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.<p><pre><code> julia> u::BigFloat = 5.0;
julia> for i=0:72
u = 6.0*cos(u)
end
julia> print(u)
2.77675840848277346491725024972478163010030433082748210024033940771878181653911</code></pre>