[math-fun] ~s highly miscellaneous hypergeometric identity
Just to shake off some rust, I fired up an old Macsyma notebook containing some matrix product rearrangements. A variable dropped out from an unusual spot, producing oo n 5 ==== (n + a) (11 n + 6 a + 1) 4 (n + a - -)! (n + 2 a - 2)! (2 n + a - 1)! \ 6
---------------------------------------------------------------------- / 2 ==== (n + a - -)! (3 n + 2 a + 1)! n = 0 3
1 3 3 (-)! (a - -)! 3 2 = ---------------, 2/3 4 an ungainly 7F6[16/27]. This is a fairly unusual z for a closed form. Also strange is the single a on the right, and the two linear factors in the summand. But oddest is the absence of n!, or (2n)!, or ..., from the summand's denominator, thanks to the dropped variable. Normally, closed forms like this require "termination to the left". So prevalent is that denominator factor that pFq notation supplies it implicitly, yet here it must be artificially (and ungainly) canceled out. A similar but lesser curiosity is oo ==== 2 2 \ 3 k + (3 a + 1) k + a > --------------------------------- = 1, / 2 k + a + 1 ==== (k + a) (k + a + 1) ( ) k = 0 k where the k! is in the numerator instead of the denominator. A slightly uglier cousin gives pi for all a (from rational terms). --rwg Holey mackerel, scombroid cobordism. PPS, the history of the Cal\Nevada border (MKleber> http://laws.findlaw.com/us/447/125.html) is indeed fascinating, The surveying uncertainties dwarfed the mathematical ones. Besides the inevitable problem with longitude, the south end was defined to be where the Colorado River crossed the 35th parallel, which turned out to be highly variable. The town of Aurora, claimed by both states, sent representatives to both legislatures, who simultaneously became house speaker (1862).
oo n 5 ==== (n + a) (11 n + 6 a + 1) 4 (n + a - -)! (n + 2 a - 2)! (2 n + a - 1)! \ 6
---------------------------------------------------------------------- / 2 ==== (n + a - -)! (3 n + 2 a + 1)! n = 0 3
1 3 3 (-)! (a - -)! 3 2 = ---------------, 2/3 4
oo ==== 2 2 \ 3 k + (3 a + 1) k + a > --------------------------------- = 1, / 2 k + a + 1 ==== (k + a) (k + a + 1) ( ) k = 0 k
It is interesting that Maple can do the second one, but for the first one gives the following answer,
sum((n + a)*(11*n + 6*a + 1)*4^n*(n + a - 5/6)!*(n + 2*a - 2)!*(2*n + a - 1)!/(n + a - 2/3)!/(3*n + 2*a + 1)!,n=0..infinity);
/ 16 7/360 (1/6)! |60 hypergeom([1/2, 1, 1, 7/6], [4/3, 4/3, 5/3], --) \ 27 16 \ + 11 hypergeom([3/2, 2, 2, 13/6], [7/3, 7/3, 8/3], --)| 27 / / 1/2 GAMMA(a - 1/2) / (Pi (1/3)!) / Comparing two answers, we get the following identity, 60*hypergeom([1/2, 1, 1, 7/6],[4/3, 4/3, 5/3],16/27)+ 11*hypergeom([3/2, 2, 2, 13/6],[7/3, 7/3, 8/3],16/27)=120*Pi*sqrt(3)/7 Alec Mihailovs http://math.tntech.edu/alec/
oo ==== 2 2 \ 3 k + (3 a + 1) k + a > --------------------------------- = 1, / 2 k + a + 1 ==== (k + a) (k + a + 1) ( ) k = 0 k
It is interesting that Maple can do the second one,
Likewise (my, at least) Macsyma. But DUH! I'm getting senile. I thought and then forgot to check: That sum merely telescopes. Telescoping series have no need of "termination to the left" (or the right).
but for the first one gives the following answer,
sum((n + a)*(11*n + 6*a + 1)*4^n*(n + a - 5/6)!*(n + 2*a - 2)!*(2*n + a - 1)!/(n + a - 2/3)!/(3*n + 2*a + 1)!,n=0..infinity);
/ 16 7/360 (1/6)! |60 hypergeom([1/2, 1, 1, 7/6], [4/3, 4/3, 5/3], --) \ 27 ...
That did a nicer job than Macsyma, but I strongly object to the gratuitous change of notation from sum to hypergeom. In Macsyma, you say makehyper, or make1hyper if you want a single one. --rwg Henry> Is there a "no man's land", where neither the CA nor the NV
highway patrol will give tickets?
More likely it's where *both* highway patrols will get you.
participants (2)
-
Alec Mihailovs -
R. William Gosper