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.
 A: 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.
