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I saw this in a pdf, and I'm wondering

Questions:

  1. How do you prove Saalschutz Theorem:

$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag{1}$$

I'm somewhat relatively new to Hypergeometrical Series. I understand that the general Hypergeometrical series takes the form$$_pF_q\left[\begin{array}{c,c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_q\end{array};x\right]=\sum\limits_{k=0}^{\infty}\dfrac {(\alpha_1)_k(\alpha_2)_k\ldots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\ldots(\beta_q)_k}\dfrac {x^k}{k!}\tag{2}$$ So therefore, by $(2)$, we should have$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\sum\limits_{k=0}^{\infty}\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}\tag{3}$$ However, I'm not sure how to manipulate the RHS of $(3)$ to get the RHS of $(1)$.

EDIT: Since $(a)_k=\Gamma(a+k)/\Gamma(a)$, the RHS of $(3)$ becomes$$\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}=\dfrac {\Gamma(k-y)\Gamma(n+1)\Gamma(k-x)\Gamma(-x-y-z-n)\Gamma)k-z)}{\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)}$$Now, I need to figure out how$$\Gamma(k-y)\Gamma(k-x)\Gamma(k-z)\Gamma(-x-y-z-n)=\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)$$$$\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)=\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)$$


Extra: I also believe that using the same general approach, we can prove$$\begin{align*} & _7F_6\left[\begin{array}{c,c}n,\frac 12n+1,-x,-y,-z,-u,x+y+z+u+2n+1\\\frac 12n,x+n+1,y+n+1,z+n+1,u+n+1,-x-y-z-u-n\end{array}\right]\\ & =\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(u+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+u+n+1)\Gamma(z+u+n+1)}\\ & \times\dfrac {\Gamma(y+z+u+n+1)\Gamma(x+u+z+n+1)\Gamma(x+y+u+n+1)}{\Gamma(x+z+n+1)\Gamma(y+u+n+1)\Gamma(x+y+z+u+n+1)}\end{align*}\tag{4}$$

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2 Answers 2

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The development of the Saalschütz's identity that I know proceeds along the following path.

Start with the known identity about the sum of the product of three binomials: $$ \bbox[lightyellow] { \begin{gathered} F(m,n,r,s)\quad \left| {\;0 \leqslant \text{integers}\,m,n} \right.\quad = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered} m - r + s \\ k \\ \end{gathered} \right)\left( \begin{gathered} n + r - s \\ n - k \\ \end{gathered} \right)\left( \begin{gathered} r + k \\ m + n \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\begin{subarray}{l} \left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\ \left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right) \end{subarray}} {\left( \begin{gathered} m - r + s \\ k \\ \end{gathered} \right)\left( \begin{gathered} n + r - s \\ n - k \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n - j \\ \end{gathered} \right)\left( \begin{gathered} k \\ j \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\begin{subarray}{l} \left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\ \left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right) \end{subarray}} {\left( \begin{gathered} m - r + s \\ j \\ \end{gathered} \right)\left( \begin{gathered} m - r + s - j \\ k - j \\ \end{gathered} \right)\left( \begin{gathered} n + r - s \\ n - k \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n - j \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered} m - r + s \\ j \\ \end{gathered} \right)\left( \begin{gathered} n + m - j \\ n - j \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n - j \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered} m - r + s \\ j \\ \end{gathered} \right)\left( \begin{gathered} n + m - j \\ m \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n - j \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered} m - r + s \\ j \\ \end{gathered} \right)\left( \begin{gathered} r \\ m \\ \end{gathered} \right)\left( \begin{gathered} r - m \\ n - j \\ \end{gathered} \right)} = \hfill \\ = \left( \begin{gathered} r \\ m \\ \end{gathered} \right)\left( \begin{gathered} s \\ n \\ \end{gathered} \right) \hfill \\ \end{gathered} \tag{1} } $$ where the steps are:
- inverse convolution on last b.;
- trinomial revision on 1st and 4th b.;
- convolution in $k$ on 2nd and 3rd b.;
- symmetry on 2nd b.;
- trinomial revision on 2nd and 3rd b.;
- convolution in $j$ on 1st and 3rd b.

