[math-fun] Scooby Dougall's thm
Dougall's thm: a hyper_f ([a, - + 1, b, c, d, n - d - c - b + 2 a + 1, - n], 7, 6 2 a [-, - b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a, n + a + 1]) 2 (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n n n = -------------------------------------------------------------------. (- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a + 1) n n n n I found an early 70s (pre matrix-calculus) printout from a secret Xerox device (a LASER printer!) implying a hyper_f ([a, b, c, d, - n, n - d - c - b + 2 a, - + 1], 7, 6 2 a [- b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a + 1, n + a + 1, -], 1) 2 = - (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n - 1 n - 1 n - 1 ((d + c + b - a) n (n - d - c - b + 2 a) + (c + b - a) (d + b - a) (d + c - a)) /((- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a) ). n n n n (Tested through n=5). The latter cannot be a variable-rename of the former, because the lower parameters minus the upper parameters total 2 in the former and 4 in the latter. Presumably, there's a q-extension. Can this be "new"? We'd need several more before we'd have enough to seed the contiguity relations and evaluate the whole mesh. --rwg
I asked Scooby Dougall's thm Dougall's thm: a hyper_f ([a, - + 1, b, c, d, n - d - c - b + 2 a + 1, - n], 7, 6 2 a [-, - b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a, n + a + 1]) 2 (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n n n = -------------------------------------------------------------------. (- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a + 1) n n n n I found an early 70s (pre matrix-calculus) printout from a secret Xerox device (a LASER printer!) implying a hyper_f ([a, b, c, d, - n, n - d - c - b + 2 a, - + 1], 7, 6 2 a [- b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a + 1, n + a + 1, -], 1) 2 = - (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n - 1 n - 1 n - 1 ((d + c + b - a) n (n - d - c - b + 2 a) + (c + b - a) (d + b - a) (d + c - a)) /((- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a) ). n n n n (Tested through n=5). The latter cannot be a variable-rename of the former, because the lower parameters minus the upper parameters total 2 in the former and 4 in the latter. Presumably, there's a q-extension. Can this be "new"? George Andrews and Dick Askey kindly inform me that they've known this for decades, and sketched a fancy derivation. I just mucked with my usual matrices, but there is an embarrassingly more obvious way: Call Dougall's lhs above D(n). Then ScoobyDougall(n) is just n (n + a) D(n) - ----------------------------------------- D(n - 1) (n - d - c - b + a) (n - d - c - b + 2 a) combined termwise. Oddly, this trick seems to fail for the q-extension ("ScoobyJackson"), and my matrices bog down in algebra. If there is a ScoobyJackson, why isn't it in BHS? --rwg TRANSSEXUAL EX-SATURNALS
I asked If there is a ScoobyJackson, why isn't it in BHS? I still don't know why, but none other than coauthor Mizan Rahman, Dr. q himself, kindly demolishes the "if": 2 n a q 1 2 k 2 k inf (a, b, c, d, -----, --; q) q (1 - a q ) ==== b c d n k \ q > --------------------------------------------- / a q a q a q b c d n + 1 ==== (q, ---, ---, ---, --------, a q ; q) k = 0 b c d n - 1 k a q --------------------------------------------------- = 1 - a 2 n a q a q a q a n a q (---, ---, ---; q) (a q; q) ((1 - -----) (1 - q ) (1 - -----) b c b d c d n - 1 n b c d b c d a a a n a q a q a q a + (1 - ---) (1 - ---) (1 - ---) q )/(---, ---, ---, -----; q) b c b d c d b c d b c d n Notice the argument of q^2 vs Jackson's q. However, this is not sufficient to explain the failure of the linear combination trick. --rwg REHABILITATE THE BILATERIA
I said
Notice the argument of q^2 vs Jackson's q. However, this is not sufficient to explain the failure of the linear combination trick.
The explanation turns out to be simple and elegant: Yours truly is an idiot. Simply repeating the computation with additional consciousness enabled, ScoobyJackson(n) = n n a (1 - q ) (1 - a q ) J(n) - ----------------------------- J(n - 1), n 2 n a q a q b c d (1 - -----) (1 - -----) b c d b c d where J(n) := Jackson(n). Furthermore, if S := ScoobyJackson, then ScooobyJackson(n) = 2 2 n n n a q a (1 - q ) (1 - a q ) (1 - -------) b c d S(n) - ---------------------------------------------------------- S(n - 1) 2 2 (n - 1) n - 1 2 n - 1 a q a q a q b c d q (1 - -------------) (1 - --------) (1 - ---------) b c d b c d b c d i.e., 2 2 n 2 2 n - 1 b c d q a q a q - a (1 - -------) (1 - -------) (1 - -----------) a b c d b c d 2 n - 1 1 a q 3 k - 1 inf (a, b, c, d, - sqrt(a) q, sqrt(a) q, --, ---------; q) q ==== n b c d k \ q ( > ----------------------------------------------------------------) / a q a q a q b c d n + 1 ==== (- sqrt(a), sqrt(a), q, ---, ---, ---, --------, a q ; q) k = 0 b c d n - 2 k a q n - 1 2 n - 1 a q a q /(b c d (1 - --------) (1 - ---------)) = b c d b c d n n n a q a q a q n a q a q a q (a q, ---, ---, ---; q) (- a (1 - q ) (1 - ----) (1 - ----) (1 - ----) b c b d c d n b c d n - 1 n - 1 n - 1 n - 1 a q a q a q a (1 - q ) (1 - --------) (1 - --------) (1 - --------) 2 n - 1 b c d a q ((- ----------------------------------------------------------- - --------- n - 1 n - 1 n - 1 b c d a q a q a q b c d (1 - --------) (1 - --------) (1 - --------) b c b d c d 2 2 n 2 2 (n - 1) 2 n - 1 a q a q a q + 1) (1 - -------)/(q (1 - -------------) (1 - ---------)) + 1) b c d b c d b c d n n n 2 n a q a q a q a q /(b c d (1 - ----) (1 - ----) (1 - ----)) - ----- + 1) b c b d c d b c d a a q a q a q /(-----, ---, ---, ---; q) , b c d b c d n where there are now seven extra powers of q downstairs, and the argument is q^3. It looks fairly easy to prove that these go on forever (reminiscent of the bottom of a Goo...ogle hits page), and indeed, Mizan says ("we" referring to George Gasper) Let me try to answer your "why" part.We were aware that a 8W7 series with an argument q^(r+1)(and appropriately altered parameter,e.g.,n+1 replaced by n+1-r),can be expressed,via Watson's and then Sears' transformation formulas,into r+1 terms. ... Apparently, both procedures (weighted difference or Watson/Sears) apply to a,b,c, and d as well as n, yielding an infinite tree of identities instead of just a sequence. --rwg MISMANAGE SINE GAMMA
participants (1)
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R. William Gosper