Wednesday, November 19, 2008

SPONTANEOUS MATH LECTURE YAY

a Dirichlet series is a difficul thing to simplify past the equation and integer boundaries, but I'm giving it a shot anyway because it gives me an oppurtunity to revise and remind myself by jotting down everything Ic an recall... So, when given a series {an}n ∈ N of complex numbers we try to consider the value of

{inf|sigma|(n=1)} (an/(n^s))

as a function of the complex variable s. In order for this to make sense, we need to consider the convergence properties of the infitite sequence stated above.

If {an}n ∈ N is a bounded sequence of complex numbers, then the corresponding Dirichlet series f converges absolutely - that is, to a whole-number point - on the open half-plane of s such that Re(s) > 1. Generally speaking if an = O(n^k), the series converges absolutely in the half plane Re(s) > k + 1.

If the set of sums an + an+1 + ... + an+k is bounded for n and k ≥ 0, then the above infinite series converges on the open half-plane of s such that Re(s) > 0.

In both cases f is an analytic function on the corresponding open half plane.

The convergence of a Dirichlet series is the intercept on the real axis, of the vertical line in the complex plane, such that there is convergence to the right of it, and divergence to the left. This is the analogue for Dirichlet series of the radius of convergence for power series. The Dirichlet series case is more complicated, though: absolute convergence and uniform convergence may occur in distinct half-planes.

On the whole, the analytic function associated with a Dirichlet series has an analytic extension to a larger domain, and it'll be rare to see one used in another fashion.

1 comment:

nicotene.razorblade said...

Delete this comment, but never try to do this again - "a Dirichlet series is a difficult thing to simplify past the equation and integer boundaries, but I'm giving it a shot anyway."
...it's why they call those people, 'lecturers'.