Then consider that the addenda in the sum are $$ \begin{gathered} t_{\,k} = \left( \begin{gathered} m - r + s \\ k \\ \end{gathered} \right)\left( \begin{gathered} n + r - s \\ n - k \\ \end{gathered} \right)\left( \begin{gathered} r + k \\ m + n \\ \end{gathered} \right) = \hfill \\ = \frac{{\left( {m - r + s} \right)^{\,\underline {\,k\,} } }} {{k!}}\frac{{\left( {n + r - s} \right)^{\,\underline {\,n - k\,} } }} {{\left( {n - k} \right)!}}\frac{{\left( {r + k} \right)^{\,\underline {\,m + n\,} } }} {{\left( {m + n} \right)!}} \hfill \\ \end{gathered} $$ where $x^{\,\underline {\,k\,} }$ indicates the falling factorial. The value of the initial term is $$ t_{\,0} = \frac{{\left( {n + r - s} \right)^{\,\underline {\,n\,} } }} {{n!}}\frac{{r^{\,\underline {\,m + n\,} } }} {{\left( {m + n} \right)!}} = \left( \begin{gathered} n + r - s \\ n \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n \\ \end{gathered} \right) $$ and the ratio of consecutive terms is a rational function in $k$ $$ \begin{gathered} \frac{{t_{\,k + 1} }} {{t_{\,k} }} = \frac{{\left( {m - r + s} \right)^{\,\underline {\,k + 1\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k - 1\,} } \left( {r + k + 1} \right)^{\,\underline {\,m + n\,} } }} {{\left( {m - r + s} \right)^{\,\underline {\,k\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k\,} } \left( {r + k} \right)^{\,\underline {\,m + n\,} } }}\frac{{k!\left( {n - k} \right)!\left( {m + n} \right)!}} {{\left( {k + 1} \right)!\left( {n - k - 1} \right)!\left( {m + n} \right)!}} = \hfill \\ = \frac{{\left( {m - r + s - k} \right)\left( {r + k + 1} \right)}} {{\left( {r - s + 1 + k} \right)\left( {r + k - m - n + 1} \right)}}\frac{{\left( {n - k} \right)}} {{\left( {k + 1} \right)}} = \hfill \\ = \frac{{\left( {r - m - s + k} \right)\left( {r + 1 + k} \right)\left( { - n + k} \right)}} {{\left( {r - s + 1 + k} \right)\left( {r - m - n + 1 + k} \right)\left( {1 + k} \right)}} \hfill \\ \end{gathered} $$ so that we can write $$ \begin{gathered} F(m,n,r,s) = t_0 \;{}_3F_2 \left[ {\left. \begin{gathered} r + 1,\;r - m - s,\; - n \hfill \\ r - s + 1,\;r - m - n + 1\; \hfill \\ \end{gathered} \right|\;1} \right] = \hfill \\ = \left( \begin{gathered} n + r - s \\ n \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + n \\ \end{gathered} \right)\;{}_3F_2 \left[ {\left. \begin{gathered} r + 1,\;r - m - s,\; - n \hfill \\ r - s + 1,\;r - m - n + 1\; \hfill \\ \end{gathered} \right|\;1} \right] \hfill \\ \end{gathered} $$ that is, changing $n$ to $q$ so not to get confused with your notation: $$ \bbox[lightyellow] { \begin{gathered} {}_3F_2 \left[ {\left. \begin{gathered} r + 1,\;r - m - s,\; - q \hfill \\ r - s + 1,\;r - m - q + 1\; \hfill \\ \end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,m,q} \right.\quad = \hfill \\ = \left( \begin{gathered} s \\ q \\ \end{gathered} \right)\left( \begin{gathered} r \\ m \\ \end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered} q + r - s \\ q \\ \end{gathered} \right)\left( \begin{gathered} r \\ m + q \\ \end{gathered} \right)} \right) = \hfill \\ = \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}} {{\Gamma \left( {s - q + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)\Gamma \left( {q + r - s + 1} \right)\Gamma \left( {r + 1} \right)}} = \hfill \\ = \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}} {{\Gamma \left( {s - q + 1} \right)\Gamma \left( {r - s + q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)}} = \hfill \\ = \frac{{s^{\,\underline {\,q\,} } \left( {r - s} \right)^{\,\underline {\, - \,q\,} } }} {{\left( {r - m} \right)^{\,\underline {\,q\,} } m^{\,\underline {\, - q\,} } }} \hfill \\ \end{gathered} \tag{2} } $$ which you can compare with your notation ( not forgetting the final $1/k!$ term) to get $$ \left\{ \begin{gathered} q = z \hfill \\ m = y + n \hfill \\ r = - x - 1 \hfill \\ s = - x - 1 - n \hfill \\ \end{gathered} \right. $$ and finally $$ \bbox[lightyellow] { \begin{gathered} {}_3F_2 \left[ {\left. \begin{gathered} - x,\; - y,\; - z \hfill \\ n + 1,\; - x - y - z - n\; \hfill \\ \end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,z,\left( {y + n} \right)} \right.\quad = \hfill \\ = \left( \begin{gathered} - x - 1 - n \\ z \\ \end{gathered} \right)\left( \begin{gathered} - x - 1 \\ y + n \\ \end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered} z + n \\ z \\ \end{gathered} \right)\left( \begin{gathered} - x - 1 \\ y + n + z \\ \end{gathered} \right)} \right) = \hfill \\ = \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}} {{\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)}} = \hfill \\ = \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,z\,} } \,n^{\,\underline {\, - \,\,z\,} } }} {{\left( { - x - 1 - n - y} \right)^{\,\underline {\,z\,} } \;\left( {n + y} \right)^{\,\underline {\, - \,z\,} } }} = \hfill \\ = \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}} {{\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)}} = \hfill \\ = \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,y\,} } \,n^{\,\underline {\, - \,\,y\,} } }} {{\left( { - x - 1 - n - z} \right)^{\,\underline {\,y\,} } \;\left( {n + z} \right)^{\,\underline {\, - \,y\,} } }} \hfill \\ \end{gathered} \tag{3} } $$

