π(PI) and the AGM: Gauss-Brent-Salamin Formula

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

Gauss-Brent-Salamin Formula for $\pi$

$$ \boxed{\displaystyle \pi = \dfrac{2\left\{M\left(1, \dfrac{1}{\sqrt{2}}\right)\right\}^{2}}{\displaystyle 1 - \sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})}}$$ where
$$ a_{0} = 1, b_{0} = \frac{1}{\sqrt{2}}, a_{n + 1} = \frac{a_{n} + b_{n}}{2}, b_{n + 1} = \sqrt{a_{n}b_{n}}$$ This formula can be easily proved by applying the following results already established in previous posts \begin{align} 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)}\tag{1}\\ I(a, b) &= \frac{1}{a}K(k) = \frac{1}{a}\int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \frac{\pi}{2aM(1, k')}\text{ where}\tag{2}\\ k^{2} &= 1 - \frac{b^{2}}{ a^{2}},\, k^{2} + k'^{2} = 1\notag\\ J(a, b) &= \int_{0}^{\pi / 2}\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}\,d\theta\notag\\ &= \left(a^{2} - \frac{1}{2}\sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})\right)I(a, b)\tag{3}\\ J(a, b) &= aE(k) = a\int_{0}^{\pi / 2}\sqrt{1 - k^{2}\sin^{2}\theta}\,d\theta\text{ where }\tag{4}\\ k^{2} &= 1 - \frac{b^{2}}{a^{2}}\notag \end{align} and Legendre's Identity $$K(k)E(k') + K(k')E(k) - K(k)K(k') = \frac{\pi}{2}$$ where $ k^{2} + k'^{2} = 1$.

Proof of Gauss-Brent-Salamin Formula

In Legendre's Identity we put $ k = k' = 1 / \sqrt{2}$ to get $$2K\left(\frac{1}{\sqrt{2}}\right)E\left(\frac{1}{\sqrt{2}}\right) - K\left(\frac{1}{\sqrt{2}}\right)^{2} = \frac{\pi}{2}$$ Also putting $ a = 1, b = 1 / \sqrt{2} \Rightarrow k = k' = 1 / \sqrt{2}$ in $(4)$ we get (and also using $(3)$) \begin{align} E\left(\frac{1}{\sqrt{2}}\right) &= J\left(1, \frac{1}{\sqrt{2}}\right)\notag\\ &= \left(1 - \frac{1}{2}\sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})\right)I\left(1, \frac{1}{\sqrt{2}}\right)\notag\\ &= X\cdot K\left(\frac{1}{\sqrt{2}}\right)\notag\end{align} where $$X = \left(1 - \frac{1}{2}\sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})\right)$$ Using this value of $ E(1, 1 / \sqrt{2})$ in Legendre's Identity we get \begin{align} (2X - 1)\cdot K\left(\frac{1}{\sqrt{2}}\right)^{2} &= \frac{\pi}{2}\notag\\ \Rightarrow (2X - 1) \cdot \dfrac{\pi^{2}}{4M\left(1, \dfrac{1}{\sqrt{2}}\right)^{2}} &= \frac{\pi}{2}\notag\\ \Rightarrow \pi = \dfrac{2M\left(1, \dfrac{1}{\sqrt{2}}\right)^{2}}{2X - 1} &= \dfrac{2M\left(1, \dfrac{1}{\sqrt{2}}\right)^{2}}{\displaystyle 1 - \sum_{n = 0}^{\infty}2^{n}(a_{n}^{2} - b_{n}^{2})}\notag \end{align} This formula has intrigued me for several years (may be 12 years since I saw it mentioned in some calculus book when I was in 11th grade) but I was able to find a complete proof very recently which led to the creation of these posts. Initially I was impressed by the rapid convergence, but did not pursue further about the proof (12 years ago I didn't have internet facility) and later in college years I had almost completely lost interest in it. Luckily I happened to read an online paper "Gauss, recurrence relations, and the agM" by Stacy G. Langton and in it I got the Gauss's original proof connecting $ AGM(1, \sqrt{2})$ and the lemniscatic integral $ \int_{0}^{1} dx / \sqrt{1 - x^{4}}$

This proof heightened my interest in AGM and elliptic integrals. In fact the proof (presented in a previous post) is so marvelous that I couldn't stop wondering about Gauss (he got the essential ideas when he was 14). The simplicity of the proof is what excites me most (it can be understood by anyone who has idea of calculus and binomial theorem and does not use anything beyond a 12th grade syllabus). Upon reading the Gauss' proof I had the feeling that I could have done it when I met the AGM concept only if I had the sense to know that it was an important concept worth studying. I believe that's what distinguishes the masters from the mere students like us. Anyway all this episode only proves that following Abel's advice "Read the masters" is always a good idea if you want to do something in the field.

I further studied many papers related to AGM available online and found proof of AGM formula for $\pi$, but none of them had the proof for the formula for $ J(a, b)$ which is essential to the proof. This was found in a somewhat obscure book "On the direct numerical calculation of elliptic functions and integrals" by Louis V. King. The proof contained there is very terse and it does not show any algebraical manipulations involved in the Landen's transformation and does away with the proof in 10 lines of mathematical symbols. I have only tried to make the terse proof available to a wider audience.

Print/PDF Version