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2022-11-09-sum-coef-factorial.md
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2022-11-09-sum-coef-factorial.md
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---
title: A polynomial whose sum of coefficients is a factorial
date: 2022-11-09 10:04:43 -0800
categories:
- math
tags:
- combinatorics
- number sequence
- from zhihu
layout: post
excerpt: 'The function $\left(1-z\right)^{n+1}\sum_{k=1}^\infty k^nz^k$ is a polynomial of degree $n$ w.r.t. $z$,
and the sum of its coefficients is $n!$.
This turns out to be properties of Eulerian numbers.'
---
*This article is translated from a
Chinese [article](https://zhuanlan.zhihu.com/p/73079049) on my Zhihu account.
The original article was posted at 2019-07-13 20:36 +0800.*
---
**Definition 1**. Let
$$f_n\!\left(z\right)\coloneqq\left(1-z\right)^{n+1}\sum_{k=1}^\infty k^nz^k,$$
where $n$ is a positive integer.
Our goal is to prove that $f_n\!\left(z\right)$ is a polynomial of degree $n$ w.r.t. $z$,
and the sum of its coefficients is $n!$.
**Lemma 1**.
$$f_{n+1}\!\left(z\right)=z\,\left(1-z\right)^{n+2}\frac{\mathrm d}{\mathrm dz}\left(\frac{f_n\!\left(z\right)}{\left(1-z\right)^{n+1}}\right).$$
*Proof.*
$$\begin{align*}
\frac{\mathrm d}{\mathrm dz}\left(\frac{f_n\!\left(z\right)}{\left(1-z\right)^{n+1}}\right)
&=\frac{\mathrm d}{\mathrm dz}\sum_{k=1}^\infty k^nz^k\\
&=\sum_{k=1}^\infty k^{n+1}z^{k-1}\\
&=\frac{f_{n+1}\!\left(z\right)}{z\,\left(1-z\right)^{n+2}}.
\end{align*}$$
{% qed %}
**Definition 2** (Eulerian numbers).
$$\left<\begin{matrix}n\\k\end{matrix}\right>\coloneqq\sum_{j=0}^{k+1}\left(-1\right)^j\binom{n+1}j\left(k-j+1\right)^n.$$
**Lemma 2**.
$$\left<\begin{matrix}n+1\\k+1\end{matrix}\right>=\left(n-k\right)\left<\begin{matrix}n\\k\end{matrix}\right>+\left(k+2\right)\left<\begin{matrix}n\\k+1\end{matrix}\right>.$$
*Proof.*
$$\begin{align*}
&\phantom{=}~\,\left(n-k\right)\left<\begin{matrix}n\\k\end{matrix}\right>+\left(k+2\right)\left<\begin{matrix}n\\k+1\end{matrix}\right>\\
&=\left(n-k\right)\sum_{j=0}^{k+1}\left(-1\right)^j\binom{n+1}j\left(k-j+1\right)^n+\left(k+2\right)\sum_{j=0}^{k+2}\left(-1\right)^j\binom{n+1}j\left(k-j+2\right)^n\\
&=\left(n-k\right)\sum_{j=0}^{k+2}\left(-1\right)^{j-1}\binom{n+1}{j-1}\left(k-j+2\right)^n+\left(k+2\right)\sum_{j=0}^{k+2}\left(-1\right)^j\binom{n+1}j\left(k-j+2\right)^n\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\left(k-j+2\right)^n\left(\left(k-n\right)\binom{n+1}{j-1}+\left(k+2\right)\binom{n+1}j\right)\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\left(k-j+2\right)^n\left(\left(k-n\right)\frac{\left(n+1\right)!}{\left(j-1\right)!\left(n-j+2\right)!}+\left(k+2\right)\frac{\left(n+1\right)!}{j!\left(n-j+1\right)!}\right)\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\left(k-j+2\right)^n\frac{\left(n+1\right)!}{j!\left(n-j+2\right)}\left(\left(k-n\right)j+\left(k+2\right)\left(n-j+2\right)\right)\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\left(k-j+2\right)^n\frac{\left(n+1\right)!}{j!\left(n-j+2\right)}\left(n+2\right)\left(k-j+2\right)\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\left(k-j+2\right)^{n+1}\frac{\left(n+2\right)!}{j!\left(n-j+2\right)}\\
&=\sum_{j=0}^{k+2}\left(-1\right)^j\binom{n+2}j\left(k-j+2\right)^{n+1}\\
&=\left<\begin{matrix}n+1\\k+1\end{matrix}\right>.
