Oh, for bleeping out loud. The recurrence c[0]:=0, c[n+1]=2 + √c[n] is identically satisfied by c[n] = 2 + 2 Cos[π/2^n] ! So limit[4^n(4 - c[n]),n->∞] = π^2. —rwg On Sun, Jan 19, 2020 at 4:40 PM Bill Gosper <billgosper@gmail.com> wrote:
I was merely proposing (mostly that some young eavesdroppers) guess the next term numerically: In[190]:= $MaxExtraPrecision = 999; Table[4^n (π^4/12 - 4^n (π^2 - 4^n (4 - c[n]))), {n, 30, 60, 10}] // N[#, 33] &
Out[190]= {2.67052553770917899134237599660117, 2.67052553770917899175060917637930, 2.67052553770917899175060956570080, 2.67052553770917899175060956570117}
In[192]:= ContinuedFraction[2.6705255377091789917506095657011663853881847333364622039593/π^6]
Out[192]= {0, 360}
So π⁶/360. A couple more and we might guess the general term. —Bill
On Sun, Jan 19, 2020 at 10:25 AM françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
Thank you Bill, Exercise: What's the next term?
if only i could know Le dimanche 19 janvier 2020 à 18:34:09 UTC+1, Bill Gosper < billgosper@gmail.com> a écrit :
François, that's an improbably simple (and rather rapid, 4^-n) recurrence for π²: c[0]=0; c[1 + n] == 2 + √c[n]
In[178]:= π^2 - Table[4^n (4 - c[n]), {n, 9}] // N
Out[178]= {1.86960440108936, 0.49702139905888, 0.126184562534061, 0.0316679675433385, 0.00792462574866804, 0.00198163386169981, 0.000495438310220919, 0.000123861437653972, 0.0000309654596399867}
In[179]:= Ratios@%
Out[179]= {0.265843083579222, 0.25388154870795, 0.250965466039401, \ 0.250241059449836, 0.250060245688306, 0.250015060701447, \ 0.250003754450763, 0.250000809182387}
But here's a "surprise": In[181]:= Table[4^n (π^2 - 4^n (4 - c[n])), {n, 9, 19}] // N
Out[181]= {8.11740945186466, 8.11746495589614, 8.11634419858456, 8.12162679433823, 8.48650717735291, -2.05397129058838, 55.7841148376465, 1247.13645935059, -27779.4541625977, -635405.816650391, -6.73592726660156*10^6}
Not really. This is an artifact of N defaulting to machine precision. In[185]:= Table[4^n (π^2 - 4^n (4 - c[n])), {n, 9, 49, 5}] // N[#, 33] &
During evaluation of In[185]:= N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating {262144 (-262144 (2-Sqrt[Plus[<<2>>]])+\[Pi]^2),268435456 (-268435456 (2-Sqrt[Plus[<<2>>]])+\[Pi]^2),274877906944 (-274877906944 (2-Sqrt[Plus[<<2>>]])+\[Pi]^2),<<4>>,309485009821345068724781056 (-309485009821345068724781056 (2-Sqrt[Plus[<<2>>]])+\[Pi]^2),316912650057057350374175801344 (-316912650057057350374175801344 (2-Sqrt[Plus[<<2>>]])+\[Pi]^2)}.
Out[185]=
{8.11741406559391897533420324219153058693403049295814616361461007478026967, 8.11742424288505356547897388334305511999233890228014150764514550352, 8.11742425282382112106005075892777291648772368657965851167370, 8.11742425283352694875742031860153490434360064813704146, 8.11742425283353642710478103713936165445769267858, 8.11742425283353643636097963159106454657039, 8.11742425283353643637001888803095879, 8.11742425283353643637002771543, 8.11742425283353643637003}
Ries pegs this as π⁴/12. Exercise: What's the next term? —rwg Hmm, Baskin-Robbins slogan: π³ colors.
On Sat, Jan 18, 2020 at 11:57 PM françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
Hi,
a [n+1]=2^(2*n+3)*(1-(1-a [n]/2^(2*n+2))^(1/2))
a [0]=4 ;
n----> infinity
a [n]----> Pi^2
We see that:
1-a [n]/2^(2*n+2)=(1-a [n+1]/2^(2*n+3))^2
Best regards
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