---- Addendum -------

Identity (1) is famous because it is one of the very few involving the product of more than two binomials, and can be found in good collections of binomial identities .

A hint on how it could be related to binomial expansion can be got from considering how it works for the last step, i.e. $$ \begin{array}{l} \left( {1 + x} \right)^{\,r} \left( {1 + y} \right)^{\,s} = \sum\limits_{\left( {0\, \le } \right)\,m} {\sum\limits_{\left( {0\, \le } \right)\,n} {\left( \begin{array}{c} r \\ m \\ \end{array} \right)\left( \begin{array}{c} s \\ n \\ \end{array} \right)x^{\,m} y^{\,n} } } = \\ = \sum\limits_{\left( {0\, \le } \right)\,m} {\;\sum\limits_{\left( {0\, \le } \right)\,n} {\sum\limits_{\left( {0\, \le } \right)\;j\,\left( { \le \,n} \right)} {\left( \begin{array}{c} m - r + s \\ j \\ \end{array} \right)\left( \begin{array}{c} r - m \\ n - j \\ \end{array} \right)z^{\,j} y^{\,n} \left( \begin{array}{c} r \\ m \\ \end{array} \right)x^{\,m} } } } = \\ = \sum\limits_{\left( {0\, \le } \right)\,m} {\;\left( {1 + yz} \right)^{\,m - r + s} \left( {1 + y} \right)^{\,r - m} \left( \begin{array}{c} r \\ m \\ \end{array} \right)x^{\,m} } = \\ = \left( {1 + yz} \right)^s \sum\limits_{\left( {0\, \le } \right)\,m} {\left( {\frac{{1 + y}}{{1 + yz}}} \right)^{\,r - m} \left( \begin{array}{c} r \\ m \\ \end{array} \right)x^{\,m} } = \\ = \left( {1 + yz} \right)^s \left( {\frac{{1 + y}}{{1 + yz}} + x} \right)^r \\ \end{array} $$ and then put $z=1$.

