# 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: 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).

• Where is this question from? – darij grinberg Sep 1 at 15:16
• @darijgrinberg this arose from a geometry problem I’ve been thinking about for a while, it’s not from any source (at least not that I know of!) – Romain S Sep 1 at 15:19
• I'm afraid I don't have anything overly nice either, but maybe you can improve on it. My ansatz is that $a_i = \left(-1\right)^i \left(\lambda \dbinom{d}{i} + \mu \dbinom{d-1}{i}\right)$ for all $i$, where $\lambda$ and $\mu$ are two constants that need to be determined. Indeed, this is the general form of $a_i$ for which $P\left(x\right)$ is linear in $x$. (This follows from some linear algebra and finite difference yoga -- let me know if you need more details.) Now you should be able to find $\lambda$ and $\mu$ from the conditions $a_m = 0$ and $P\left(x\right) = c-x$. – darij grinberg Sep 1 at 15:24
• What is the geometry problem, though? – darij grinberg Sep 1 at 15:24
• @darijgrinberg If you want to write that up, if it is a complete solution, I will accept it since I doubt there is a much sleeker way to do it... This problem actually came up when considering the intersection of planes along a curve, but the source of the problem is not helpful to solving this, in this case, unfortunately... – Romain S Sep 1 at 16:53

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.

• (+1) Your (4) is verified by Maple. – River Li Sep 5 at 14:22
• Thanks for writing this up! I've made some simplifications and corrections. – darij grinberg Sep 5 at 23:00

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.