BBP Formula for Pi

\displaystyle \pi=\sum_{i=0}^{\infty} \frac{1}{16^{i}}\left(\frac{4}{8 i+1}-\frac{2}{8 i+4}-\frac{1}{8 i+5}-\frac{1}{8 i+6}\right)

This formula due to Bailey-Borwein-Plouffe is discovered by using integer relation algorithm PSLQ. They searched for integer relations between the quantities

{\pi}, {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+1}\quad,} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+2},\quad} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+3},\quad} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+4},\quad} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+5},\quad} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+6},\quad} {\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}}\frac{1}{8 i+7}.}

and found the above relation. Finding the relation is the harder part, proving it is easy.

Proof:

\displaystyle \sqrt{2}^{n} \sum_{i=0}^{\infty} \int_{0}^{\frac{1}{\sqrt{2}}} x^{n-1+8 i} d x=\sqrt{2}^{n} \sum_{i=0}^{\infty}\left[\frac{x^{n+8 i}}{8 i+n}\right]_{0}^{\frac{1}{\sqrt{2}}}= \sum_{i=0}^{\infty} \frac{1}{16^{i}(8 i+n)}

\displaystyle \sqrt{2}^{n} \sum_{i=0}^{\frac{1}{\sqrt{2}}} \int_{0}^{x-1+8 i} d x= \sqrt{2}^n \int_{0}^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^{8}} d x

Therefore,

\displaystyle \sum_{i=0}^{\infty} \frac{1}{16^{i}(8 i+n)}= \sqrt{2}^n \int_{0}^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^{8}} d x.

We get

\displaystyle \begin{aligned} &\sum_{i=0}^{\infty} \frac{1}{16^{i}}\left(\frac{4}{8 i+1}-\frac{2}{8 i+4}-\frac{1}{8 i+5}-\frac{1}{8 i+6}\right)\\ &=\int_{0}^{\frac{1}{\sqrt{2}}} \frac{4 \sqrt{2}-8 x^{3}-4 \sqrt{2} x^{4}-8 x^{5}}{1-x^{8}} d x\\ &= 16 \int_{0}^{1} \frac{y-1}{y^{4}-2 y^{3}+4 y-4} d y\\ &=4 \int_{0}^{1} \frac{2-y}{y^{2}-2 y+2} d y+4 \int_{0}^{1} \frac{y}{y^{2}-2} d y\\ &=\int_{0}^{1} \frac{4-4 y}{y^{2}-2 y+2} d y+4 \int_{0}^{1} \frac{1}{1+(y-1)^{2}} d y+4 \int_{0}^{1} \frac{y}{y^{2}-2} d y\\ &=\left[-2 \ln \left(y^{2}-2 y+2\right)+4 \arctan (y-1)+2 \ln \left(2-y^{2}\right)\right]_{0}^{1}\\ &=\pi\\ \end{aligned}

Similar formula for {\pi^2:}

\displaystyle \begin{aligned} \pi^{2}=& \sum_{i=0}^{\infty} \frac{1}{16^{i}}\left[\frac{16}{(8 i+1)^{2}}-\frac{16}{(8 i+2)^{2}}-\frac{8}{(8 i+3)^{2}}-\frac{16}{(8 i+4)^{2}}\right.\\ &\left.-\frac{4}{(8 i+5)^{2}}-\frac{4}{(8 i+6)^{2}}+\frac{2}{(8 i+7)^{2}}\right] \end{aligned}

Quest for Pi: https://www.davidhbailey.com//dhbpapers/pi-quest.pdf

PSLQ Algorithm: https://www.davidhbailey.com/dhbpapers/pslq-comp-alg.pdf
BBP Formula: https://www.experimentalmath.info/bbp-codes/bbp-alg.pdf

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