skip to content
Aymen Hafeez

A cool formula for π/8

/ 1 min read

View more blogs with the tag mathematics

I came across this expression whilst writing the π\pi and primes post. In that post we showed that

πxcot⁡(πx)=1−2∑n=1∞ζ(2n)x2n\pi x\cot(\pi x)=1-2\sum_{n=1}^{\infty}\zeta(2n)x^{2n}

πx2cot⁡(πx)=12−∑n=1∞ζ(2n)x2n\frac{\pi x}{2}\cot(\pi x)=\frac{1}{2}-\sum_{n=1}^{\infty}\zeta(2n)x^{2n}

Letting x=14x=\frac{1}{4}

π8=12−∑n=1∞ζ(2n)42n\frac{\pi}{8}=\frac{1}{2}-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{4^{2n}}

We’ve shown that the Riemann zeta function at even positive integers is given by the following expression:

ζ(2n)=(−1)k+1(2π)2nβ2n2(2n)!\zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}\beta_{2n}}{2(2n)!}

The first few even valued Bernoulli numbers, β2n\beta_{2n}, are

β2=16;  β4=−130;  β6=142;  β8=−130\beta_2=\frac{1}{6}; \ \ \beta_4=-\frac{1}{30}; \ \ \beta_6=\frac{1}{42}; \ \ \beta_8=-\frac{1}{30}

And so, ζ(2n)\zeta(2n) for the first few values of nn is

ζ(2)=π26;  ζ(4)=π490;  ζ(6)=π6945;  ζ(8)=π89450\zeta(2)=\frac{\pi^2}{6}; \ \ \zeta(4)=\frac{\pi^4}{90}; \ \ \zeta(6)=\frac{\pi^6}{945}; \ \ \zeta(8)=\frac{\pi^8}{9450}

Expanding the sum out:

π8=12−π26⋅42−π490⋅44−π6945⋅46−π89450⋅48−. . .\frac{\pi}{8}=\frac{1}{2}-\frac{\pi^2}{6\cdot4^2}-\frac{\pi^4}{90\cdot4^4}-\frac{\pi^6}{945\cdot4^6}-\frac{\pi^8}{9450\cdot4^8}-.\ .\ .

We can simplify it visually by defining a constant ana_n:

an=(−1)n+14nβ2n2(2n)!a_n=\frac{(-1)^{n+1}4^n\beta_{2n}}{2(2n)!}

So the expression becomes

π8=12−a1π242−a2π444−a3π646−a4π848−. . .\frac{\pi}{8}=\frac{1}{2}-a_1\frac{\pi^2}{4^2}-a_2\frac{\pi^4}{4^4}-a_3\frac{\pi^6}{4^6}-a_4\frac{\pi^8}{4^8}-.\ .\ .

which is just a really nice formula.