Gauss–Legendre algorithm: Difference between revisions

Revision as of 04:40, 29 May 2009

Revision as of 04:41, 29 May 2009

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If <math>c_0 = \sin\varphi\!</math>, <math>c_{i+1} = a_i - a_{i+1}\!</math>. then

:<math>\sum_{i=0}^{\infty} 2^{i-1} c_i^2 = 1 - {E(\sin\varphi)\over K(\sin\varphi)}\!</math>

where <math>E(k)\!</math> be the [[Elliptic integral#Complete elliptic integral of the second kind|complete elliptic integral of the second kind]]:

:<math>E(k) = \int_0^{\frac{\pi}{/2}}\sqrt {1-k^2 \sin^2\theta}\, d\theta.\!</math>

Gauss knew of both of these results.<ref name="brent">{{Citation

| last=Brent