Negative arguments of hypergeometric functions

Question

I am writing this question because I get bits and pieces of information from hypergeometric function, but it is a bit too hard to learn from scratch.

The hypergeometric function, $_2F_1(a,b;c;z)$ is defined as

$$ _2F_1(a,b;c;z) = \sum_{i=0}^\infty \frac{(a)_i(b)_i}{(c)_i}\frac{z^i}{i!} $$

Wikipedia says that if $a$ or $b$ is a negative integer, it truncates into

$$ _2F_1(-a,b;c;z) = \sum_{i=0}^a \binom{a}{i}\frac{(b)_i}{(c)_i}z^i, $$

most possibly due to $(a)_n = 0$ if $0>a>-n$. $i!$ is also moved into $\binom{-a}{i}$, because

$$ \binom{a}{i} = \frac{(a)_i}{i!}. $$

My questions are

• Does this extend to all $_PF_Q$s? I see no reason it shouldn't extend at all, because it is just adding extra Pochhammer terms here and there which are all independent of each other. But I want to confirm.

• What happens if both $a$ and $b$ are negative integers? Can I claim

$$ _2F_1(-a,-b;c;z) = \sum_{i=0}^{\min(a,b)} \binom{a}{i}\binom{b}{i} \frac{i!}{(c)_i}{z^i}? $$

Thanks in advance.

Answer 1

I think the confussion is due to the notation.

For Hypergeometric functions the most common notation is to define the rising factorial or Pochhammer polynomial as:

$$(x)_{n} = \prod_{j=0}^{n-1}(j+x) = x(x+1)(x+2)\cdots(x+n-1) = \frac{\Gamma(x+n)}{\Gamma(x)}$$

Note that if $x = -a$ with $a>0$

\begin{align*} (-a)_{n} = (-a)(-a+1)\cdots(-a+n-1) =& (-1)^na(a-1)(a-2)\cdots(a-n+1)\\ =& (-1)^n(a-n+1)_{n} \tag{1} \end{align*}

$(a-n+1)_{n}$ is the falling factorial. Take a lot of care because some authors and websites use the notation $(a)_{n}$ for the falling factorial.

With this definition note that

$$ (a-n+1)_{n} = \frac{\Gamma(a+1)}{\Gamma(a-n+1)} = \frac{a!}{(a-n)!}$$

So the correct definition of the binomial coefficient is

$$ \binom{a}{n} = \frac{a!}{(a-n)!n!} = \frac{(a-n+1)_{n}}{n!}$$

not

$$ \binom{a}{n} = \frac{(a)_{n}}{n!} $$

So if we have the series

$$ {}_pF_q(a_1,...,a_q;b_1,...,b_q;z) = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}\cdots(a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{z^n}{n!}$$

and at least of the $a_{i} = -m$ with $m\in \mathbb{N}$

\begin{align*} {}_pF_q(a_1,...,a_q;b_1,...,b_q;z) = &\sum_{n=0}^{\infty} \frac{(a_{1})_{n}\cdots(-m)_{n}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{z^n}{n!}\\ =& \sum_{n=0}^{\infty} \frac{(a_{1})_{n}\cdots(-1)^n(m-n+1)_{n}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{z^n}{n!} \quad (\textrm{from (1)})\\ =& \sum_{n=0}^{m} \frac{(a_{1})_{n}\cdots(m-n+1)_{n}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{(-z)^n}{n!} \quad ((m-n+1)_{n}=0 \quad \textrm{ if } n>=m+1)\\ =& \sum_{n=0}^{m} \binom{m}{n} \frac{(a_{1})_{n}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} (-z)^n \end{align*}

From this is easy to conclude that if we have two negative parameters $-m,-v$ with $m,v \in \mathbb{N}$

$$\sum_{n=0}^{\infty} \frac{(a_{1})_{n}\cdots(-m)_{n}(-v)_{n}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{z^n}{n!}\\ = \sum_{n=0}^{\min(m,v)} \frac{(a_{1})_{n}\cdots(-m)_{n}(-v)_{m}\cdots (a_p)_{n}}{(b_1)_{n}\cdots (b_q)_{n}} \frac{z^n}{n!}$$

Answer 2

I published something about hypergeometric series with negative integers at both numerator and denominator a few years ago. You may be interested to have a look at it : Hypergeometric series with negative integer at the denominator