[Calculation 9] Simple Corollaries from Gauss and Bailey Formula

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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; – \frac{w}{1-w}\right) … Continue reading

[Calculation 8] Gauss’s Quadratic Trasformation

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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 be expanded in … Continue reading

[Calculation 6] Kummer’s Theorem

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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 \left( … Continue reading

[Calculation 3] Euler’s Transformation Formula

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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} \right)\\ &=& … Continue reading

[Calculation 2] Pfaff’s Transformation Formula

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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 … Continue reading