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?—rwgHmm, 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 Envoyé depuis Yahoo Mail pour Android