**Abstract.** We prove the following approximation of $ \pi $

which is derived from the arithmetic-geometric mean of $ 1 $ and $ \sqrt{2}/2 $.

**Note.** Throughout this article tedious steps are often skipped to illustrate the main picture; in particular, computing integral substitutions and power series expansions.

## Table of Contents

## 1. History

The methods used for approximating $ \pi $ span millennia and severely vary in complexity; we will describe a few of the popular methods. The reference here is [1]. Around 200 BC, Archimedes approximated the circumference $ C $ and radius $ r $ of a circle by inscribing it in a polygon with $ n $ sides. It is not hard to see that as $ n $ approaches infinity, we get $ \pi $ using the circumference formula $ C = 2 \pi r $. Then during the invention of calculus in the 1600s, Newton and others used integrals and power series expansions to calculate $ \pi $. For instance, the identity

\[\arctan(x) = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+1}\]evaluated at $ x = 1 $ gives us

\[\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}+\frac{1}{7}-\cdots.\]Furthermore, in the 1700s, Euler calculated the values of the Riemann zeta function. Famously, for $ x=2 $, we get that

\[\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.\]The underlying problem is this family of methods are slow, often taking hundreds of iterations to even yield a couple of digits.

In contrast, the Gauss-Legendre algorithm has quadratic convergence. Let us say we want to calculate 512 decimal places of $ \pi $. Then the Gauss-Legendre algorithm only needs 9 iterations, while almost all of the older methods (normally) need at least 800 iterations, if not significantly more. As an aside, Ramanujan's equation for $ \pi $

\[\frac{1}{\pi} = \frac{2 \sqrt{2}}{9801} \sum_{n=0}^{\infty} \frac{(4n)! (1103+26390n)}{(n!)^4 396^{4n}}.\]only need around $ 65 $ iterations to converge [3]. His equations are now used for large approximations of $ \pi $ due to computational complexity optimization and storage restrictions.

## 2. Elliptic integrals

Let $ (a \cos \theta, b \sin \theta) $ be an ellipse parameterized by $ \theta \in [0, 2 \pi] $. Then it's arc-length is given by

\[\int_{0}^{2 \pi} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} d \theta. \tag{1}\]This integral is not easy to compute in itself, and it was generalized to the study of so-called elliptic integrals in the early 1700s. Now notice that in (1) all of the data concerning our specific ellipse's arc-length is contained within the interval $ 0 \leq \theta \leq \pi/2 $. Hence, it would make sense to reduce our study to complete elliptic integrals, meaning those with amplitude $ \pi/2 $. Furthermore, we will restrict ourselves to complete elliptic integrals of the first and second kind. For the proof that every elliptic integral is of the first, second, or third kind, see [2].

**Definition.** Let

Then $ F(k) $ and $ E(k) $ are called *complete elliptic integrals of the first and second kinds*, respectively, with *symmetric forms* $ F_S(a, b) $ and $ E_S(a, b) $. We refer to $ k $ as the modulus of our integral.

From here on we will assume that by elliptic integral we mean a complete elliptic integral of the first or second kind. We see that the ordinary and symmetric forms of elliptic integrals are related by the following proposition.

**Proposition 2.1.** *Let $ k^2 = 1 - b^2/a^2 $. Then the identities*

*are true.*

*Proof.* Let us prove the second equation with the first following similarly. Substituting, we get

and hence our result. $ \blacksquare $

We also recall a result regarding the symmetry of these integrals originally discovered by Legendre.

**Proposition 2.2.** (Legendre's Identity). *Suppose $ k_1^2 + k_2^2 = 1 $. Then*

*holds.*

*Proof.* We leave the details of this proof to the reader. Taking the derivative with respect to $ k_1 $ shows that the L.H.S. is constant. To see that this value is $ \pi/2 $, we take the limit as $ k_1 $ goes to $ 0 $. $ \blacksquare $

## 3. Main results

We now prove the Gauss-Legendre algorithm. We will not discuss error analysis, which is done in [4].

**Definition.** Let $ a_0, b_0 \in \mathbb{R} $. Let $ a_{n+1} = (a_n + b_n)/2 $ and $ b_{n+1} = (a_n b_n)^{1/2} $ be the arithmetic and geometric means, respectively, of the $ n $th terms. Then we call their common limit

the *arithmetic-geometric (AM-GM) mean* of $ a_0 $ and $ b_0 $.

**Theorem** (Gauss-Legendre). *Set $ a_0 = 1 $ and $ b_0 = \sqrt{2}/2 $. Then the series*

*converges to $ \pi $.*

We need two lemmas in order to prove our result.