Note: the bounds put in brackets are to indicate that, from an algebraic point of view, they are superfluous because inherent to the binomial coeff. (which is null outside them); this is important to carry on in order to correctly apply the convolution.

A combinatorial proof can be found in this interesting paper

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  • $\begingroup$ How do you derive the identity for the three binomials you mentioned at the beginning of your answer? Is there a way to relate that to the coefficient of $x^n$ in some sort of binomial identity? $\endgroup$
    – Frank W
    Jun 10, 2018 at 20:37
  • $\begingroup$ @FrankW.: I put and Addendum to the answer , in reply to your request. $\endgroup$
    – G Cab
    Jun 11, 2018 at 1:11
  • $\begingroup$ @FrankW.: a) well yes, $0 \le n,m$ was a shortcut to indicate the sum over $m$ and $n$, both non-negative b) I changed the symbology, and also amended for a typo. Hope now it is clear $\endgroup$
    – G Cab
    Jun 11, 2018 at 13:55
  • $\begingroup$ If I may, how are you getting the second line with the three separate sums? I see that the $\binom rm$ is preserved, but I’m not understanding where the extra $z$ factor comes in. Could you elaborate that a bit for me? Also doesn’t the sum have to be infinite for it to fit the hypergeometric function? Thanks. $\endgroup$
    – Frank W
    Jun 12, 2018 at 2:31
  • $\begingroup$ I hope you don’t mind, I added a question here: math.stackexchange.com/questions/2816488/…. Perhaps you can give me a more detailed answer explaining each step? I’m having difficulty understanding. $\endgroup$
    – Frank W
    Jun 12, 2018 at 3:37
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Prove$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}$$


Proof: Begin with the identity$$(1-z)^{a+b-c}\space_2F_1(a,b;c;z)=_2F_1(c-a,b-a;c;z)\tag1$$This can be easily proven by setting both solutions of the second order linear differential equation$$(z-z^2)\frac {d^2y}{dz^2}+\bigr\{c-(a+b+1)z\bigr\}\frac {dy}{dx}-aby=0$$equal to each other, and changing the dependent variables. Starting with $(1)$, rewrite it as a summation, and then find the coefficient of $z^n$.$$\sum\limits_{k=0}^\infty\frac {(a+b-c)_k}{k!}(-z)^k\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r}{(c)_r}\frac {z^r}{r!}=\sum\limits_{l=0}^{\infty}\frac {(c-a)_l(b-a)_l}{(c)_l}\frac {z^l}{l!}\tag2$$The coefficient of $z^n$ of $(2)$ is therefore$$\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r(c-a-b)_{n-r}}{(n-r)!(c)_rr!}=\frac {(c-a)_n(b-a)_n}{(c)_nn!}\tag3$$And from $(3)$, it follows that the left-hand side is equal to$$\sum\limits_{r=0}^{\infty}\frac {(a)_r(b)_r}{(c)_r}\frac {(c-a-b)_n(-n)_r}{(1+a+b-c-n)_rn!}=_3F_2\left[\begin{array}{c c}a,b,-n\\c,1+a+b-c-n\end{array}\right]\frac 1{n!(c-a-b)_n}$$Equating to the right-hand side of $(3)$, and simplifying, we get the identity$$_3F_2\left[\begin{array}{c c}a,b,-n\\c,1+a+b-c-n\end{array}\right]=\frac {\Gamma(c)\Gamma(c-a-b)\Gamma(n-a+b)\Gamma(n-a+c)}{\Gamma(b-a)\Gamma(c-a)\Gamma(c+n)\Gamma(n+c-a-b)}\tag4$$Replacing $a=-x,b=-y,n=z,$ and $c$ with $n+1$, we deduce Saalschutz's theorem.

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  • $\begingroup$ nice alternative way (+1), although I am a bit perplex about whether the convergency criteria for $F$ are preserved . $\endgroup$
    – G Cab
    Jun 11, 2018 at 14:07

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