# Senior thesis

1. Mar 27, 2006

### benorin

I am writing my senior thesis (I am an undergrad math major at UCSB) on Dirichlet Series, which are, in the classical sense, series of the form

$$\xi (s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$​

where $$a_n,s\in\mathbb{C}$$ and $$a_n$$ is multiplicative, hence

$$\forall n,m\in\mathbb{N}, \, a_{nm}=a_{n}a_{m}$$​

I have begun this bit on analytic continuation for such series, here it goes:

$$\xi (s)+\sum_{n=1}^{\infty}(-1)^{n}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n^s}+\sum_{n=1}^{\infty}(-1)^{n}\frac{a_n}{n^s}=2\sum_{n=1}^{\infty}\frac{a_{2n}}{(2n)^s}=2^{1-s}a_2\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$​

so that

$$\xi (s)=(1-a_22^{1-s})^{-1}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a_n}{n^s}$$​

which is the first first stage of analytic continuation. Now, to the above series apply Euler's series transformation, which, if you don't recall, is

$$\sum_{n=1}^{\infty}(-1)^{n-1}b_n=\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{m=0}^{n}(-1)^{m}\left( \begin{array}{c}n\\m\end{array}\right) b_{m+1}$$​

to get the the second stage, namely

$$\boxed{\xi (s)=(1-a_22^{1-s})^{-1}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{m=0}^{n}(-1)^{m}\left( \begin{array}{c}n\\m\end{array}\right)\frac{a_{m+1}}{(m+1)^s}}$$​

when this same process of continuation is applied to the Riemann zeta it produces a series for the zeta function that converges for all s in the complex plane except s=1 (see prior thread for details.) My trouble is proving convergence in the present, more general case. Any thoughts?

Last edited: Mar 27, 2006
2. Mar 27, 2006

### shmoe

Not all dirichlet series have an analytic continuation.

In the case of zeta, the 1-2^(1-s) is cancelling the pole at s=1. If the coefficients of $$\xi(s)$$ are real and non-negative then you have a pole at the real point on the line of convergence. I don't see anything that will gurantee this pole to be canceled in your case (could be of higher order, or at a different location, or...)- if this doesn't happen that sum cannot possibly converge.

3. Mar 27, 2006

### benorin

So I should then ask, For what sequences $$a_n$$ does this favorable condition occur?

4. Mar 27, 2006

### benorin

I see the cancelation by the factor of 1-2^(1-s) at s=1, since

$$1-2^{1-s}=0 \Leftrightarrow s=1+i\frac{2k\pi}{\log 2}, k\in\mathbb{Z},$$

but we may assume principle values so that s=1 is the only point of interest. But for the factor of 1-a22^(1-s), s=1 would not be of interest unless a_2=1.

5. Mar 27, 2006

### shmoe

Happily, all the interesting ones. (under the right definition of "interesting") The 'naturally' occuring arithmetic sequences typically do, e.g. dirichlet L-functions. I don't know that there's any nice conditions on the coefficients to guarantee analytic continuation, but I can't say that I've looked to hard. The Selberg class has some nice assumptions like an Euler product and the Ramanujan hypothesis, yet it still includes analytic continuation as an assumption for this class and I'm pretty confidant it doesn't follow from these coefficient conditions.

The obvious thing to try after success with this method on the riemann zeta function is to head towards dirichlet L-functions. Sandow mentions this (and more) was in the future in the intro of that '94 paper. Did you try to track this work down?

6. Apr 12, 2006

### benorin

Analytic continuation via Euler's Series Transformation

I wish to consider when exactly does Euler's series transformation provide an analytic continuation of a function defined by an alternating series. I will use a modified version of the transformation given above:

For a known convergent alternating series $\sum (-1)^k b_k ,$ Euler's series transformation is given by

$$\sum_{k=0}^{\infty}(-1)^{k}b_k=\sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) b_{m}$$​

An example: the result is trivial, yet the concept of continuation by the series transformation is rather at hand.

Let z be complex. Consider the function f(z) defined by the alternating series $$f(z) = \sum_{k=0}^{\infty}(-1)^kz^k$$ which converges to $$\frac{1}{1+z}$$ on the unit disk $$|z|<1$$. Applying Euler's series transformation to f(z) we obtain

$$f(z) = \sum_{k=0}^{\infty}(-1)^kz^k = \sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) z^{m}$$​

and since the binomial theorem gives $$\sum_{m=0}^{k}(-1)^{m}\left( \begin{array}{c}k\\m\end{array}\right) z^{m} = (1-z)^{k}$$ we may simplify this to obtain

$$f(z) = \sum_{k=0}^{\infty}\frac{1}{2^{k+1}}\cdot (1-z)^{k} = \frac{1}{2}\sum_{k=0}^{\infty}\left( \frac{1-z}{2} \right) ^{k}$$​

where the last series is a geometric series which converges to $$\frac{1}{1+z}$$ on the disk $$\left| \frac{1-z}{2} \right| <1 \Rightarrow |z-1|<2$$. Notice that the series thus obtained converges everywhere the given series did and on a disk twice as big!

If one applies the transformation yet again, the series $$\frac{1}{4}\sum_{k=0}^{\infty}\left( \frac{3-z}{4} \right) ^{k}$$ is obtained, which is a geometric series converging to $$\frac{1}{1+z}$$ on the disk $$\left| \frac{3-z}{4} \right| <1 \Rightarrow |z-3|<4$$.

I suspect that successive applications of the transformation would produce series with circles of convergence having radii that grow as powers of 2 whose left most point is z=-1. Can this process be carried out indefinely to give a series which converges in the half-plane $$\Re z >-1$$ ?

But how often will it happen that an analytic continuation is obtained (necessary and sufficient conditions)? What is the maximal region of convergence thereby obtained? In "Theory and Application of Infinte Series," Knopp discusses sufficient conditions that a greater rapidity of convergence be obtained by an application of the series transformation, but I have yet to find a discussion of continuation. Any thoughts?

Last edited: Apr 12, 2006
7. Apr 12, 2006

### benorin

Found it! It even in that very text (Knopp) under the title Ep-transformation, (the p-fold application of Euler's series transformation to not necessarily alternating series.) The maximal region of continuation is determined by the placement of the singular points of the function.

8. Apr 13, 2006

### shmoe

Knopp should have some theorems about composing these Euler transformations, yes? I know they're in Hardy's "Divergent Series" book, and iirc they were essentially due to Knopp. If so, you've probably already realized your conjecture about how the radius of convergence expands in your example (doubling each time) was correct!