What is the geometric meaning of ∫10x4n(1−x)4n1+x2dx=rn+(−1)n4nπ4∫01x4n(1−x)4n1+x2dx=rn+(−1)n4nπ4?
rn=∫10Qn(x)dxrn=∫01Qn(x)dx with QnQn the quotient in the euclidean division of X4n(1−X)4nX4n(1−X)4n by X2+1X2+1.
So we can approximate ππ by (−14)n−1rn(−14)n−1rn, which is a rationnal. It yields 227227 for n=1n=1.
The absolute error is bounded above by (and below by half this):
14n−1∫10x4n(1−x)4ndx=(4n)!24n−1(8n+1)!∼2π−−√2−10n14n−1∫01x4n(1−x)4ndx=(4n)!24n−1(8n+1)!∼2π2−10n.
So when we increase order by one, we gain approximately 10log10(2)≃310log10(2)≃3 decimals.
