# Coefficients of $(x-1)(x-2)\cdots(x-k)$

I'm interested in the coefficients of $x$ in the expansion of,

$$(x-1)(x-2)\cdots(x-k) = x^k + P_1(k) x^{k-1} + P_2(k)x^{k-2} + \cdots + P_k(k),$$

Where $k$ is an integer. In particular I am interested in thinking of these coefficients as polynomials in $k$.

Its not to hard to show that,

$$P_1(k) = -\sum_{i=1}^k i =-k(k-1)/2$$ $$P_2(k) = \sum_{i=2}^k i \sum_{j=1}^{i-1} j = k^4/8 + k^3/12-k^2/8-k/12$$

And I am pretty sure that $$P_n(k) = (-1)^k\sum_{i_1=n}^k i_1 \sum_{i_2=1}^{i_1-1}i_2 \sum_{i_3=1}^{i_2-1}i_3\cdots i_{n-1}\sum_{i_n=1}^{i_{n-1}-1}i_n$$

I haven't gotten around to proving it but it works for $P_1$, $P_2$ and $P_3$ which gives me some confidence in the formula. For the purpose of this question assume that the formula works in general.

The last polynomial is a bit awkward because $P_k(k) = (-1)^kk!$ meaning that the coefficients are heavily dependent upon $k$ and somewhat ill defined. However I am primarily interested in $P_n$ when $n<k$.

My questions are the following,

• Is there a simple formula for the coefficients of $P_n(k)$.
• Is there a tight upper bound $M_k \geq P_n(x)$ for $x=1,2,\ldots,k$ which holds for all $n$.
• I would also be interested in an upper bound on the coefficients if their explicit form is unavailable.

It is the Stirling numbers of the first kind.

By definition they are the coefficients in the expansion

$(x)_n = \sum_{k=0}^n s(n,k) x^k,$

where $(x)_n$ is the falling factorial

$(x)_n = x(x-1)(x-2)\cdots(x-n+1).$

So $P_n(k)=s(n+1,k).$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\sum_{k = 0}^{n}a_{k}x^{k} = \sum_{k = 0}^{n}\left[k!\,a_{k}\right]\,{x^{k} \over k!}$$ Then, $$a_{k} = \left.{1 \over k!}\totald[k]{\pars{\sum_{k = 0}^{n}a_{k}x^{k}}}{x} \right\vert_{x\ =\ 0}$$ In your case, the 'simple formula' you are looking for is given by: $$P_{i} = \left. {1 \over \pars{k - i}!}\totald[k - i]{\bracks{\pars{x - 1}\ldots\pars{x - k}}}{x} \right\vert_{x\ =\ 0}$$