Generalization of identity for terminating hypergeometric function

Question

Let ${}_2F_1(a,b;c;z)$ be the ordinary hypergeometric function for $z \in \mathbb{C}$ \begin{equation} {}_2F_1(a,b;c;z) = \sum_{k=0}^{\infty} \frac{z^k}{k!} \frac{(a)_{k} (b)_k}{(c)_k}\,, \end{equation} where the coefficients $(a)_k, (b)_k,(b)_k$ are the usual Pochhammer symbols or rising factorials. For the special terminating case of $a = -n$ for $n \in \mathbb{N}$, this series reduces to a polynomial. Let now $n \in \mathbb{N}$, $c,1+b-c-n \neq 0,-1,\dots,-n+1$, then this review gives the identity \begin{equation} {}_2F_1(-n,b;c;1-z) = \frac{(c-b)_{n}}{(c)_n} {}_2F_1(-n,b;1+b-c-n;z)\,. \end{equation} They mention that this can be proven by induction. My question now is if there is a generalization of this identity to more general terminating hypergeometric functions like the generalized hypergeometric function ${}_pF_q$ as it is given here (or of course other functions)? Or maybe anyone sees how the above can be proven without induction, such that I can maybe try to generalize it with the help of it? Thank you in advance.

Answer 1

I assume that the reason you want a non-inductive proof is that induction requires you to know your goal before you start, but there's actually a simple way to find out what the result could be. Assume that the general identity is of the form

$$F(-n,\vec{b};\vec{c};1-z) = \alpha_n(\vec{b},\vec{c}) F(-n,\vec{u};\vec{v};z)$$ where $\alpha_n$ is independent of $z$. Then we have an identity of polynomials, which is just an identity of coefficients. Expanding

$$\sum_{k=0}^n \frac{(-n)_k \prod_{b_i \in \vec{b}}(b_i)_k}{\prod_{c_i \in \vec{c}}(c_i)_k} \frac{(1-z)^k}{k!} = \alpha_n(\vec{b},\vec{c}) \sum_{j=0}^n \frac{(-n)_j \prod_{u_i \in \vec{u}}(u_i)_j}{\prod_{v_i \in \vec{v}}(v_i)_j} \frac{z^j}{j!}$$

we get that for $0 \le j \le n$,

$$(-1)^j \sum_{k=j}^n \frac{(-n+j)_{k-j} \prod_{b_i \in \vec{b}}(b_i)_k}{(k-j)! \prod_{c_i \in \vec{c}}(c_i)_k} = \alpha_n(\vec{b},\vec{c}) \frac{\prod_{u_i \in \vec{u}}(u_i)_j}{\prod_{v_i \in \vec{v}}(v_i)_j}$$

From the case $j=n$ we derive

$$\alpha_n(\vec{b},\vec{c}) = (-1)^n \frac{\prod_{b_i \in \vec{b}}(b_i)_n}{\prod_{c_i \in \vec{c}}(c_i)_n} \frac{\prod_{v_i \in \vec{v}}(v_i)_n}{\prod_{u_i \in \vec{u}}(u_i)_n}$$

From the case $j=n-1$ we derive

$$(-1)^{n-1} \frac{\prod_{b_i \in \vec{b}}(b_i)_{n-1}}{\prod_{c_i \in \vec{c}}(c_i)_{n-1}} \left(1 - \frac{\prod_{b_i \in \vec{b}} (b_i+n-1)}{\prod_{c_i \in \vec{c}} (c_i+n-1)} \right) = \alpha_n(\vec{b},\vec{c}) \frac{\prod_{u_i \in \vec{u}}(u_i)_{n-1}}{\prod_{v_i \in \vec{v}}(v_i)_{n-1}}$$

Putting them together,

$$\frac{\prod_{v_i \in \vec{v}} (v_i+n-1)}{\prod_{u_i \in \vec{u}} (u_i+n-1)} = 1 - \frac{\prod_{c_i \in \vec{c}} (c_i+n-1)}{\prod_{b_i \in \vec{b}} (b_i+n-1)}$$

In the case that $\vec{b} = b$ and $\vec{c} = c$, the RHS is simply $1 - \frac{c+n-1}{b+n-1} = \frac{b-c}{b+n-1}$ and identifying $v+n-1 = b-c$, $u+n-1=b+n-1$ gives the parameters of the quoted theorem. If we have e.g. $\vec{b} = (b_1,b_2)$, $\vec{c} = (c_1,c_2)$ then the RHS is $$\frac{(b_1+n-1)(b_2+n-1)-(c_1+n-1)(c_2+n-1)}{(b_1+n-1)(b_2+n-1)} = \frac{(b_1 + b_2 - c_1 - c_2)n + (b_1-1)(b_2-1) - (c_1-1)(c_2-1)}{(b_1+n-1)(b_2+n-1)}$$

which is suggestive that the most general identity possible requires $b_1 + b_2 - c_1 - c_2 = 1$. I haven't considered also the case $j=n-2$, but it would add a further constraint which would probably tell you exactly what is worth trying to prove. In any case, it seems that $\vec{u} = \vec{b}$.