Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Incidentally I've obtained a hypergeometric identity that I've not seen before:

$${}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2$$

So, I wonder if it is well-known and possibly represents a particular case of something more general?

P.S. I've tried to simplify() the l.h.s. in Maple but it did not succeed, giving a hope that the identity is not completely trivial. ;)

EDIT: There seems to be a bug in formula rendering, so I'm repeating it below in plain LaTeX:

{}_3F_2(-m,-n,m+n; 1, 1; 1) = \frac{m^2+n^2+mn}{(m+n)^2} {\binom{m+n}{m}}^2

share|improve this question
2  
Have you tried to see if it follows from the methods in Petovsek, Wilf, and Zeilberger? math.upenn.edu/~wilf/AeqB.html –  Qiaochu Yuan Jun 28 '10 at 20:27
1  
I don't need a proof. I just wonder if it is known. –  Max Alekseyev Jun 28 '10 at 20:31
4  
I think Qiaochu's point may be that his reference actually provides a Maple program to do this sort of computation. If their program can do this computation (and it likely can) then the identity you've found is known in the sense that anyone curious could derive it in a couple minutes by computer. Of course, if you have a combinatorial proof that's very cool. –  Daniel Litt Jun 28 '10 at 21:23
    
Thank you for suggestions! I've found that such programs are already incorporated in Maple as packages 'sumtools' and 'SumTools'. And sumtools[hypersum]() was able to evaluate the closed form. So, there is an automated way to prove this identity that makes it somewhat less interesting. –  Max Alekseyev Jun 28 '10 at 23:16

1 Answer 1

up vote 11 down vote accepted

Your relation is a particular case of the Karlsson--Minton relations (see Section 1.9 in the $q$-Bible by Gasper and Rahman). It's also a contiguous identity to Pfaff--Saalschütz.

EDIT. First of all I apologise for giving insufficient comments on the problem. I learned from Max a very nice graph-theoretical interpretation of the identity which makes good reasons for not burring it in the list of "ordinary" problems.

The hypergeometric series (function) $$ {}_ {p+1}F_ p\biggl(\begin{matrix} a_ 0,\ a_ 1,\ \dots,\ a_ p \cr b_ 1,\ \dots ,\ b_ p\end{matrix};x\biggr) = \sum_ {k=0}^\infty \frac{(a_ 0)_ k(a_ 1)_ k\dots (a_ p)_ k}{(b_ 1)_ k\dots (b_ p)_ k}\frac{x^k}{k!}, $$ where $$ (a)_ 0=1 \quad\text{and}\quad (a)_ k =\frac{\Gamma(a+k)}{\Gamma(a)}= a(a+1)\dots (a+k-1) \quad\text{for } k\in \mathbb Z_ {>0} $$ (I consider the ones with finite domain of convergence $|z|<1$), have very nice history and links to practically everything in mathematics. There are many transformation and summation theorems for them, both classical and contemporary. There are very efficient algorithms and packages for proving them, like the algorithm of creative telescoping (due to W. Gosper and D. Zeilberger) and the package HYP which allows one to manipulate and identify binomial and hypergeometric series (due to C. Krattenthaler). An example of classical summation theorem is the Pfaff--Saalschütz sum $$ {}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr c,\ 1+a+b-c-m\end{matrix};1\biggr) =\frac{(c-a)_ m(c-b)_ m}{(c)_ m(c-a-b)_ m} $$ where $m$ is a negative integer, with a generalisation $$ {}_ {p+1}F_ p\biggl(\begin{matrix} a,\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=0 \quad\text{if } \operatorname{Re}(-a)>m_ 1+\dots+m_ p $$ and $$ {}_ {p+1}F_ p\biggl(\begin{matrix} -(m_ 1+\dots+m_ p),\ b_ 1+m_ 1,\ \dots,\ b_ p+m_ p \cr b_ 1,\ \dots ,\ b_ p\end{matrix};1\biggr)=(-1)^{m_ 1+\dots+m_ p} \frac{(m_ 1+\dots+m_ p)!}{(b_ 1)_ {m_ 1}\dots (b_ p)_ {m_ p}} $$ due to B. Minton and Per W. Karlsson (here $m_ 1,\dots,m_ p$ are nonnegative integers). Max's original identity is not a straightforward particular case but a linear combination of three contiguous Pfaff--Saalschütz-summable hypergeometric series. (Two hypergeometric functions are said to be contiguous if they are alike except for one pair of parameters, and these differ by unity.) Because of having three hypergeometric functions, I do not see any fun in writing the corresponding details but indicate a simpler hypergeometric derivation.

Applying Thomae's transformation $$ {}_ 3F_ 2\biggl(\begin{matrix} -m,\ a,\ b \cr c,\ d\end{matrix};1\biggr) =\frac{(d-b)_ m}{(d)_ m}\cdot{}_ 3F_ 2\biggl(\begin{matrix} -m,\ c-a,\ b \cr c,\ 1+b-d-m\end{matrix};1\biggr) $$ the problem reduces to evaluation of the series $$ {}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr 1,\ n\end{matrix};1\biggr). $$ Writing $$ \frac{(n+1)_ k}{(n)_ k}=\frac{n+k}{n}=1+\frac kn $$ the latter series becomes $$ {}_ 3F_ 2\biggl(\begin{matrix} -m,\ n+1,\ m+n \cr 1,\ n\end{matrix};1\biggr) ={}_ 2F_ 1\biggl(\begin{matrix} -m,\ m+n \cr 1 \end{matrix};1\biggr) +\frac{(-m)(m+n)}{n} {}_ 2F_ 1\biggl(\begin{matrix} -m+1,\ m+n+1 \cr 2 \end{matrix};1\biggr) $$ and the latter two series are summed with the help of the Chu--Vandermonde summation (a particular case of the Gauss summation theorem).

As for general forms of Max's identity, I can mention that there is no use of the integrality of $n$ in the last paragraph, and I could even expect something a la Minton--Karlsson in general.

share|improve this answer
2  
Wadim, what does "contiguous identity" mean? Also, would you mind stating KM relations for those of us not doing our daily prayers? –  Victor Protsak Jun 28 '10 at 23:44
1  
From Slater's book: "Two hypergeometric functions are said to be contiguous if they are alike except for one pair of parameters, and these differ by unity." What I'm saying about PS is that a linear combination of two contiguous hypergeometrics summed by PS gives the OP. As for the KM identities, I know see that they don't quite include the OP. In any case, after rereading of the question, I'd assume that the author wonders whether the cyclotomic $m^2+mn+n^2$ can be generalized to a higher degree, at least to explain its appearance. I don't believe so, but this isn't in hypergeometry. –  Wadim Zudilin Jun 29 '10 at 0:39
1  
Thanks a lot! I suspected that it should be well-known. –  Max Alekseyev Jun 29 '10 at 0:43
2  
Wadim, thank you for clarifying the term "contiguous". I don't know what made you angry, maybe you need to take a break for a few days? Your contributions to MO are certainly valuable, but this answer is a bit cryptic, please, fill in some details for the benefit of educated non-specialists like myself when you get a chance. –  Victor Protsak Jun 29 '10 at 2:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.