[math-fun] what the heck is this?
sum((i-a)!*(i-b)!*(i-c)!/((i+a+1)!*(i+b+1)!*(i+c+1)!),i,0,inf) =((-a-1)!*(-b-1)!*(-c-1)!/2)*(((c+b+a)!/((b+a)!*(c+a)!*(c+b)!))-1/(a!*b!*c!)) inf ==== \ (i - a)! (i - b)! (i - c)!
-------------------------------------- = / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0
(- a - 1)! (- b - 1)! (- c - 1)! (c + b + a)! 1 -------------------------------- (-------------------------- - --------). 2 (b + a)! (c + a)! (c + b)! a! b! c! E.g., (c78) minfactorial(limit(subst([c = a,b = a],%),a,0)) inf ==== \ 1 (d78) > -------- = zeta(3) / 3 ==== (i + 1) i = 0 Looks like there's a bilateral version, too. --rwg PS: An easy sudoku with a blank row, (auto)morphed from 18 Nov LIFE "magazine": [[8,3,_;_,_,1;9,_,_],[_,7,_;8,4,_;_,_,_],[_,_,6;_,9,_;_,1,3]; [7,9,_;_,_,_;_,_,_],[1,_,_;_,_,_;_,_,2],[_,_,_;_,_,_;_,3,4]; [4,5,_;_,7,_;2,_,_],[_,_,_;_,6,4;_,8,_],[_,_,8;9,_,_;_,7,5]] [ [ 8 3 _ ] [ _ 7 _ ] [ _ _ 6 ] ] [ [ ] [ ] [ ] ] [ [ _ _ 1 ] [ 8 4 _ ] [ _ 9 _ ] ] [ [ ] [ ] [ ] ] [ [ 9 _ _ ] [ _ _ _ ] [ _ 1 3 ] ] [ ] [ [ 7 9 _ ] [ 1 _ _ ] [ _ _ _ ] ] [ [ ] [ ] [ ] ] [ [ _ _ _ ] [ _ _ _ ] [ _ _ _ ] ] [ [ ] [ ] [ ] ] [ [ _ _ _ ] [ _ _ 2 ] [ _ 3 4 ] ] [ ] [ [ 4 5 _ ] [ _ _ _ ] [ _ _ 8 ] ] [ [ ] [ ] [ ] ] [ [ _ 7 _ ] [ _ 6 4 ] [ 9 _ _ ] ] [ [ ] [ ] [ ] ] [ [ 2 _ _ ] [ _ 8 _ ] [ _ 7 5 ] ]
Yesterday I said,
Looks like there's a bilateral version, too. sum((i-a)!*(i-b)!*(i-c)!/((i+a+1)!*(i+b+1)!*(i+c+1)!),i,minf,inf) =(-a-1)!*(-b-1)!*(-c-1)!*(c+b+a)!/((b+a)!*(c+a)!*(c+b)!)
oo ==== \ (i - a)! (i - b)! (i - c)!
-------------------------------------- =
/ (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i =-oo (- a - 1)! (- b - 1)! (- c - 1)! (c + b + a)! ---------------------------------------------. (b + a)! (c + a)! (c + b)! Alternatively, as c -> -oo, sum((-1)^i*(i-a)!*(i-b)!/((i+a+1)!*(i+b+1)!),i,0,inf) =box((-a-1)!*(-b-1)!/2)*(1/(a!*b!)-1/((b+a)!)) inf ==== i \ (- 1) (i - a)! (i - b)! (- a - 1)! (- b - 1)! 1 1
------------------------- = --------------------- (----- - --------). / (i + a + 1)! (i + b + 1)! 2 a! b! (b + a)! ==== i = 0
These things are seriously weird. There are many *two term*, *inhomogeneous* relations among the contiguous sums, but none, so far, that would yield a closed form for one in isolation. --rwg
Dr. q himself, Mizan Rahman, gets the bilateral version as a limiting case of Bailey's 6psi6, saying
This bilateral sum looked too familiar to overlook.It's q-analogue is a special case of the well-known 6psi6 summation formula that you will find,for instance in our book,(5.3.1).Just let a go to q^(2),then set e=a^(1/2),let b,c,d,go to 1/a,1/b,1/c,respectively,then simplify.Note that a q-analogue of your sum is not a 3psi3,rather a 4psi4..In fact the relevant formula is (5.3.3),which is a q-analogue of Dixon's formula.In the simplification process,you have to remember that the limit of (q/a;q)_inf/(q/a^(1/2);q)_inf,as a goes to q^(2),exists,and equals 2(1-q^(-2)).
