[Calculation 7] Bailey’s Formulas for Hypergeometric Functions

Edited by Leun Kim

Theorem. (Bailey) The followings are valid:
\begin{align*}
&\text{(i) } (1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right) = {}_2 F_1 (a,c-b;c;z),\quad|z|<1,\;\text{Re}z< \frac{1}{2},\\
&\text{(ii) } {}_2 F_1 \left( a,b; \frac{a+b+1}{2}; \frac{1}{2}\right) = \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+a+b}{2})}{\Gamma(\frac{1+a}{2})\Gamma(\frac{1+b}{2})},\\
&\text{(iii) } {}_2 F_1 \left(a,1-a;c;\frac{1}{2}\right) = \frac{\Gamma(\frac{1}{2}c)\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{1+c-a}{2})}.
\end{align*}

 
Proof. (i) By the direct calculations, the left hand side of (i) becomes
\begin{eqnarray*}
(1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right)
&=&
\sum_{k=0}^\infty \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} z^k (1-z)^{-a-k}\\
&=&
\sum_{k=0}^\infty \sum_{r=0}^\infty \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} \frac{(a+k)_r}{r!} z^{k+r}
\end{eqnarray*}
for $|z|< \frac{1}{2}$. Calculating the coefficient of $z^n$, we get
\begin{eqnarray*}
\sum_{k=0}^n \frac{(a)_k (b)_k (-1)^k}{(c)_k k!} \frac{(a+k)_{n-k}}{(n-k)!}
&=&
\frac{(a)_n}{n!} \sum_{k=0}^n \frac{(b)_k (-n)_k}{k!(c)_k}\\
&=&
\frac{(a)_n}{n!} \frac{(c-b)_n}{(c)_n}
\end{eqnarray*}
where we applied the identities
$$
(a)_k (a+k)_{n-k} = (a)_n, \qquad
\frac{(-1)^k}{(n-k)!} = \frac{(-n)_k}{n!}
$$
for the first equality, and the Chu-Vandermonde identity (in [2]) for the second equality. Finally we obtain
\begin{eqnarray*}
(1-z)^{-a} {}_2 F_1 \left( a,b;c; – \frac{z}{1-z}\right)
&=&
\sum_{n=0}^\infty \frac{(a)_n}{n!}\frac{(c-b)_n}{(c)_n} z^n\\
&=&
{}_2 F_1 (a,c-b;c;z),
\end{eqnarray*}
which proves (i) for $|z|< \frac{1}{2}$. Complete result follows by analytic continuation.$\square$
 

(ii) Letting $z \to -1$ in (i), we find that
\begin{equation}\tag{1}
{}_2 F_1 \left(a,b;c;\frac{1}{2}\right) =
2^a {}_2 F_1 (a,c-b;c;-1).
\end{equation}
Setting $c=\frac{1}{2}(a+b+1)$, we obtain
\begin{eqnarray*}
{}_2 F_1 \left( a,b; \frac{a+b+1}{2}; \frac{1}{2}\right)
&=&
2^a {}_2 F_1 \left( a, \frac{1}{2}(1+a-b); \frac{1}{2}(1+a+b);-1\right)\\
&=&
2^a \frac{\Gamma(\frac{1+a+b}{2})\Gamma(1+\frac{1}{2}a)}{\Gamma(1+a)\Gamma(\frac{1+b}{2})}\\
&=&
\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+a+b}{2})}{\Gamma(\frac{1+a}{2})\Gamma(\frac{1+b}{2})}
\end{eqnarray*}
where we used Kummer’s Theorem for the second equality, and the Duplication Formula of Gamma functions for the last equality.$\square$
 
(iii) Setting $b=1-a$ in the equation (1), we have
\begin{eqnarray*}
{}_2 F_1 \left(a,1-a;c;\frac{1}{2}\right)
&=&
2^a {}_2 F_1 (a, c+a-1; c; -1)\\
&=&
2^a {}_2 F_1 (c+a-1,a; c; -1)\\
&=&
2^a \frac{\Gamma(c)\Gamma(\frac{c+a+1}{2})}{\Gamma(c+a)\Gamma(\frac{c-a+1}{2})}\\
&=&
\frac{\Gamma(\frac{1}{2}c)\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{1+c-a}{2})}
\end{eqnarray*}
where we used Kummer’s Theorem for the third equality, and the Duplication Formula of Gamma functions twice for the last equality.$\square$
 
References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.10-11.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Leun Kim, [Calculation 6] Kummer’s Theorem.

 
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.



One comment