$\displaystyle \frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2}}}} \cdots$
$\displaystyle \frac{2}{\varpi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{\frac{1}{2}}{\sqrt{\frac{1}{2}}}} \cdot \sqrt{\frac{1}{2} + \frac{\frac{1}{2}}{\sqrt{\frac{1}{2} + \frac{\frac{1}{2}}{\sqrt{\frac{1}{2}}}}}} \cdots ,$
Unfortunately, Levin's proof of this formula gets into Weierstrass with complex arguments to arrive at the result. This post is attempt to give a simpler proof with just trigonometric identities. We'll use the lemniscate sine and cosine functions but will reparametrize them in such a way that the proof remains simple.
The lemniscate elliptic functions, $\text{sl}(s)$ and $\text{cl}(s)$ are close analogues of the classical sine and cosine functions as can be seen in the figures in Relation to geometric shapes. The lemniscate curve was made immortal when Fagnano first found that they obey a doubling identity. That is,
$\displaystyle \text{sl}(2s)^2=\frac{4\text{sl}(s)^2(1-\text{sl}(s)^4)}{(1+\text{sl}(s)^4)^2}$
It can also easily seen that $\text{sl}'(s)^2=1-\text{sl}(s)^4$ where prime denotes differential w.r.t. $s$.
Surprisingly, this is all we need for our proof from the lemniscate. To begin, let's parametrise the lemniscate sine function with a parameter $t$ such that
$\text{sl}(s/2)=\sqrt{\tan(t)}$
Then using Fagnano's formula and double angle formulas of sine and cosine using tangents, we can quickly see that
$\text{sl}(s)=\sqrt{\sin(4t)}$
From here, it is straightforward that $\text{sl}'(s)=\cos(4t)$ and $\text{sl}'(s/2)=\sqrt{1-\tan^2(t)}$
Now with simple trigonometric identities, we have
$\displaystyle \frac{1}{\text{sl}'(s/2)}=\frac{\cos(t)}{\sqrt{\cos(2t)}}=\sqrt{\frac{1+\cos(2t)}{2\cos(2t)}}=\sqrt{\frac{1+\sec{2t}}{2}}$
But we know that $\text{sl}'(s)=\cos(4t)$ and therefore,
$\displaystyle \cos(2t)=\sqrt{\frac{1+\cos(4t)}{2}}=\sqrt{\frac{1+\text{sl}'(s)}{2}}$
Using this expression, we have
$\displaystyle\frac{1}{\text{sl}'(s/2)}=\sqrt{\frac{1+\left(\sqrt{\frac{1+\text{sl}'(s)}{2}}\right)^{-1}}{2}}$
To avoid this nested square roots, we introduce a new function such that $\hat{\text{sl}}(s)=\cos(2t)$. We then have,
$\displaystyle\text{sl}'(s/2)^{-1}=\sqrt{\frac{1+\hat{\text{sl}}(s)^{-1}}{2}}$ and $\displaystyle\hat{\text{sl}}(s)=\sqrt{\frac{1+\text{sl}'(s)}{2}}$
Now comes the interesting part. From our definitions of the lemniscate functions in terms of $t$,
$\displaystyle \frac{\text{sl}(s)}{2\text{sl}(s/2)}=\frac{\sqrt{\sin(4t)}}{2\sqrt{\tan(t)}}=\cos(t)\sqrt{\cos(2t)}$
Also,
$\displaystyle\frac{\hat{\text{sl}}(s)}{\text{sl}'(s/2)}=\frac{\cos(2t)}{\sqrt{1-\tan^2(t)}}=\cos(t)\sqrt{\cos(2t)}$
Therefore, we see that
$\displaystyle \frac{\text{sl}(s)}{2\text{sl}(s/2)}=\frac{\hat{\text{sl}}(s)}{\text{sl}'(s/2)}$
Repeated application of this relation $n$ times,
$\displaystyle \frac{\text{sl}(s)}{2^n\text{sl}(s/2^n)}=\frac{\hat{\text{sl}}(s)}{\text{sl}'(s/2)}\frac{\hat{\text{sl}}(s/2)}{\text{sl}'(s/4)}\cdots\frac{\hat{\text{sl}}(s/2^{n-1})}{\text{sl}'(s/2^n)}$
As $n \to \infty$ and noting that $\lim_{n \to \infty}2^n\text{sl}(s/2^n)=s$, we have
$\displaystyle \frac{\text{sl}(s)}{s}=\hat{\text{sl}}(s)\cdot\text{sl}'(s/2)^{-1}\cdot\hat{\text{sl}}(s/2)\cdot\text{sl}'(s/4)^{-1}\cdot\hat{\text{sl}}(s/4)\cdot\text{sl}'(s/8)^{-1}\cdots$
Based on the relations we saw before, if $a$ and $b$ are two consecutive terms on the RHS above, we see that
$\displaystyle b=\sqrt{\frac{1+a^{-1}}{2}}$
We have $\text{sl}(\varpi/2)=1$, $\text{sl}'(\varpi/2)=0$ and $\hat{\text{sl}}(\varpi/2)=1/\sqrt{2}$. Using these values in the infinite product above, we get the lemniscatic analogue of Viete's formula.
Hope you enjoyed this. See ya later.
Until then
Yours Aye
Me
