# An identity on $\small{}_pF_q\left(\left.\begin{array}{c} a_1+1,a_2+1,\dots ,a_p+1\\ b_1+1,b_2+1,\dots ,b_q+1\end{array}\right| z\right)$

I stumbled upon this relation while trying to answer this post. I was trying to find a relation between the two generalized hypergeometric functions,

$$A=\,_3F_2\left(\color{blue}{\tfrac12,\tfrac12},\tfrac12;\color{red}{\tfrac32,\tfrac32};\color{fuchsia}{\tfrac12}\right)$$

$$B=\,_3F_2\left(\tfrac32,\tfrac32,\tfrac32;\tfrac52,\tfrac52;\tfrac12\right)$$

It seems,

$$A+\tfrac1{18}B = \,_2F_1\left(\tfrac12,\tfrac12;\tfrac32;\tfrac12\right) =\frac{\pi}{2\sqrt2}$$

Note that from a $$_3F_2$$, the sum reduces to a $$_2F_1$$, and $$\tfrac1{18}= \color{blue}{\tfrac12\tfrac12} \color{red}{\tfrac23\tfrac23} \color{fuchsia}{\tfrac12}$$.

Question: In general, let

$$p=q+1\\c_n = a_n+1\\d_n = b_n+1$$

where $$a_n, b_n$$ are arbitrary but the last pair must satisty $$a_p+1=b_q$$. Is it true that,

$${}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)+z\,\frac{a_1a_2\dots a_{p-1}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q \end{array}\right| z\right)\\={}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| z\right)\\ {} \\$$

(Note: The pair $$a_p,b_q$$ disappears in the $$\text{RHS}$$.)

## 1 Answer

We first use the differentiation formula for the generalized hypergeometric function $$$$\frac{a_1a_2\dots a_{p}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q \end{array}\right| z\right)=\frac{d}{dz}{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)$$$$ Then, the LHS of the proposed identity can be written as $$$$_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)+z\,\frac{a_1a_2\dots a_{p-1}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q \end{array}\right| z\right)=\left( 1+\frac{z}{a_p}\frac{d}{dz} \right){} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\tag{1}\label{eq1}$$$$ To differentiate the hypergeometric function, we use the Euler's integral transform \begin{align} & _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\\ &=\frac{\Gamma(b_q)}{\Gamma(a_p)\Gamma(b_q-b_p)} \int_0^1t^{a_p-1}\left( 1-t \right)^{b_q-a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| t\right)\,dt \end{align} Here $$b_q=a_p+1$$, then \begin{align} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)&= a_p \int_0^1t^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| zt\right)\,dt\\ &=\frac{a_p}{z^{a_p}} \int_0^zu^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| u\right)\,du \end{align} Then \begin{align} \frac{d}{dz}&\,{} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right)\\ &=\frac{a_p}{z}\,{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1} \end{array}\right| z\right)-\frac{a_p}{z} \,{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q \end{array}\right| z\right) \end{align} Plugging this expression in eq. \eqref{eq1} we find theRHS of the proposed identity.

• Thanks! From the $c_n = a_n+1$ etc, I knew it had to involve differentiation. While forming the identity, I fortunately noticed the condition $b_q=a_p+1$ and wondered why it was necessary. – Tito Piezas III Jan 24 at 2:27
• You're welcome. The proof does not make use of the assumption $p=q+1$, I think it can be removed. – Paul Enta Jan 24 at 9:01