[Calculation 7] Bailey’s Formulas for Hypergeometric Functions

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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.

 
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Leun Kim

Ph.D Candidate at The University of Tokyo
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.



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