Primes, Probabilities, Products

Over at Pro-Science, Kris noticed a news item, that researchers were reporting the discovery of a new mathematical object, a ‘third degree transcendental L-function’. And he was wondering just what this means, and why it is important.

The best answer is actually given in the article, that

“It’s a big step towards our understanding the ‘world of L’, which is where most of the secrets of number theory are kept.”

But I think I can elaborate a little further.

A very interesting number theoretic question is this – given two random integers, a and b, what is the probability that they are relatively prime? That is, what is the probability that the only number that divides both a and b is 1?

Notice that, as a matter of probability, we only must concern ourselves with -prime- potential divisors of a and b. We’re only interested in the probability of there being no divisors period, and if there is a composite divisor, then there must be prime divisors. So in our probability calculations, it suffices to consider only the probability of prime divisors.

To begin with, given a prime number p, what is the probability that p divides a? This is equal to, in effect, the ratio of integers divisible by p to all integers. Of course, each of those values is infinite, but starting at 0, every p’th number is divisible by p. So the ratio would come out to 1/p. Thus the probability of a being divisible by p is 1/p.

As such, the probability of a not being divisible by p is 1 – 1/p.

So, the probability of a being divisible by p is 1/p, and likewise for b. If p divides them both, then p is a common divisor. The probability of this occurring is then 1/p².

But, of course, we would like a and b to have no common divisors. So the probability of p not being a common divisor of a and b is thus

But, if we want the probability of no prime being a common divisor, we must multiply the above probability over all primes p. If we let p_k be the k’th prime (p₁ being 2, of course), the probability that no prime is a common divisor, and thus a and b have a common divisor of 1, is given by

The question then becomes, what does that infinite product equal, if anything?

For now, we’ll assume it comes out to a finite value. As such, we can calculate the following,

A sort of basic series to remember, though, is that

As such, we can make the substitution to yield,

Now, this is the part that’s always difficult to convey, at least without a blackboard present.

But let’s expand the first couple of sums in that product.

And so on.

To determine the product of all these sums, we can do the following. Pick a single term from a given sum, and multiply it but one term chosen from all the other sums. The product of the original sums then becomes the sum of these constructed products.

Choosing the 1 term from each sum, they multiply together to give a 1.

Choosing the 1/2² term from the first sum, and 1’s from all the successive sums, they multiply to give a 1/2². Similarly we can construct 1/3², 1/4², and 1/5², choosing those terms from their sums, and 1’s from all the other sums.

Choosing the 1/2² from the first sum, and 1/3² from the second, and 1’s from all the others, that multiplies to give 1/6².

In this way, choosing terms appropriately from every sum, we can construct 1/n², for any integer n. And because prime factorizations are unique, the 1/n² term can only be constructed in one way, and will occur once and only once in the final sum.

Multiplying out the sums in this way, the final sum of the products is the sum of 1/n² over the positive integers. As such, we get

And quite remarkably, through many and sundry methods, it can be shown that

As such, taking reciprocals, the probability of two randomly selected integers being relatively prime is given by

And that is a fascinating result by itself.

But as it applies to the matter at hand, far more interesting is generalized identity that results from that same logic as above, that

And this is a fascinating result. We start with a product where the terms of the product are indexed off of the prime numbers, a set of no known order or behavior, and we’ve transformed it into a sum indexed strictly off the natural numbers, the most well behaved set you could imagine. Somehow, implicit in the structure and nature of the summation power series function, the nature of the set of prime numbers is unraveled and ordered. And perhaps, if we could understand the structure of that power series, we could somehow understand the structure of the prime numbers.

Now, as this relates to the news item, the power series

Is an L-function, specifically a first order algebraic L-function, if I understand correctly. But there is a whole world, a whole class of other L-functions and series with similar properties to explore. For example, the Generalized Riemann Hypothesis states that if f(s) = 0, for some L-series f, allowing that s may be complex, then the real part of s is less than 1/2. But as the example I laid out shows, L-functions carry a deep connection to the set, and specifically the distribution of, prime numbers.

Now, I don’t specifically know what good the first 3rd order transcendental L-function is, but it seems like discovering an off-shoot of man’s evolutionary ancestors, something new, but related, offering a different perspective on an old problem. Apparently as well, the geometry and computation that went into the analysis of this critter was quite impressive, and included methods generally applicable to the analysis of a wide range of L-functions, perhaps opening further venues of study.

It’s all very exciting.

Anyway. I must sleep now. For I return to Pittsburgh in the morning, for much hilarity and merry-making.

I wish.