[math-fun] telescoping triggish products
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
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 bob baillie>in any case, video like this:http://www.guardian.co.uk/science/2013/feb/15/meteorite-explodes-over-russia... is pretty amazing! And just what I was looking for. http://www.youtube.com/watch?feature=player_embedded&v=gQ6Pa5Pv_io#! Flash: 4:37; Bang: 7:00 But can someone explain the multiplicity of bangs that characterize these bolides? Is it just sonic booms from individual fragments on different trajectories, or maybe similar trajectories with different speeds? And what's with the bifid symmetry? Maybe this is all just a promotion from the Beijing Cruller and Fireworks Collective. Or is it a natural, convective consequence of heating a horizontally cylindrical region of a fluid? And speaking of heat, I haven't heard anyone report feeling the heat of the flash. "Finally", in this heavily viewed video, http://www.youtube.com/watch?v=d5xMYRBpLSI , the driver hangs a right and seems to chase the meteor at high speed. But notice where the cloud disappears behind his(?) roof edge. It's drifting way faster in that same direction. I thought there was little wind in the stratosphere. Did the meteor impart a bunch of momentum? Energy must be fairly cheap in Chelyabinsk to justify so many single pane windows in such a frigid climate. A harbinger of prosperity.
Bill Gosper:
But can someone explain the multiplicity of bangs that characterize these bolides? Is it just sonic booms from individual fragments on different trajectories, or maybe similar trajectories with different speeds?
Here is what Warren Murray (a co-editor for The Guardian) said about this: "Put simply, a body going supersonic creates multiple shock waves and therefore can make multiple booms: one in front, one behind, plus others potentially. This partly depends on its shape, changes in trajectory etc. The space shuttle was always known for two booms. Also, consider how very loud those booms in Russia would have been, and the fact that echoes off the sides of mountains etc would have in turn travelled a very long way and been heard some time afterwards."
Energy must be fairly cheap in Chelyabinsk to justify so many single pane windows in such a frigid climate. A harbinger of prosperity.
On the other hand, they effectively halved the cost of replacement. ;)
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
bob baillie>in any case, video like this:http://www.guardian.co.uk/science/2013/feb/15/meteorite-explodes-over-russia... is pretty amazing!
And just what I was looking for. http://www.youtube.com/watch?feature=player_embedded&v=gQ6Pa5Pv_io#! Flash: 4:37; Bang: 7:00
But can someone explain the multiplicity of bangs that characterize these bolides? Is it just sonic booms from individual fragments on different trajectories, or maybe similar trajectories with different speeds?
And what's with the bifid symmetry? Maybe this is all just a promotion from the Beijing Cruller and Fireworks Collective. Or is it a natural, convective consequence of heating a horizontally cylindrical region of
a fluid? And speaking of heat, I haven't heard anyone report feeling the heat of the flash.
"Finally", in this heavily viewed video, http://www.youtube.com/watch?v=d5xMYRBpLSI ,
the driver hangs a right and seems to chase the meteor at high speed. But notice where the cloud disappears behind his(?) roof edge. It's drifting way faster in that same direction. I thought there was little wind in the stratosphere. Did
the meteor impart a bunch of momentum?
Energy must be fairly cheap in Chelyabinsk to justify so many single pane windows in such a frigid climate. A harbinger of prosperity.
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...]
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
These bilateral products invariant in the unit disk may be common: Product[1 + (-1/2 + (I*Sqrt[7])/2)*z^2^n + z^(3*2^n) + (-1/2 - (I*Sqrt[7])/2)*z^2^(1 + n), {n, -oo,oo}] == I/Sqrt[7] [...]
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.
--rwg
On Sun, Feb 24, 2013 at 11:21 PM, Bill Gosper <billgosper@gmail.com> wrote:
These bilateral products invariant in the unit disk may be common: Product[1 + (-1/2 + (I*Sqrt[7])/2)*z^2^n + z^(3*2^n) + (-1/2 - (I*Sqrt[7])/2)*z^2^(1 + n), {n, -oo,oo}] == I/Sqrt[7]
Product[((I + z^2^k)*(-I + z^2^(1 + k)))/(1 + z^2^k), {k, -oo, oo}] == -((-1)^(3/4)/(2*Sqrt[2])), Product[1 + z^2^k*(-1 + z^2^k)*((-1)^(1/3) + z^2^k),{k, -oo, oo}] == -(-1)^(2/3)/3 |z|<1. ==rwg
[...]
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.
--rwg
On Mon, Feb 25, 2013 at 10:18 AM, Bill Gosper <billgosper@gmail.com> wrote:
On Sun, Feb 24, 2013 at 11:21 PM, Bill Gosper <billgosper@gmail.com>wrote:
These bilateral products invariant in the unit disk may be common: Product[1 + (-1/2 + (I*Sqrt[7])/2)*z^2^n + z^(3*2^n) + (-1/2 - (I*Sqrt[7])/2)*z^2^(1 + n), {n, -oo,oo}] == I/Sqrt[7]
Product[((I + z^2^k)*(-I + z^2^(1 + k)))/(1 + z^2^k), {k, -oo, oo}] == -((-1)^(3/4)/(2*Sqrt[2])),
Product[1 + z^2^k*(-1 + z^2^k)*((-1)^(1/3) + z^2^k),{k, -oo, oo}] == -(-1)^(2/3)/3
Product[1 + I*Sqrt[2]*z^3^k - z^(2*3^k) + z^(4*3^k) - I*Sqrt[2]*z^3^(1 + k), {k, -oo, oo}] == I/Sqrt[2]
|z|<1.
There seem(s) to be about a dozen of these ^3^k flavor. I've been so amused at the z-invariance that I've neglected to give the underlying telescoping forms. Here's this latest: Product[2*Cos[(2*t)/3^k] + 2*Sqrt[2]*Sin[t/3^k] - 1, {k, 1, Infinity}] == 1 + Sqrt[2]*Sin[t] (∀t) rwg
[...]
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.
--rwg
participants (2)
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Bill Gosper -
Hans Havermann