If $P(x)=\sum_{i=0}^da_i\left(\prod_{j=i}^{d+i-1}(x+j)\right)$ is linear, what is its constant term?

Question

Question: Fix $d,m\in\mathbb{N}$ with $0\leq m\leq d$ and define $$P(x)=\sum_{i=0}^da_i\left(\prod_{j=i}^{d+i-1}(x+j)\right),$$ where each $a_i$ is a constant, $a_m=0$. Suppose that, after expansion, $P(x)=c-x$ for some constant $c$. Show that $c=\frac{m}{d}-d$.

I obtained a rough solution by evaluating $P\left(-(d+k)\right)$ for each $0\leq k\leq m$, which yields $m+1$ linear relations on $a_0,\dots, a_{m-1}$ and $c$, from which one can then solve by scaling and subtracting. However, I am hoping for a cleaner, more succinct answer (in fact, one might even be able to use the above approach in a neater way than I did).

Answer 1

Here is a detailed write-up of the solution sketched in @darijgrinberg 's comment.

Summary of solution: the problem can be restated as saying that the $a_i$'s are the coordinates of $P(x)=c-x$ in a certain basis. So we only need to compute the coordinates of $1$ and $x$ in this basis to obtain the $a_i$ in terms of $c$.

Detailed solution : let $\beta_{d,k}(x)=\prod_{j=k}^{d+k-1}(x+j)$ (so that $P=\sum_{k=0}^d a_k\beta_{d,k}(x)$) and ${\cal B}_d=(\beta_{d,0},\beta_{d,1},\ldots,\beta_{d,d})$.

Lemma 1. ${\cal B}_d$ forms a basis of ${\mathbb R}_d[x]$, the space of polynomials of degree $\leq d$.

Proof of lemma 1. It will suffice to show that the members of ${\cal B}_d$ are linearly independent. So suppose that $\sum_{k=0}^d \lambda_k \beta_{d,k}=0$ for some scalars $\lambda_0,\lambda_1,\ldots,\lambda_k$. Evaluating at $-d$, we see that $\lambda_0=0$. Next, evaluating at $-(d+1)$, we see that $\lambda_1=0$, etc.

Our goal is now to compute the coordinates of $1$ and $x$ in the basis ${\cal B}_d$. The idea is to iterate the difference operator $\Delta$ defined by $\Delta(Q)=Q(x+1)-Q(x)$ for a polynomial $Q$. We will use two well-known facts on $\Delta^{i}(Q)$ which are straightforward to check by induction on $i$ once stated.

Fact 1. $\Delta^{i}(Q)=\sum_{k=0}^{i}(-1)^{i-k}\binom{i}{k}Q(x+k)$.

Fact 2. If the two leading monomials of $Q$ are $ax^d+bx^{d-1}$ and $i\leq d-1$, then the two leading monomials of $\Delta^i(Q)$ are $(i!\binom{d}{i}a)x^{d-i}+(\frac{i}{2}(i+1)!\binom{d}{i+1}a+i!\binom{d-1}{i}b)x^{d-i-1}$.

Combining the two facts for $i=d$, we deduce

$$ (d!)a=\Delta^d(Q)=\sum_{k=0}^{d}(-1)^{d-k}\binom{d}{k}Q(x+k). \label{1}\tag{1} $$

And for $i=d-1$, we deduce similarly

\begin{align} &((d!)a)x+\bigg(\frac{d-1}{2}d!a+(d-1)!b\bigg)=\Delta^{d-1}(Q) \\ &= \sum_{k=0}^{d-1}(-1)^{d-1-k}\binom{d-1}{k}Q(x+k). \label{2}\tag{2} \end{align}

Notice that in the LHS of \eqref{2}, the constant term can be rewritten as $(d!)a \times \rho$ where $\rho=\frac{d-1}{2}+\frac{b}{da}$. Subtracting $\rho$ times \eqref{1} from \eqref{2}, we deduce :

\begin{align} ((d!)a)x= \sum_{k=0}^{d}(-1)^{d-1-k}\bigg(\binom{d-1}{k}+\rho\binom{d}{k}\bigg)Q(x+k) \label{3}\tag{3} \end{align}

(since $\dbinom{d-1}{d}=0$). We now apply this context to $Q=\beta_{d,0}$. Then we have $a=1,b=\frac{d(d-1)}{2}$ and hence $\rho=d-1$, so that \eqref{1} becomes

$$ 1=\frac{(-1)^d}{d!} \sum_{k=0}^{d}(-1)^k \binom{d}{k}\beta_{d,k} \label{1'}\tag{1'} $$

and \eqref{3} becomes

$$ x= \sum_{k=0}^{d}\frac{(-1)^{d-1-k}}{d!}\bigg(\binom{d-1}{k}+(d-1)\binom{d}{k}\bigg)\beta_{d,k} \label{3'}\tag{3'} $$

Note that $\binom{d-1}{k}+(d-1)\binom{d}{k}=\big(d-\frac{k}{d}\big)\binom{d}{k}$, so that \eqref{3'} simplifies to

$$ x=\frac{(-1)^d}{d!}\sum_{k=0}^{d}(-1)^{k+1}\big(d-\frac{k}{d}\big)\binom{d}{k}\beta_{d,k} \label{3''}\tag{3''} $$

Combining \eqref{1'} and \eqref{3''}, we deduce

$$ a_k=\frac{(-1)^d}{d!}(-1)^{k}\binom{d}{k}\bigg(c-\big(\frac{k}{d}-d\big)\bigg) \ (0\leq k\leq d) \label{4}\tag{4} $$ and your claim immediately follows.

