[math-fun] What do you mean, "Fake news"?
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16) In[372]:= N[%364,105] Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 ==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg
Whoa! On Tue, Aug 4, 2020 at 9:40 AM Bill Gosper <billgosper@gmail.com> wrote:
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg
As someone late to the party, what trick is being use to create these conspiracy theories? My guess is that one tries to express the log of the target number (Gamma[1/4]) as an approximate rational combination of the logs of “the usual suspects” (e.g. golden mean), in the same way as subset-sum, where the approximation is set up as finding a short vector in a lattice using LLL. -Veit
On Aug 4, 2020, at 12:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg
*Grin* I guess the question is, are there enough expressions of the complexity of this one, to make it plausible that every 106-significant-figure constant can be the value of one? Just eyeballing it, the complexity looks about right. I would echo Greg Huber's "Whoa!" if the expression were, say, half as long. I don't suppose Gosper is willing to show us N[%364,200]. Of course the *real* question is Veit's; just because a "representation" exists doesn't mean that it's easy to find, and I have no clue what tricks Gosper is employing. On Tue, Aug 4, 2020 at 1:28 PM Veit Elser <ve10@cornell.edu> wrote:
As someone late to the party, what trick is being use to create these conspiracy theories? My guess is that one tries to express the log of the target number (Gamma[1/4]) as an approximate rational combination of the logs of “the usual suspects” (e.g. golden mean), in the same way as subset-sum, where the approximation is set up as finding a short vector in a lattice using LLL.
-Veit
On Aug 4, 2020, at 12:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
—rwg
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Bill's 'identity' is, of course, very clever. The right-hand-side of the 'identity' is in the algebraic closure of pi and exp(pi), whereas Yuri Nesterenko proved in 1996 that pi, exp(pi), and Gamma(1/4) are all algebraically independent of each other. My suspicion is that Bill deliberately chose a lattice generated by a large bunch of simple expressions in the algebraic closure of pi and exp(pi) and looked for small linear relations (using LLL, as Viet Elser mentioned) to engineer a 'counter-example' to Nesterenko's result. Best wishes, Adam P. Goucher
Sent: Tuesday, August 04, 2020 at 5:39 PM From: "Bill Gosper" <billgosper@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] What do you mean, "Fake news"?
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
In[382]:= N[Gamma[1/3], 85] Out[382]= 2.678938534707747633655692940974677644128689377957301100950428327590417610167743819541 In[383]:= N[(2^(14/9) 3^(55/108) (π Sinh[9 √3 π])^(2/3))/(6^(1/3) (1 + 2^(2/3))/(1 + 2^(1/3)) - 3^(2/3))^(2/9)/√E^(13 √3 π), 85] Out[383]= 2.67893853470774763365569294097467764412868937795730110095042832759041\7610167743819541 On Tue, Aug 4, 2020 at 9:39 AM Bill Gosper <billgosper@gmail.com> wrote:
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg
Okay good, thanks Bill, we only need about 60 digits anyways, and some of us are having trouble with the gamma function. —Brad
On Aug 6, 2020, at 10:46 AM, Bill Gosper <billgosper@gmail.com> wrote:
In[382]:= N[Gamma[1/3], 85]
Out[382]= 2.678938534707747633655692940974677644128689377957301100950428327590417610167743819541
In[383]:= N[(2^(14/9) 3^(55/108) (π Sinh[9 √3 π])^(2/3))/(6^(1/3) (1 + 2^(2/3))/(1 + 2^(1/3)) - 3^(2/3))^(2/9)/√E^(13 √3 π), 85]
Out[383]= 2.67893853470774763365569294097467764412868937795730110095042832759041\7610167743819541
On Tue, Aug 4, 2020 at 9:39 AM Bill Gosper <billgosper@gmail.com> wrote:
Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16)
In[372]:= N[%364,105]
Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147
==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg
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participants (6)
-
Adam P. Goucher -
Allan Wechsler -
Bill Gosper -
Brad Klee -
Greg Huber -
Veit Elser