Theorem. The following holds: $$ \pi \;{}_2 F_1 \left( \frac{1}{2}, \frac{1}{2}; 1; 1-x\right) = \log \left( \frac{16}{x}\right) {}_2 F_1 \left( \frac{1}{2}, \frac{1}{2};1;x\right) - 4 \sum_{k=1}^\infty \frac{(\frac{1}{2})_k^2}{(k!)^2} \sum_{j=1}^k \frac{x^k}{(2j-1)(2j)}. $$ Proof. First, we recall the Corollary …
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 …
Here we introduce some examples from the Dixon Theorem. By setting suitable coefficients, we can obtain simple formulas of infinite series, which are related to the Gamma functions. Example. From the …
Here we note Dixon's theorem, which gives some special values of ${}_3 F_2$, since the proof is almost automatic by using Gauss and Kummer's formulas which we've shown before. Theorem. (Dixon's …
Corollary 1. For $\frac{1}{2} < z < 2$, $$ {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-\frac{1}{z}\right) = \sqrt{z} {}_2 F_1 \left(\frac{1}{2}, \frac{1}{2};1;1-z\right). $$ Proof. We recall Bailey's Formula ((i) in [2]) for $w\in\mathbb R$: \begin{equation}\tag{1} (1-w)^{-a} {}_2 F_1 \left( a,b;c; - …
Theorem. (Gauss's Quadratic Transformation) \begin{equation}\tag{1} (1+z)^{-2a} {}_2 F_1 \left(a,b; 2b; \frac{4z}{(1+z)^2}\right) = {}_2 F_1 \left(a, 1+\frac{1}{2}-b;b+\frac{1}{2};z^2\right). \end{equation} Proof. The proof is almost similar to that of [2]. Note that the left hand side of (1) can …
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) } …
Theorem. (Kummer's Theorem) $$ {}_2 F_1 (a,b;1+a-b;-1) = \frac{\Gamma(1+a-b)\Gamma\left(1+\frac{1}{2}a\right)}{\Gamma(1+a)\Gamma\left(1+\frac{1}{2}a-b\right)} $$ To prove Kummer's theorem, we introduce the following lemma, which is called Kummer's quadratic transformation: Lemma. (Kummer's Quadratic Transformation) \begin{equation}\tag{1} {}_2 F_1 (a,b;1+a-b;z) = (1-z)^{-a} {}_2 F_1 …
Theorem. (Gauss's Summation Formula) For $\text{Re}c>\text{Re}b>0$, $$ {}_2 F_1 (a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} $$ holds. Proof. We remember that the Euler Integral Representation for the hypergeometric function is $$ _2 F_1 (a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-tz)^a}\,dt. $$ Taking the limit $z\to 1$ …
Theorem. (Saalschütz's Theorem) $$ {}_3 F_2 (a,b,-n;c,1+a+b-c-n;1) = \frac{(c-a)_n (c-b)_n}{(c)_n (c-a-b)_n} $$ holds for $n\in \mathbb N_0$. Proof. We recall the Euler Transformation Formula: $$(1-z)^{a+b-c} {}_2 F_1 (a,b;c;z) = {}_2 F_1 (c-a,c-b;c;z).$$ Equating the coefficients of $z^n$ …
Theorem (Euler's Transformation Formula) $${}_2 F_1 (a,b;c;z) = (1-z)^{c-a-b} {}_2 F_1 (c-a,c-b;c;z)$$ Proof. Applying Pfaff’s Transformation Formula twice, we obtain \begin{eqnarray*} {}_2 F_1(a,b;c;z) &=& (1-z)^{-a} {}_2 F_1 \left(a,c-b;c; \frac{z}{z-1} \right)\\ &=& (1-z)^{-a} \left(1-\frac{z}{z-1}\right)^{b-c} {}_2 F_1 \left(c-a,c-b;c;\frac{\frac{z}{z-1}}{\frac{z}{z-1} -1} …
Theorem. (Pfaff’s Transformation Formula) $${}_2 F_1(a,b;c;z) = (1-z)^{-a} {}_2 F_1 \left(a,c-b;c; \frac{z}{z-1} \right)$$ Proof. We remember the Euler Integral Representation for the hypergeometric function: $$ {}_2 F_1 (a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1 t^{b-1}(1-t)^{c-b-1} (1-tz)^{-a}\,dt. $$ Substitution $t=1-s$ yields \begin{eqnarray*} {}_2 F_1 …
The classical hypergeometric function ${}_{2}F_1$ is defined by $$ {}_{2}F_1(a,b;c;z) = \sum_{k=0}^\infty \frac{(a)_k(b)_k}{(c)_k} \frac{z^k}{k!} $$ where $(\cdot)_k$ is Pochhammer symbol, that is, $$ (q)_k = \frac{\Gamma(q+k)}{\Gamma(q)} $$ provided that $q+k$ is not a negative integer, with the convention $1/{\pm\infty} = …