# Generalization $\zeta_\varphi(s)=\sum_{k=0}^\infty {\exp(I\varphi*k) \over (1+k)^s}$

This is more a reference-request for some fiddling/exploration with the $\zeta$-function. In expressing the $\zeta$ and the alternating $\zeta$ (="$\eta$") in terms of matrixoperations I asked myself, what we get, if we generalize the idea of the alternating signs to cofactors from the complex unit-circle.

$$\zeta_\varphi(s)=\sum_{k=0}^\infty {\exp(I\varphi*k) \over (1+k)^s}$$

With this the usual "alternating $\zeta$" (or "$\eta$")- function were identified with $\varphi=\pi$, so each second term of the $\zeta$-series has the cofactor of $-1$. I looked at $\varphi=\pi/2$ and $\varphi=\pi/4$ so far. The $\zeta_{\pi/2}(s)$ for $s=[0,-1,-2,-3,-4,\ldots ]$ are for instance

$[1/2*(1+I), 1/2I, -1/2, -1I, 5/2, 8I, -61/2, -136I, 1385/2, \ldots]$.

But after this,... well a better approach is first to ask here for known results/discussion because involving $\zeta$ means usually, that very likely someone has looked at something like this before....(For instance, it seems we have analogues to the bernoulli-polynomials and possibly there is also an analogon to the Euler-Maclaurin-formula)

Since Gerry asked for the computation and I lost him at $M_\varphi$ I'll put some more explanation here.

I express the problem in my matrix-notation (which is admittedly a very private notation lacking rigor but hopefully gives a clue of what I'm doing). For this the main ingredient is the ubiquituous occurence of a vector V(x) (ideally of infinite size as all matrices involved) which is meant to denote the reference to the argument x in a taylor-series of a function f(x). Here the coefficients of the formal powerseries are collected in some vector A , V(x) means $\small [1,x,x^2,x^3,...x^n]$ (ideally $n\to\infty$ and f(x) is principally expressed as dot- or matrixproduct $\small f(x) = A \cdot V(x)$. Then this notation allows to formulate the required manipulations on the formal powerseries by vector-operations, matrixproducts, matrixpowers and inversion.

Let P be the lower triangular Pascalmatrix, containing the binomial-coefficients. Then the binomial-theorem allows to write $$\small P \cdot V(x) = V(x+1)$$ $$\small P^2 \cdot V(x) = V(x+2)$$ and so on. Then we can do the linear combination
$$\small (P^0 + P + P^2 + ... + P^k) \cdot V(1) = V(1) + V(2) + .... + V(1+k) = S(1,k)$$

Here we get in all rows of S the sums-of-like-powers of exponents 0,1,2,3,... and so on, simultaneously.

Using the alternating signed sum we can extend the sum to infinite index k getting

$$\small \begin{eqnarray} AS(1) &=& V(1)-V(2)+V(3)-...\\\ &=& (P^0-P+P^2-P^3+...-...)\cdot V(1) \\\ &=& (Id + P)^{-1} \cdot V(1) \\\ &=& H \cdot V(1) \end{eqnarray}$$

which involves the closed-form-formula for geometric series for a matrix-argument.

In each row of the result AS(1) we get now the Dirichlet $\eta$ at the nonpositive integer index according to the rowindex. Also the matrix $\small H = (Id+P)^{-1}$ contains that $\eta$'s and moreover the rows describe a modification of the bernoulli-polynomials adapted to the problem of alternating summing of like powers.
The non-alternating sum, resulting in $\zeta$-values (usually expressed in terms of bernoulli-numbers) cannot be taken this way because $\small Id - P$ cannot be inverted (but there is a workaround such that we still get a solution).

