The final scheme was In[193]:= FullSimplify[%] Out[193]= (1 - (-1)^(1/3) x^2 + (-1)^(2/3) x^4)/((-1)^(2/3) - (-1)^(1/3) x + x^2) (Note FullSimplify blowing it here, because) In[194]:= PolynomialQuotientRemainder[Numerator[%], Denominator[%], x] Out[194]= {-(-1)^(1/3) - x + (-1)^(2/3) x^2, 0} This fairly directly gives Product[(1/2)*((-1)^(1/3) + x^(-2)^(-k) - (-1)^(2/3)*x^(2^(1 - k)/(-1)^k)), {k, Infinity}] == -(((3 + I*Sqrt[3])*(-(-1)^(2/3) + ((-1)^(1/3) - x)*x))/(6*x^(4/3)*Log[x])) But Mathematica 9.01 seems utterly incapable of reducing this to Product[Sin[Pi/6 + ((-1)^k*z)/2^k] + 1/2, {k, 1, Infinity}] == (2*Sin[Pi/6 + z] - 1)/(Sqrt[3]*z) (after x->E^(I*z)). If anybody feels like step-by-step deriving this in his favorite computer algebra system, I'd like to see it. So after all that, it's already (d114) in http://www.tweedledum.com/rwg/idents.htm ! Now I'm seriously puzzled as to what scheme leads to the not quite identical identity in a paper Mourad is reviewing. Minor consolation: My idents.mfe notebook has the quotient of (d114) by (d113) in the somewhat startling form Product[Csc[Pi/6 + theta/(-2)^n]/4 + 1/2, {n, 1, Infinity}] == Tan[Pi/6 + theta]/theta - Sec[Pi/6 + theta]/(2*theta) So we can similarly divide the apparently new ("penultimate") result by (d113) to get Product[Csc[Pi/6 + (-1/2)^k*t] - 1, {k, 0, Infinity}] == Cot[Pi/6 + t]/Sqrt[3] which I think is the nicest of the lot. (Way easier to prove than to find.) I need a "grad student" to convert these da*ned mfes to nbs and then to html pages. (Which could be a chore. Yesterday, 9.01 ate my entire swap space, froze and da*ned near crashed my OS when I tried to save a graphicsy nb as a pdf.) Can Maxima convert mfes to html? Macsyma can't do displayed formulas. --rwg What? You people don't read minds? I neglected to mention that all the sliding block puzzles in Neil's current tabulation are diagonal traverses: upper-leftmost piece (goalpiece) to its lower right extreme (which might exclude the lower right square if the goalpiece isn't convex). (Sorry. I'd be curious who of you *did* read my mind here.) On Thu, Feb 21, 2013 at 1:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
The penultimate scheme was E^(-2*I*Pi/3) + 2*x + I*E^(I*Pi/6)*x^2 == (E^(-2*I*Pi/3) + 2*x^2/E^(I*Pi/3) + x^4)/ (1 + 2*x/E^(I*Pi/3) - E^(I*Pi/3)*x^2)
leading to
Product[2 - 2*Sin[Pi/6 + (-1/2)^k*z], {k, Infinity}] == 2*(1 + Sin[Pi/6 + z])/3
Aha, a new one. At least for me. --rwg
On Sun, Feb 17, 2013 at 4:59 AM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Feb 14, 2013 at 12:48 PM, Bill Gosper <billgosper@gmail.com>wrote:
Decades ago in an MIT talk, I gave a method to enumerate a family of telescoping products akin to Product[(1/2)*(1 + x^2^(-n)), {n, Infinity}] == (-1 + x)/Log[x] . (Which telescopes because you can shift the index by changing the x variable.)
Recently, Mourad Ismail sent a paper reminding me to carry out the enumeration, on which Neil and I started yesterday. Neil found
Product[(1/2)*(z^(-2)^(-k) + 1), {k, 1, Infinity}] == (z - 1)/(z^(2/3)*Log[z])
and fears it to be centuries old, after getting severely scooped on a continued fraction identity last week.
I found
Product[-(-1)^(2/3) + (-1)^(1/3)*z^(-2)^(-k), {k, Infinity}] == (-1 + (-1)^(2/3)*z)/((-1 + (-1)^(2/3))*z^(2/3))
which looks oscillatory and nonconvergent, but is really just equivalent to
Product[2*Sin[Pi/6 + (-1/2)^k*t], {k, Infinity}]==(2*Cos[Pi/6 + t])/Sqrt[3]
from http://www.tweedledum.com/rwg/idents.htm . (Nice(?) exercise.)
Two of the above identities come from the scheme (a*x^4 + b*x^2 + c)/(c*x^2 + b*x + a) reducing to a polynomial, which Reduce promises contains three more. And there are other schemes... --rwg
The antepenultimate scheme is (1+x^2+x^4)/(1+x+x^2) == 1-x+x^2 . But this is just (1+x^3)/(1+x), so we can divide Neil's result into itself with z cubed: Product[1 - z^(-2)^(-k) + z^(2^(1 - k)/(-1)^k), {k, 1, Infinity}] == (1 + z + z^2)/(3*z^(4/3))
and this is equivalent to Product[-1 + 2*Cos[t/(-2)^k], {k, 1, Infinity}] == (1/3)*(1 + 2*Cos[t])
which is already in idents.htm . Only two more shots at novelty. In this scheme. --rwg
[...Meteor mumblings...]