# 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
$$\tan(\phi - \theta) = \frac{b}{a}\tan\theta$$
In fact we will study the effect of a sequence of these transformations applied to the defining integral for $J(a, b)$. Setting the stage with the following initial values
$$a_{0} = a,\, b_{0} = b,\, c_{1} = \frac{a - b}{2}$$
and
$$a_{n+1} = \frac{a_{n} + b_{n}}{2},\, b_{n + 1} = \sqrt{a_{n}b_{n}},\, c_{n + 1} = \frac{a_{n} - b_{n}}{2}$$
$$\tan(\theta_{n + 1} - \theta_{n}) = \frac{b_{n}}{a_{n}}\tan\theta_{n}$$
$$\Delta_{n} = \sqrt{a_{n}^{2}\cos^{2}\theta_{n} + b_{n}^{2}\sin^{2}\theta_{n}}$$
(note that unlike sequences $\{a_{n}\}, \{b_{n}\}, \{c_{n}\}$, the sequence $\{\theta_{n}\}$ is a sequence of variable amplitudes and not of constant real numbers) we are going to establish the following in order
1. $\Delta_{n + 1} + c_{n + 1}\cos\theta_{n + 1} = \Delta_{n}$
2. $\Delta_{n + 1} - c_{n + 1}\cos\theta_{n + 1} = a_{n}b_{n} / \Delta_{n}$
3. $2\Delta_{n + 1} = \Delta_{n} + \dfrac{a_{n}b_{n}}{\Delta_{n}}$
4. $\dfrac{d\theta_{n}}{\Delta_{n}} = \dfrac{1}{2}\cdot\dfrac{d\theta_{n + 1}}{\Delta_{n + 1}}$
5. $\Delta_{n}\Delta_{n + 1} = \dfrac{\Delta_{n}^{2} + a_{n}b_{n}}{2}$
6. $\Delta_{n}d\theta_{n} - \Delta_{n + 1}d\theta_{n + 1} = c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} - \dfrac{a_{n}b_{n}}{\Delta_{n}}d\theta_{n}$

