Gauss: Arithmetic-Geometric Mean, Elliptic Functions, Approximations to Pi

Gauss-Legendre algorithm:

$\displaystyle a_{0}=1 \quad b_{0}=\frac{1}{\sqrt{2}} \quad c_{0}=\frac{1}{4} \quad p_{0}=1$

$\displaystyle \begin{aligned} &a_{n+1}=\frac{a_{n}+b_{n}}{2}\\ \\&\begin{array}{l} b_{n+1}=\sqrt{a_{n} b_{n}} \\ \\t_{n+1}=t_{n}-p_{n}\left(a_{n}-a_{n+1}\right)^{2} \\\\ p_{n+1}=2 p_{n} \end{array} \end{aligned}$

$\displaystyle \pi = \lim_{n \rightarrow \infty} \frac{\left(a_{n+1}+b_{n+1}\right)^{2}}{4 t_{n+1}}$

The validity of the above algorithm can be seen as a consequence of the following fact:

$\displaystyle \pi= \lim_{n \rightarrow \infty} \frac{\left(a_{n+1}+b_{n+1}\right)^{2}}{4 t_{n+1}}=\dfrac{2 M\left(1, \frac{1}{\sqrt{2}}\right)^{2}}{1-\displaystyle \sum_{n=0}^{\infty} 2^{n}\left(a_{n}^{2}-b_{n}^{2}\right)}$

Proof of this formula: It will follows from the relation between Arithmetic-Geometric mean and the elliptic integrals. In fact, it it closely related to Legendre relation between elliptic integrals of first and second kind.

Details:

Arithmetic-Geometric Mean: ${M(a,b)}$

$\displaystyle \begin{aligned} &a_{0}=a\\ &b_{0}=b \end{aligned}$

$\displaystyle \begin{array}{l} a_{n+1}=\frac{1}{2}\left(a_{n}+b_{n}\right) \\ b_{n+1}=\sqrt{a_{n} b_{n}} \end{array}$

$\displaystyle a_n, b_n \rightarrow_{n\rightarrow \infty} M(a,b)$

The convergence is very fast- It’s a quadratic convergence

${|a_{n+1} -M(a,b)| =O\left(|a_n-M(a.b)|^2\right)}$

Consider

$\displaystyle c_{n}^{2}=a_{n}^{2}-b_{n}^{2}$

We see

$\displaystyle c_{n+1}= \frac{1}{2}\left(a_{n}-b_{n}\right)=a_{n}-a_{n+1}=a_{n+1}-b_n$

2. AGM and Elliptic Integrals (First kind):

$\displaystyle I(a,b)=\int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta}}$

$\displaystyle I(a,b)=\frac{\pi}{2 M(a, b)}$

Proof: The integral ${I(a,b)}$ can be seen to be invariant under the AM, GM transformation ${(a,b) \rightarrow (\frac{a+b}{2}, \sqrt{ab}).}$ That is

$\displaystyle I(a,b)=I(a_n, b_n)$

Hence it should be equal to the quantity obtained by setting both coordinates equal to the limit ${M(a,b).}$

(Invariance): Change of variables

$\displaystyle \sin \theta=\frac{2 a \sin t}{(a+b)+(a-b) \sin ^{2}t}$ gives

$\displaystyle I(a, b)=\int_{0}^{\pi / 2} \frac{d t}{\sqrt{\left(\frac{1}{2}(a+b)\right)^{2} \cos ^{2} t+(\sqrt{a b})^{2} \sin ^{2} t}}= I (\frac{a+b}{2}, \sqrt{ab})$

And it’s easy to see that

$\displaystyle I(x, x) =\frac{\pi}{2x}$

Therefore

$\displaystyle I(a, b)=I(M(a,b), M(a,b)) =\frac{\pi}{2M(a,b)}$

Unsymmetric Form of the elliptic integral:

$\displaystyle K(k)=\int_{0}^{\pi / 2} \frac{1}{\sqrt{1-k^{2} \sin ^{2} \theta}} d \theta= \int_{0}^{1} \frac{\mathrm{d} t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}$

The two forms are related by

$\displaystyle K(k) =I(1, \sqrt{1-k^2})$

The AGM relation looks like

$\displaystyle K(k)=\frac{\pi}{2 M(1, \sqrt{1-k^{2}})}$

$\displaystyle M(1, \cos \phi)=\frac{\pi}{2K(\sin \phi)}$

3. Elliptic Integrals of the second kind:

$\displaystyle J(a, b)=\int_{0}^{\pi / 2} \sqrt{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta} d \theta$

$\displaystyle E(k)=\int_{0}^{\pi / 2} \sqrt{1-k^{2} \sin ^{2} \theta} d \theta=\int_{0}^{1} \frac{\sqrt{1-k^{2} t^{2}}}{\sqrt{1-t^{2}}} d t$

$\displaystyle J(a, b)=a E(k), k^{2}=1-\frac{b^{2}}{a^{2}}$

4. Legendre Identity is a relation between first and second elliptic integrals which says:

if ${k}$ and ${k'}$ satisfy $k^{2}+k^{\prime 2}=1$ , then we have

$\displaystyle K(k) E\left(k^{\prime}\right)+K\left(k^{\prime}\right) E(k)-K(k) K\left(k^{\prime}\right)=\frac{\pi}{2}$

To prove this identity, differentiate ${K(k) E\left(k^{\prime}\right)+K\left(k^{\prime}\right) E(k)-K(k) K\left(k^{\prime}\right)}$ with respect to ${k}$ and show that it is zero when ${k^2+k'^2 =1.}$

5. AGM iterations and the elliptic integrals:

First integral is invariant

$\displaystyle I(a_1, b_1) = I(a,b)$

as we have seen before.

