

-----------message from Christian Krathenthaler about a classical derivation
-----------of Lemma of the paper kirillov.tex-------------------------

From KRATT@Pap.UniVie.AC.AT Wed Nov 29 09:26:25 1995
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 Wed, 29 Nov 1995 15:25:55 +0100 (MET)
Date: Wed, 29 Nov 1995 15:25:55 +0100 (MET)
Subject: the irrelevant identity
To: zeilberg@euclid.math.temple.edu
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Dear Doron,

Thanks a lot for your nice "shorty". I know that it is "irrelevant". 
But maybe you are still interested to know a "classical" proof of
your identity in Lemma 2. It is a special case of one of the "strange
evaluations", namely of the identity (this is Plain-\TeX\ compatible)
$$
\displaylines{
\sum_{j = 0}^{\infty}{q^j} {{(1-{\let \over / a {q^{4 j}}})} \over
      {({\let \over / a})}}  
     {{({\let \over / e, {b\over e}}; q) _{j}} \over {({\let \over /
      {{a {q^3}}\over e}, {{a e {q^3}}\over b}}; {q^3}) _{j}}}  
{{({\let \over / {{a q}\over b}, {{a {q^2}}\over b}}; {q^2}) _{j}}
      \over {({\let \over / b q, b}; {q^2}) _{j}}}
     {{({\let \over / a, b}; {q^3}) _{j}} \over {({\let \over / q,
      {{a q}\over b}}; q) _{j}}}  
\cr       = 
  { {(\let \over / {b\over a}, {q\over b}, {{{q^2}}\over b}, {e\over a},
   {{a {q^3}}\over b}, e q, e {q^2}, {{a e {q^3}}\over {{b^2}}} ;{q^3})
   _\infty} \over {(\let \over / {1\over a}, q, {q^2}, {{b e}\over a},
   {{a {q^3}}\over {{b^2}}}, {{e q}\over b}, {{e {q^2}}\over b},
   {{a e {q^3}}\over b} ;{q^3}) _\infty} },
}
$$
whenever the series terminates
(this is somewhere in my unpublished paper "Some quadratic, cubic,
quartic very well-poised summation formulas for basic hypergeometric
series", one of the reasons that it is unpublished being that, for
instance, the above identity follows from (3.8.19) in Gasper and
Rahman's book by doing the replacements $a\to e$, $b\to b/e$, $d\to
a$, $c\to a/e$, in this order, then only the first term on the
left-hand side of (3.8.19) survives if we assume that $aq/b$ is of
the form $q^-N$ with $N$ a nonnegative integer, and the surviving
term is an $_8\phi_7$-sum which can be evaluated by means of
Jackson's sum). In this identity do the replacements $b\to aq^{1+m}$,
$a\to a^2$, $e\to a$, to obtain
$$
\displaylines{
\sum_{j = 0}^{\infty}{q^j} {{({\let \over / {a^2} {q^{4 j}}})}
      \over {({\let \over / {a^2}})}}  
     {{({\let \over / {a^2}, {a^2} {q^{1 + m}}};
      {q^3}) _{j}} \over {({\let \over / q,
      {q^{-m}}}; q) _{j}}}  
{{({\let \over / {q^{-m}}, {q^{1 - m}}}; {q^2}) _{j}} \over
      {({\let \over / {a^2} {q^{2 + m}}, {a^2} {q^{1 + m}}}; {q^2}) _{j}}}
{{({\let \over / a, a {q^{1 + m}}}; q) _{j}}
    \over {({\let \over / a {q^3}, a {q^{2 - m}}}; {q^3})
      _{j}}}    
\cr   = 
{ {(\let \over / {q^{1 + m}}, {1\over {{a^2} {q^m}}}, {{{q^{1 -
   m}}}\over {{a^2}}}, {1\over a}, {q^{2 - m}}, a q, a {q^2}, {{{q^{1 -
   2 m}}}\over a} ;{q^3}) _\infty} \over {(\let \over / {a^{-2}}, q, {q^2},
   a {q^{1 + m}}, {{{q^{1 - 2 m}}}\over {{a^2}}}, {1\over {a {q^m}}},
   {{{q^{1 - m}}}\over a}, a {q^{2 - m}} ;{q^3}) _\infty} },
}
$$
and then let $a\to\infty$. There are some manipulations to do on the
right-hand side, depending on whether $m$ is $\equiv 0$ or $\equiv 1$
mod 3. In case $m\equiv 2$ mod 3 it is obvious from the term
$(q^{2-m};q^3)_\infty$ that the right-hand side vanishes.

   All the best,

      Christian

