4 ms·
Not sure what the poster is getting at. It seems to be 'normal' case of precision with floating point. In Julia for example, you can define u to be BigFloat (de
by npalli 3y ago
Not sure what the poster is getting at. It seems to be 'normal' case of precision with floating point. In Julia for example, you can define u to be BigFloat (defaulting to 32 bytes for me below). It gets you the correct answer. The python and other default versions of other software are definitely not the "right" answer.
julia> u::BigFloat = 5.0;
julia> for i=0:72
u = 6.0*cos(u)
end
julia> print(u)
2.77675840848277346491725024972478163010030433082748210024033940771878181653911