Second integral satisfies

$\displaystyle 2 J\left(a_{1}, b_{1}\right)=a b I(a, b)+J(a, b)$

Let $\displaystyle S=\sum_{n=0}^{\infty} 2^{n-1} c_{n}^{2}$

Iterating this formula gives

$\displaystyle J(a, b)=\left(a^{2}-S\right) I(a, b)$

Proof: Introduce

$\displaystyle L(a, b)=\left(a^{2}-b^{2}\right) \int_{0}^{\pi / 2} \frac{\sin ^{2} \theta d \theta}{\sqrt{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta}}$

Note that

$\displaystyle L(a, b)=a^{2} I(a, b)-J(a, b)$

We can check that

$\displaystyle L(a, b)=(1 / 2)\left(a^{2}-b^{2}\right) I(a, b)+2 L\left(a_{1}, b_{1}\right)$

Thus we get

$\displaystyle 2 J\left(a_{1}, b_{1}\right)=a b I(a, b)+J(a, b)$

Iterating it

$\displaystyle L(a, b)=\frac{1}{2}\left(c_{0}^{2}+2 c_{1}^{2}+\cdots+2^{n-1} c_{n-1}^{2}\right){I(a, b)}+2^{n} L\left(a_{n}, b_{n}\right)$

$\displaystyle 2^{n} L\left(a_{n}, b_{n}\right) \rightarrow 0$

$\displaystyle \implies L(a, b)=\left(\frac{1}{2} \sum_{n=0}^{\infty} 2^{n} c_{n}^{2}\right) I(a, b) =SI(a,b)$

Therefore

$\displaystyle J(a, b)=\left(a^{2}-S\right) I(a, b)$

Final proof of the formula for ${\pi}$ :

Legendre identity witj ${k=k' =\frac{1}{\sqrt2}}$ gives

$\displaystyle 2 K\left(\frac{1}{\sqrt{2}}\right) E\left(\frac{1}{\sqrt{2}}\right)-K\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{\pi}{2}$

We have from before that

$\displaystyle E\left(\frac{1}{\sqrt{2}}\right) =J\left(1, \frac{1}{\sqrt{2}}\right) \\ =\left(1-\frac{1}{2} \sum_{n=0}^{\infty} 2^{n}c_n^2\right) I\left(1, \frac{1}{\sqrt{2}}\right)= \left(1-\frac{1}{2} \sum_{n=0}^{\infty} 2^{n}c_n^2\right) K\left(\frac{1}{\sqrt{2}}\right)$

$\displaystyle \implies \left(1- \sum_{n=0}^{\infty} 2^{n}c_n^2\right) K\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{\pi}{2}$

But we have

$\displaystyle K\left(\frac{1}{\sqrt{2}}\right)= \frac{\pi}{2 M\left(1, \frac{1}{\sqrt{2}}\right)}$

Hence

$\displaystyle \pi=\dfrac{2 M\left(1, \frac{1}{\sqrt{2}}\right)^{2}}{\displaystyle 1-\sum_{n=0}^{\infty} 2^{n}c_n^2}$

$\displaystyle = \lim_{n \rightarrow \infty}\frac{\left(a_{n+1}+b_{n+1}\right)^{2}}{4 t_{n+1}}$

Miscellaneous formulae:

$\displaystyle K(0)=\frac{\pi}{2}$

$\displaystyle K(k)=\int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=\frac{\pi}{2 M(1, \sqrt{1-k^{2}})}=\frac{\pi}{2 M\left(1, k^{\prime}\right)}$

$\displaystyle K(k)=\sum_{n=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2^{n} n !} k^{2 n} \int_{0}^{\pi / 2} \sin ^{2 n} \theta d \theta$

$\displaystyle \int_{0}^{\pi / 2} \sin ^{2 n} \theta d \theta=\frac{2 n-1}{2 n} \int_{0}^{\pi / 2} \sin ^{2 n-2} \theta d \theta= \frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n} \cdot \frac{\pi}{2}$

$\displaystyle K(k)= \sum_{n=0}^{\infty}\left(\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n}\right)^{2} k^{2 n}$

$\displaystyle E(k)=\frac{\pi}{2}\left(1-\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)} {2 \cdot 4 \cdot 6 \cdots 2 n} \frac{k^{2 n}}{2n-1}\right)$