Every polynomial sequence satisfies a linear recurrence?

Question

I noticed an interesting pattern, which is that the sequence $a_n = n^2$ satisfies the recurrence relation $1a_{n-3} - 3 a_{n-2} + 3 a_{n-1} - 1a_n = 0.$ This is an alternating weighted sum of 4 consecutive terms in the sequence, with the weights taken from Pascal's triangle [1,3,3,1].

I noticed that the pattern seems to extend to other polynomials, such as $a_n=n$ (with $1a_{n-2} - 2 a_{n-1} + 1a_n = 0$) and $a_n = n^3$ (with $1a_{n-4} - 4a_{n-3} + 6a_{n-2}-4a_{n-1}+1a_n = 0)$.

I am looking for a (hopefully elegant) proof that this pascal-weighted alternating sum recurrence holds for any polynomial sequence $a_n = n^p$, with an explanation for why the pascal weights in particular appear. I know several facts that are potentially useful, but I haven't seen how to put them together.

1. The integer points of every degree $d$ polynomial have constant order-$d$ successive differences, for the same reason that in a continuous setting, the $d$th derivative is constant.
2. This means that some fixed number of previous terms in the sequence convey enough data about the various-order differences to construct the next term in the sequence.
3. The terms from Pascal's triangle represent the coefficients of the binomial expansion of $(x+y)^n$. Presumably this alternating sum causes a bunch of nice cancellations, but I don't yet see how.
4. The terms from Pascal's triangle are also recursive, in that each row is generated as the sum of the two neighboring terms in the previous row.

In short, why does it seem that:

$$f(x)=\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} (x-k)^p \stackrel{?}{\equiv} 0. $$

Answer 1

Okay, I managed to construct a satisfyingly elegant answer.

1. We know from the binomial theorem that $(x+1)^p$ can be written as a linear combination of powers of $x^0,x^1,\ldots,x^{p}$. The coefficients of the linear combination come from Pascal's triangle.

2. Hence if we represent each term of the sequence $a_n \equiv n^p$ as a vector of powers $\begin{bmatrix}n^0,n^1,n^2,\ldots,n^p\end{bmatrix}^\top$, then we can generate successive terms in the sequence by applying a linear operator (matrix) made from Pascal's triangle:

   $$\begin{bmatrix}(x+1)^0\\(x+1)^1\\\vdots\\(x+1)^p\end{bmatrix} = \begin{bmatrix}1 & &&\\1 & 1 &\\1&2&1&\\&\vdots&\\{p \choose 0}&{p \choose 1}&\ldots&{p \choose p}\end{bmatrix}\begin{bmatrix}x^0\\x^1\\\vdots\\x^p\end{bmatrix}$$

   Let's call this matrix $P$. By repeatedly applying $P$, we generate the successive terms of the sequence: $a_{n+k} = P^k a_n$.

3. $P$ is a $p\times p$ triangular matrix with ones along the diagonal, which means that its characteristic polynomial is $\chi(\lambda)=(1-\lambda)^p = \sum_{i=0}^p (-1)^i {p \choose i}\lambda^i$. This is exactly the alternating sum of binomial coefficients we were looking for.

4. Every matrix satisfies its own characteristic polynomial. (This is the Cayley–Hamilton theorem.) For example, when $p=3$, we know that

$$P^3 -3P^2 + 3P - I = \mathbf{0}^{4\times 4}$$

5. But then, if $\mathbf{x}=\begin{bmatrix}n^0,n^1,n^2,\ldots,n^p\end{bmatrix}^\top$ is any term in the sequence, then $P^0\mathbf{x}=I\mathbf{x} = \mathbf{x}$ represents that term and $P^1\mathbf{x}, P^2\mathbf{x}, \ldots P^p\mathbf{x}$ represents the $p$ following terms. From this, the characteristic polynomial of $P$ shows that the alternating sum of $(p+1)$ successive terms in the polynomial sequence, weighted by binomial coefficients, vanishes— which is what we wanted to show.

