# [Calculation 10] Dixon Theorem

Edited by Leun Kim

Here we note Dixon’s theorem, which gives some special values of ${}_3 F_2$, since the proof is almost automatic by using Gauss and Kummer’s formulas which we’ve shown before.

 Theorem. (Dixon’s Theorem) $${}_3 F_2 (a,b,c;1+a-b,1+a-c;1) = \frac{\Gamma(1+\frac{a}{2})\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+a)\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)\Gamma(1+a-b-c)}$$

Proof. First, we recall the Gauss Summation Formula ([2]) and Kummer’s Theorem ([3]) as below:
\begin{align*}
&{}_2 F_1 (\alpha,\beta;\gamma;1) = \frac{\Gamma(\gamma) \Gamma(\gamma-\alpha-\beta)}{\Gamma(\gamma-\alpha)\Gamma(\gamma-\beta)}\quad\text{(Gauss)}\\
\end{align*}
Using these formulas, we obtain
\begin{align*}
&{}_3 F_2 (a,b,c;1+a-b,1+a-c;1)\\
&=
\sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c+k)\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)\Gamma(1+a-b+k)\Gamma(1+a-c+k)k!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c+k){}_2 F_1 (b+k, c+k;1+a+2k;1)}{\Gamma(1+a+2k)\Gamma(1+a-b-c)k!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{k=0}^\infty \sum_{m=0}^\infty \frac{\Gamma(a+k)\Gamma(b+k+m)\Gamma(c+k+m)}{k!m!\Gamma(1+a+2k+m)\Gamma(1+a-b-c)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}\sum_{p=0}^\infty \sum_{k=0}^p \frac{\Gamma(a+k)\Gamma(b+p)\Gamma(c+p)}{k!(p-k)!\Gamma(1+a+k+p)\Gamma(1+a-b-c)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\sum_{p=0}^\infty \frac{\Gamma(a)\Gamma(b+p)\Gamma(c+p)}{p!\Gamma(1+a-b-c)\Gamma(1+a+p)} {}_2 F_1(a,-p;1+a+p;-1)\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\sum_{p=0}^\infty \frac{\Gamma(a)\Gamma(b+p)\Gamma(c+p)\Gamma(1+\frac{a}{2})}{p!\Gamma(1+a-b-c)\Gamma(1+a)\Gamma(1+\frac{a}{2}+p)}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)} \sum_{p=0}^\infty \frac{(b)_p (c)_p}{(1+\frac{a}{2})_p p!}\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)} {}_2 F_1 (b,c; 1+ \frac{a}{2};1)\\
&=
\frac{\Gamma(1+a-b)\Gamma(1+a-c)}{\Gamma(1+a)\Gamma(1+a-b-c)}
\cdot
\frac{\Gamma(1+\frac{a}{2}) \Gamma(1+\frac{a}{2}-b-c)}{\Gamma(1+\frac{a}{2}-b)\Gamma(1+\frac{a}{2}-c)}
\end{align*}
where we applied Gauss’s Summation Formula ([2]) for the second equality and the last equality, Kummer’s Theorem ([3]) for the 6th equality.$\square$

References.
[1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, 1964, pp. 13-14.
[2] Leun Kim, [Calculation 5] Gauss’s Summation Formula.
[3] Leun Kim, [Calculation 6] Kummer’s Theorem.

#### Leun Kim

Ph.D Candidate at The University of Tokyo
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.

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