Equation numbers refer to his (& Gasper's) book, Basic Hypergeometric Series. Without even knowing this, I should have gone after the more general sum with the index shifted by an arbitrary phase parameter. Mizan then prophetically admonished
I hope this will help realize that many of the summation formulas that may appear new are really special or limiting cases of some long-known formulas in different notations.Or in some other disguise.
Indeed, retracing my Macsyma steps reveals that the unilateral version, inf ==== \ (i - a)! (i - b)! (i - c)!
-------------------------------------- = / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0
(- a - 1)! (- b - 1)! (- c - 1)! (c + b + a)! 1 -------------------------------- (-------------------------- - --------), 2 (b + a)! (c + a)! (c + b)! a! b! c! is just the limit a->0 of Dougall's very-well-poised 5F4 (2.1.7). The same trick applied to Jackson's q-extension (2.7.1) (which erroneously claims to extend (2.1.6)) gives stringpoch(intosum(multthru(q, sum(q^n*(q^(n+1)+1)*qpoch(b*q,c*q,a*q,q,n)/(a^n*b^n*c^n*qpoch(q^2/b,q^2/c,q^2/a,q,n)),n,0,inf) =(a-q)*(b-q)*(c-q)/((a-1)*(b-1)*(c-1)*q) -(1-q/a)*(1-q/b)*(1-q/c)*qpoch(q,q/(b*c),q/(a*b),q/(a*c)),q,inf) /(qpoch(1/b,1/c,1/a,q/(a*b*c),q,inf)*q))) oo n + 1 n + 1 ==== (b q, c q, a q; q) q (q + 1) \ n
--------------------------------------- = / 2 2 2 ==== n n n q q q n = 0 a b c (--, --, --; q) b c a n
(a - q) (b - q) (c - q) ----------------------- (a - 1) (b - 1) (c - 1) q q q q q q (q, ---, ---, ---; q) (1 - -) (1 - -) (1 - -) b c a b a c oo a b c - ------------------------------------------------. 1 1 1 q (-, -, -, -----; q) b c a a b c oo What makes these peculiar is their failure to "terminate on the left". I.e. the summand's denominator lacks the "obligatory" i! or (q;q)_n present in nearly all sums that come out in Gamma (or q-Gamma) functions. Also mysterious is the difficulty of recurrence-relating them to their contiguous neighbors. OK, so they weren't news. If you want not so well known, try www.tweedledum.com/rwg/idents.htm . I'll bet with enough tawa salmon and masala chai, I could "q" that 4F3[125/128]. --rwg PS: Why didn't I try this earlier?: (c1) makelist(sum((i-a)!*(i-b)!*(i-c)!/((i+a+1)!*(i+b+1)!*(i+c+1)!),i,0,inf),c,c,c+2) inf ==== \ (i - a)! (i - b)! (i - c)! (d1) [ > --------------------------------------, / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c - 1)! > --------------------------------------, / (i + a + 1)! (i + b + 1)! (i + c + 2)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c - 2)! > --------------------------------------] / (i + a + 1)! (i + b + 1)! (i + c + 3)! ==== i = 0 (c2) substpart(factor(piece),contiguate(%),1,1) 2 (d2) (c + 1) (c + 2) (c + a + 2) (c + b + 2) inf ==== \ (i - a)! (i - b)! (i - c - 2)! ( > --------------------------------------)/(c + b + a + 1) / (i + a + 1)! (i + b + 1)! (i + c + 3)! ==== i = 0 2 + (c + 1) (2 c + 2 b c + 2 a c + 6 c + a b + 3 b + 3 a + 5) inf ==== \ (i - a)! (i - b)! (i - c - 1)! ( > --------------------------------------)/(c + b + a + 1) / (i + a + 1)! (i + b + 1)! (i + c + 2)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c)! + > -------------------------------------- = 0 / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0 Answer: Because this isn't very useful--it doesn't tell us any values of the contiguous series because we only know the last of the three.
