# I'm writing my senior thesis on Dirichlet Series

1. Feb 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

$$\zeta (s)=\sum_{k=1}^{\infty}\frac{a_k}{k^s}$$​

where $$a_k,s\in\mathbb{C}$$ and are more generally given by (whence the name generalized Dirichlet series)

$$\xi (s)=\sum_{k=1}^{\infty}a_k e^{-\lambda_k s}$$​

where $$\{ \lambda_k\}$$ is a sequence of real numbers such that $$\lambda_k < \lambda_{k+1},\forall k\in\mathbb{Z} ^+$$ and such that $$\lambda_k\rightarrow\infty\mbox{ as }k\rightarrow\infty .$$ I note that the usual power series and classical Dirichlet series are both special cases of generalized Dirichlet series (the sufficiently curious will check this by putting $$\lambda_k = \log k$$ in $$\xi$$ to obtain the later, and put $$a_k = b_n$$ if $$k=2^{n}$$ and $$a_n = 0$$ otherwise in $$\zeta$$ to obtain the former.)

As is well known, the formula for multipling two absolutely convergent classical Dirichlet series, say f and g given by

$$f(s)=\sum_{k=1}^{\infty}\frac{a_k}{k^s}\mbox{ and }g(s)=\sum_{n=1}^{\infty}\frac{b_n}{n^s},$$​

is the so-called Dirichlet product or convolution given by

$$h(s):=f(s)g(s)=\left( \sum_{k=1}^{\infty}\frac{a_k}{k^s}\right) \left( \sum_{n=1}^{\infty}\frac{b_n}{n^s}\right) = \sum_{k=1}^{\infty} \left( \sum_{n|k}a_{n}b_{\frac{k}{n}}\right) \frac{1}{k^s} ,$$​

where the inner sum is over all positive divisors n of k (the symbol n|k is read "n divides k".) This is an analog of the Cauchy product of power series, namely

$$\left( \sum_{k=0}^{\infty}a_{k}z^{k}\right) \left( \sum_{n=0}^{\infty}b_{n}z^{n}\right) = \sum_{k=0}^{\infty} \left( \sum_{n=0}^{k}a_{n}b_{k-n}\right) z^{k} .$$​

I am attempting to formulate an analog of this multiplication formula for generalized Dirichlet series, this is what I've got:

Question: is the following correct?

The formula for multipling two absolutely convergent general Dirichlet series, say F and G given by

$$F(s)=\sum_{k=1}^{\infty}a_{k}e^{-\alpha_k s}\mbox{ and }G(s)=\sum_{n=1}^{\infty} b_{n}e^{-\beta_n s},$$​

is

$$H(s):=F(s)G(s)=\left( \sum_{k=1}^{\infty}a_{k}e^{-\alpha_k s}\right) \left( \sum_{n=1}^{\infty} b_{n}e^{-\beta_n s}\right) = \sum_{k=2}^{\infty} \left( \sum_{\substack{p+q=k\\p,q\geq 1}}a_{p}b_{q}e^{-\left(\alpha_p + \beta_q \right) s}\right) ,$$​

where the inner sum is over all integer partitions of k into exactly two summands p and q.

Note that this should reduce to the Dirichlet product and/or the Cauchy product under the respective special cases mentioned above.

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

### shmoe

The nice thing about the product form for a power series or dirichlet series is you get a power series or dirichlet series back owing to the way you've grouped the pairs of indicies (k,n) in your double infinite sum as either k*n=constant or k+n=constant.

This doesn't happen with how you've written the product of generalized dirichlet series. In fact you've just used the cauchy product formula on it. So I'd ask what the point is? Is this more natural than a convolution or just leaving it as a double sum?

3. Feb 27, 2006

### benorin

I haven't yet stumbled across a formula (not that the isn't one, for perhaps there is one) for such a product: what I have posted is what I managed by writing out the first few products of the partial sums. Any thoughts on how to give a more meaningful formula?

Also, ( I may as well go for it) I am charged with coming up with some original content :surprised: ! Any suggestions?

4. Feb 27, 2006

### benorin

Do not think that the above is to be the content of my paper: I have just begun studding the topic.

5. Feb 27, 2006

### shmoe

You do see that your product is just the usual cauchy product for series though, right? I don't think you can give a nice form for the product in the general case. By 'nice form' here I mean one that is a generalized dirichlet series as it's written (with increasing lambdas), and an infinite sum over only one index. In the special cases, the exponent lambdas are from a nice set, like log's of natural numbers, and this is preserved in the product. You won't get this in general.

I don't know if this will lead anywhere at all, but it's in the direction of your product idea. For the special dirichlet case, we get a nice product by grouping the pairs (k,n) over an infinite family of hyperbolas, each with a finite number of points. For the power series case, we group over an infinite family of lines with slope -1, each with a finite number of points. If you tried different curves, say circles (which requires some of the coefficients to be zero), will there be restrictions on the lambdas that make this into a nice product form? I haven't given this much thought, so it might not lead anywhere interesting.