Forgotten functions used to calculate $\pi$
Today we will show the proof of the following result:
\[ 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2\cdot 4}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4 \cdot 6 }\right)^2 + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8 }\right)^2 + ... = \frac{4}{\pi} \]
We show the proof of this result using an elliptic integrals approach (the result can also be obtained trough the Vandermonde's identity). However, following the path of elliptic integrals turned out very interesting given that it needs a tour through some of the main results of the theory of this sepcial functions to prove it.
This is the outline of the proof:
- Obtain the series expansion for $E(p), K(p)$ (the elliptic integrals of the first and second kind)
- Prove a result know as the Landen transformation
- Obtain the following series expansion as corollary of 1 and 2 (our result is a particular case of this expansion)
\[ E(p) =\frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \] \[ \quad p=\sqrt{1-q^2}\]
where $\displaystyle E(p)$ is the elliptic integral of the second kind, $\displaystyle p$ is known as the modulus and $q$ is known as de complementary modulus. If $E(p)=1$ when $p=1$ we get our result.
Proof
Consider the complete elliptic integral of the first kind:
\[K(p) = \int_{0}^{\frac{\pi}{2}} \frac{d \theta}{\sqrt{1-p^2\sin^2 \theta}} \]
where $\displaystyle p$ is known as the modulus.
We will obtain its series expansion:
Proposition.
\[K(p) = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} p^{2j}\] Proof
First, recall that
\[\arcsin (x) = \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} \frac{x^{2j+1}}{2j+1} \quad |x|\leq 1\] Hence \[ \frac{1}{\sqrt{1-x^2}} = \arcsin' (x)= \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} x^{2j} \quad |x|\leq 1 \] Now, given that $\displaystyle -1<\sin \theta < 1 $ then $ 0< p^2 \sin^2 \theta < p^2$
Therefore
If $\displaystyle p^2 \leq 1$ \[\frac{1}{\sqrt{1-p^2\sin^2 \theta}} = \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \sin^{2j} \theta \] Integrating \begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{d\theta }{\sqrt{1-p^2\sin^2 \theta}} =& \int_{0}^{\frac{\pi}{2}} \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \sin^{2j} \theta d\theta \\ =& \sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j} \int_{0}^{\frac{\pi}{2}} \sin^{2j} \theta d\theta \end{align*} It is easy to show that (proof will be added in the Appendix) \[\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{2j} \theta d\theta = \frac{(2j-1)!!}{(2j)!!} \frac{\pi}{2}\] Hence \[ K(p)= \int_{0}^{\frac{\pi}{2}}\frac{d\theta }{\sqrt{1-p^2\sin^2 \theta}} = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} p^{2j} \tag{1}\]
Now we are going to prove the expansion series for the complete elliptic integral of the second kind:
Proposition
\[ E(p)=\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta = \frac{\pi}{2} \sum_{j=0}^{\infty } \frac{1}{1-2j} \left[\frac{(2j-1)!!}{(2j)!!} p^j \right]^2\]
Proof
Lets apply the Feynman's trick: \[ \frac{d}{dp} E(p) = -\int_{0}^{\frac{\pi}{2}} \frac{p\sin^2 \theta}{\sqrt{1-p^2\sin^2 \theta }}d\theta \] Hence \[-\frac{p \sin^2 \theta }{\sqrt{1-p^2\sin^2 \theta}} = -\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \sin^{2j+2} (\theta) \] Therefore \begin{align*} -\int_{0}^{\frac{\pi}{2}}\frac{p \sin^2 \theta }{\sqrt{1-p^2\sin^2 \theta}}d\theta = & -\int_{0}^{\frac{\pi}{2}}\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \sin^{2j+2} (\theta) d\theta \\ =& -\sum_{j=0}^{\infty} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \int_{0}^{\frac{\pi}{2}}\sin^{2j+2} (\theta) d\theta \\ =& -\sum_{j=0}^{\infty} \frac{\pi}{2} \frac{(2j-1)!!}{(2j)!!} p^{2j+1} \frac{(2j+1)!!}{(2j+2)!!}\\ =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j+2)!!^2} \frac{(2j+1)}{(2j+2)} p^{2j+1} \end{align*}