Answer 2

Some thoughts

Clearly, $d\ge 1$. Let $x = -(m-1), -m, -(m+1), \cdots, -(d-1)$ respectively to get \begin{align} P(1-m) &= c + m - 1, \\ P(-m) &= c + m, \\ P(-m - 1) &= c + m+1, \\ &\cdots\cdots\\ P(-d+1) &= c + d - 1. \end{align} Then we have (weighted sum of the equations above) $$\sum_{k=0}^{d-m} P(-m - k + 1)(-1)^k\binom{d+1}{k} = \sum_{k=0}^{d-m} (c + m + k - 1) (-1)^k\binom{d+1}{k}. \tag{1} $$

Claim 1: It holds that $$\sum_{k=0}^{d-m} P(-m - k + 1)(-1)^k\binom{d+1}{k} = 0.$$ (The proof is given at the end.)

By (1) and Claim 1, we have $$\sum_{k=0}^{d-m} (c + m + k - 1) (-1)^k\binom{d+1}{k} = 0$$ which results in $$c = -m + 1 - \frac{\sum_{k=0}^{d-m} k (-1)^k\binom{d+1}{k}}{\sum_{k=0}^{d-m} (-1)^k\binom{d+1}{k}} = -m + 1 - (d+1)\frac{d-m}{d} = \frac{m}{d} - d$$ where we have used the identity (see 26.3.10 in https://dlmf.nist.gov/26.3) $$(-1)^N \binom{M}{N} = \sum_{k=0}^N (-1)^k \binom{M+1}{k}, \quad 0\le N \le M$$ to get $$\sum_{k=0}^{d-m} (-1)^k\binom{d+1}{k} = (-1)^{d-m}\binom{d}{d-m}$$ and $$\sum_{k=0}^{d-m} k (-1)^k\binom{d+1}{k} = (d+1)\frac{d-m}{d}(-1)^{d-m}\binom{d}{d-m}. \tag{2}$$ (The proof of (2) is given at the end.)

$\phantom{2}$

Proof of Claim 1: We have \begin{align} &\sum_{k=0}^{d-m} P(-m - k + 1)(-1)^k\binom{d+1}{k}\\ =\ & \sum_{k=0}^{d-m} \sum_{i=0}^d a_i\left(\prod_{j=i}^{d+i-1}(-m - k + 1+j)\right)(-1)^k\binom{d+1}{k}\\ =\ & \sum_{i=0}^d a_i \sum_{k=0}^{d-m} \left(\prod_{j=i}^{d+i-1}(-m - k + 1+j)\right)(-1)^k\binom{d+1}{k}\\ =\ & \sum_{i=0}^d a_i A_i \end{align} where $$A_i = \sum_{k=0}^{d-m} \left(\prod_{j=i}^{d+i-1}(-m - k + 1+j)\right)(-1)^k\binom{d+1}{k}.$$ It suffices to prove that $A_i = 0$ for all $i \ne m$.

We split into three cases:

1. $m = d$: For $0\le i < m$, we have $$A_i = \prod_{j=i}^{d+i-1}(-d + 1+j) = 0.$$

2. $m = 0$: For $1\le i\le d$, noting that $\prod_{j=i}^{d+i-1}(-m - k + 1+j) = 0$ for $i + 1 \le k \le d$, we have \begin{align} A_i &= \sum_{k=0}^{d} \left(\prod_{j=i}^{d+i-1}( - k + 1+j)\right)(-1)^k\binom{d+1}{k}\\ &= \sum_{k=0}^i \left(\prod_{j=i}^{d+i-1}( - k + 1+j)\right)(-1)^k\binom{d+1}{k}\\ &= \sum_{k=0}^i \frac{(d+i-k)!}{(i-k)!}(-1)^k\binom{d+1}{k}\\ &= d! \sum_{k=0}^i (-1)^k \binom{d+1}{k} \binom{d+i-k}{i-k}\\ &= 0 \end{align} where we have used the identity (see @arindam mitra's answer:
   Prove combinatorial identity using inclusion/exclusion principle) $$\sum_{k=0}^M (-1)^k \binom{N}{k}\binom{N + r - k}{M - k} = 0, \quad 0 \le r \le M-1$$ to get (let $M = i$, $N = d + 1$, $r = i - 1$) $$\sum_{k=0}^i (-1)^k \binom{d+1}{k} \binom{d+i-k}{i-k} = 0.$$

3. $1 \le m \le d - 1$: If $0\le i < m$, clearly $\prod_{j=i}^{d+i-1}(-m - k + 1+j) = 0$ and hence $A_i = 0$.

If $m < i \le d$, I $\color{blue}{\textrm{GUESS}}$ $A_i = 0$.

Remark: With the help of Maple, $\color{blue}{\textrm{it appears that}}$ $$\sum_{k=0}^{d-m} \Big(\prod_{j=i}^{d+i-1}(-m - k + 1+j)\Big)(-1)^k\binom{d+1}{k} = (-1)^{d-m}\binom{d}{m} \prod_{0\le k \le d, \, k\ne m} (i-k). \tag{2}$$ How to prove it?

$\phantom{2}$

Proof of (2): If $d-m = 0$, it is obvious. If $d-m\ge 1$, we have \begin{align} \sum_{k=0}^{d-m} k (-1)^k\binom{d+1}{k} &= \sum_{k=1}^{d-m} k (-1)^k\binom{d+1}{k}\\ &= (d+1) \sum_{k=1}^{d-m} (-1)^k \binom{d}{k-1}\\ &= -(d+1) \sum_{j=0}^{d-m-1} (-1)^j \binom{d}{j}\\ &= -(d+1)(-1)^{d-m-1}\binom{d-1}{d-m-1}\\ &= (d+1)\frac{d-m}{d}(-1)^{d-m}\binom{d}{d-m}. \end{align} We are done.