### Initialization

And now come the laborious manipulations which must be appreciated for the fact they lead to such simple looking results in spite of their inherent complexity. The starting point here is to express $\theta_{n + 1}$ in terms of $\theta_{n}$ and calculate the derivative $d\theta_{n + 1} / d\theta_{n}$ as follows
$$\tan(\theta_{n + 1} - \theta_{n}) = \frac{b_{n}}{a_{n}}\tan\theta_{n}$$ $$\Rightarrow a_{n}(\tan\theta_{n + 1} - \tan\theta_{n}) = b_{n}\tan\theta_{n}(1 + \tan\theta_{n + 1}\tan\theta_{n})$$ $$\Rightarrow (a_{n} - b_{n}\tan^{2}\theta_{n})\tan\theta_{n + 1} = (a_{n} + b_{n})\tan\theta_{n}$$ $$\Rightarrow \tan\theta_{n + 1} = \frac{(a_{n} + b_{n})\tan\theta_{n}}{a_{n} - b_{n}\tan^{2}\theta_{n}}$$ \begin{aligned}\Rightarrow \cos\theta_{n + 1} &= \frac{a_{n} - b_{n}\tan^{2}\theta_{n}}{\sqrt{(a_{n} - b_{n}\tan^{2}\theta_{n})^{2} + (a_{n} + b_{n})^{2}\tan^{2}\theta_{n}}}\\ &= \frac{a_{n} - b_{n}\tan^{2}\theta_{n}}{\sqrt{a_{n}^{2} + b_{n}^{2}\tan^{4}\theta_{n} + (a_{n}^{2} + b_{n}^{2})\tan^{2}\theta_{n}}}\\ &= \frac{a_{n} - b_{n}\tan^{2}\theta_{n}}{\sqrt{(1 + \tan^{2}\theta_{n})(a_{n}^{2} + b_{n}^{2}\tan^{2}\theta_{n})}}\\ &= \frac{a_{n} - b_{n}\tan^{2}\theta_{n}}{\sec^{2}\theta_{n}\sqrt{a_{n}^{2}\cos^{2}\theta_{n} + b_{n}^{2}\sin^{2}\theta_{n}}}\\ &= \frac{a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n}}{\Delta_{n}}\end{aligned} Similarly
$$\sin\theta_{n + 1} = \frac{(a_{n} + b_{n})\sin\theta_{n}\cos\theta_{n}}{\Delta_{n}}$$ Also we have (from the defining equation for $\theta_{n + 1}$)
\begin{aligned}\frac{d\theta_{n + 1}}{d\theta_{n}} &= (a_{n} + b_{n})\cos^{2}\theta_{n + 1}\frac{(a_{n} - b_{n}\tan^{2}\theta_{n})\sec^{2}\theta_{n} + 2b_{n}\tan^{2}\theta_{n}\sec^{2}\theta_{n}}{(a_{n} - b_{n}\tan^{2}\theta_{n})^{2}}\\ &= (a_{n} + b_{n})\cos^{2}\theta_{n + 1}\frac{a_{n} + b_{n}\tan^{2}\theta_{n}}{\cos^{2}\theta_{n}(a_{n} - b_{n}\tan^{2}\theta_{n})^{2}}\\ &= (a_{n} + b_{n})\cos^{2}\theta_{n + 1}\frac{a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n}}{(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})^{2}}\end{aligned} $$\Rightarrow \frac{d\theta_{n + 1}}{d\theta_{n}} = \frac{(a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})}{\Delta_{n}^{2}}$$ Now that the preliminary setup is done we can prove the desired results
Proof for 1) $$\Delta_{n + 1} + c_{n + 1}\cos\theta_{n + 1} = \Delta_{n}$$ We have
\begin{aligned}\Delta_{n + 1} &= \sqrt{a_{n + 1}^{2}\cos^{2}\theta_{n + 1} + b_{n + 1}^{2}\sin^{2}\theta_{n + 1}}\\ &= \frac{\sqrt{(a_{n} + b_{n})^{2}\cos^{2}\theta_{n + 1} + 4a_{n}b_{n}\sin^{2}\theta_{n + 1}}}{2}\\ &= \frac{(a_{n} + b_{n})\sqrt{(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})^{2} + 4a_{n}b_{n}\sin^{2}\theta_{n}\cos^{2}\theta_{n}}}{2\Delta_{n}}\\ &= \frac{(a_{n} + b_{n})\sqrt{(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})^{2}}}{2\Delta_{n}}\\ &= \frac{(a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})}{2\Delta_{n}}\end{aligned} Therefore (from definition of $c_{n + 1}$ and value of $\cos\theta_{n + 1}$)
\begin{aligned}\Delta_{n + 1} + c_{n + 1}\cos\theta_{n + 1} &= \frac{(a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})}{2\Delta_{n}}\\ &\,\,\,\,\,\,\,\, +\, \frac{(a_{n} - b_{n})(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})}{2\Delta_{n}}\\ &= \frac{2(a_{n}^{2}\cos^{2}\theta_{n} + b_{n}^{2}\sin^{2}\theta_{n})}{2\Delta_{n}} = \frac{\Delta_{n}^{2}}{\Delta_{n}} = \Delta_{n}\end{aligned} Proof for 2) $$\Delta_{n + 1} - c_{n + 1}\cos\theta_{n + 1} = a_{n}b_{n} / \Delta_{n}$$ Similarly
\begin{aligned}\Delta_{n + 1} - c_{n + 1}\cos\theta_{n + 1} &= \frac{(a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})}{2\Delta_{n}}\\ &\,\,\,\,\,\,\,\, -\, \frac{(a_{n} - b_{n})(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})}{2\Delta_{n}}\\ &= \frac{2a_{n}b_{n}(\cos^{2}\theta_{n} + \sin^{2}\theta_{n})}{2\Delta_{n}} = \frac{a_{n}b_{n}}{\Delta_{n}}\end{aligned} Proof for 3) is obtained from adding 1) and 2).
