[Calculation 12] A Basic Formula for Hypergeometric Functions

Edited by Leun Kim

Theorem. Let $n \notin \mathbb Z$. Then we have
&{}_2 F_1 \left( a+n+1, b+n+1; a+b+n+2; 1-z\right)\\
\qquad\qquad\qquad=\frac{\Gamma(a+b+n+2)\Gamma(-n)}{\Gamma(a+1)\Gamma(b+1)}\; {}_2 F_1 (a+n+1,b+n+1;n+1;z)\\
&\, \qquad\qquad\qquad\qquad+\frac{\Gamma(a+b+n+2)\Gamma(n)z^{-n}}{\Gamma(a+n+1)\Gamma(b+n+1)} \;{}_2 F_1 (a+1,b+1;-n+1;z).

Proof. We consider the following ODE, which is called hypergemoetric differential equation:
z(1-z)y”(z) + (c-(a+b+1)z) y'(z) – ab y(z)=0.
Then we see that the general solution for (2) is the form of
y= A_1\; {}_2 F_1 (a,b;c;z) + A_2 z^{1-c} {}_2 F_1 (a+1-c,b+1-c;2-c;z)
valid for $|z|<1$, where $A_1, A_2$ are constants. Changing $z$ into $1-\zeta$ in (2), we obtain
\zeta (1-\zeta) y”(\zeta) + ((a+b+1-c) – (a+b+1)\zeta)y'(\zeta)-aby(\zeta)=0
which has the solutions
{}_2 F_1 (a,b;a+b+1-c;\zeta)\quad \text{and}\quad \zeta^{c-a-b} {}_2 F_1 (c-a,c-b;1+c-a-b;\zeta).
Thus comparing the coefficients of (3) and (5), we obtain the relation
&{}_2 F_1 (a,b;c;z) \\
&= C_1 \; {}_2 F_1 (a,b;a+b+1-c;1-z) + C_2 (1-z)^{c-a-b} \, {}_2 F_1 (c-a,c-b;1+c-a-b;1-z)
valid for $|z|<1$ and $|1-z|<1$, where $C_1, C_2$ are constants. The constant $C_1, C_2$ can be found by limiting $z \to 0$ and $z \to 1$ with Gauss Summation Formula [2]. Then we obtain
\frac{\Gamma(a+b+1-c)\Gamma(1-c)}{\Gamma(a+1-c)\Gamma(b+1-c)} C_1 +
\frac{\Gamma(1+c-a-b)\Gamma(1-c)}{\Gamma(1-a)\Gamma(1-b)}C_2 =1,
C_1 = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.
Therefore we find that
&{}_2 F_1 (a,b;c;z) =
\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\; {}_2 F_1 (a,b;a+b+1-c;1-z)\\
\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)} (1-z)^{c-a-b} {}_2 F_1 (c-a,c-b;1+c-a-b;1-z).
Replacing $a \to a+n+1$, $b \to b+n+1$ and $c \to a+b+n+2$ yields (1).$\square$


Corollary. If $n$ is a nonnegative integer, then
&{}_2 F_1 \left( a+n+1, b+n+1; a+b+n+2; 1-z\right)\\
&\, \qquad
= \frac{\Gamma(a+b+n+2)\Gamma(n)z^{-n}}{\Gamma(a+n+1)\Gamma(b+n+1)} \sum_{k=0}^{n-1} \frac{(a+1)_k (b+1)_k}{(-n+1)_k k!} z^k
&\, \qquad
– \frac{(-1)^n \Gamma(1+b+n+2)}{\Gamma(a+1)\Gamma(b+1)\Gamma(n+1)} \sum_{k=0}^\infty \frac{(a+n+1)_k (b+n+1)_k}{(n+1)_k k!}\\
&\, \qquad
\times \left( \psi(a+n+k+1) + \psi (b+n+k+1) – \psi (n+k+1) – \psi (k+1) + \log x \right) x^k
where $\psi$ is Digamma function. If $n=0$ then the first expression on the right side above is understood to be equal to $0$.

For the proof of this corollary, see for example Andrews [4]. The method in [4] can be applied to this case also.

[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp.1-4.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, pp. 77-78.
[4] G. E. Andrews, Special Functions, Cambridge University Press, pp. 82-84.

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.