Hello, two formulas for number Pi Formula 1 Pi=2^(n+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/120)) ; approximative value for n=8 ; 2^(8+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/120)); Formula 2 a(n)=(2^(n)/sqrt(271))*sqrt(A(n)+B(n)-330); A(n)=378900*(u(n))^(1/45)/(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000); B(n)=(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000)/(u(n))^(1/45); u(n+1)=sqrt(2*u(n)+2)/2 ; u(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi this formula becomes: a(n)=sqrt(-(378900*v(n)^(1/45))/(-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)+ (-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)/v(n)^(1/45)-330)/(2^(-n)*sqrt(271)); V(n+1)=sqrt(2*v(n)+2)/2 ; v(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi Best regards
Formula 2 omission of power (1/3) in formula 2 a(n)=(2^(n)/sqrt(271))*sqrt(A(n)+B(n)-330); A(n)=378900*(u(n))^(1/45)/(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000)^(1/3); B(n)=(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000)^(1/3)/(u(n))^(1/45); u(n+1)=sqrt(2*u(n)+2)/2 ; u(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi Le Jeudi 8 février 2018 23h19, françois mendzina essomba2 <m_essob@yahoo.fr> a écrit : Hello, two formulas for number Pi Formula 1 Pi=2^(n+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/120)) ; approximative value for n=8 ; 2^(8+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/120)); Formula 2 a(n)=(2^(n)/sqrt(271))*sqrt(A(n)+B(n)-330); A(n)=378900*(u(n))^(1/45)/(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000); B(n)=(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000)/(u(n))^(1/45); u(n+1)=sqrt(2*u(n)+2)/2 ; u(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi this formula becomes: a(n)=sqrt(-(378900*v(n)^(1/45))/(-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)+ (-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)/v(n)^(1/45)-330)/(2^(-n)*sqrt(271)); V(n+1)=sqrt(2*v(n)+2)/2 ; v(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi Best regards
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françois mendzina essomba2