Proof for 4) $$\frac{d\theta_{n}}{\Delta_{n}} = \frac{1}{2}\cdot\frac{d\theta_{n + 1}}{\Delta_{n + 1}}$$ First we note that
\begin{aligned}\frac{d\Delta_{n}}{d\theta_{n}} &= \frac{(b_{n}^{2} - a_{n}^{2})\sin\theta_{n}\cos\theta_{n}}{\Delta_{n}}\\ &= -\frac{c_{n}^{2}\sin\theta_{n}\cos\theta_{n}}{\Delta_{n}}\end{aligned} Differentiating 3) with respect to $\theta_{n}$ we get
$$2\cdot\frac{d\Delta_{n + 1}}{d\theta_{n + 1}}\cdot\frac{d\theta_{n + 1}}{d\theta_{n}} = \frac{d\Delta_{n}}{d\theta_{n}} - \frac{a_{n}b_{n}}{\Delta_{n}^{2}}\frac{d\Delta_{n}}{d\theta_{n}}$$ $$\Rightarrow 2\cdot\frac{d\Delta_{n + 1}}{d\theta_{n + 1}}d\theta_{n + 1} = \left(\frac{d\Delta_{n}}{d\theta_{n}} - \frac{a_{n}b_{n}}{\Delta_{n}^{2}}\frac{d\Delta_{n}}{d\theta_{n}}\right)d\theta_{n}$$ $$\Rightarrow \frac{2c_{n + 1}^{2}\sin\theta_{n + 1}\cos\theta_{n + 1}}{\Delta_{n + 1}}\,d\theta_{n + 1} = \frac{c_{n}^{2}\sin\theta_{n}\cos\theta_{n}}{\Delta_{n}}\left(\frac{\Delta_{n}^{2} - a_{n}b_{n}}{\Delta_{n}^{2}}\right)d\theta_{n}$$ Now putting the values of $\cos\theta_{n + 1}$ and $\sin\theta_{n + 1}$ and noting that
$$c_{n + 1} = \frac{a_{n} - b_{n}}{2},\,\, c_{n}^{2} = a_{n}^{2} - b_{n}^{2}$$ \begin{aligned}\Delta_{n}^{2} - a_{n}b_{n} &= \Delta_{n}^{2} - a_{n}b_{n}(\cos^{2}\theta_{n} + \sin^{2}\theta_{n})\\ &= (a_{n} - b_{n})(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})\end{aligned} we get
$$\frac{(a_{n} - b_{n})^{2}(a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})\sin\theta_{n}\cos\theta_{n}}{2\Delta_{n}^{2}}\frac{d\theta_{n + 1}}{\Delta_{n + 1}}$$ $$= \frac{(a_{n}^{2} - b_{n}^{2})\sin\theta_{n}\cos\theta_{n}(a_{n} - b_{n})(a_{n}\cos^{2}\theta_{n} - b_{n}\sin^{2}\theta_{n})}{\Delta_{n}^{2}}\frac{d\theta_{n}}{\Delta_{n}}$$ $$\Rightarrow \frac{1}{2}\cdot\frac{d\theta_{n + 1}}{\Delta_{n + 1}} = \frac{d\theta_{n}}{\Delta_{n}}$$ Proof for 5) $$\Delta_{n}\Delta_{n + 1} = \frac{\Delta_{n}^{2} + a_{n}b_{n}}{2}$$ We have
\begin{aligned}\Delta_{n}^{2} + a_{n}b_{n} &= \Delta_{n}^{2} + a_{n}b_{n}(\cos^{2}\theta_{n} + \sin^{2}\theta_{n})\\ &= (a_{n} + b_{n})(a_{n}\cos^{2}\theta_{n} + b_{n}\sin^{2}\theta_{n})\\ &= \Delta_{n}^{2}\cdot\frac{d\theta_{n + 1}}{d\theta_{n}}\\ &= \Delta_{n}^{2}\cdot\frac{2\Delta_{n + 1}}{\Delta_{n}} = 2\Delta_{n}\Delta_{n + 1}\end{aligned} (from (4) above)
Proof for 6) $$\Delta_{n}d\theta_{n} - \Delta_{n + 1}d\theta_{n + 1} = c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} - \frac{a_{n}b_{n}}{\Delta_{n}}d\theta_{n}$$ The result in 5) can be rewritten as
$$\Delta_{n} = \frac{\Delta_{n}^{2} + a_{n}b_{n}}{2\Delta_{n + 1}}$$ Replacing $\Delta_{n + 1}$ using 4) we get
$$\Delta_{n} =\frac{\Delta_{n}^{2} + a_{n}b_{n}}{\Delta_{n}}\frac{d\theta_{n}}{d\theta_{n + 1}}$$ $$\Rightarrow \Delta_{n}d\theta_{n + 1} = \left(\Delta_{n} + \frac{a_{n}b_{n}}{\Delta_{n}}\right)d\theta_{n}$$ Replacing $\Delta_{n}$ on L.H.S. using 1) we get
$$\Delta_{n + 1}d\theta_{n + 1} + c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} = \Delta_{n}d\theta_{n} + \frac{a_{n}b_{n}}{\Delta_{n}}d\theta_{n}$$ $$\Rightarrow \Delta_{n}d\theta_{n} - \Delta_{n + 1}d\theta_{n + 1} = c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} - \frac{a_{n}b_{n}}{\Delta_{n}}d\theta_{n}$$
Having established these results it may still not be clear to the reader as to the overall goal of proving these results. The last result gives some indication about the path to reach the goal. It tries to estimate the change in the value of the quantity $\Delta d\theta$ as we apply the Landen transformation. Upon integrating these differentials we can find the change in the integral defining $J(a, b)$ effected due to the Landen transformation. And we hope to add these changes involved in the series of Landen transformations to finally get the value of $J(a, b)$.

