Euler's Transformation

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Theorem

Let $a, b, c \in \C$.

Let $\size x < 1$

Let $\map \Re c > \map \Re b > 0$.

Then:

$\ds \map F {a, b; c; x} = \paren {1 - x}^{c - a - b} \map F {c - a, c - b; c; x}$


where $\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$.


Proof

First, we observe:

\(\ds \dfrac {\dfrac x {x - 1} } {\dfrac x {x - 1} - 1}\) \(=\) \(\ds \dfrac {\dfrac x {x - 1} } {\dfrac x {x - 1} -1} \times \dfrac {\paren {x - 1} } {\paren {x - 1} }\) multiplying by $1$
\(\ds \) \(=\) \(\ds \dfrac x {x - \paren {x - 1} }\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds x\)


Applying Pfaff's Transformation twice, we obtain:

\(\ds \map F {a, b; c; x}\) \(=\) \(\ds \paren {1 - x}^{-a} \map F {a, c - b; c; \dfrac x {x - 1} }\) First application of Pfaff's Transformation
\(\ds \) \(=\) \(\ds \paren {1 - x}^{-a} \map F {c - b, a; c; \dfrac x {x - 1} }\) symmetry in first two terms
\(\ds \) \(=\) \(\ds \paren {1 - x}^{-a} \paren {1 - x}^{c - b} \map F {c - b, c - a; c; \dfrac {\dfrac x {x - 1} } {\dfrac x {x - 1} - 1} }\) Second application of Pfaff's Transformation
\(\ds \) \(=\) \(\ds \paren {1 - x}^{-a} \paren {1 - x}^{c - b} \map F {c - a, c - b; c; x}\) symmetry in first two terms and $(1)$ above
\(\ds \) \(=\) \(\ds \paren {1 - x}^{c - a - b} \map F {c - a, c - b; c; x}\) Product of Powers


Therefore, after two applications of Pfaff's Transformation, we have:

$\map F {a, b; c; x} = \paren {1 - x}^{c - a - b} \map F {c - a, c - b; c; x}$

$\blacksquare$


Also see


Source of Name

This entry was named for Leonhard Paul Euler.


Sources