# Lerch-$\small \zeta(\varphi,0,-n)$ of integer *n* purely real and imaginary ($\small \zeta_\varphi (-n)^2$ is real) for $\small n \ge 2$?

Are the Lerch-$\zeta(\varphi,0,-n)$ of integer n (for shortness I use the notation of my earlier question $\small \zeta_\varphi(-n)$) periodically purely real and imaginary: $\zeta_\varphi (-n)^2$ is real, ($n \ge 2$) ? And how to prove this?

I've a strange observation of periodicity which I would like to explain/derive/prove analytically.

Consider the triangle of eulerian numbers E (ideally of infinite size, row and col-indices beginning at zero, with elements $\small e_{r,c}$)

$\qquad E = \small \begin{array} {rrrrr} 1 & . & . & . & . & . \\ 1 & 0 & . & . & . & . \\ 1 & 1 & 0 & . & . & . \\ 1 & 4 & 1 & 0 & . & . \\ 1 & 11 & 11 & 1 & 0 & . \\ 1 & 26 & 66 & 26 & 1 & 0 \end{array}$

Assume some angular parameter $\varphi$, the associated complex number from the unit-circle $z=z_\varphi= \exp(i \varphi ) \qquad z \ne 1$, $\small {\varphi \over \pi}$ not necessarily rational.

Now consider the Eulerian polynomials, whose coefficients are taken from a row of E, $\small E(z,r) = \sum_{c=0}^\infty ( e_{r,c} \cdot z^c )$

Then compute $\small \zeta_\varphi(-n) = {z \over (1-z)} \cdot E(z,n) \cdot (1-z)^{-n}$ .

Observation: I observe, that for $n>2$ the $\small \zeta_\varphi(-n)$ lie either on the real or on the imaginary axis, in other words $\small \zeta_\varphi(-n) ^2$ are real..

Q: I seem to be missing just the key idea, so my question here is, how I could derive that behave analytically, given the description via the Eulerian triangle.

Addendum/Generalization: While it was already surprising, that this works with rational roots z of the complex unit, it seems to me even more interesting, that the observation holds for arbitrary $\small \varphi \ne 0$ and $\small z_\varphi = \exp(I \cdot \varphi) \ne 1$ .

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Using your notations, I am getting $\zeta_\phi(-3)^2 = (z^2+4z+1)^2/(1-z)^8$. For $z=e^{i \pi/8}$, that value is $(1-i) \xi$, where $\xi \approx 45,013.022$, hence $\zeta_\phi(-3)^2$ is not real – Sasha Jul 26 '11 at 19:28
@Sasha: thanks for the crosscheck; I've reproduced your result. You missed that leading $z$ in the formula; you needed to write $\small \ldots = z^2 \cdot (z^2 +4z + 1)^2/(1-z)^8$ – Gottfried Helms Jul 26 '11 at 19:42
Thanks, in this case you deal with $\zeta(\varphi, -n, 0)$ in wikipedia's convention, I used $a=1$ as stated in the title. – Sasha Jul 26 '11 at 19:51
I see; I'll update the title. Unfortunately, in wikipedia the formula begins then with a division by zero, but this might go as minor notation problem then... – Gottfried Helms Jul 26 '11 at 19:59
given that $s = -n$, the first term is $z^0*0^n$ which is perfectly defined in the case at hand. – Sasha Jul 26 '11 at 20:13

By the symmetry of the triangle of Eulerian numbers the rational function $$R_{n}(z) = {z \over (1-z)} \cdot E(z,n) \cdot (1-z)^{-n}$$ satisfies$$R_{n}(z)=(-1)^{n+1}R_{n}(\frac{1}{z}).$$ Since the numerator and denominator polynomials of $R_{n}(z)$ have real (in fact, integer) coefficients we have$$\overline{R_{n}(z)}=R_{n}(\bar z).$$ Your observation should now be explained (there is no need to restrict to $n>2$).

To clarify further:

Let $z = \exp(i \cdot \varphi) \ne 1$ lie on the unit circle so $\bar z = \frac {1}{ z}$ and let $n \geq 1$. Suppose $R_{n}(z) = a+ib$.