**Lemma 3.1.** *Let $ a_0 = a $ and $ b_0 = b $ as in definition 3.1. Set*

*Then $ E_S(a, b) = S F_S(a, b) $.*

*Proof.* This proof is taken from [5]. Consider the integral

Explicitly, this expands to

\[L(a, b) = (a^2 - b^2) \int_0^{\pi/2} \frac{\sin^2 \theta}{\sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta}} d \theta.\]Substituting $ x^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta $,

\[L(a, b) = \int_b^a \sqrt{\frac{a^2-x^2}{x^2-b^2}} dx.\]Now substituting $ y = (x+ab/x)/2 $ and considering the associated AM-GM sequence,

\[\begin{aligned} L(a, b) & = \frac{1}{2} \int_{b_1}^{a_1} \frac{(a^2 - b^2) + 4(a_1^2 - y^2)}{\sqrt{(a_1^2 - y^2)(y^2 - b_1^2)}} dy \\ & = \frac{1}{2} (a^2 - b^2) F_S(a, b) + 2 L(a_1, b_1), \end{aligned}\]and thus

\[\frac{L(a, b)}{F_S(a, b)} = \frac{1}{2}(a_0^2 - b_0^2) + 2 \frac{L(a_1, b_1)}{F_S(a, b)}.\]Since $ 2^n(a_n^2 - b_n^2) \to 0 $, we get $ 2^n L(a_n, b_n) \to 0 $, and hence repeatedly applying this identity we get

\[L(a, b) = \sum_{n=0}^{\infty} 2^{n-1} (a_n^2 - b_n^2) F_S(a, b),\]or equivalently

\[L(a, b) = - S F_S(a, b) + a^2 F_S(a, b).\]Combining this equation with (2) gives us the result. $ \blacksquare $

**Lemma 3.2.** *We have*

*In particular,*

*Proof.* It is not hard to calculate the power series expansion

(The "!!" means double factorial.) Calculating the AM and GM means gives us

\[M(1+k, 1-k) = M(1, \sqrt{1-k^2}).\]Hence, applying proposition 2.1 for $ b/a = \sqrt{1-k^2} $, we get

\[a F_S(a, b) = \frac{\pi}{2 M(1, b/a)} \\ F_S(a, b) = \frac{\pi}{2 M(a, b)},\]with the last equality following from the identity $ M(ca, cb) = cM(a, b) $. $ \blacksquare $

We finally have the tools we need to prove the Gauss-Legendre Algorithm does indeed converge to $ \pi $.

*Proof (Gauss-Legendre).* Set $ k = \sqrt{2}/2 $ to be our modulus. Then we notice $ 2k^2 = 1 $, and hence we can apply proposition 2.2 to get

Now let us evaluate these integrals by first converting them into symmetric form then applying our lemmas. Since $ k^2 = 1 - 1/k^2 $, we can set $ a = 1 $ and $ b = k $ in proposition 2.1 to get

\[2 F_S(1, k) E_S(1, k) - F_S(1, k)^2 = \frac{\pi}{2}.\]Then applying lemma 3.1 gives us an equation only dependent on $ F_S(1, k) $

\[(2 S - 1) F_S(1, k)^2 = \frac{\pi}{2}.\]Finally, we apply lemma 3.2 to write our equation in terms of the AM-GM mean

\[\begin{aligned} \frac{\pi}{2} & = (2S-1) \left( \frac{\pi}{2 M(1, k)} \right)^2 \\ \pi & = \frac{2M(1, k)^2}{2S-1}. \end{aligned}\]Plugging in $ S $ gives us our result. $ \blacksquare $

## 4. References

- David H. Bailey, Simon M. Plouffe, Peter B. Borwein, and Jonathan M. Borwein,
*The quest for $ \pi $*, The Mathematical Intelligencer**19**(1997), no. 1, 50–56. - Gosta Mittag-Leffler,
*An introduction to the theory of elliptic functions*, Annals of Mathematics**24**(1923), no. 4, 271–351. - Srinivasa Ramanujan,
*Modular equations and approximations to $ \pi $*, The Quarterly Joural of Pure and Applied Mathematics**45**(1914), 350372. - Eugene Salamin,
*Computation of $ \pi $ using arithmetic-geometric mean*, Mathematics of Computation**30**(1976), no. 135, 565–570. - Paramanand Singh,
*$ \pi $ and the AGM: Evaluating elliptic integrals*, 2009. Link.