A closed form for the sum $\sum_{i=0}^j (-1)^i \binom{j}{i} (j+c-i)^k$

Question

Let $c\in \mathbb{Q}$ be a constant, and let $k\in \mathbb{N}$ be fixed.

Is there a closed form for the sum $$ L_j := \sum_{i=0}^j (-1)^i \binom{j}{i} (j+c-i)^k, $$ for $j \leq k$ ?

Answer 1

Here is an attempt. Note that $$L_j=\sum_{i=0}^j(-1)^i\binom{j}{i}\sum_{r=0}^k\binom{k}{r}c^{k-r}(j-i)^{r}.$$ Therefore, $$L_j=\sum_{r=0}^kc^{k-r}\binom{k}{r}\sum_{i=0}^j(-1)^i\binom{j}{i}(j-i)^r.$$ Let $s=j-i$. Then, $$L_j=\sum_{r=0}^kc^{k-r}\binom{k}{r}\sum_{s=0}^j(-1)^{j-s}\binom{j}{s}s^r.$$ Note that $\displaystyle\frac{1}{j!}\sum_{s=0}^j(-1)^{j-s}\binom{j}{s}s^r$ is the Stirling number $\begin{Bmatrix}r\\j\end{Bmatrix}$. So, $$L_j=j!\sum_{r=0}^k\binom{k}r\begin{Bmatrix}r\\j\end{Bmatrix}c^{k-r}=j!\sum_{r=j}^k\binom{k}r\begin{Bmatrix}r\\j\end{Bmatrix}c^{k-r}.$$ One particular result is when $j=k$, where we have $L_k=k!$. Also, $L_j=0$ if $j>k$. For $j\leq k$, $L_j$ is a polynomial in $c$ with non-negative integer coefficients of degree $k-j$. Some other trivial answers are $L_0=c^k$ (for $k\geq 0$, and $L_1=(c+1)^k-c^k$ (for $k\geq 1$).

Answer 2

You can assume that there is no "closed" form.

With a Generating function:

$\displaystyle L_{k,j}(x):=\frac{d^k}{dz^k}e^{xz}\left(e^z-1\right)^j |_{z=0}=\sum\limits_{i=0}^j (-1)^{j-i}{\binom j i}(i+x)^k = {\Delta}^j x^k \enspace$

$\hspace{7cm}$ ($j^{\,\text{th}}$ forward difference of $x^k$)

$\Rightarrow \enspace L_j = L_{k,j}(c)$

Answer 3

A closed form seems to be out of reach for general $c\in\mathbb{Q}$. We derive an expression which makes this plausible. We use the coefficient of operator $[z^k]$ to denote the coeffcient of $z^k$ in a series. This way we can write e.g. \begin{align*} n^k=k![z^k]e^{nz}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{i=0}^j}&\color{blue}{(-1)^i\binom{j}{i}(j+c-i)^k}\\ &=\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}(c+i)^k\tag{2}\\ &=\sum_{i=0}^j(-1)^{i+j}\binom{j}{i}k![z^k]e^{(c+i)z}\tag{3}\\ &=(-1)^jk![z^k]e^{cz}\sum_{i=0}^j\binom{j}{i}(-e^z)^i\tag{4}\\ &=(-1)^jk![z^k]e^{cz}\left(1-e^z\right)^j\tag{5}\\ &\color{blue}{=k![z^k]e^{cz}\left(e^z-1\right)^j}\tag{6} \end{align*} We conclude from the representation (6) that the expression $e^{cz}\left(e^z-1\right)^j$ cannot be simplified for general $c\in\mathbb{Q}$. Taking the coefficient $[z^k]$ from it will typically lead to more than one term, because we have to do a Cauchy multiplication of the corresponding series. So, we won't get a closed form in general.

Comment:

• In (2) we exchange the order of summation $i \to j-i$.

• In (3) we apply the coefficient of operator as we did in (1).

• In (4) we do some rearrangements as preparation for the next step.

• In (5) we apply the binomial theorem.

• In (6) we do some final simplifications.