A Quest for Pi

For $m<n$, it is easy to show that,

$\displaystyle\frac{x^m}{(x-a_0)(x-a_1)\cdots(x-a_{n-1})}=\sum_{k=0}^{n-1}\frac{a_k^m}{x-a_k}\prod_{i=0\\i \neq k}^{n-1}(a_k-a_i)^{-1}$

Let's choose, $a_k=\eta^k$ where $\eta$ is the $n^{\text{th}}$ root of unity. Then we have,

$\displaystyle \frac{x^m}{1-x^n}=\sum_{k=0}^{n-1}\frac{1}{n\eta^{k(n-1)}}\frac{\eta^{km}}{\eta^k-x}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{\eta^{km}}{1-x/\eta^k}$

Integrating $x$ in the above expression between $0$ and $t$ and comparing the real part, we get,

$\begin{align}
\displaystyle \sum_{j \geq 0}\frac{t^{jn+m}}{jn+m}&=\frac{1}{n}\sum_{k=0}^{n-1}\cos(2\pi km/n)\log(1-2\cos(2\pi k/n)t+t^2)^{-1/2}\\ &\quad +\frac{1}{n}\sum_{k=0}^{n-1}\sin(2\pi km/n)\tan^{-1}\left(\frac{\sin(2\pi k/n)}{t^{-1}-\cos(2\pi k/n)}\right)
\end{align}$

Comparing the Imaginary parts, we also get,

$\displaystyle \sum_{k=0}^{n-1}\cos(2\pi km/n)\tan^{-1}\left(\frac{\sin(2\pi k/n)}{t^{-1}-\cos(2\pi k/n)}\right)=\sum_{k=0}^{n-1}\sin(2\pi km/n)\log(1-2\cos(2\pi k/n)t+t^2)^{-1/2}$

Using the properties of Conjugates, we also have,

$\begin{align}
\displaystyle \sum_{j \geq 0}\frac{t^{jn+m}}{jn+m}&=\frac{1}{n}\log(1-t)^{-1}+\mathbf{1}_{n \text{ even}}\frac{(-1)^m}{n}\log(1+t)^{-1}\\ & \quad +\frac{2}{n}\sum_{k=1}^{\lceil n/2 \rceil-1}\sin(2\pi km/n)\tan^{-1}\left(\frac{\sin(2\pi k/n)t}{1-\cos(2\pi k/n)t}\right)\\ & \quad +\frac{2}{n}\sum_{k=1}^{\lceil n/2 \rceil-1}\cos(2\pi km/n)\log(1-2\cos(2\pi k/n)t+t^2)^{-1/2}
\end{align}$

On the other hand, using Vandermonde matrix or otherwise,

$\displaystyle \frac{1}{1-x/\eta^m}=\sum_{k=0}^{n-1}\eta^{-km}\frac{x^k}{1-x^n}$

Integrating the above,

$\displaystyle \log(1-\eta^{-m}t)^{-1}=\sum_{k=1}^n\eta^{-mk}P_{n,k}(t)$ where $\displaystyle P_{n,k}(t)=\sum_{j \geq 0}\frac{t^{nj+k}}{nj+k}$

Comparing real and imaginary parts, we have,

$\displaystyle \log(1-2\cos(2\pi m/n)t+t^2)^{-1/2}=\sum_{k=1}^n\cos(2\pi km/n)P_{n,k}(t)$

$\displaystyle \tan^{-1}\left(\frac{\sin(2\pi m/n)}{t^{-1}-\cos(2\pi m/n)}\right)=\sum_{k=1}^n\sin(2\pi km/n)P_{n,k}(t)$

The second one here is more interesting. Using $t=\cos(2\pi m/n)$, we get infinitely many representations for $\pi$.

$\displaystyle (n-4m)\frac{\pi}{2n}=\sum_{k=1}^n\sin(2\pi km/n)P_{n,k}(\cos(2\pi m/n))$

For example, with $m=1$ and $n=3$, we get a Bailey–Borwein–Plouffe type formula (BBP formula) for $\pi$ in Base-64 which can be verified with Wolfram Alpha.

$\displaystyle \frac{\pi}{(\sqrt{3}/2)^3}=\sum_{k=0}^\infty\left[\frac{1}{64^k}\left(\frac{4}{6k+1}+\frac{2}{6k+2}-\frac{2^{-1}}{6k+4}-\frac{4^{-1}}{6k+5}\right)\right]$

(or)

$\displaystyle \frac{32\pi}{3\sqrt{3}}=\sum_{k=0}^\infty\left[\frac{1}{64^k}\left(\frac{16}{6k+1}+\frac{8}{6k+2}-\frac{2}{6k+4}-\frac{1}{6k+5}\right)\right]$


Until then
Yours Aye
Me