Kummer's Hypergeometric Theorem/Examples/2F1(0.5,0.5;1;-1)

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Example of Use of Kummer's Hypergeometric Theorem

$1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$


Proof

From Kummer's Hypergeometric Theorem:

$\ds \map F {n, -x; x + n + 1; -1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {\dfrac n 2 + 1} } { \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {n + 1} } $

where:

$\ds \map F {n, -x; x + n + 1; -1}$ is the Gaussian hypergeometric function of $-1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} } { \paren {x + n + 1}^{\overline k} } \dfrac {\paren {-1}^k} {k!}$
$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
$\map \Gamma {n + 1} = n!$ is the Gamma function.


We have:

\(\ds \map F {\dfrac 1 2, \dfrac 1 2; 1; -1}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {\dfrac 1 2}^{\overline k} \paren {\dfrac 1 2}^{\overline k} } { 1^{\overline k} } \dfrac {\paren {-1}^k} {k!}\) Definition of Gaussian Hypergeometric Function
\(\ds \) \(=\) \(\ds 1 - \dfrac { \paren {\dfrac 1 2}^2 } { \paren {1!}^2 } + \dfrac { \paren {\dfrac 1 2 \times \dfrac 3 2 }^2 } { \paren {2!}^2 } - \dfrac { \paren {\dfrac 1 2 \times \dfrac 3 2 \times \dfrac 5 2 }^2 } { \paren {3!}^2 } + \cdots\) One to Integer Rising is Integer Factorial, $1^k = 1$, Number to Power of Zero Rising is One
\(\ds \) \(=\) \(\ds 1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots\)

and:

\(\ds \map F {\dfrac 1 2, \dfrac 1 2; 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {-\dfrac 1 2 + \dfrac 1 2 + 1} \map \Gamma {\dfrac {\dfrac 1 2} 2 + 1} } { \map \Gamma {-\dfrac 1 2 + \dfrac {\dfrac 1 2} 2 + 1} \map \Gamma {\dfrac 1 2 + 1} }\) Kummer's Hypergeometric Theorem
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {1} \map \Gamma {\dfrac 5 4} } { \map \Gamma {\dfrac 3 4} \map \Gamma {\dfrac 3 2 } }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {\dfrac 1 4 \map \Gamma {\dfrac 1 4} } { \map \Gamma {\dfrac 3 4 } \dfrac 1 2 \map \Gamma {\dfrac 1 2} }\) Definition of Gamma Function and $\map \Gamma {1} = 1$
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 4} } { 2 \map \Gamma {\dfrac 3 4 } \sqrt \pi }\) Gamma Function of $\dfrac 1 2$


Recall from the Euler Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Therefore:

\(\ds \map \Gamma {\dfrac 1 4} \map \Gamma {1 - \dfrac 1 4}\) \(=\) \(\ds \dfrac \pi {\map \sin {\dfrac \pi 4} }\)
\(\ds \map \Gamma {\dfrac 1 4} \map \Gamma {\dfrac 3 4}\) \(=\) \(\ds \dfrac \pi {\dfrac {\sqrt 2} 2 }\) Sine of $45 \degrees$
\(\text {(2)}: \quad\) \(\ds \map \Gamma {\dfrac 1 4}\) \(=\) \(\ds \dfrac {\sqrt 2 \pi} {\map \Gamma {\dfrac 3 4} }\)

Substituting this result back into our equation above:

\(\ds \map F {\dfrac 1 2, \dfrac 1 2; 1; -1}\) \(=\) \(\ds \dfrac {\map \Gamma {\dfrac 1 4} } { 2\map \Gamma {\dfrac 3 4 } \sqrt \pi }\) from $\paren {1}$ above
\(\ds \) \(=\) \(\ds \dfrac {\dfrac {\sqrt 2 \pi} {\map \Gamma {\dfrac 3 4} } } { 2 \map \Gamma {\dfrac 3 4 } \sqrt \pi }\) substituting $\paren {2}$ into $\paren {1}$
\(\ds \) \(=\) \(\ds \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }\)


Therefore:

$1 - \paren {\dfrac 1 2}^2 + \paren {\dfrac {1 \times 3} {2 \times 4} }^2 - \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^2 + \cdots = \dfrac {\sqrt \pi} {\sqrt 2 \paren {\map \Gamma {\dfrac 3 4} }^2 }$

$\blacksquare$


Sources

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