Who would have thought that area and circumference of a circle were related to factorising numbers and to infinite series?
Yet the probability that two integers m and n picked at random are co-prime (have no common factors) is:
ζ(z) is the Riemann zeta function, which for z=2 is
sum from r=1 to infinity of 1/r2.
Let N be a positive whole number, and p1, p2, p3…. pr be the primes less than N
Let Π(N) be the number of coprime pairs of numbers ≤N
⌊x⌋ is the greatest whole number ≤x, e.g. ⌊2.9⌋=2
Then divide up all the N2 pairs of numbers≤N into the Π(N) which have no common divisor, the (⌊N/p1⌋)2 which have p1 as a common divisor, the (⌊N/p2⌋)2 which have p2 as a common divisor, etc.
We will have double-counted those which have p1p2 (etc.) as a common divisor. Subtract (⌊N/p1p2⌋)2 (etc.) Then we will have double-subtracted those which have p1p2p3 as a common divisor. And so on…
With all the corrections
N2 = Π(N) + sum over all primes≤N of (⌊N/p⌋)2 −
sum over all pairs of primes≤N of (⌊N/pipj⌋)2 +
sum over all groups of three primes≤N of (⌊N/pipjpk⌋)2 −…
Π(N)/N2 = 1 − [sum (1/N)⌊(N/p)⌋]2 + sum [(1/N)⌊N/pipj⌋]2 − sum [(1/N)⌊N/pipjpk⌋]2 + …
Take the limit as N tends to infinity
Then the probability of two numbers≤N being coprime approaches
1 − sum (1/p)2 + sum (1/pipj)2 − ….
which is the infinite product over all the primes of
But ζ(2) is defined as the sum from r=1 to infinity of 1/r2. Factorising r into prime factors, that is:
the sum over all powers r1, r3, r3… ≥0 of 1/[22r1.32r2.52r3….]
and that is this infinite product of infinite sums:
(sum for all r1≥0 of 1/22r1).(sum for all r2≥0 of 1/32r2)…
The infinite sums are geometric series adding to
1/[1−(1/2)2], 1/[1−(1/3)2], 1/[1−(1/5)2], ….
so their infinite product, ζ(2), is equal to 1/(infinite product ), which is equal to the probability of any two numbers, however big, being coprime.
It remains to prove that ζ(2)=π2/6
Euler proved it this way:
sin x = x − x3/3! + x5/5! − x7/7! − …. (Maclaurin series)
But the roots of sin x = 0 are 0, ±2π, ±3π, ±π…
So (this is a daring step, but possible to justify rigorously) for some constant A
x − x3/3! + x5/5! − x7/7! −…. = A x(1−x2/π2)(1−x2/(2π)2)(1−x2/(3π)2)…
Equating the coefficients of x, A=1
Equating the coefficients of x3:
−(1/3!) = −1/π2−1/(2π)2−1/(3π)2…
1/12 + 1/22 + 1/32 +… = π2/6 ▇
Thanks for all that to Julian Havil’s brilliant book Gamma.
For more about primes, watch Terry Tao’s video:
And for more about the zeta (ζ) function and the Riemann hypothesis, watch Ed Frenkel’s video: