How to prove Vandermonde's identity in complex numbers without using the binomial theorem?

Question

This exercise came from Complex Analysis by Freitag and Busam 1.2.11.

$$\binom{a}{n}:= \prod_{j=1}^n\frac{a-j+1}{j} $$ Show: $\sum\limits_{\nu=0}^{\infty}\dbinom{\alpha}{\nu} z^\nu$ is absolutely convergent for all $z \in \mathbb{C}$ with $|z|<1$. Let $b_\alpha(z):=\sum\limits_{\nu=0}^{\infty}\dbinom{\alpha}{\nu} z^\nu$. Show: For all $z \in \mathbb{C}$ with $|z|<1$ and arbitrary $\alpha, \beta \in \mathbb{C}$ we have $$ b_{\alpha+\beta}(z)=b_\alpha(z) b_\beta(z) . $$Remark. We shall see later that for $z \in \mathbb{C}$ with $|z|<1$ there holds $$ b_\alpha(z)=(1+z)^\alpha:=\exp (\alpha \log (1+z)) . $$ For $\alpha=n \in \mathbb{N}_0$, one obtains the binomial formula $$ (1+z)^n=\sum_{\nu=0}^n\binom{n}{\nu} z^\nu $$

I think most people will prove $\sum\limits_{k=0}^n \binom{a}{k}\binom {b}{n-k}= \binom{a+b}{n}$ by using the coefficient of $(1+z)^a (1+z)^b $ but my book give this exercise before mentioning this fact as stated "Remark. We shall see later that for $z \in \mathbb{C}$ with $|z|<1$ there holds $ b_\alpha(z)=(1+z)^\alpha:=\exp (\alpha \log (1+z)) . $" (so the book asks for different approach).

I wasn't successful in finding anything useful after a lot if time, so I decided to ask here.

This question is in the second section of the book so Complex integration isn't introduced yet or any other advanced tools.

The hint given by the book is use induction which I have no idea how to do so.

Answer 1

If you want a combinatorical proof: On the right hand Side you choose $n$ elemnts out of $a + b$ Elements. You can visualize a set of cardinality $a+b$ as a union of a set $A$ of cardinality $a$ and a set $B$ of cardinality $b$ which are disjunct.

Now you can count how many of the $n$ elements where in set $A$ and you must chose the missing ones out of set $B$, or in formulas if you choose $k$ Elements of $A$ there are $\binom{a}{k}$ possibilites to do so. To get $n$ elements you have to chose the missing $n-k$ out of $B$ and there are $\binom{b}{n-k}$ possible ways to do so.

Together we have $$\sum_{k=0}^n \binom{a}{k} \binom{b}{n-k} = \binom{a+b}{n}$$ because if you chose $n$ elements of $A \uplus B$ there will be somewhere between $0$ to $n$ elements out of $A$ and it is unique for every choice, meaning you can have 2 elements out of $A$ and 5 elements out of $A$ at the same time.

Answer 2

Here is one possible proof outline.

1. Think of $\binom{x+b}{n}$ as a function of $x$. Prove that this is a polynomial function, and the degree of the polynomial is $n$.

2. Now, consider the list of polynomial functions $\binom{x}{0},\binom{x}1,\dots,\binom{x}n$. Prove that this list is a basis for the vector space of polynomials in $x$ with degree at most $n$.

3. Combining points $1$ and $2$, we know that there exist coefficients $c_0,c_1,\dots,c_n$ such that $$\binom{x+b}{n}=\sum_{k=0}^n c_k\binom{x}k\tag1$$As long as we can prove $c_k=\binom{b}{n-k}$, we will be done. We prove that this is true by induction on $n$. The base case $n=0$ is straightforward, so let us take $n\ge 1$, assume this is true for $n-1$, and use that to prove it for $n$.

4. Let $\Delta$ be the operator which given a polynomial $f$ in the variable $x$, returns the polynomial $\Delta f$, defined by $(\Delta f)(x)=f(x+1)-f(x)$. Prove that $$ \Delta \binom{x}{n}=\binom x{n-1} $$ This is really just Pascal's identity, $\binom{x+1}{n}=\binom xn+\binom x{n-1}$, which can be proven algebraically by manipulating the definition $\binom xn=\prod_{j=1}^n \frac{x-j+1}j$.

5. Now, apply $\Delta$ to both sides of $(1)$. Using point $4$, the result is$$\binom{x+b}{n-1}=\sum_{k=1}^nc_k \binom{x}{k-1}=\sum_{k=0}^{n-1}c_{k+1}\binom{x}{k}$$Now, applying the induction hypothesis, we know that $c_{k+1}=\binom{b}{n-1-k}$ for each $k\in \{0,1,\dots,n-1\}$. This means that $c_k=\binom{b}{n-k}$ for each $k\in \{1,\dots,n\}$, so we have proved $$ \binom{x+b}n=c_0\binom x0+\sum_{k=1}^n \binom{b}{n-k}\binom{x}{k}\tag2 $$ We will be done once we prove $c_0=\binom{b}n$. This follows by subsituting $x\gets 0$ in $(2)$.