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
Using z^5 instead of z^3 gives the rather kludgy Product[1 - 4*Sin[t/2^k]*Sin[(3*t)/2^k], {k, 1, Infinity}] == (2*Cos[4*t] + 2*Cos[2*t] + 1)/5 The simpler scheme (1+a*x^2+b^x4)/(1+a*x+b*x^2) includes (1 + x^2 + x^4)/(1 + x + x^2) == 1 - x + x^2 which gives two product identities: Product[1 + x^2^(1 - k) - x^2^(-k), {k, Infinity}] == (1/3)*(1 + x + x^2) and, if |x|<1, Product[1 - x^2^k + x^2^(1 + k), {k, 0, Infinity}] == 1/(1 + x + x^2) . Note that the latter product converges for a dense set on |x|=1, but the identity fails, giving, e.g., 3==1 for x=-1. Multiplying the identities gives the curious, bilateral product Out[331]=Product[1 - x^2^k + x^2^(1 + k), {k, -Infinity, Infinity}] == 1/3 when |x|<1. Unlike with the unilaterally infinite products, you can't recover the telescoping identity from identities like %331. Attempting to test it for x=.5, say: In[368]:= N[List @@ % /. x -> .5] Out[368]= {1.15476, 0.333333} This appears to be some sort of bug in Mathematica's bilateral product numerics, since if we break it in two someplace, In[339]:= detour[%331, -6] Out[339]= Product[1 - x^2^k + x^2^(1 + k), {k, -5, Infinity}]* Product[1 - x^2^k + x^2^(1 + k), {k, -Infinity, -6}] == 1/3 In[341]:= N[List @@ %339[[1]] /. x -> .5] Out[341]= {0.340579, 0.978725} In[369]:= Times @@ %341 Out[369]= 0.333333 I'm guessing this bug relates to Product's egregious "procedural" semantics, e.g., In[370]:= Product[f[n], {n, 1, -1}] Out[370]= 1 which, with mathematical semantics would give 1/f[0]. This is like saying Integrate[f[x],{x,1,0}] should give 0 because the limits are backwards. Macsyma errs by default, but gives 1/f(0) if prodhack:true. It NEVER smashes it to the empty product. This leads to blatant contradictions when Mathematica comes up with a closed form (even when you wish it wouldn't), e.g.: In[371]:= Product[k, {k, a}] Out[371]= a! In[373]:= Product[k, {k, -1/2}] -> (-1/2)! Out[373]= 1 -> Sqrt[\[Pi]] --rwg