Hypergeometric functions: why is $\frac{P(n)}{Q(n)} = \frac{(n+a_1)(n+a_2)...(n+a_p)}{(n+b_1)(n+b_2)...(n+b_q)(n+1)}$

Question

$$\frac{a_{n+1}}{a_n} =\frac{P(n)}{Q(n)} = \frac{(n+a_1)(n+a_2)...(n+a_p)}{(n+b_1)(n+b_2)...(n+b_q)(n+1)}$$

So, this is what I think I understand thus far:

A geometric series has a constant ratio between every term. It is thus a special case of the hypergeometric series, in which there is a constant ratio between the functions applied to every term. So, the terms themselves don't necessarily have a constant ratio (if they do, it's a geometric series), but when the terms have been taken through their respective functions, the outputs have a constant ratio.

How this fact however, leads to the rightmost expression however, is beyond me. Also, another confusing bit is that $(n+1)$ is included for historical reasons, and that if $Q(n)$ doesn't include $(n+1)$, then $a_p = 1$, so as to not lose generality. I don't get this either.

Answer 1

I'm not sure what "a constant ratio between the functions applied to every term" means, but one possible definition of a hypergeometric function is that it's a series $\displaystyle \sum_{n \ge 0} c_n z^n$ where $\frac{c_{n+1}}{c_n}$ is a rational function of $n$. Often we'll assume $c_0 = 1$, but that doesn't really matter.

Being a rational function means we can write as a ratio of two polynomials; you can make it $\frac{c_{n+1}}{c_n} = \frac{P(n)}{Q(n)}$, or $\frac{c_{n+1}}{c_n} = \frac{P(n+1)}{Q(n)}$, it doesn't really matter. (If $P(n)$ is a polynomial, then $P(n+1)$ is a polynomial, and vice versa).

For the "traditional form" of the hypergeometric function - we'll get to why it is traditional in a bit - let's assume that $\frac{c_{n+1}}{c_n} = \frac{P(n)}{Q(n)(n+1)}$. How do we know this exists? Well, if $\frac{c_{n+1}}{c_n}$ is a rational function of $n$, then so is $\frac{c_{n+1}}{c_n} \cdot (n+1)$, and that rational function can be written as $\frac{P(n)}{Q(n)}$ for some polynomials $P$ and $Q$.

Finally, we can factor these polynomials into linear factors over the complex numbers. If $\frac{c_{n+1}}{c_n} = \frac{P(n)}{Q(n)(n+1)}$, where $P(n)$ has degree $p$ and $Q(n)$ has degree $q$, then let:

• $-a_1, -a_2, \dots, -a_p$ be the roots of $P(n)$;
• $-b_1, -b_2, \dots, -b_q$ be the roots of $Q(n)$.

This means that $P(n)$ factors as $(n+a_1)(n+a_2)\dotsm (n+a_p)$, and $Q(n)$ factors as $(n+b_1)(n+b_2) \dotsm (n+b_q)$, giving us $$ \frac{c_{n+1}}{c_n} = \frac{(n+a_1)(n+a_2)\dotsm (n+a_p)}{(n+b_1)(n+b_2) \dotsm (n+b_q)(n+1)}. $$ Actually, this is not fully general. In general, there could also be a constant factor in $P$ and in $Q$. That constant factor will be absorbed in the power $z^n$, though, so by replacing $z^n$ by $(rz)^n$ for some constant $r$, we can forget about the constant factor.

From here, we can write down $c_n$ in terms of $c_0$, which will give us the hypergeometric function in its more common form. We have $$ c_n = c_0 \cdot \frac{c_1}{c_0} \cdot \frac{c_2}{c_1} \cdots \frac{c_n}{c_{n-1}}. $$ From each factor $\frac{c_{i+1}}{c_i}$ for $i=0,\dots,n-1$, we pick up:

• A factor of $a_1+i$ in the numerator. These multiply together to $a_1(a_1+1)\dotsm (a_1+n-1)$, which we write as the rising power $a_1^{\overline n}$.
• Similarly, we get $a_2^{\overline n}, \dots, a_p^{\overline n}$ in the numerator.
• A factor of $b_1+i$ in the denominator. These multiply together to another rising power $b_1^{\overline n}$, but now in the denominator.
• Similarly, we get $b_2^{\overline n}, \dots, b_q^{\overline n}$ in the denominator.
• Finally, the factors of $1+i$ for $i=0,1,\dots,n-1$ in the denominator multiply together to get $n!$.

So the final result is that $$ \sum_{n\ge 0} c_n z^n = \sum_{n\ge 0}c_0 \frac{a_1^{\overline n} a_2^{\overline n} \dotsm a_p^{\overline n}}{b_1^{\overline n} b_2^{\overline n} \dotsm b_q^{\overline n}} \cdot \frac{z^n}{n!}. $$ When we also set $c_0 = 1$, this expression is the definition of the hypergeometric function $${}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right).$$ (If we don't want $c_0 = 1$, then just multiply this ${}_pF_q$ expression by whatever you want $c_0$ to be.)

The reason for the "traditional" factor of $n+1$ in the factorization of $\frac{c_{n+1}}{c_n}$ is to get the $n!$ in the denominator, because this is very common for many hypergeometric functions that we encounter. (It should remind you of the Taylor series.) It also makes taking derivatives of ${}_pF_q$ nicer.

Answer 2

A power series $$ S(z) = \sum\limits_{0 \le n} {c_n z^n } $$

is a (generalized) hypergeometric function when the ratio of consecutive coefficients is a rational function in $n$, that is the ratio of two polynomials in $n$, which can be expressed in terms of the respective roots as $$ \frac{{c_{n + 1} }}{{c_n }} = \frac{{P(n)}}{{Q(n)}} = \frac{{\left( {n + a_1 } \right)\left( {n + a_2 } \right) \cdots \left( {n + a_p } \right)}} {{\left( {n + b_1 } \right)\left( {n + b_2 } \right) \cdots \left( {n + b_q } \right)}} $$

Then $$ S(z) = c_0 \sum\limits_{0 \le n} {\frac{{a_1 ^{\overline {\,n\,} } a_2 ^{\overline {\,n\,} } \cdots a_p ^{\overline {\,n\,} } }} {{b_1 ^{\overline {\,n\,} } b_2 ^{\overline {\,n\,} } \cdots b_q ^{\overline {\,n\,} } }}z^n } \quad \left| {\;x^{\overline {\,n\,} } = \frac{{\Gamma \left( {x + n} \right)}}{{\Gamma \left( x \right)}}} \right. $$

But for historical reasons instead of $z^n$ we want to sum in $z^n/n!$, and since $n! = 1^{\overline {\,n\,} } $ we put $$ S(z) = c_0 \sum\limits_{0 \le n} {\frac{{a_1 ^{\overline {\,n\,} } a_2 ^{\overline {\,n\,} } \cdots a_p ^{\overline {\,n\,} } 1^{\overline {\,n\,} } }}{{b_1 ^{\overline {\,n\,} } b_2 ^{\overline {\,n\,} } \cdots b_q ^{\overline {\,n\,} } }}\frac{{z^n }}{{n!}}} $$ which can be simplified if one of the $b$'s is equal to $1$.