BBP Formula for Pi Proof

$$\pi =\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{16}^k}(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6})$$

The BBP formula was found using an integer relation algorithm. Proofs were found afterward and do not give insight into how the formula was found.

First consider the integral ${\int }_0^{1/\sqrt{2}}\frac{x^{r-1}dx}{1-x^8}$ for some positive integer $r$. Then we can evaluate the integral:

$$\begin{array}{rl}& {\int }_0^{1/\sqrt{2}}\frac{x^{r-1}dx}{1-x^8}={\int }_0^{1/\sqrt{2}}{x}^{r-1}\frac{1}{1-x^8}dx={\int }_0^{1/\sqrt{2}}{x}^{r-1}(1+x^8+x^{16}+x^{24}+\dots )dx\\ & ={\int }_0^{1/\sqrt{2}}\left(\sum _{k=0}^{\mathrm{\infty }}{x}^{8k+r-1}\right)dx={\left[\sum _{k=0}^{\mathrm{\infty }}\frac{x^{8k+r}}{8k+r}\right]}_0^{2^{-1/2}}=\sum _{k=0}^{\mathrm{\infty }}\frac{2^{(-1/2)(8k+r)}}{8k+r}=2^{-r/2}\sum _{k=0}^{\mathrm{\infty }}\frac{2^{-4k}}{8k+r}=2^{-r/2}\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+r}\end{array}$$

Now let $S(r)=2^{-r/2}\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+r}={\int }_0^{1/\sqrt{2}}\frac{x^{r-1}dx}{1-x^8}$. We can rewrite the BBP summation and turn it into an integral.

$$\begin{array}{rl}& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{{16}^k}(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6})=4\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+1}-2\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+4}-\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+5}-\sum _{k=0}^{\mathrm{\infty }}\frac{{16}^{-k}}{8k+6}\\ & =4\cdot 2^{1/2}S(1)-2\cdot 2^{4/2}S(4)-2^{5/2}S(5)-2^{6/2}S(6)=4\sqrt{2}S(1)+8S(4)-4\sqrt{2}S(5)-8S(6)\\ & ={\int }_0^{1/\sqrt{2}}\frac{4\sqrt{2}-8x^3-4\sqrt{2}{x}^4-8x^5}{1-x^8}dx\end{array}$$

Next, make the substitution $y=\sqrt{2}x$. This changes the integration bounds to $[0,1]$.

$$\begin{array}{rl}& {\int }_0^{1/\sqrt{2}}\frac{4\sqrt{2}-8x^3-4\sqrt{2}{x}^4-8x^5}{1-x^8}dx={\int }_0^1\frac{4\sqrt{2}-2\sqrt{2}{y}^3-\sqrt{2}{y}^4-\sqrt{2}{y}^5}{1-y^8/16}\frac{dy}{\sqrt{2}}\\ & ={\int }_0^1\frac{4-2y^3-y^4-y^5}{1-y^8/16}dy=16{\int }_0^1\frac{4-2y^3-y^4-y^5}{16-y^8}dy=16{\int }_0^1\frac{y^5+y^4+2y^3-4}{y^8-16}dy\end{array}$$

Now for some factoring, we can do the following for the denominator. The factoring of $y^4+4$ is not the easiest to do.

$$\begin{array}{rl}& y^8-16=(y^4+4)(y^4-4)=(y^4+4)(y^2+2)(y^2-2)\\ & =(y^2+2y+2)(y^2-2y+2)(y^2+2)(y^2-2)\end{array}$$

Then for the numerator, first we can find the factor $y-1$ with the rational root theorem. The other part of factoring is not quite easy.

$$y^5+y^4+2y^3-4=(y-1)(y^4+2y^3+4y^2+4y+4)=(y-1)(y^2+2)(y^2+2y+2)$$

Now the integral can be simplified.

$$\begin{array}{rl}& 16{\int }_0^1\frac{(y-1)(y^2+2)(y^2+2y+2)}{(y^2+2y+2)(y^2-2y+2)(y^2+2)(y^2-2)}dy=16{\int }_0^1\frac{y-1}{(y^2-2y+2)(y^2-2)}dy\end{array}$$

The next step is to use partial fraction decomposition. We need constants $A,B,C,D$ such that

$$\begin{array}{rl}& \frac{Ay+B}{y^2-2y+2}+\frac{Cy+D}{y^2-2}=\frac{y-1}{(y^2-2y+2)(y^2-2)}\\ & (Ay+B)(y^2-2)+(Cy+D)(y^2-2y+2)=y-1\\ & A{y}^3+B{y}^2-2Ay-2B+C{y}^3+D{y}^2-2C{y}^2-2Dy+2Cy+2D=y-1\\ & (A+C)y^3+(B-2C+D)y^2+(-2A+2C-2D)y+(-2B+2D)=y-1\end{array}$$

This gives us the following linear system.

$$\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& -2& 1\\ -2& 0& 2& -2\\ 0& -2& 0& 2\end{array}\right]\left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\\ -1\end{array}\right]$$

After solving it, the constants are $(A,B,C,D)=(-1/4,1/2,1/4,0)$. The integral can be simplified further.

$$\begin{array}{rl}& 16{\int }_0^1(\frac{Ay+B}{y^2-2y+2}+\frac{Cy+D}{y^2-2})dy={\int }_0^1\frac{-4y+8}{y^2-2y+2}dy+{\int }_0^1\frac{4y}{y^2-2}dy\\ & ={\int }_0^1\frac{4}{y^2-2y+2}dy+{\int }_0^1\frac{-4y+4}{y^2-2y+2}dy+{\int }_0^1\frac{4y}{y^2-2}dy\\ & ={\int }_0^1\frac{4}{y^2-2y+2}dy-2{\int }_0^1\frac{d(y^2-2y+2)}{y^2-2y+2}+2{\int }_0^1\frac{d(y^2-2)}{y^2-2}\\ & ={\int }_0^1\frac{4}{y^2-2y+2}dy-2{[\ln |y^2-2y+2|]}_0^1+2{[\ln |y^2-2|]}_0^1\\ & ={\int }_0^1\frac{4}{y^2-2y+2}dy-2(\ln (1)-\ln (2))+2(\ln (1)-\ln (2))={\int }_0^1\frac{4}{y^2-2y+2}dy\end{array}$$

Now we have one last integral to evaluate.

$$\begin{array}{rl}& 4{\int }_0^1\frac{dy}{y^2-2y+2}=4{\int }_0^1\frac{dy}{(y-1{)}^2+1}=4{\int }_0^1\frac{d(y-1)}{(y-1{)}^2+1^2}\\ & =4\arctan (y-1){|}_0^1=4(\arctan (0)-\arctan (-1))=4(0+\pi /4)=\pi \end{array}$$

This completes the proof. The BBP formula proof does not require any very advanced techniques. It is accessible to those who understand material in typical college calculus. The main difficulties are rewriting the summation as an integral using the $S(r)$ function at the beginning and factoring the high degree polynomials after the change of variables to $y$.