[Apologies, Funsters, for the repetition. The formulas in my PS, while true, were tautological and belied the text. "Luckily", I also neglected to bcc the eavesdroppers, so this stone in the head kills two in the bush. --Louis B. Mayer. (Not really.)] Dr. q himself, Mizan Rahman, gets the bilateral version as a limiting case of Bailey's 6psi6, saying
This bilateral sum looked too familiar to overlook.It's q-analogue is a special case of the well-known 6psi6 summation formula that you will find,for instance in our book,(5.3.1).Just let a go to q^(2),then set e=a^(1/2),let b,c,d,go to 1/a,1/b,1/c,respectively,then simplify.Note that a q-analogue of your sum is not a 3psi3,rather a 4psi4..In fact the relevant formula is (5.3.3),which is a q-analogue of Dixon's formula.In the simplification process,you have to remember that the limit of (q/a;q)_inf/(q/a^(1/2);q)_inf,as a goes to q^(2),exists,and equals 2(1-q^(-2)).
Equation numbers refer to his (& Gasper's) book, Basic Hypergeometric Series. Without even knowing this, I should have gone after the more general sum with the index shifted by an arbitrary phase parameter. Mizan then prophetically admonished
I hope this will help realize that many of the summation formulas that may appear new are really special or limiting cases of some long-known formulas in different notations.Or in some other disguise.
Sure enough, retracing my Macsyma steps reveals that the unilateral version, inf ==== \ (i - a)! (i - b)! (i - c)!
-------------------------------------- = / (i + a + 1)! (i + b + 1)! (i + c + 1)! ==== i = 0
(- a - 1)! (- b - 1)! (- c - 1)! (c + b + a)! 1 -------------------------------- (-------------------------- - --------), 2 (b + a)! (c + a)! (c + b)! a! b! c! is just the limit a->0 of Dougall's very-well-poised 5F4 (2.1.7). The same trick applied to Jackson's q-extension (2.7.1) (which erroneously claims to extend (2.1.6)) gives stringpoch(intosum(multthru(q, sum(q^n*(q^(n+1)+1)*qpoch(b*q,c*q,a*q,q,n)/(a^n*b^n*c^n*qpoch(q^2/b,q^2/c,q^2/a,q,n)),n,0,inf) =(a-q)*(b-q)*(c-q)/((a-1)*(b-1)*(c-1)*q) -(1-q/a)*(1-q/b)*(1-q/c)*qpoch(q,q/(b*c),q/(a*b),q/(a*c)),q,inf) /(qpoch(1/b,1/c,1/a,q/(a*b*c),q,inf)*q))) oo n + 1 n + 1 ==== (b q, c q, a q; q) q (q + 1) \ n
--------------------------------------- = / 2 2 2 ==== n n n q q q n = 0 a b c (--, --, --; q) b c a n
(a - q) (b - q) (c - q) ----------------------- (a - 1) (b - 1) (c - 1) q q q q q q (q, ---, ---, ---; q) (1 - -) (1 - -) (1 - -) b c a b a c oo a b c - ------------------------------------------------. 1 1 1 q (-, -, -, -----; q) b c a a b c oo What makes these peculiar is their failure to "terminate on the left". I.e. the summand's denominator lacks the "obligatory" i! or (q;q)_n present in nearly all sums that come out in Gamma (or q-Gamma) functions. Also mysterious is the difficulty of recurrence-relating them to their contiguous neighbors. OK, so they weren't news. If you want not so well known, try www.tweedledum.com/rwg/idents.htm . I'll bet with enough tawa salmon and masala chai, I could "q" that 4F3[125/128]. --rwg PS: Why didn't I try this earlier?: (c1) makelist(sum((i-a)!*(i-b)!*(i-c)!/((i+a+1)!*(i+b+1)!*(i+d)!),i,0,inf),d,d,d+2) inf ==== \ (i - a)! (i - b)! (i - c)! (d1) [ > ----------------------------------, / (i + a + 1)! (i + b + 1)! (i + d)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c)!
--------------------------------------, / (i + a + 1)! (i + b + 1)! (i + d + 1)! ==== i = 0
inf ==== \ (i - a)! (i - b)! (i - c)!
--------------------------------------] / (i + a + 1)! (i + b + 1)! (i + d + 2)! ==== i = 0
(c2) (sign_uses_hashing:false,tlimswitch:true,substpart(factor(piece),contiguate(%),1,1)) (d2) (d + a + 1) (d + b + 1) (d + c + 1) inf ==== \ (i - a)! (i - b)! (i - c)! ( > --------------------------------------) / (i + a + 1)! (i + b + 1)! (i + d + 2)! ==== i = 0 2 /(d + c + 2 b + 2 a + 1) - (2 d + 2 c d + 3 b d + 3 a d + 3 d + b c inf ==== \ (i - a)! (i - b)! (i - c)! + a c + c + b + a + 1) ( > --------------------------------------) / (i + a + 1)! (i + b + 1)! (i + d + 1)! ==== i = 0 inf ==== \ (i - a)! (i - b)! (i - c)! /(d + c + 2 b + 2 a + 1) + > ---------------------------------- = / (i + a + 1)! (i + b + 1)! (i + d)! ==== i = 0 (a + 1) (- a)! (b + 1) (- b)! (- c)! (d + 2) -------------------------------------------------- (a + 1)! (b + 1)! (d + c + 2 b + 2 a + 1) (d + 2)! Answer: Because this isn't very useful--it doesn't tell us any values of the contiguous series because [with d = c+1, e.g.] we only know the last of the three.
