How circles, prime numbers, and infinite series all link up

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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(2)(-1)=6/(π2)=0.60792…

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 −…

so

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

(1−(1/2)2).(1−(1/3)2).(1−(1/5)2).(1−(1/7)2)…. [1]

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 [1]), 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−x22)(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

and so

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: