A curious identity involving the Appell hypergeometric series.

By playing around with exact solutions to ordinary differential equations from 1 (here I mean with entry 1.4.5.31 in page 180 to be exact) we have discovered a following curious identity involving the Appell hypergeometric function $F_1() $.

Let $\xi_1,\xi_2,\xi_3,\xi_4,\xi_5 $ and $t$ be parameters. Then we define another parameters $\chi_1,\chi_2,\chi_3$ as follows:

\begin{array}{lll} \chi_1 &=& -\frac{2 \left(\xi_1^2-\xi_4 \xi_1-\xi_5 \xi_1+\xi_4 \xi_5\right)} {(\xi_1-\xi_2) (\xi_1-\xi_3)}\\ \chi_2 &=& \frac{2 \left(\xi_2^2-\xi_4 \xi_2-\xi_5 \xi_2+\xi_4 \xi_5\right)}{(\xi_1-\xi_2) (\xi_2-\xi_3)}\\ \chi_3 &=& \frac{2 \left(\xi_3^2-\xi_4 \xi_3-\xi_5 \xi_3+\xi_4 \xi_5\right)} {(\xi_1-\xi_3) (\xi_3-\xi_2)} \end{array}

Then we have checked numerically that the following identity holds true:

\begin{eqnarray} &&\frac{1}{t-\xi_2} \cdot F_1\left( 1, \chi_1+1,\chi_3+2,2, \frac{\xi_1-\xi_2}{t-\xi_2}, \frac{\xi_3-\xi_2}{t-\xi_2} \right) + \\ && \frac{(1+\chi_1)(\xi_1-\xi_2)}{2(t-\xi_2)^2} \cdot F_1\left( 2, \chi_1+2,\chi_3+2,3, \frac{\xi_1-\xi_2}{t-\xi_2}, \frac{\xi_3-\xi_2}{t-\xi_2} \right) + \\ && \frac{(2+\chi_3)(\xi_3-\xi_2)}{2(t-\xi_2)^2} \cdot F_1\left( 2, \chi_1+1,\chi_3+3,3, \frac{\xi_1-\xi_2}{t-\xi_2}, \frac{\xi_3-\xi_2}{t-\xi_2} \right) = \\ % &&\left(t-\xi_1\right)^{-\chi_1-1} \left(t-\xi_2\right)^{-\chi_2} \left(t-\xi_3\right)^{-\chi_3-2} \end{eqnarray}

A snippet of Mathematica code that verifies that identity numerically is included below:

t =.; tt =.;
{xi[1], xi[2], xi[3], xi[4], xi[5]} = 
  RandomReal[{-1, 0}, 5, WorkingPrecision -> 50];
{chi[1], chi[2], 
   chi[3]} = {-((
    2 (xi[1]^2 - xi[1] xi[4] - xi[1] xi[5] + xi[4] xi[5]))/((xi[1] - 
       xi[2]) (xi[1] - xi[3]))), (
   2 (xi[2]^2 - xi[2] xi[4] - xi[2] xi[5] + xi[4] xi[5]))/((xi[1] - 
      xi[2]) (xi[2] - xi[3])), (
   2 (xi[3]^2 - xi[3] xi[4] - xi[3] xi[5] + xi[4] xi[5]))/((xi[1] - 
      xi[3]) (-xi[2] + xi[3]))};

remm1 := 2 a ((1/(t - xi[2])
         AppellF1[1, chi[1] + 1, chi[3] + 2, 
         2, (xi[1] - xi[2])/(t - xi[2]), (xi[3] - xi[2])/(t - 
           xi[2])] + ( (1 + chi[1]) (xi[1] - xi[2]))/(2 (t - xi[2])^2)
         AppellF1[2, 2 + chi[1], 2 + chi[3], 3, (xi[1] - xi[2])/(
         t - xi[2]), (-xi[2] + xi[3])/(
         t - xi[2])] + ((2 + chi[3]) (-xi[2] + xi[3]))/(
        2 (t - xi[2])^2)
         AppellF1[2, 1 + chi[1], 3 + chi[3], 3, (xi[1] - xi[2])/(
         t - xi[2]), (-xi[2] + xi[3])/(t - xi[2])] ) - (t - 
        xi[1])^(-chi[1] - 
       1) (t - xi[2])^-chi[2] (t - xi[3])^(-chi[3] - 2));

