# Computing single summands of a zeta function

Given a zeta function

$$\zeta(s)=\sum_{n=1}^\infty |\lambda_n |^{-s},$$

I can do many tricks to get certain information. For example $\zeta'(0)$ might relate to the determinant of the operator where $\lambda_n$ are the eigenvalues.

Say I don't start with the $\lambda_n$'s but have an expression for which I know or postulates that it is a function of $\zeta(s)$,

is it possible to reconstruct $\lambda_n$ for chosen $n$ from my $\zeta(s)$?

If I have a graph of a function, I can in principle derive the coefficients of the Taylor expansion by measuring the curvature and the curvatue of the curvature and so on. Can something like this be done with an infinite sum of exponentiated numbers like in this case? At least I remember there was some technique which involved phase summation where only the $n=1$ survives.

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Depending on what exactly you mean by "reconstruct", but: as $\Re s\to\infty$, the function $\zeta(s)$ is asymptotic to $|\lambda_1|^{-s}$. (I'm ignoring the possibility of many $\lambda_n$ with the same modulus - easy to modify this comment.) Then, the function $\zeta(s) - |\lambda_1|^{-s}$ is asymptotic to $|\lambda_2|^{-s}$. And so on.... –  Greg Martin Feb 14 '13 at 22:05
@GregMartin: Thanks for the comment, by reconstruct I mean compute the specific value $\lambda_n$ form $\zeta(s)$ in case it's not given in terms of all the $\lambda_n$. Like say I have a conventional power series and I know that this object is the zeta function, i.e. I know that it can be written as summ of numbers to the power $-s$, then how do I know the $n$'th number. You procedure, subtracting $\lambda_1$ would need $\lambda$ as an input. If there is a way to get the first lambda, then I guess you answered my question. Are you saying I know $\lambda_1$ from asymptoic behaviour exactly? –  NiftyKitty95 Feb 14 '13 at 22:19
Depending on what measurements you're allowed to make of your function, yes. There is a unique positive real number $r$ such that $\zeta(s) \sim r^{-s}$ as $\Re s \to \infty$, and $\lambda_1$ satisfies $|\lambda_1|=r$. (By the way, in your formulation, there's no hope of finding the exact values of $\lambda_n$ as complex numbers, since $\zeta$ depends only upon their moduli and not their phase. So you can only hope to find $|\lambda_1|$, $|\lambda_2|$, etc. and not $\lambda_1$, $\lambda_2$, etc.) –  Greg Martin Feb 14 '13 at 23:17
@GregMartin: Okay, if you want to write this as an answer, maybe more in manual style, then I'll be happy to accept. The absolute-value thing is clear, yes. –  NiftyKitty95 Feb 15 '13 at 11:18
If you're allowing yourself to make any measurement you want of $\zeta(s)$, then it is possible to recursively reconstruct any finite number of the $|\lambda_n|$. (As has been mentioned, $\zeta(s)$ depends only on the moduli of the $\lambda_n$ and not on their phases, so reconstructing $|\lambda_n|$ is the best we can hope for.)
Let's rewrite the function slightly: let $0 < m_1 < m_2 < m_3 < \cdots$ be the distinct moduli in the set $\{|\lambda_n|\colon n\in\mathbb N\}$, and let $\nu_j\ge1$ equal the number of integers $n$ such that $|\lambda_n| = m_j$. This allows us to write $$\zeta(s) = \sum_{j=1}^\infty \nu_j m_j^{-s}.$$
We may now calculate the following quantities, in the order given, using only the values of $\zeta(s)$ (for $s$ real, for that matter): \begin{align*} m_1 &= \lim_{s\to\infty} \frac{\log\zeta(s)}{-s} \\ \nu_1 &= \lim_{s\to\infty} \frac{\zeta(s)}{m_1^{-s}} \\ m_2 &= \lim_{s\to\infty} \frac{\log\big(\zeta(s)-\nu_1m_1^{-s}\big)}{-s} \\ \nu_2 &= \lim_{s\to\infty} \frac{\zeta(s)-\nu_1m_1^{-s}}{m_2^{-s}} \\ m_3 &= \lim_{s\to\infty} \frac{\log\big(\zeta(s)-\nu_1m_1^{-s}-\nu_2m_2^{-s}\big)}{-s} \end{align*} and so on.