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)\\
&=&
(1-z)^{c-a-b} {}_2 F_1 (c-a,c-b;c;z),
\end{eqnarray*}
which proves the theorem.$\square$
References.
[1] aw.twi.tudelft.nl/~koekoek/onderw1112/specfunc_en.html
[2] Leun Kim, Pfaff’s Transformation Formula.
Calculation, Euler, formula, Hypergeometric, HypergeometricFunction, HypergeometricSeries, SpecialFunction, Transformation, 超幾何級数, 超幾何関数, 계산, 초기하급수, 초기하함수, 특수함수
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