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

Question

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.

Answer 1

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

Answer 2

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