t = RandomReal[{0, 1}, WorkingPrecision -> 50];
{remm1}

1 Polyanin, A. D.; Zaitsev, Valentin F., Handbook of exact solutions for ordinary differential equations., Boca Raton, FL: CRC Press. xxvi, 787 p. (2003). ZBL1015.34001.

The question is now can we prove that identity? I tried to use the recurrence relations from the Wikipedia website Appell hypergeometric function but they do not involve a source term that we have in the right hand side above. How do we go about proving that relation?

Solution 1:

By substituting for $ x = 1/(t-\xi_2) $ it amounts to show that both sides of the aforementioned identity have the same expansion in powers of $x$. But now we can easily find coefficients of those expansions, the one on the right hand side is a convolution of two binomial expansions whereas the one on the left hand side is given from the definitions of the Appell series. If we compare those and meticulously simplify their difference proving the identity is equivalent to proving the following:

\begin{eqnarray} -\frac{1+\chi_2}{n+\chi_3} \cdot (\xi_1-\xi_2) \cdot F_{2,1} \left[ 2-n, 2+\chi_1,1-n-\chi_3, \frac{\xi_2-\xi_1}{\xi_2-\xi_3}\right] = (-\xi_2+\xi_3) \left( F_{2,1} \left[2-n,1+\chi_1,-n-\chi_3 , \frac{\xi_2-\xi_1}{\xi_2-\xi_3}\right] - F_{2,1} \left[1-n,1+\chi_1,-n-\chi_3 , \frac{\xi_2-\xi_1}{\xi_2-\xi_3}\right] \right) \quad (i) \end{eqnarray}

for $n=1,2,3,\dots $.

Now let us simplify the right hand side of $(i)$. We have:

\begin{eqnarray} rhs &=& (-\xi_2+\xi_3) \cdot\sum\limits_{m=0}^\infty \underbrace{\left[ (2-n)^{(m)} - (1-n)^{(m)} \right]}_{(2-n)^{(m-1)} \cdot m} \cdot \frac{(1+\chi_1)^{(m)}}{(-n-\chi_3)^{(m)}} \cdot \frac{(\frac{\xi_2-\xi_1}{\xi_2-\xi_3})^m}{m!} \\ &=& (-\xi_2+\xi_3) \cdot\sum\limits_{m=1}^\infty (2-n)^{(m-1)} \cdot \frac{(1+\chi_1)^{(m)}}{(-n-\chi_3)^{(m)}} \cdot \frac{(\frac{\xi_2-\xi_1}{\xi_2-\xi_3})^m}{(m-1)!} \\ &\underbrace{=}_{m-1\rightarrow m}& (-\xi_2+\xi_3) \cdot \frac{(1+\chi_1)}{(-n-\chi_3)} \cdot \frac{\xi_2-\xi_1}{\xi_2-\xi_3} \cdot \sum\limits_{m=0}^\infty (2-n)^{(m)} \cdot \frac{(2+\chi_1)^{(m)}}{(-n+1-\chi_3)^{(m)}} \cdot \frac{(\frac{\xi_2-\xi_1}{\xi_2-\xi_3})^m}{(m)!} \\ &=& \frac{1+\chi_1}{n+\chi_3} \cdot (\xi_2-\xi_1) \cdot F_{2,1} \left[ 2-n,2+\chi_1,-n+1-\chi_3,\frac{\xi_2-\xi_1}{\xi_2-\xi_3}\right] \\ &=& lhs \end{eqnarray} This completes the proof.