It turns out that the series so obtained does not converge (because of the simple fact that $\Delta_{n}$ does not tend to zero as $n \rightarrow \infty$). So we modify our strategy a little bit and instead study the changes in the quantity $(\Delta_{n} - a_{n}^{2} / \Delta_{n})$ (this does tend to zero as $n \rightarrow \infty$).

Armed with this insight we can see that
\begin{aligned}a_{n}^{2}\frac{d\theta_{n}}{\Delta_{n}} - a_{n + 1}^{2}\frac{d\theta_{n + 1}}{\Delta_{n + 1}} &= (a_{n}^{2} - 2a_{n + 1}^{2})\frac{d\theta_{n}}{\Delta_{n}}\\ &= \left(\frac{a_{n}^{2} - b_{n}^{2}}{2} - a_{n}b_{n}\right)\frac{d\theta_{n}}{\Delta_{n}}\end{aligned} Thus
$$a_{n}^{2}\frac{d\theta_{n}}{\Delta_{n}} - a_{n + 1}^{2}\frac{d\theta_{n + 1}}{\Delta_{n + 1}} = \frac{1}{2}c_{n}^{2}\frac{d\theta_{n}}{\Delta_{n}} - \frac{a_{n}b_{n}}{\Delta_{n}}d\theta_{n}$$ Using this result with 6) we get
\begin{aligned}\left(\Delta_{n} - \frac{a_{n}^{2}}{\Delta_{n}}\right)d\theta_{n} - \left(\Delta_{n + 1} - \frac{a_{n + 1}^{2}}{\Delta_{n + 1}}\right)d\theta_{n + 1} &= c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} - \frac{1}{2}c_{n}^{2}\frac{d\theta_{n}}{\Delta_{n}}\\ &= c_{n + 1}\cos\theta_{n + 1}d\theta_{n + 1} - \frac{1}{2}\cdot 2^{n}c_{n}^{2}\frac{d\theta_{o}}{\Delta_{o}}\end{aligned} (using (4) to get from $d\theta_{n}$ to $d\theta_{o}$)
Putting $n = 0, 1, 2, \ldots$ and on adding and integrating the result we get (here we use the fact that $(\Delta_{n} - a_{n}^{2} / \Delta_{n})$ tends to zero as $n \rightarrow \infty$)
$$\int_{0}^{\theta_{0}}\Delta_{0}d\theta - a_{0}^{2}\int_{0}^{\theta_{0}}\frac{d\theta}{\Delta_{0}} = c_{1}\sin\theta_{1} + c_{2}\sin\theta_{2} + c_{3}\sin\theta_{3} + \cdots$$ $$- \frac{1}{2}(c_{0}^{2} + 2c_{1}^{2} + 4c_{2}^{2} + 8c_{3}^{2} + \cdots)\int_{0}^{\theta_{0}}\frac{d\theta}{\Delta_{0}}$$ Putting $\theta_{0} = \pi / 2$ and noting that $\theta_{1} = \pi, \theta_{2} = 2\pi \ldots$ (and in general $\theta_{n} = 2^{n - 1}\pi$) we get
$$J(a, b) = \left(a^{2} - \frac{1}{2}\sum_{n = 0}^{\infty}2^{n}c_{n}^{2}\right)I(a, b)$$
and finally the elliptic integral of second kinds gets expressed in terms of AGM and the associated sequence.

