^m=k^m+mk^{m-1}+\cdots-k^m.$$

Illustration ($n=3$):

$$\Delta^4 k^m=((4^m-3^m)-(3^m-2^m))-((3^m-2^m)-(2^m-1^m)) \\-((3^m-2^m)-(2^m-1^m))-((2^4-1^m)-(1^4-0^m)) \\=4^m-4\cdot3^m+6\cdot2^m-4\cdot1^m+0^m.$$ and $$\begin{matrix} 1&&1&&1&&1&&1 \\&0&&0&&0&&0 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&2&&3&&4 \\&1&&1&&1&&1 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&4&&9&&16 \\&1&&3&&5&&7 \\&&2&&2&&2 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&8&&27&&64 \\&1&&7&&19&&37 \\&&6&&12&&18 \\&&&6&&6 \\&&&&0 \end{matrix}$$

Final remark:

On can show that $\Delta_{n+1}k^{n+1}=(-1)^nn!$, so that for a polynomial of degree $n+1$ the sum is

$$(-1)^nn!p_{n+1}b^{n+1},$$ independently of $a$.

Solution 3:

This result is known and was demonstrated by a student named Ruiz.

Here's the reference :

Sebastián Martín Ruiz, An Algebraic Identity Leading to Wilson's Theorem, The Mathematical Gazette, 80 (489) 579-582 (Nov. 1996).

You accesss to it at JSTOR : http://www.jstor.org/stable/3618534 or on the arXiv: arXiv:math/0406086.