where we make use of the fact that (the proof will be added to the appendix)
\[\int_{0}^{\frac{\pi}{2}}\sin^{2j+2} (\theta) d\theta = \frac{(2j+1)!!}{(2j+2)!!}\] Hence, integrating with respect to $p$ \begin{align*} \int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} \frac{(2j+1)}{(2j+2)^2} p^{2j+2} + C \\ =& -\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j+1)!!^2}{(2j+2)!!^2} \frac{1}{(2j+1)} p^{2j+2} + C \end{align*} Now if $\displaystyle p=0$ then $\displaystyle C=\frac{\pi}{2}$ therefore \[\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =\frac{\pi}{2}-\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(2j+1)!!^2}{(2j+2)!!^2} \frac{1}{(2j+1)} p^{2j+2} \] If $m=j+1$ \[\int_{0}^{\frac{\pi}{2}} \sqrt{1-p^2\sin^2 \theta }d\theta =\frac{\pi}{2}-\frac{\pi}{2}\sum_{m=1}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(2m-1)} p^{2m} = \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} p^{2m} \] Therefore \[\boxed{E(p)= \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(2m-1)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} p^{2m}}\tag{2} \]
Now we have to make use of the Landen transformation (the proof is very extensive and will be added in the appendix later)
Proposition (Landen transformation) \[\boxed{K\left(\frac{1-q}{1+q}\right) = \frac{1+q}{2} K(p)} \tag{3}\] \[\boxed{E\left(\frac{1-q}{1+q} \right) = \frac{E(p) +qK(p)}{1+q}} \tag{4}\]
Finally, we are going to show the proof for the following expansion of $E(p)$
Proposition \[E(p) =\frac{\pi(1+q)}{4} \sum_{n=0}^{\infty} \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\]
Proof
From (3) we can write: \[ E(p) =E\left(\frac{1-q}{1+q} \right)-qK(p)\] Substituting (4) in (3) we can write \[ E(p) =E\left(\frac{1-q}{1+q} \right)-qK(p) =E\left(\frac{1-q}{1+q} \right)-\frac{2q}{1+q} K\left(\frac{1-q}{1+q}\right) \] Then, using the power series for $E,K$ in (1) and (2): \[K\left(\frac{1-q}{1+q} \right) = \frac{\pi}{2} \sum_{j=0}^{\infty} \frac{(2j-1)!!^2}{(2j)!!^2} \left(\frac{1-q}{1+q} \right)^{2j} \] \[E\left(\frac{1-q}{1+q} \right)= \frac{\pi}{2}\sum_{m=0}^{\infty} \frac{(1-2m)!!^2}{(2m)!!^2} \frac{1}{(1-2m)} \left(\frac{1-q}{1+q} \right)^{2m} \] Hence \begin{align*} E(p) = & \frac{\pi(1+q)}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n} -\frac{2q}{1+q} \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n}\\ =& \frac{\pi(1+q)}{2} \sum_{n=0}^{\infty}\left\{ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-q}{1+q}\right)^{2n} \left(\frac{1}{1-2n} - \frac{2q}{(1+q)^2}\right)\right\}\\ =& \frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\left[\frac{2}{1-2n} - 1 + \left(\frac{1-q}{1+q}\right)^2\right] \\ =&\frac{\pi(1+q)}{4} \left\{ 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2(n+1)}\right\} \\ =&\frac{\pi(1+q)}{4} \left\{ 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n-2)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \right\} \\ =& \frac{\pi(1+q)}{4} \left\{1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \left[(2n)^2 - \frac{2n+1}{2n-1} (2n-1)^2\right] \right\}\\ =& \frac{\pi(1+q)}{4} \left\{ 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \right\}\\ =& \frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} \end{align*} Hence \[\boxed{ E(p) =\frac{\pi(1+q)}{4} \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-q}{1+q}\right)^{2n} }\] Note that if $\displaystyle p=1$ then $q=0$. Moreover, \[E(1) = \int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin^2 }\theta d\theta = \int_{0}^{\frac{\pi}{2}} cos \theta d\theta = 1 \] Therefore, we can conclude that \[\boxed{ 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2\cdot 4}\right)^2 + \left(\frac{1\cdot 3}{2\cdot 4 \cdot 6 }\right)^2 + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8 }\right)^2 + ... = \frac{4}{\pi} }\]
Appendix I will be adding the omitted results the following days