Approximating pi

I was recently interviewed by my city’s local newspaper about $\pi$ Day and the general fascination with memorizing the digits of $\pi$ . I was asked by the reporter if the only constraint in our knowledge of the digits of $\pi$ was the ability of computers to calculate the digits, and I answered in the affirmative.

Here’s the current state-of-the-art for calculating the digits of $\pi$ . Amazingly, this expression was discovered 1995… in other words, very recently.

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

Because of the term $16^n$ in the denominator, this infinite series converges very quickly.

Proof: If $k < 8$ , then we calculate the integral $I_k$ , defined below:

$I_k = \displaystyle \int_0^{1/\sqrt{2}} \frac{x^{k-1}}{1-x^8} dx$

$= \displaystyle \int_0^{1/\sqrt{2}} x^{k-1} \sum_{n=0}^\infty x^{8n} dx$

$= \displaystyle \int_0^{1/\sqrt{2}} \sum_{n=0}^\infty x^{8n+k-1} dx$

$= \displaystyle \sum_{n=0}^\infty \int_0^{1/\sqrt{2}} x^{8n+k-1} dx$

$= \displaystyle \sum_{n=0}^\infty \left[ \frac{x^{8n+k}}{8n+k} \right]^{1/\sqrt{2}}_0$

$= \displaystyle \sum_{n=0}^\infty \frac{1}{8n+k} \left[ \left( \frac{1}{\sqrt{2}} \right)^{8n+k} - 0 \right]$

$= \displaystyle \sum_{n=0}^\infty \frac{1}{2^{k/2}} \frac{1}{16^n (8n+k)}$

We now form the linear combination $P = 4\sqrt{2} I_1 - 8 I_4 - 4\sqrt{2} I_5 - 8 I_6$ :

$P = \displaystyle \sum_{n=0}^\infty \left( \frac{4\sqrt{2}}{2^{1/2}} \frac{1}{16^n (8n+1)} - \frac{8}{2^{4/2}} \frac{1}{16^n (8n+4)} - \frac{4\sqrt{2}}{2^{5/2}} \frac{1}{16^n (8n+5)} - \frac{8}{2^{6/2}} \frac{1}{16^n (8n+6)} \right)$

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

Also, from the original definition of the $I_k$ ,

$P = \displaystyle \int_0^{1/\sqrt{2}} \frac{4\sqrt{2} - 8x^3 -4\sqrt{2} x^4 - 8x^5}{1-x^8} dx$ .

Employ the substitution $x = y/\sqrt{2}$ :

$P = \displaystyle \int_ 0^1 \frac{4\sqrt {2} - 2\sqrt {2} y^3 - \sqrt {2} y^4 - \sqrt {2} y^5}{1 - y^8/16}\frac {dy} {\sqrt {2}}$

$P = \displaystyle \int_ 0^1 \frac{16 (4 - 2 y^3 - y^4 - y^5)}{16 - y^8} dy$

$P = \displaystyle \int_0^1 \frac{16(y-1)(y^2+2)(y^2+2y+2)}{(y^2-2)(y^2+2)(y^2+2y+2)(y^2-2y+2)} dy$

$P = \displaystyle \int_0^1 \frac{16y-16}{(y^2-2)(y^2-2y+2)} dy$

Using partial fractions, we find

$P = \displaystyle \int_ 0^1\frac{4 y}{y^2 - 2} dy - \int_ 0^1 \frac{4 y - 8}{y^2 - 2 y + 2} dy$

The expression on the right-hand side can be simplified using standard techniques from Calculus II and is equal to $\pi$ .

So that’s the proof… totally accessible to a student who has mastered concepts in Calculus II. But this begs the question: how in the world did anyone come up with the idea of starting with the integrals $I_k$ to develop an infinite series that leads to $\pi$ ? Let me quote from page 118 of J. Arndt and C. Haenel, $\pi -$ Unleashed (Springer, New York, 2000):

Certainly not by chance, even if luck played some part in the discovery. All three parties [David Bailey, Peter Borwein and Simon Plouffe] are established mathematicians who have been working with the number $\pi$ for a considerable time… Yet the series was not discovered through mathematical deduction or inference. Instead, the researchers used a tool called Computer Algebra System and a particular procedure called the “PSQL algorithm” to generate their series. They themselves write that they found their formula “through a combination of inspired testing and extensive searching.”

The original paper that announced the discovery of this series can be found at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P123.pdf.

Approximating pi

Published by John Quintanilla

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