\end{align*}$$
{% qed %}
**Lemma 3**.
$$\left<\begin{matrix}n\\0\end{matrix}\right>=1,\quad\left<\begin{matrix}n\\n\end{matrix}\right>=0.$$
*Brief proof.*
Easily proved by Definition 2. {% qed %}
**Lemma 4**.
$$f_n\!\left(z\right)=\sum_{k=1}^n\left<\begin{matrix}n\\n-k\end{matrix}\right>z^k.$$
*Proof.*
By mathematical induction.
When $n=1$,
$$f_n\left(z\right)=z=\sum_{k=1}^n\left<\begin{matrix}n\\n-k\end{matrix}\right>z^k,$$
the result stands.
Suppose the result stands when $n=n_0$, and then when $n=n_0+1$,
$$\begin{align*}
&\phantom{=}~\,f_n\!\left(z\right)\\
&=z\,\left(1-z\right)^{n_0+2}\frac{\mathrm d}{\mathrm dz}\left(\frac{f_{n_0}\left(z\right)}{\left(1-z\right)^{n_0+1}}\right)
&\text{(Lemma 1)}\\
&=z\,\left(1-z\right)^{n_0+2}\frac{\frac{\mathrm df_{n_0}\left(z\right)}{\mathrm dz}\left(1-z\right)^{n_0+1}-f_{n_0}\!\left(z\right)\frac{\mathrm d\left(\left(1-z\right)^{n_0+1}\right)}{\mathrm dz}}{\left(1-z\right)^{2n_0+2}}\\
&=z\,\left(1-z\right)^{n_0+2}\frac{\frac{\mathrm df_{n_0}\left(z\right)}{\mathrm dz}\left(1-z\right)^{n_0+1}+\left(n_0+1\right)f_{n_0}\!\left(z\right)\left(1-z\right)^{n_0}}{\left(1-z\right)^{2n_0+2}}\\
&=z\left(\frac{\mathrm df_{n_0}\!\left(z\right)}{\mathrm dz}\left(1-z\right)+\left(n_0+1\right)f_{n_0}\left(z\right)\right)\\
&=z\left(\left(1-z\right)\frac{\mathrm d}{\mathrm dz}\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k+\left(n_0+1\right)\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k\right)
&\text{(supposed)}\\
&=z\left(\left(1-z\right)\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>kz^{k-1}+\left(n_0+1\right)\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k\right)\\
&=z\left(\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>kz^{k-1}-\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>kz^k+\left(n_0+1\right)\sum_{k=1}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k\right)\\
&=z\left(\sum_{k=0}^{n_0}\left<\begin{matrix}n_0\\n_0-k-1\end{matrix}\right>\left(k+1\right)z^k-\sum_{k=0}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>kz^k+\left(n_0+1\right)\sum_{k=0}^{n_0}\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k\right)
&\text{(Lemma 3)}\\
&=z\sum_{k=0}^{n_0}\left(\left<\begin{matrix}n_0\\n_0-k-1\end{matrix}\right>\left(k+1\right)z^k-\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>kz^k+\left(n_0+1\right)\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>z^k\right)\\
&=\sum_{k=0}^{n_0}\left(\left<\begin{matrix}n_0\\n_0-k-1\end{matrix}\right>\left(k+1\right)-\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>k+\left(n_0+1\right)\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>\right)z^{k+1}\\
&=\sum_{k=1}^{n_0+1}\left(k\left<\begin{matrix}n_0\\n_0-k\end{matrix}\right>+\left(n_0-k+2\right)\left<\begin{matrix}n_0\\n_0-k+1\end{matrix}\right>\right)z^k\\
&=\sum_{k=1}^{n_0+1}\left<\begin{matrix}n_0+1\\n_0-k+1\end{matrix}\right>z^k
&\text{(Lemma 2)}\\
&=\sum_{k=1}^n\left<\begin{matrix}n\\n-k\end{matrix}\right>z^k.
\end{align*}$$
Then we can derive that the result is true by mathematical induction. {% qed %}
**Lemma 5**.
$$\sum_{j=0}^n\left(-1\right)^{n-j}\binom njj^n=n!.$$
*Proof.*
Because $\mathrm e^x-1\sim x$ (in terms of infinitesimal quantity),
$\left(\mathrm e^x-1\right)^n\sim x^n$, i.e.
$$\left(\mathrm e^x-1\right)^n=x^n+o\!\left(x^n\right)$$
(where $o$ denotes higher order of infinitesimal quantity).
