4 Oct
2018
4 Oct
'18
10:17 a.m.
Hello I searched for such a formula with integers, but it may be impossible, a small consolation for me. 1/4*Pi = (arctan(1/3)^2+arctan(2)*(arctan(2/9)^2+arctan(13/16)*(arctan(4/33)^2+arctan(49/128)*(arctan(8/129)^2+arctan(193/1024)*sqrt((a[4])^2+v[4]*sqrt((a[5])^2+v[5]*cdots)))^(1/2))^(1/2))^(1/2))^(1/2); a[n]=arctan(1/(2^n)/(2+1/(2^n)^2))^2 ; v[n]=arctan(1/2/(2^n)^3+3/2/(2^n)); For an exact value, we can associate v [n] with c[n] = arctan (1/2 ^ (n + 1)) Pi/4=(arctan(1/3)^2+arctan(2)*(arctan(2/9)^2+arctan(13/16)*(arctan(4/33)^2+arctan(49/128)*(arctan(8/129)^2+arctan(193/1024)*arctan(1/16))^(1/2))^(1/2))^(1/2))^(1/2); FME...