On 2016-02-18 23:27, Simon Plouffe wrote:
Hello, I think this is Vieta formula for Pi in disguise,
99 terms of it and then multiply by 2,
and it would correspond to about 100 terms of it, http://keisan.casio.com/exec/system/1354861725
Best regards, Simon Plouffe
Interesting! But I think not. (And therefore am not?) Listing just a few terms reveals, e.g. the 6th to be In[842]:= 64 Sqrt[2 - Nest[Sqrt[2 + #] &, 0, 5]] Out[842]= 64 Sqrt[2 - Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2]]]]]] In[843]:= N@% Out[843]= 3.14128 This would have made a much better puzzle, had I noticed it. More rapidly convergent: In[848]:= 96*Sqrt[2 - Nest[Sqrt[2 + #] &, 1, 5]] Out[848]= 96 Sqrt[2 - Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[3]]]]]] In[849]:= N@% Out[849]= 3.14145 This corresponds to In[804]:= 3 Nest[#/(Sqrt[1 - #] + Sqrt[1 + #]) &, .5`33, 52] 2^53 // tim During evaluation of In[804]:= 0.000608 Out[804]= 3.14159265358979323846264338327950 In[805]:= Sin@% Out[805]= 0.*10^-33 This shames my original puzzle, In[787]:= Nest[#/(Sqrt[1 - #] + Sqrt[1 + #]) &, 1`33, 98] 2^99 Out[787]= 3.141592653589793 by not losing significance (which I now almost understand), and, mysteriously requiring only half of the 33/.30103 iterations I'd expect. --rwg