### Formula for J(a, b): An Alternative Approach

The above proof of the formula for $J(a, b)$ is somewhat cumbersome and indirect. By studying a closely related integral $L(a, b)$ defined by
$$L(a, b) = (a^{2} - b^{2}) \int_{0}^{\pi / 2} \frac{\sin^{2}\theta \,\,d\theta}{\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}}$$
one can arrive at the formula for $J(a, b)$ established earlier.

As can be easily seen, this integral is related to $J(a, b)$ and $I(a, b)$ as follows
$$L(a, b) = a^{2}I(a, b) - J(a, b)$$
The idea here is to find a relation between $L(a_{n}, b_{n})$ and $L(a_{n + 1}, b_{n + 1})$. To simplify writing it is sufficient to consider the relationship between $L(a, b)$ and $L(a_{1}, b_{1})$. We start with the substitution
$$t^{2} = a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta$$
so that
$$t\,\,dt / d\theta = (b^{2} - a^{2})\sin\theta\cos\theta = -\sqrt{(a^{2} - t^{2})(t^{2} - b^{2})}$$
and therefore the integral gets transformed into
$$L(a, b) = \int_{b}^{a}\sqrt{\frac{a^{2} - t^{2}}{t^{2} - b^{2}}}\,\,dt$$
Now the substitution $x = (1 / 2)(t + ab / t)$ leads us to
$$a_{1}^{2} - x^{2} = (a^{2} - t^{2})(t^{2} - b^{2}) / 4t^{2}$$
$$x^{2} - b_{1}^{2} = (t^{2} - ab)^{2} / 4t^{2}$$
$$dx / dt = (t^{2} - ab) / 2t^{2}$$
so that
$$\frac{dt}{\sqrt{(a^{2} - t^{2})(t^{2} - b^{2})}} = \pm \frac{1}{2}\frac{dx}{\sqrt{(a_{1}^{2} - x^{2})(x^{2} - b_{1}^{2})}}$$
The ambiguous sign on the right can be resolved if we observe that when $t$ increases from $b$ to $b_{1}$, $x$ decreases from $a_{1}$ to $b_{1}$ (so that we select the - sign here) and when $t$ increases from $b_{1}$ to $a$, $x$ increases from $b_{1}$ to $a_{1}$ (so that we select the + sign here).

