Umbral calculus/Pochhammer – invert falling factorial of binomial in term of falling factorial of monomial

Question

Consider the variables $x,n \in \mathbb{Z}^+$ and define the following falling factorial operator:

\begin{equation}
L[x^n] = (x)_n = \prod_{k=0}^{n-1}(n-k)
\end{equation}

now from consider the following function with $c \in \mathbb{R}^+$

\begin{equation}
q_n(x)=L[(x-c)^n] = \sum_{m=0}^{n}{n \choose m}(-c)^{m}(x)_{n-m}
\end{equation}

I have found but not understood the following expression:

\begin{equation}
(x)_m = \sum_{n=0}^m {m \choose n} c^{m-n}q_n(x)
\end{equation}

In other word the falling factorial is inverted in terms of the falling factorial of a function of a binomial expansion.

I have verified the expression handily, and it is true, but I'm unable to understand how this has been computed.

Does anybody know how I could proceed?

Answer 1

The stated formulas need to be somewhat revised. Here we use D. E. Knuth's notation $x^{\underline{n}}=x(x-1)\cdots(x-n+1)$ to denote falling factorials.

We consider the operator $L[x^n]=x^{\underline{n}}$. We have for $c\in \mathbb{R}^+$: \begin{align*} L[(x-c)^n] = (x-c)^{\underline{n}}=\sum_{m=0}^{n}{n \choose m}(-c)^{\underline{m}}\,x^{\underline{n-m}}\tag{1} \end{align*} which can be found for instance in section 2.6 of The Umbral Calculus by Steven Roman.

Note that in (1) we have the factor $(-c)^{\underline{m}}$ and not $(-c)^m$ which is stated in the definition of $q_n(x)$ by OP. A plausibility check $n=2$ results in \begin{align*} (x-c)^{\underline{2}}&=(x-c)(x-c-1)\\ &=\color{blue}{x^2-(2c+1)x+c^2+c}\\ \sum_{m=0}^{n}{n \choose m}(-c)^{\underline{m}}\,x^{\underline{n-m}} &=\binom{2}{0}(-c)^{\underline{0}}x^{\underline{2}}+\binom{2}{1}(-c)^{\underline{1}}x^{\underline{1}} +\binom{2}{2}(-c)^{\underline{2}}x^{\underline{0}}\\ &=1\cdot1\cdot x(x-1)+2(-c)x+1\cdot c^2\cdot 1\\ &=\color{blue}{x^2-(2c+1)x+c^2+c}\\ \end{align*} showing equality of (1) in case of $n=2$.

Substituting $c$ with $-c$ in (1) we obtain \begin{align*} (x+c)^{\underline{n}}=\sum_{m=0}^{n}{n \choose m}c^{\underline{m}}\,x^{\underline{n-m}}\tag{2} \end{align*} and substituting $x$ with $x-c$ in (2) we obtain \begin{align*} \color{blue}{x^{\underline{n}}} &=\sum_{m=0}^{n}{n \choose m}c^{\underline{m}}\,(x-c)^{\underline{n-m}}\\ &\,\,\color{blue}{=\sum_{m=0}^{n}{n \choose m}c^{\underline{n-m}}\,(x-c)^{\underline{m}}}\tag{3}\\ \end{align*} which is OP's stated formula with $c^{n-m}$ replaced with $c^{\underline{n-m}}$.

In (3) we have changed the order of summation by replacing $m$ with $n-m$.