That peculiar series I've been hacking boils down to (c43) (hyper_f[4,3]([1,a,b,c],[1-a,1-b,1-c]),%%=multthru(hypersimp(%%))) Time= 50 msec. (d43) hyper_f ([1, a, b, c], [1 - a, 1 - b, 1 - c]) = 4, 3 gamma(1 - a) gamma(1 - b) gamma(1 - c) gamma(- c - b - a + 1) 1 ------------------------------------------------------------- + - 2 gamma(- b - a + 1) gamma(- c - a + 1) gamma(- c - b + 1) 2 which, as you can see, I've now automated, in spite of the absence of a contiguous system. Thus, you typically get a gamma product + a rational term + a canonical 4F3 with parameters in the form [1,x,y,z; 1-x,1-y,2-z] : (c54) (hyper_f[4,3]([1,a,b,c],[-a,-b,-c]),%%=hypersimp(%%)) Time= 200 msec. (d54) hyper_f ([1, a, b, c], [- a, - b, - c]) = 4, 3 (a b c (c + b + a + 2) (2 c + 2 b + 2 a + 3) gamma(- a) gamma(- b) gamma(- c - b - a - 2) gamma(- c) (-------------------------------------------------------- + 1) gamma(- b - a - 1) gamma(- c - a - 1) gamma(- c - b - 1) /((b + a + 1) (c + a + 1)) + (b + 1) (c + 1) 2 2 2 2 2 (4 b c + 2 a c + 2 c + 4 b c + 6 a b c + 8 b c + 2 a c + 5 a c + 3 c 2 2 2 2 + 2 a b + 2 b + 2 a b + 5 a b + 3 b + a + 2 a + 1) /((b + a + 1) (c + a + 1)) - 2 hyper_f ([1, c + 1, b + 1, a + 1], 4, 3 [- c, - b, 1 - a], 1) b c (2 c + 2 b + 2 a + 3)) /((2 b + 1) (c + b + 1) (2 c + 1)) This isn't much use unless you luck out with a zero coefficient on the 4F3: (c50) (sum(i^2*(i-a-1)!*(i-b-1)!*(i-c-1)!/((i+a)!*(i+b)!*(i+c)!),i,1,inf), resimplify(%%=make1hyper(%%))=factcomb(factor(maxfactorial(makefact(hypersimp(%%)))))) Time= 34879 msec. inf ==== 2 \ i (i - a - 1)! (i - b - 1)! (i - c - 1)! (d50) > ----------------------------------------- = / (i + a)! (i + b)! (i + c)! ==== i = 1 3 %pi hyper_f ([2, 2, 1 - a, 1 - b, 1 - c], [1, a + 2, b + 2, c + 2], 1) 5, 4 /((a - 1)! (a + 1)! sin(%pi a) (b - 1)! (b + 1)! sin(%pi b) (c - 1)! (c + 1)! 3 sin(%pi c)) = %pi (c + b + a - 1)!/(2 (a - 1)! sin(%pi a) (b - 1)! sin(%pi b) (b + a)! (c - 1)! sin(%pi c) (c + a)! (c + b)!) Note that this took half a minute and extensive post-simplification, and needs a bit more: (c51) ((-a)!*(-b)!*(-c)!,(1=%%)=%%,factcomb(negfactorial(%/%%)*%%)) Time= 60 msec. inf ==== 2 \ i (i - a - 1)! (i - b - 1)! (i - c - 1)! (d51) > ----------------------------------------- = / (i + a)! (i + b)! (i + c)! ==== i = 1 hyper_f ([2, 2, 1 - a, 1 - b, 1 - c], [1, a + 2, b + 2, c + 2], 1) (- a)! 5, 4 (- b)! (- c)!/((a + 1)! (b + 1)! (c + 1)!) = (- a)! (- b)! (- c)! (c + b + a - 1)! -------------------------------------, 2 (b + a)! (c + a)! (c + b)! a monomial! This is strange because the base case is a binomial, but due to the lack of "left termination", the three term contiguity relations are inhomogeneous, which adds in rational terms. Getting this calculation to even fit in memory, let alone in 34 seconds, was a bear to program. I need another 4F3[1] identity to recoup the effort! Apropos surprisingly messy results, one can easily sum certain hybrid trig/hypergeometric series by, e.g., inf inf ==== ==== %i n x \ sin(n x) \ %e (d60) > ---------------- = Im( > ----------------) / binomial(2 n, n) / binomial(2 n, n) ==== ==== n = 0 n = 0 but the answer, which is real, will look complex. Macsyma's IMAGPART can nearly always eliminate the "%i"s, but leaves a big mess. Extensive chiropractic got it down to inf ==== 2 \ sin(n x) 2 cos (x) - 76 cos(x) + 47 (d158) > ---------------- = sqrt(2) sqrt(-------------------------- + 1) / binomial(2 n, n) 3/2 ==== (17 - 8 cos(x)) n = 0 sqrt(4 cos(x) + sqrt(17 - 8 cos(x)) - 1) 3/4 asin(----------------------------------------)/(17 - 8 cos(x)) 2 sqrt(2) 2 2 cos (x) - 76 cos(x) + 47 + sqrt(2) sqrt(1 - --------------------------) 3/2 (17 - 8 cos(x)) sqrt(- 4 cos(x) + sqrt(17 - 8 cos(x)) + 1) 3/4 asinh(------------------------------------------)/(17 - 8 cos(x)) 2 sqrt(2) 4 sin(x) + ------------- 17 - 8 cos(x) (Seventeen? Forty seven??) Symbolically testing: c159) taylor(%,x,0,5) inf ==== 3 \ n 3 ( > ----------------) x inf / binomial(2 n, n) ==== ==== \ n n = 0 (d159)/T/ ( > ----------------) x - --------------------------- / binomial(2 n, n) 6 ==== n = 0 inf ==== 5 \ n 5 ( > ----------------) x / binomial(2 n, n) ==== n = 0 + --------------------------- + . . . = 120 3 (2 %pi sqrt(3) + 18) x (9 sqrt(6) sqrt(3) sqrt(2) + 37 %pi sqrt(3) + 351) x ---------------------- - ----------------------------------------------------- 27 729 5 (445 sqrt(6) sqrt(3) sqrt(2) + 1407 %pi sqrt(3) + 12639) x + ----------------------------------------------------------- + . . . 43740 Doing the series: (c162) block([algebraic:true],expand(ratsimp(hypersimp(%)))) 5 5 3 3 469 sqrt(3) %pi x 7 x 37 sqrt(3) %pi x 5 x 2 sqrt(3) %pi x (d160) ------------------ + ---- - ----------------- - ---- + --------------- 14580 20 729 9 27 5 5 3 3 2 x 469 sqrt(3) %pi x 7 x 37 sqrt(3) %pi x 5 x 2 sqrt(3) %pi x + --- = ------------------ + ---- - ----------------- - ---- + --------------- 3 14580 20 729 9 27 2 x + ---, so it has my vote. 3 With x = pi/2, we get another one of those screwy hypergeometrics with a 1 upstairs to cancel the familiar n! out of the denominator: 3 3 5 1 hyper_f ([1, 1, -], [-, -], - --) 3, 2 2 4 4 16 (d171) ------------------------------------ = 2 47 sqrt(sqrt(17) + 1) sqrt(2) sqrt(sqrt(17) - --) asinh(------------------) 17 2 sqrt(2) ----------------------------------------------------- 17 47 sqrt(sqrt(17) - 1) sqrt(2) sqrt(sqrt(17) + --) asin(------------------) 17 2 sqrt(2) 4 + ---------------------------------------------------- + -- 17 17 (c172) dfloat(%) (d172) 0.45369611887698d0 = 0.45369611887698d0 I was anxious to try Mma's big hammer simplifiers on (d158), but couldn't figure out how to get past the first step of syntactically eliminating the imaginary units. Can some Mma wizard explain how to get Re[ArcSin[2]], e.g., (c61) realpart(asin(2)) %pi (d61) --- 2 Maple also does this case, but my V.4 can't seem to do the mess fom the sum, nor even compute the mess in the first place. Meanwhile, I remain convinced that simplifications such as (d158) are an open AI problem. Computer algebra and AI parted ways with the Risch and Berlekamp algorithms. Maybe it's time they got back together.
participants (1)
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R. William Gosper