Now the generalization (for which I ask for references) is $$\small S_\varphi (1) = V(1) + z V(2) + z^2 V(3) + z^3 V(4)+ z^4 V(5) + ...$$ where $\small z = \exp(I \cdot \varphi)$ lies on the complex unit-circle. Thus $\small AS(1) = S_{\pi}(1)$ and generally $$\small \begin{eqnarray} S_{\varphi}(1) &=& (P^0 + z \cdot P^1 + z^2 \cdot P^2 + z^3 \cdot P^3 + ...) \cdot V(1) \\\ & =& (Id - z \cdot P)^{-1} \cdot V(1) \\\ &=&M_\varphi \cdot V(1) \end{eqnarray}$$ where the matrix $\small M_\varphi$ can be computed as long as $\varphi \ne 0$

The $\zeta_\varphi(k)$ for k=[0,-1,-2,-3,...] can now be taken from the according row of $\small S_\varphi(1)$. I assume that using the matrix $\small M_\varphi$ we can construct analogues of the bernoulli-polynomials to compute the $\small \zeta_\varphi(s)$ at real or complex s, which is what I was referring to in my above question.

[update] the idea of taking the "geometric series" for some matrix as I did it here $\small M_\varphi = (Id - z \cdot P)^{-1}$ is known as "Neumann-series"

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How are you getting values where the series doesn't converge? Have you worked out an analytic continuation? a functional equation? –  Gerry Myerson Jul 5 '11 at 10:36
For a first impression I used the Pari/GP-sumalt-function (which implements sort of Abel-summation, I think). After that I defined the lower triangular pascalmatrix P, and computed, using $z=\exp(I\varphi)$ the expression for the geometric series of zP $M_{\varphi} = (Id - z*P)^{-1}$ .Then, according to the similar process for the case of $z=1$ and $z=-1$ we get in the first column the required $\zeta_{\varphi}(s)$ for s=0,-1,-2,... (note, that for $z=1$ the geometric series-formula is inapplicable and needs a workaround) –  Gottfried Helms Jul 5 '11 at 11:11
You lost me at $M_{\phi}$. –  Gerry Myerson Jul 5 '11 at 13:04
Hmmm. Technically this is easy ( Pari/GP, could not format this properly): z=I \\ which is exp(I*Pi/2) ; dim=8; P=matpascal(dim-1); M=(matid(dim)-z*P) ^ -1 ; print(M[,1]) –  Gottfried Helms Jul 5 '11 at 13:16
For the inversion of the (Id + P)-matrix there is a nice article which covers the proof and some generalization to powers of P in Inverting the Pascal Matrix Plus One Rita Aggarwala and Michael P. Lamoureux The American Mathematical Monthly Vol. 109, No. 4 (Apr., 2002), pp. 371-377 Published by: Mathematical Association of America Article Stable URL: jstor.org/stable/2695500 –  Gottfried Helms Jul 6 '11 at 11:59

jax and Gerry Myerson are correct, $\mathrm{Li}_s ( \omega) \omega^{-1}$ (which is the formal expression you are taking about, taken as a function of $s$), for a root of unity $\omega$, can be written as a linear sum of Dirichlet L-functions. Let $\omega$ be an $n$-th root of unity. Hence $f(k) = \omega^k$ can be interpreted as a function on $\mathbb{Z} / n \mathbb{Z}$. The group of all (Dirichlet) characters on $\mathbb{Z} / n \mathbb{Z}$, called $\hat{G}$, actually forms an orthonormal basis for the vector space of all functions $\mathbb{Z} / n \mathbb{Z} \to \mathbb{C}$, with the inner product $\langle f, g \rangle = \phi(n)^{-1}\sum f(a) \bar{g}(a)$. Hence we can write $f(k)$ as a linear combination of elements from $\hat{G}$:

$$\omega^k = \sum_{\chi \in \hat{G}} a_{\chi} \chi(k).$$

We can deduce the coefficients via what are called the Gauss sums (this is essentially a finite version of an inverse Fourier transform),

$$a_{\chi} = \frac{1}{\phi(n)} \sum_{m=0}^{n-1} \omega^m \bar{\chi}(m).$$

When applying this decomposition of $\omega^k$ to your original zeta expansion, we get

$$\sum_{n=1}^\infty \frac{\omega^{n-1}}{n^s} = \frac{1}{\omega \phi(n)} \sum_{\chi \in \hat{G}} \left( \sum_{m=0}^{n-1} \omega^{m} \bar{\chi}(m)\right) L(s,\chi).$$

