On Mon, Nov 28, 2016 at 3:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
I.e., equally spaced over a period. What's the term? (Special case of sum|product over a root system.)
I once convinced myself that *all* of them, even such nasties as
Sum[Cot[f + (j*Pi)/n]*Cot[g + (j*Pi)/n]*Cot[h + (j*Pi)/n], {j, 1, n}]/n == -Cot[f*n] - Cot[g*n] - Cot[h*n] - (1/2)*Csc[f - g]*Csc[f - h]*Csc[g - h]* ((-Cot[h*n])*Sin[2*f - 2*g] + Cot[g*n]*Sin[2*f - 2*h] - Cot[f*n]*Sin[2*g - 2*h])
come out in closed form as direct consequences of nothing more than x^n-1 == Product[x-I^(4*k/n), {k, 0,n -1}].
Computer algebra junkies may "enjoy" starting here and plowing through to Sum[1/(a + b*Cos[c + (2*k*Pi)/n]), {k, 0, -1 + n}] == ((b^(2*n) - (-a + Sqrt[a^2 - b^2])^(2*n))*n)/(Sqrt[a^2 - b^2]* (b^(2*n) + (-a + Sqrt[a^2 - b^2])^(2*n) - 2*b^n*(-a + Sqrt[a^2 - b^2])^n*Cos[c*n]))
("I ask a simple question; I get a pageant." --Stan Freberg) --rwg
Pageantry unbridled: Sum[(a + b*Sin[(2*Pi*k)/n + x])/ (c + d*Sin[(2*Pi*k)/n + y]), {k,n}] == (1/d)* (n*((a*d*(d^(2*n) - (Sqrt[c^2 - d^2] - c)^(2*n)) + b*Cos[x - y]*(c*((Sqrt[c^2 - d^2] - c)^(2*n) - d^(2*n)) + Sqrt[(c - d)*(c + d)]* ((Sqrt[c^2 - d^2] - c)^(2*n) + d^(2*n)) - 2*Sqrt[(c - d)*(c + d)]*d^n* (Sqrt[c^2 - d^2] - c)^n*Cos[n*(y - Pi/2)]))/ (Sqrt[(c - d)*(c + d)]*((Sqrt[c^2 - d^2] - c)^ (2*n) - 2*d^n*(Sqrt[c^2 - d^2] - c)^n* Cos[n*(y - Pi/2)] + d^(2*n))) - (2*b*Sin[x - y])/(2*Cot[n*(y - Pi/2)] - ((((Sqrt[c^2 - d^2] - c)/d)^(2*n) + 1)* Csc[n*(y - Pi/2)])/((Sqrt[c^2 - d^2] - c)/d)^ n))) reminiscent of those old pre-computer-algebra integral tables whose formulas had to be as encyclopedic as possible. But strangely, Gradshteyn & Ryzhik's table http://fisica.ciens.ucv.ve/~svincenz/TISPISGIMR.pdf 1.34-1.39 seems to have mostly weird LHSs that make nice RHSs. Has anyone seen tabulated or derived even so simple a summand as Sum[1/(a + b Cos[(2 π k)/n + y]), {k, n}] == ((b^(2*n) - (-a + Sqrt[a^2 - b^2])^(2*n))*n)/(Sqrt[a^2 - b^2]* (b^(2*n) + (-a + Sqrt[a^2 - b^2])^(2*n) - 2*b^n*(-a + Sqrt[a^2 - b^2])^n*Cos[c*n])) ? G&R even has some sums over quadratic residues, e.g., Sum[Sin[(2*k^2*Pi)/n], {k, n}] == (1/2)*Sqrt[n]*(1 + Cos[(n*Pi)/2] - Sin[(n*Pi)/2]) == Sqrt[2]*Sqrt[n]*Cos[(n*Pi)/4]* Sin[Pi/4 - (n*Pi)/4] so there are probably lots of nice closed forms involving Jacobi symbols. There are even exotic claims like Sum[Tanh[x/(n Sin[k \[Pi]/2/n]^2)]/(1+Tanh[x]^2/Tan[k \[Pi]/2/n]^2),{k,1,n-1}] ==Coth[2 n x]-(Tanh@x+Coth@x)/2/n , but this seems to be false, e.g., for n=3, x = 9. --rwg
Note the messed up ordering of the nasty cot cot cot result. Is any Computer Algebra able to, e.g., convert Sin[x - y] - Sin[x - z] + Sin[y - z] to Sin[x - y] + Sin[y - z] + Sin[z - x] ? At least Macsyma had trigsign:false to partially inhibit the reverse transformation. But then the sum would be reversed. But it also had ordergreat and orderless, which aliased symbols to have different display orderings. It only accepted individual symbols, not relations between them, but it might barely suffice here. In Mathematica, you can Inactivate all the trigs and then run around manually flipping the anticyclic ones. I'll bet people have built such tools for themselves, but if anybody has a really good solution, WRI should have heard about it.