Binomial coefficient - upper factorials

Question

Could you help me proof the formula with upper factorials: $$ \left(a+b\right)^{\left(n\right)}=\sum_{k=0}^{n}{\binom{n}{k}a^{\left(n-k\right)}b^{\left(k\right)}} $$ I am able to prove an analogous equation with ordinary coefficient binomial, but I have problems with this. Obviously I tried simple induction, but I failed.

Answer 1

Another argument is via generating functions.

$$\frac1{(1-x)^a}=\sum_{n\ge 0}\binom{a+n-1}nx^n=\sum_{n\ge 0}\frac{a^{\overline{n}}}{n!}x^n\;,$$

so

$$\begin{align*} \sum_{n\ge 0}\frac{(a+b)^{\overline{n}}}{n!}x^n&=\frac1{(1-x)^{a+b}}\\ &=\frac1{(1-x)^a}\cdot\frac1{(1-x)^b}\\ &=\left(\sum_{n\ge 0}\frac{a^{\overline{n}}}{n!}x^n\right)\left(\sum_{n\ge 0}\frac{b^{\overline{n}}}{n!}x^n\right)\\ &=\sum_{n\ge 0}\left(\sum_{k=0}^n\frac{a^{\overline{n-k}}}{(n-k)!}\cdot\frac{b^{\overline{k}}}{k!}\right)x^n\\ &=\sum_{n\ge 0}\frac1{n!}\sum_{k=0}^n\binom{n}ka^{\overline{n-k}}b^{\overline{k}}x^n\;, \end{align*}$$

and equating coefficients of $x^n$ yields the result.

Answer 2

Note that the upper factorial is defined as : $$ \left(\forall n\in\mathbb{N}^{*}\right)\left(\forall x\in\mathbb{R}\right),\ x^{\left(n\right)}=\prod_{k=0}^{n-1}{\left(x+k\right)} $$

So what we'll need to do is to prove by induction that : $$ \left(\forall k\in\mathbb{N}^{*}\right),\ \prod_{j=0}^{k-1}{\left(a+b+j\right)}=\sum_{i=0}^{k}{\binom{k}{i}\prod_{j=0}^{n-i-1}{\left(a+j\right)}\prod_{j=0}^{i-1}{\left(b+j\right)}} $$

The statement holds for $k=1\cdot$

Let $ n $ be a positive integer, assuming that the statement holds for $k=n$, let's prove that the statement holds for $k=n+1 $ :

\begin{aligned} \small\prod_{k=0}^{n}{\left(a+b+k\right)}&\small=\left(a+b+n\right)\prod_{k=0}^{n-1}{\left(a+b+k\right)}\\ &\small=\left(a+b+n\right)\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{n-k-1}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}\\ &\small=\sum_{k=0}^{n}{\binom{n}{k}\left(a+n-k\right)\prod_{i=0}^{n-k-1}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\sum_{k=0}^{n}{\binom{n}{k}\left(b+k\right)\prod_{i=0}^{n-k-1}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}\\ &\small=\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{n-k-1}{\left(a+i\right)}\prod_{i=0}^{k}{\left(b+i\right)}} \\ &\small=\prod_{i=0}^{n}{\left(a+i\right)}+\sum_{k=1}^{n}{\binom{n}{k}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\sum_{k=0}^{n-1}{\binom{n}{k}\prod_{i=0}^{n-k-1}{\left(a+i\right)}\prod_{i=0}^{k}{\left(b+i\right)}}+\prod_{i=0}^{n}{\left(b+i\right)}\\ &\small=\prod_{i=0}^{n}{\left(a+i\right)}+\sum_{k=1}^{n}{\binom{n}{k}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\sum_{k=1}^{n}{\binom{n}{k-1}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\prod_{i=0}^{n}{\left(b+i\right)}\\ &\small=\prod_{i=0}^{n}{\left(a+i\right)}+\sum_{k=1}^{n}{\left[\binom{n}{k}+\binom{n}{k-1}\right]\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\prod_{i=0}^{n}{\left(b+i\right)}\\ &\small=\prod_{i=0}^{n}{\left(a+i\right)}+\sum_{k=1}^{n}{\binom{n+1}{k}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}+\prod_{i=0}^{n}{\left(b+i\right)}\\ \small\prod_{k=0}^{n}{\left(a+b+k\right)}&\small=\sum_{k=0}^{n+1}{\binom{n+1}{k}\prod_{i=0}^{n-k}{\left(a+i\right)}\prod_{i=0}^{k-1}{\left(b+i\right)}}\end{aligned}

Hence, the statement holds for every natural number $k\geq 1 \cdot$