Functional equations of L-functions with primitive Dirichlet characters, either odd or even (which are separate cases), can be found using theta functions, but I'm not sure if we know of good formulas otherwise. I wish I could be of more help in knowing how the specific values for $s= 0 , -1, -2, - 3, \dots$ are found, but I'm not aware of the literature there. But it does appear we do know how to calculate them somehow, because Wolfram Alpha is quite able to provide the analytic values of PolyLog[n,z] for various negative integers $n$ and roots of unity $\omega$. At any rate, here are a couple good references for Dirichlet characters:

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very nice -thanks! I just took a look at Pari/GP' polylog-function which provides the results at least numerically. I'll try your online-links then. –  Gottfried Helms Jul 6 '11 at 13:37
I've found a very nice article of L.C.Maximon in 2003 ("The dilogarithm function for complex arguments"), where he collect references to articles on generalizations of the dilog - some of them just in the manner like I asked it here. The nearest seem to be an article of A.Jonquiere (1889) and M. Lerch(1887). This shall give me some food to chew on. Very likely I'll continue this very interesting subject with some specific follow-up questions. Thanks to all respondents so far. (I'll accept this answer above to "close the case" as it is very near to what I'm doing myself). –  Gottfried Helms Jul 23 '11 at 6:29
On the matter of dilogarithms, @Gottfried, you'll also want to peer at Lewin's books; you should be able to find something useful there. –  Ｊ. Ｍ. Jul 23 '11 at 6:48

It is the polylogarithm function $\mathrm{Li_s}(z)=\sum_{n=1}^\infty \frac{z^n}{n^s}$. In your notation $\zeta_\varphi(s)=\mathrm{Li_s}(e^{i \varphi})e^{-i \varphi}$.

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I don't think so - you've got a function of $z$, what's wanted is a function of $s$. –  Gerry Myerson Jul 5 '11 at 10:33
@Andrew: the link to the wikipedia gives excessive very good informations, thanks for the hint. As Gerry mentions, my focus is indeed the indexed zeta as in my question, where the role of z and s as parameter and argument are exchanged; but there is such a lot of information there that I'll get enough stuff to review and to look into up to the weekend. –  Gottfried Helms Jul 6 '11 at 8:03

It sounds like you'd want to look into Dirichlet characters and their corresponding $L$-series.

(edit: Oops, the sequences are in a different order. Sorry.)

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I think Gottfried's functions can be expressed as linear combinations of Dirichlet $L$-functions, so I think you're right to suggest looking at them. –  Gerry Myerson Jul 5 '11 at 10:37
Hmm, one difference that I see is that dirichlet-characters have periodically zeros, but here $\text{abs}(\exp(I*\varphi*k))$ is always 1... –  Gottfried Helms Jul 5 '11 at 12:37
The trivial Dirichlet character is the exception, it's always 1. –  Gerry Myerson Jul 5 '11 at 13:01

Even though it won't be considered 'a proper answer' - I will elaborate on the connection to the Bernoulli polynomials.

Lets assume the bernoulli polynomials are periodic - they repeat their values between $[0,1]$ on the real line. The coefficients of the fourier series will then be given by the following integrals with

$$-\cos\left(\frac{\pi m}{2}\right)\frac{(2\pi n)^m}{m!}\int_0^1B_m(x)\cos(2\pi n x)\textrm{d}x = \left\{\begin{array}{cl} 1& \mbox{if } m \mbox{ even} \\0 & \mbox{otherwise} \end{array}\right.$$

and $$-\sin\left(\frac{\pi m}{2}\right)\frac{(2\pi n)^m}{m!}\int_0^1B_m(x)\sin(2\pi n x)\textrm{d}x = \left\{\begin{array}{cl} 0& \mbox{if } m \mbox{ even} \\1 & \mbox{otherwise} \end{array}\right.$$

Where $B_m(x)$ is the $m$th bernoulli polynomial.

So $$B_m(x)=-\cos(\frac{\pi m}{2})\frac{m!}{(2\pi)^m}\sum_{n=0}^\infty \frac{\cos(2\pi n x)}{n^m}$$ for even $m$ and otherwise the same with sine on interval $[0,1]$.

So if we insert $x=0$ we get $\zeta(m)$ (for even $m$). If you insert the appropriate $x = 1/4, 2/4, 3/4$ etc you get $\beta(m)$ and $\eta(m)$.