Theorem. Let $n \notin \mathbb Z$. Then we have\begin{align}\tag{1} &{}_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). \end{align} |

**Proof.** We consider the following ODE, which is called hypergemoetric differential equation:

\begin{equation}\tag{2}

z(1-z)y”(z) + (c-(a+b+1)z) y'(z) – ab y(z)=0.

\end{equation}

Then we see that the general solution for (2) is the form of

\begin{equation}\tag{3}

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)

\end{equation}

valid for $|z|<1$, where $A_1, A_2$ are constants. Changing $z$ into $1-\zeta$ in (2), we obtain

\begin{equation}\tag{4}

\zeta (1-\zeta) y”(\zeta) + ((a+b+1-c) – (a+b+1)\zeta)y'(\zeta)-aby(\zeta)=0

\end{equation}

which has the solutions

\begin{equation}\tag{5}

{}_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).

\end{equation}

Thus comparing the coefficients of (3) and (5), we obtain the relation

\begin{align*}

&{}_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)

\end{align*}

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

\begin{align*}

&{}_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).

\end{align*}

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\begin{align*} &{}_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 \end{align*} 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.

**References.**

[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.

#### Leun Kim

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## 3 comments

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[…] Proof. First, we recall the Corollary in [2] with and […]

Interesting! I’ve been thinking to study the subjects like hypergeometric functions… but recently I feel my energy and passion vaporized away.

I know that you’re really good at these kind of things! Actually I’m studying basics as I would like to figure out some formulas by Ramanujan.