[Calculation 11] Simple Examples from the Dixon Theorem

Edited by Leun Kim

Here we introduce some examples from the Dixon Theorem. By setting suitable coefficients, we can obtain simple formulas of infinite series, which are related to the Gamma functions.
 

Example. From the Dixon Theorem, we have
\begin{align*}
&\text{(i) } 1 + \frac{1}{5} \left( \frac{1}{2}\right)^2 + \frac{1}{9} \left( \frac{1\cdot 3}{2\cdot 4}\right)^2 +
\frac{1}{13} \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^2+
\cdots =\frac{\pi^2}{4\Gamma^4(\frac{3}{4})},
\\
&\text{(ii) } 1 + \frac{1}{5^2} \left( \frac{1}{2}\right) + \frac{1}{9^2} \left( \frac{1\cdot 3}{2\cdot 4}\right) +
\frac{1}{13^2} \left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)
+\cdots =\frac{\pi^{5/2}}{8\sqrt{2}\Gamma^2(\frac{3}{4})},
\\
&\text{(iii) } 1 + \left( \frac{1}{2}\right)^3 + \left( \frac{1\cdot 3}{2\cdot 4}\right)^3
+
\left( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^3
+\cdots =\frac{\pi}{\Gamma^4(\frac{3}{4})}.
\end{align*}

 
Proof. First, we recall the Dixon Theorem [2]:
\begin{align}\tag{1}
&{}_3 F_2 (a,b,c;1+a-b,1+a-c;1) \\
&=
\frac{\Gamma(1+\frac{a}{2})\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+a)\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)\Gamma(1+a-b-c)}.
\end{align}
(i) Setting $a=b=\frac{1}{2}$ and $c=\frac{1}{4}$ in the equation (1), we obtain
\begin{eqnarray*}
{}_3 F_2 \left(\frac{1}{2},\frac{1}{2},\frac{1}{4};1,\frac{5}{4};1\right)
&=&
\frac{\Gamma(\frac{5}{4})\Gamma(1)\Gamma(\frac{5}{4})\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2})\Gamma(\frac{3}{4})\Gamma(1)\Gamma(\frac{3}{4})}\\
&=&
\frac{2\Gamma^2(\frac{5}{4})}{\Gamma^2 (\frac{3}{4})}\\
&=&
\frac{\pi^2}{4\Gamma^4(\frac{3}{4})},
\end{eqnarray*}
where we used the Legendre Duplication Formula:
$$
\Gamma \left( \frac{3}{4}\right) \Gamma \left( \frac{5}{4}\right)
=
2^{1-2\cdot \frac{3}{4}}\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)
=
2^{-\frac{3}{2}}\pi
$$
for the last equality.

(ii) Setting $a=\frac{1}{2}$ and $b=c=\frac{1}{4}$ in the equation (1) yields the result by the similar steps.

(iii) Setting $a=b=c=\frac{1}{2}$ in the equation (1) yields the result.$\square$
 
References.
[1] Bruce C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, p.24.
[2] Leun Kim, [Calculation 10] Dixon Theorem.

 
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.



0 comments