After a heavy dose of elliptic integral theory in the previous posts we can now prove the celebrated AGM formula for $\pi$ given independently by Gauss, Richard P. Brent and Eugene Salamin. So here we go
π(PI) and the AGM: Evaluating Elliptic Integrals contd.
Complete Elliptic Integrals of Second Kind
After getting familiar with the AGM sequences and Landen Transformation, it is time to apply these concepts to evaluate elliptic integrals. Here we are going to focus on elliptic integrals of the second kind. To be more specific we are going to deal with $ J(a, b)$ defined by
$$ J(a, b) = \int_{0}^{\pi / 2}\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}\,d\theta $$
Our strategy (well actually Landen's and Legendre's) here will be to
analyze the defining integral under the Landen transformation
$$ \tan(\phi - \theta) = \frac{b}{a}\tan\theta $$
By
Paramanand Singh
Friday, August 14, 2009
π(PI) and the AGM: Evaluating Elliptic Integrals
Complete Elliptic Integrals of First Kind
In my earlier post I described the method for calculating complete elliptic integrals of first kind namely $ K(k)$ and $ I(a, b)$. To summarize the results we have$$ K(k) = \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \frac{\pi}{2M(1, \sqrt{1 - k^{2}})} = \frac{\pi}{2M(1, k^{\prime})}$$ $$ I(a, b) = \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}} = \frac{\pi}{2M(a, b)}$$ where $ M(a, b)$ denotes the Arithmetic-Geometric Mean of two numbers $ a$ and $ b$ and $ k^{\prime}$ is the complementary modulus related to $ k$ by the following relation $$ k^{2} + k^{\prime 2} = 1$$
By
Paramanand Singh
Thursday, August 13, 2009
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