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Euler's zeta function, which forms the basis for Riemann's Hypothesis, is the sum of the integers from 1 to infinity raised to a power. It is written:

This converges for complex s such that the real part of s is greater than 1. Riemann's zeta function extends this function into the whole complex plane.

Bernoulli numbers are the zeta functions of integers greater than 1. They are notable for the results obtained, for example zeta(2)=(pi^2)/6, and zeta(4)=(pi^4)/90.

We shall be looking at zeta(2), and so here is a proof that zeta(2) does indeed relate to pi, first discovered by Euler.

If the polynomial:

has roots at:

then this equation is true.

Note that the Maclaurin expansion of sinx/x gives:

and this expression has roots at:

Using:

it is simple to deduce that the roots of this are:

Using the first result, we get:

and this gives the desired equation:

Euler was also the first to notice that his zeta function can also be expressed in terms of the primes. This observation is the main motivating force behind solving Riemann's hypothesis - a solution of Riemann's hypothesis would allow the prime numbers to be untangled.

This paradox is concerned with Euler's prime product, and demonstrates a potential loop-hole with the prime product.

The denominator of the prime product is factorizable, and so we may wonder if the denominator contains every prime as a factor. By factorizing the denominator, we can possibly gain an insight into the construction of pi.

By considering zeta(2), we can factorize the denominator to get:

This doesn't look easy, the factors of n-1 given the factors of n is one of the hardest prime problems. But Dirichlet has proved that given an arithmetic progression, an+b, where a and b are co-prime [gcd(a,b)=1], then an+b contains an infinite number of primes.

So, if we create an arithmetic progression using a prime as n, then ap+1 will contain an infinite number of primes. For every prime q that this sequence contains, then q-1 contains p as a factor.

Or, the denominator contains every prime an infinite number of times. And this is the paradox - if this was true, then the prime product would tend to zero, and so we have proved that pi=0.

The solution to this paradox is in how we calculate the prime product. The prime product is only valid if we have evaluate the same number of primes in the numerator as we do with the denominator.

So, for example:

in valid, but:

is not.

Another point is that in any valid expansion of the prime product, the number contains P, where P is the last prime evaluated. The denominator however will never contain P as a factor.

The first 10 values of the prime product when calculated like this are:

1.3333333333333332
1.5
1.5625
1.5950520833333334
1.608344184027778
1.6179176613136576
1.6235354309709968
1.628045251612583
1.6311286706497278
1.6330704904957396
1.6347716055900061

To see more values, click [here]

Finally, the only two primes in the expansion of p²-1 are 2 and 3, which conviently make the denominator for the LHS of the prime product, i.e. (pi^2)/6.

Please address questions/comments/suggestions to : Jon Perry