Watson's Hypergeometric Theorem

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Theorem

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

Then:

$\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1} = \dfrac {\map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }$

where:

${}_3 \operatorname F_2$ is the generalized hypergeometric function
$x^{\overline k}$ denotes the $k$th rising factorial power of $x$.


Proof

From Thomae's Transformation, we have:

\(\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {e, f} } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma e \map \Gamma f \map \Gamma s } {\map \Gamma a \map \Gamma {s + b} \map \Gamma {s + c} } \map { {}_3 \operatorname F_2} { { {e - a, f - a, s} \atop {s + b, s + c} } \, \middle \vert \, 1}\) Thomae's Transformation: $s = e + f - a - b - c$
\(\ds \leadsto \ \ \) \(\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } } } {\map \Gamma a \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + b} \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + c} } \map { {}_3 \operatorname F_2} { { {\dfrac 1 2 \paren {a + b + 1} - a, 2 c - a, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } \atop {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + b, \paren {\dfrac 1 2 \paren {1 - a - b} + c } + c} } \, \middle \vert \, 1}\) $s = \paren {\dfrac 1 2 \paren {1 - a - b} + c }$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } } \map { {}_3 \operatorname F_2} { { {\dfrac 1 2 \paren {-a + b + 1}, 2 c - a,\dfrac 1 2 \paren {1 - a - b} + c } \atop {\dfrac 1 2 \paren {1 - a + b} + c, \dfrac 1 2 \paren {1 - a - b} + 2 c } } \, \middle \vert \, 1}\) simplifying

The generalized hypergeometric function on the right hand side can be summed using Dixon's Hypergeometric Theorem.

From Dixon's Hypergeometric Theorem, we have:

$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $

Let:

\(\ds n\) \(=\) \(\ds 2 c - a\)
\(\ds -x\) \(=\) \(\ds \dfrac 1 2 \paren {-a + b + 1}\)
\(\ds -y\) \(=\) \(\ds \dfrac 1 2 \paren {1 - a - b} + c\)

Then:

\(\ds x + n+ 1\) \(=\) \(\ds \dfrac 1 2 \paren {1 - a - b} + 2 c\)
\(\ds y + n + 1\) \(=\) \(\ds \dfrac 1 2 \paren {1 - a + b} + c\)
\(\ds \dfrac n 2 + 1\) \(=\) \(\ds c - \dfrac a 2 + 1\)
\(\ds x + y + \dfrac n 2 + 1\) \(=\) \(\ds \dfrac a 2\)
\(\ds n + 1\) \(=\) \(\ds 1 - a + 2 c\)
\(\ds x + y + n + 1\) \(=\) \(\ds c\)
\(\ds x + \dfrac n 2 + 1\) \(=\) \(\ds \dfrac 1 2 \paren {1 - b} + c\)
\(\ds y + \dfrac n 2 + 1\) \(=\) \(\ds \dfrac 1 2 \paren {1 + b}\)


Making these substitutions to the generalized hypergeometric function on the right hand side, we have:

\(\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }\) Dixon's Hypergeometric Theorem: Before substitution
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map { {}_3 \operatorname F_2} { { {2 c - a, \dfrac 1 2 \paren {-a + b + 1}, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } \atop {\paren {\dfrac 1 2 \paren {1 - a - b} + 2 c }, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) Dixon's Hypergeometric Theorem: After substitution

We now have:

\(\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } } \dfrac {\map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) from $(1)$ and $(2)$ above
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a } \dfrac {\map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) $\map \Gamma {c + \dfrac 1 2 \paren {1 - a + b} }$ and $\map \Gamma {2 c + \dfrac 1 2 \paren {1 - a - b} }$ cancel
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {2 c} } {\map \Gamma a \map \Gamma {2 c - a + 1} } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) simplifying

From Legendre's Duplication Formula, we have:

\(\ds \map \Gamma {2 c}\) \(=\) \(\ds \dfrac {2^{2 c - 1} \map \Gamma c \map \Gamma {c + \dfrac 1 2} } {\map \Gamma {\dfrac 1 2} }\)
\(\ds \map \Gamma a\) \(=\) \(\ds \dfrac {2^{a - 1} \map \Gamma {\dfrac a 2} \map \Gamma {\dfrac 1 2 \paren {1 + a} } } {\map \Gamma {\dfrac 1 2} }\)
\(\ds \map \Gamma {2 c - a + 1}\) \(=\) \(\ds \dfrac {2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {c - \dfrac a 2 + 1} } {\map \Gamma {\dfrac 1 2} }\)

Therefore:

\(\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {2 c} } {\map \Gamma a \map \Gamma {2 c - a + 1} } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) From $(3)$ above: Before substitution
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {2^{2 c - 1} \map \Gamma c \map \Gamma {c + \dfrac 1 2} } {\map \Gamma {\dfrac 1 2} } } } {\paren {\dfrac {2^{a - 1} \map \Gamma {\dfrac a 2} \map \Gamma {\dfrac 1 2 \paren {1 + a} } } {\map \Gamma {\dfrac 1 2} } } \paren {\dfrac {2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {c - \dfrac a 2 + 1} } {\map \Gamma {\dfrac 1 2} } } } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }\) After substitution
\(\ds \) \(=\) \(\ds \dfrac {2^{2 c - 1} \map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {2^{a - 1} 2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }\) simplifying: $\map \Gamma c$, $\map \Gamma {c - \dfrac a 2 + 1}$, $\map \Gamma {\dfrac 1 2}$ and $\map \Gamma {\dfrac a 2}$ cancel
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }\) simplifying further: Product of Powers, Quotient of Powers


$\blacksquare$


Also known as

Some sources refer to Watson's Hypergeometric Theorem as Watson's Theorem.


Also see


Source of Name

This entry was named for George Neville Watson.


Sources

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