Apply Newton's binomial theorem to the left-hand side, and we have
$$\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\mathrm e^{jx}=x^n+o\!\left(x^n\right).$$
Take $n$th derivative of the equation, and we have
$$\sum_{j=0}^n\left(-1\right)^{n-j}\binom njj^n\mathrm e^{jx}=n!+o\!\left(1\right).$$
Take $x=0$, and we have
$$\sum_{j=0}^n\left(-1\right)^{n-j}\binom njj^n=n!.$$
{% qed %}
**Lemma 6**.
$$\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\left(j+1\right)^n=n!.$$
*Proof.*
$$\begin{align*}
\left(n+1\right)!&=\sum_{j=0}^n\left(-1\right)^{n-j+1}\binom{n+1}jj^{n+1}&
\text{(Lemma 5)}\\
&=\sum_{j=1}^n\left(-1\right)^{n-j+1}\binom{n+1}jj^{n+1}\\
&=\sum_{j=1}^n\left(-1\right)^{n-j+1}\frac{\left(n+1\right)!}{j!\left(n-j+1\right)!}j^{n+1}
\\
&=\sum_{j=1}^n\left(-1\right)^{n-j+1}\frac{\left(n+1\right)n!}{\left(j-1\right)!\left(n-j+1\right)!}j^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\frac{\left(n+1\right)n!}{j!\left(n-j\right)!}\left(j+1\right)^n\\
&=\left(n+1\right)\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\left(j+1\right)^n.
\end{align*}$$
{% qed %}
**Lemma 7**.
$$\sum_{k=0}^n\left<\begin{matrix}n\\k\end{matrix}\right>=n!.$$
*Proof.*
$$\begin{align*}
&\phantom{=}~\,\sum_{k=0}^n\left<\begin{matrix}n\\k\end{matrix}\right>\\
&=\sum_{k=0}^n\sum_{j=0}^{k+1}\left(-1\right)^j\binom{n+1}j\left(k-j+1\right)^n\\
&=\sum_{k=0}^n\sum_{j=0}^k\left(-1\right)^j\binom{n+1}j\left(k-j+1\right)^n\\
&=\sum_{j=0}^n\sum_{k=j}^n\left(-1\right)^j\binom{n+1}j\left(k-j+1\right)^n\\
&=\sum_{j=0}^n\left(-1\right)^j\binom{n+1}j\sum_{k=j}^n\left(k-j+1\right)^n\\
&=\sum_{j=0}^n\left(-1\right)^j\binom{n+1}j\sum_{k=1}^{n-j+1}k^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\binom{n+1}{n-j}\sum_{k=1}^{j+1}k^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\binom{n+1}{j+1}\sum_{k=1}^{j+1}k^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\left(\binom nj+\binom n{j+1}\right)\sum_{k=1}^{j+1}k^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\sum_{k=1}^{j+1}k^n+\sum_{j=0}^n\left(-1\right)^{n-j}\binom n{j+1}\sum_{k=1}^{j+1}k^n\\
&=\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\left(j+1\right)^n+\sum_{j=0}^n\left(-1\right)^{n-j}\binom nj\sum_{k=1}^jk^n+\sum_{j=1}^n\left(-1\right)^{n-j+1}\binom nj\sum_{k=1}^jk^n\\
&=n!+\sum_{j=1}^n\left(-1\right)^{n-j}\binom nj\sum_{k=1}^jk^n-\sum_{j=1}^n\left(-1\right)^{n-j}\binom nj\sum_{k=1}^jk^n
&\text{(Lemma 6)}\\
&=n!.
\end{align*}$$
{% qed %}
*Proof of the original proposition.*
By Lemma 4, $f_n\!\left(z\right)$ is a polynomial of degree $n$ in $z$
(Lemma 3 guarantees that the coefficient of the $n$th degree term is not $0$).
The sum of coefficients
$$\begin{align*}
\sum_{k=1}^n\left<\begin{matrix}n\\n-k\end{matrix}\right>
&=\sum_{k=0}^{n-1}\left<\begin{matrix}n\\k\end{matrix}\right>\\
&=\sum_{k=0}^n\left<\begin{matrix}n\\k\end{matrix}\right>
&\text{(Lemma 3)}\\
&=n!.
&\text{(Lemma 7)}
\end{align*}$$
{% qed %}