Also note that we have
$$t = x \pm \sqrt{x^{2} - b_{1}^{2}}$$
\begin{aligned}L(a, b) &= \int_{b}^{a}\frac{(a^{2} - t^{2})}{\sqrt{(a^{2} - t^{2})(t^{2} - b^{2})}}\,\,dt\\ &= \int_{b}^{a} \frac{\phi(t)}{\sqrt{(a^{2} - t^{2})(t^{2} - b^{2})}}\,\,dt\\ &= \frac{1}{2}\int_{b_{1}}^{a_{1}}\frac{\psi(x)}{\sqrt{(a_{1}^{2} - x^{2})(x^{2} - b_{1}^{2})}}\,\,dx\end{aligned} where $\phi(t) = a^{2} - t^{2}$ and
\begin{aligned}\psi(x) &= \phi(x + \sqrt{x^{2} - b_{1}^{2}}) + \phi(x - \sqrt{x^{2} - b_{1}^{2}})\\ &= a^{2} - (x + \sqrt{x^{2} - ab})^{2} + a^{2} - (x - \sqrt{x^{2} - ab})^{2}\\ &= 2a^{2} + 2ab - 4x^{2} = a^{2} - b^{2} + (a^{2} + 2ab + b^{2}) - 4x^{2}\\ &= (a^{2} - b^{2}) + 4(a_{1}^{2} - x^{2})\end{aligned} We thus get
$$L(a, b) = \frac{a^{2} - b^{2}}{2}\int_{b_{1}}^{a_{1}}\frac{dx}{\sqrt{(a_{1}^{2} - x^{2})(x^{2} - b_{1}^{2})}} + 2\int_{b_{1}}^{a_{1}}\sqrt{\frac{a_{1}^{2} - x^{2}}{x^{2} - b_{1}^{2}}}\,\,dx$$
The first integral on the right hand side is $I(a_{1}, b_{1})$ which is the same as $I(a, b)$ (use the substitution $x^{2} = a_{1}^{2}\cos^{2}\theta + b_{1}^{2}\sin^{2}\theta$) so we have the recursive relation for $L(a,b)$ as follows
$$L(a, b) = (1 / 2)(a^{2} - b^{2})I(a, b) + 2L(a_{1}, b_{1})$$
Dividing both sides by $I(a, b)$ we get
$$\frac{L(a_{0}, b_{0})}{I(a, b)} = \frac{1}{2}c_{0}^{2} + 2\frac{L(a_{1}, b_{1})}{I(a, b)}$$
and a repeated application of this identity gives us
$$\frac{L(a, b)}{I(a, b)} = \frac{1}{2}(c_{0}^{2} + 2c_{1}^{2} + \cdots + 2^{n - 1}c_{n - 1}^{2}) + 2^{n}\frac{L(a_{n}, b_{n})}{I(a, b)}$$
Now $2^{n}L(a_{n}, b_{n}) \rightarrow 0$ as $n \rightarrow \infty$ (this is clear from the fact that $2^{n}(a_{n}^{2} - b_{n}^{2}) \rightarrow 0$ as $n \rightarrow \infty$).
Therefore on letting $n \rightarrow \infty$ we get
$$L(a, b) = \left(\frac{1}{2}\sum \limits_{n = 0}^{\infty}2^{n}c_{n}^{2}\right)I(a, b)$$
Using the relation between $L(a, b)$ and $J(a, b)$ we get
$$J(a, b) = \left(a^{2} - \frac{1}{2}\sum\limits_{n = 0}^{\infty}2^{n}c_{n}^{2}\right)I(a, b)$$
We are now in a position to state and prove the Brent-Salamin AGM formula for $\pi$ and this will be done in the next post.

#### 3 comments :: π(PI) and the AGM: Evaluating Elliptic Integrals contd.

1. Hi there,

I really like this blog, I’ve found it fascinating. I was wondering if you could tell me where you found the alternative proof above?

Thanks.

2. Hi Bob,
I read the alternate proof in some paper online. It’s an old entry so I don’t remember the source now. Sorry.

3. Hi, Paramanand,
I found the limits in this post interesting.

From the Landen transformation section:

$\theta_{n+1} = \arctan(\frac{b_n}{a_n} tan(\theta_n)) + \theta_n$

Near $\pi / 2$, the $\tan()$ approaches $\infty$. Therefore, it doesn't matter much what $b_n$ and $a_n$ are, the $\arctan()$ is still $\pi / 2$, and thus $\theta_{n+1} = 2 \theta_n$.

Later on, it took me a while to see the double sign reversal with $\psi(x)$ .

As noted above, corresponding to $t$ going up from $b \to a$ , $x$ goes down $a_1 \to b_1$ and then up again from $b_1 \to a_1$ . This is reflected in the 2 versions of $\psi(x)$ .
Since the overall integral is stated as $\int_{b_1}^{a_1}$ , the result of the first part of the integral range above must be sign reversed. This is the 2nd $\psi(x)$ in the expansion above. This corresponds to the part of the integral where the bottom $\sqrt{..} \space$ is negative, resulting in a positive $\psi(x)$ component. The other $\psi(x)$ has no sign reversals.

Thanks for the http://math.stackexchange.com/questions/ask hint for previewing. It was very helpful for checking the formatting. I hope it worked ok.

Thanks, a very informative section. --gjk