6. Incidentally, this shows that the recurrence holds for any polynomial sequence $a_n = c_0 + c_1 n + c_2 n^2 + \ldots + c_p n^p$ if we left-multiply the characteristic equation by $\mathbf{c}^\top = [c_0, c_1, \ldots, c_p]$, because then

$$\mathbf{c}^\top(I-P)^p \mathbf{x} = \mathbf{c}^\top \sum_{i=0}^p (-1)^i {p \choose i}P^i \mathbf{x} = \sum_{i=0}^p (-1)^i {p \choose i} a_i$$

Answer 2

We use the coefficient of operator $[z^p]$ to denote the coefficient of $z^p$ in a series.

We obtain \begin{align*} \color{blue}{\sum_{k=0}^{p+1}}&\color{blue}{\binom{p+1}{k}(-1)^k(x-k)^p}\\ &=\sum_{k=0}^{p+1}\binom{p+1}{k}(-1)^kp![z^p]e^{(x-k)z}\tag{1}\\ &=p![z^p]e^{xz}\sum_{k=0}^{p+1}\binom{p+1}{k}(-1)^k\left(e^{-z}\right)^k\\\ &=p![z^p]e^{xz}\left(1-e^{-z}\right)^{p+1}\tag{2}\\ &=p![z^p]e^{xz}\left(z-\frac{z^2}{2!}+\frac{z^3}{3!}-\cdots\right)^{p+1}\tag{3}\\ &\,\,\color{blue}{=0} \end{align*} and the claim follows.

Comment:

• In (1) we use the exponential series expansion $e^{(x-k)z}=\sum_{q=0}^\infty (x-k)^q\frac{z^q}{q!}$ and extract the coefficient of $z^p$.

• In (2) we apply the binomial theorem.

• In (3) we observe the smallest power in $\left(1-e^{-z}\right)^{p+1}$ is $z^{p+1}$, so that the coefficient of $z^p$ of the expression is zero.

Answer 3

I would like to provide an answer from the perspective of induction / combinations.

We can induct on $p$, and obviously the base case where $p=0$ is true. Now we suppose $f(x)=0$ for $m<p$, and we try to prove that $f(x)=0$ for $p$.

First, we can show that $f(p+1)=0$. Consider a team of $p+1$ people. We perform the following operations:

Choose $0$ people, and award $0$ points per person,

Choose $1$ person (in every possible way), and deduct $1^{p-1}$ point per person,

Choose $2$ people (in every possible way), and award $2^{p-1}$ points per person...

After all the operations, the team should end up with exactly $$\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} k^p=\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} (p+1-k)^p=f(p+1)$$ points. We can also consider the team's points by considering how many points have each person gained.

Each person would've first lost $1^{p-1}{p \choose 0}$ points, then gain $2^{p-1}{p \choose 1}$ points, then lose $3^{p-1}{p \choose 2}$ points, etc., which can be written as $$\sum_{k=0}^{p} {p\choose k} (-1)^{k+1} (k+1)^{p-1}=\sum_{k=0}^{p} {p\choose k} (-1)^{k+1} (p+1-k)^{p-1}$$ which we assumed to be $0$. Therefore, $f(p+1)=(p+1)\cdot 0=0$.

For $x≠p+1$, we can let $x=p+1-x_1$, so $$f(x)=\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} (p+1-x_i-k)^p=\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} \sum^p_{i=0}{p\choose i}(p+1-k)^i(-x_1)^{p-i}$$ $$=\sum^p_{i=0}{p\choose i}(-x_1)^{p-i}\sum_{k=0}^{p+1} {p+1\choose k} (-1)^{k} (p+1-k)^i$$ We have either assumed or proved that every term in this sum is $0$, so $f(x)=0$ for $p$.

In fact, using the same method, we can show that we don't even need $a_n=n^p$, $a_n=(x_1+n)(x_2+n)···(x_p+n)$ would also satisfy the recurrence. $a_n=n^p$ would be the special case where $x_1=x_2=...=x_p=0$