Pi
Number theory and Riemann zeta function

The Riemann zeta function ζ (s) is used in many areas of mathematics. When evaluated at s = 2 it can be written as ζ (2) = 1 1 2 + 1 2 2 + 1 3 2 + ⋯ {\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots }

Finding a simple solution for this infinite series was a famous problem in mathematics called the Basel problem. Leonhard Euler solved it in 1735 when he showed it was equal to π /6. Euler's result leads to the number theory result that the probability of two random numbers being relatively prime (that is, having no shared factors) is equal to 6/π. This probability is based on the observation that the probability that any number is divisible by a prime p is 1/ p (for example, every 7th integer is divisible by 7.) Hence the probability that two numbers are both divisible by this prime is 1/ p, and the probability that at least one of them is not is 1 − 1/ p. For distinct primes, these divisibility events are mutually independent; so the probability that two numbers are relatively prime is given by a product over all primes: ∏ p ∞ (1 − 1 p 2) = (∏ p ∞ 1 1 − p − 2) − 1 = 1 1 + 1 2 2 + 1 3 2 + ⋯ = 1 ζ (2) = 6 π 2 ≈ 61 %. {\displaystyle {\begin{aligned}\prod _{p}^{\infty }\left(1-{\frac {1}{p^{2}}}\right)&=\left(\prod _{p}^{\infty }{\frac {1}{1-p^{-2}}}\right)^{-1}\\&={\frac {1}{1+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\cdots }}\\&={\frac {1}{\zeta (2)}}={\frac {6}{\pi ^{2}}}\approx 61\%.\end{aligned}}}

This probability can be used in conjunction with a random number generator to approximate π using a Monte Carlo approach.

The solution to the Basel problem implies that the geometrically derived quantity π is connected in a deep way to the distribution of prime numbers. This is a special case of Weil's conjecture on Tamagawa numbers, which asserts the equality of similar such infinite products of arithmetic quantities, localized at each prime p, and a geometrical quantity: the reciprocal of the volume of a certain locally symmetric space. In the case of the Basel problem, it is the hyperbolic 3-manifold SL (R) / SL (Z).

The zeta function also satisfies Riemann's functional equation, which involves π as well as the gamma function: ζ (s) = 2 s π s − 1 sin (π s 2) Γ (1 − s) ζ (1 − s). {\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\ \sin \left({\frac {\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s).}

Furthermore, the derivative of the zeta function satisfies exp (− ζ ′ (0)) = 2 π. {\displaystyle \exp(-\zeta '(0))={\sqrt {2\pi }}.}

A consequence is that π can be obtained from the functional determinant of the harmonic oscillator. This functional determinant can be computed via a product expansion, and is equivalent to the Wallis product formula. The calculation can be recast in quantum mechanics, specifically the variational approach to the spectrum of the hydrogen atom.
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