Then from the above$$\overline{R_{n}(z)}=R_{n}(\bar z) = R_{n}(\frac{1}{z})=(-1)^{n+1}R_{n}(z).$$ Hence $$a-ib = (-1)^{n+1}(a+ib).$$ If $n$ is even we get $a = 0,$ and so $R_{n}(z)$ is pure imaginary; if $n$ is odd we get $b= 0,$ and $R_{n}(z)$ is real.

A particularly interesting case is when $z = i$. The sequence $(2 \cdot R_{n}(i))n \geq 1$ equals $[-1, -1 \cdot i, 2, 5 \cdot i, -16, -61 \cdot i,...]$, which is the sequence of up-down numbers multiplied by powers of $i$. For more on this see my comment of Jan 2011 in A000111 in the OEIS.

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thanks! For $R_n(z)$ n=1 this can even be seen when z is explicitely written as $z=\alpha + \beta i$ and then multiplied out (exploiting $\alpha^2 + \beta^2 = 1$ ) - the imaginary part vanishes then. But we still cannot allow n=0 ; we have always $-0.5 + x i$ which is not on the real or imaginary axis. – Gottfried Helms Jul 27 '11 at 7:00
Peter, what you define as $R_{n}(z)$ is $Li_{-n}(z)$ according to the last formula in link. – Bruce Arnold Jul 27 '11 at 21:30
Nice. Or even simpler Table[I^(n+1)*2*PolyLog[-n,-I],{n,1,11}]. – Bruce Arnold Jul 28 '11 at 22:14
@Peter: again thanks; that clarification is nice and concise, so I think, I'd accept that answer. Btw, in my previous question I showed that same sequence (for $\varphi=\pi/2$ ) only I did not yet expect, that it would be better to let the series begin with $z^1$ at the first term. (But I think it makes no sense to adapt that old question&answer) – Gottfried Helms Jul 29 '11 at 8:36

Unfortunately, I couldn't make Peter's answer fully transparent to me. So I tried two things; both led to a solution.
1) rewrite the $\small \zeta _\varphi(-n)$ - expression based on the description via the Eulerian polynomials as taylor-series in terms of $\small \varphi$, which is real. We can factor out $\small i^n$ and stay with a series in $\small \varphi ^2$, thus the result is $\small i^n \cdot x$ where x is real.

2) Rewrite the expression using the Eulerian polynomials using $\small z=exp(i \varphi)$
in $\small \zeta _\varphi(-n) = {z \over 1-z} E(z,n) \cdot (1-z)^{-n}$

Factor out $\small {z \over (1-z)^2}$ to get as first cofactor
$\qquad \small {z \over (1-z)^2} = {1 \over (1/\sqrt{z} - \sqrt z )^2} ={1 \over (\exp(- i \varphi/2) - \exp(i \varphi /2))^2 } = {1 \over (-2 i\sin(\varphi /2))^2 } = -1/4 \cdot \sin(\varphi /2)^{-2}$.
This value is always real, and is also $\zeta_\varphi(-1)$.

Then , for instance, for n=3 we get for the remaining part of the formula:

$\qquad\small \begin{array} {ll} (1 + 4z + 1z^2) \cdot (1-z)^{-2} &=& (1/z + 4 + 1z) \cdot (1/\sqrt{z} - \sqrt z )^{-2} \\\ & = &(1 \cdot (1/z + z) + 4 ) \cdot -1/4 \cdot \sin(\varphi /2)^{-2} \\\ & = & - ( {1 \over 2} \cos(\varphi)) + 1)\cdot \sin(\varphi /2)^{-2} \\\ \end{array}$

Together this gives
$\qquad \small \zeta_\varphi(-3)= 1/4 \cdot \sin(\varphi /2)^{-4} \cdot ( {1 \over 2} \cos(\varphi)) + 1)$

Well, this is only for one n, but I begin to understand, how this works and why we get these purely imaginary/real results (and I even begin to see Peter's mentioned symmetries here.... )

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