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To investigate the relations between Appell / Jensen polynomials and the real-valued function for real $t$

$$\Omega(t) = \xi(1/2+it)/\xi(1/2),$$

where $\xi(s)$ is the Landau Riemann xi function, I need reasonable approximations, say to 3 significant digits, of

$$ Tr_{2n}= \sum_{k=0}^\infty 1/(z_k)^{2n}$$

for $n=1,2,3,4$ where the $z_k$ are the imaginary parts of the nontrivial zeros of the Riemann zeta function above the real axis. (Assume the RH is true, of course.)

I don't have access to Mathematica nor Maple, so help would be appreciated.

Edit (June 5, 2020):

To allay any further doubts about the convergence of $Tr_2$:

Titchmarsh in his classic book On the Theory of the Riemann Zeta Function has, on p. 18, Eqn 2.1.14

$$\Xi(z)= \xi(1/2+iz),$$

and on p. 30 he states that it is an even integral function of order 1, whose exponent of convergence is 1. "Hence $\Xi(z)$ has an infinity of zeros, whose exponent of convergence is 1. The same is true of $\xi(s).$" In his Theory of Functions on p. 249 is

Theorem 8.22: If $r_1, r_2...$ are the moduli of the roots of $f(z)$, then the series $\sum 1/r^{\alpha}$ is convergent if $\alpha > \rho.$

$\rho$ in an earlier paragraph is called the order of the integral function $f(z)$.

The absolute contribution of a zero of $\Omega$, $a+ib$, and its complex conjugate to the sum of the inverse squares of the zeros is $2(a^2-b^2)/(a^2+b^2)^2=2\cos(2\theta)/r^2$ with $\theta=0$ for the real zeros. This is less than $2/r^{\alpha}$ for $0< \alpha < 2$, so the trace of the paired inverse squares, even including any complex zeros if they were to be found, is absolutely convergent

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  • $\begingroup$ Do you know for certain that those series converge? $\endgroup$ – Gerry Myerson May 27 '20 at 6:25
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    $\begingroup$ $(s-1)\zeta$ is entire order $1$ (and maximal type of course) so $\sum 1/|z_k|^{1+\epsilon}$ will converge for any $\epsilon>0$ (which is the same as the OP sum since the real parts of the non-trivial zeroes are bounded regardless of RH) while of course $\sum 1/|z_k|= \infty$ and only the symmetric sum of the critical zeroes reciprocals converges conditionally (here we can consider trivial roots as they do not affect much, or take the symmetric from of $\zeta$ which gets rid of those) $\endgroup$ – Conrad May 27 '20 at 14:10
  • $\begingroup$ @Conrad, I was a little more careful in deciphering my old notes. Everything seems consistent now. Thanks for the comments. $\endgroup$ – Tom Copeland May 27 '20 at 17:43
  • $\begingroup$ excellent - and thanks for pointing to that article by Coffey which seems interesting $\endgroup$ – Conrad May 27 '20 at 17:50
  • $\begingroup$ @Conrad, all related to the fascinating history of ideas presented in "Zeros of entire Fourier transforms" by Dimitrov and Rusev. $\endgroup$ – Tom Copeland May 27 '20 at 18:23
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Just to make things visible and to help towards a hypothese (well aware of the comment of Gerry Myerson): I've somewhere found online the set of first 100 000 values of the imaginary parts to 8 digits. Partial sums in increasing segments of $2^k$ are the following:

   partial sums of sum(i=1,2*2^k-1, 1/(imag(root_i))^(2n)   

   k     n=1             n=2                 n=3                  n=4
  -----------------------------------------------------------------------------------------
   0  0.00500524412341  0.0000250524687349  0.000000125393721912  0.000000000627626189713
   1  0.00886667676944  0.0000327283874871  0.000000141065528135  0.000000000660375162896
   2   0.0121739737744  0.0000356030857633  0.000000143677509519  0.000000000662837761448
   3   0.0150045293315  0.0000366728779872  0.000000144108076453  0.000000000663021051125
   4   0.0172998726364  0.0000370263290034  0.000000144166228966  0.000000000663031185306
   5   0.0190794372367  0.0000371333200692  0.000000144173144661  0.000000000663031660997
   6   0.0203971381635  0.0000371628546828  0.000000144173860457  0.000000000663031679531
   7   0.0213326190435  0.0000371703457908  0.000000144173925660  0.000000000663031680139
   8   0.0219725965445  0.0000371721091123  0.000000144173930966  0.000000000663031680156
   9   0.0223966100709  0.0000371724981445  0.000000144173931358  0.000000000663031680157
  10   0.0226699376045  0.0000371725793391  0.000000144173931384  0.000000000663031680157
  11   0.0228420530479  0.0000371725955004  0.000000144173931386  0.000000000663031680157
  12   0.0229482871375  0.0000371725985893  0.000000144173931386  0.000000000663031680157
  13   0.0230127402289  0.0000371725991596  0.000000144173931386  0.000000000663031680157
  14   0.0230512683394  0.0000371725992617  0.000000144173931386  0.000000000663031680157
  15   0.0230740041036  0.0000371725992795  0.000000144173931386  0.000000000663031680157

update corrected wrong upper index in sum-expression
update2 corrected the mention in the partial sums protocol, that the sums are of course from the powers of the imaginary part of root_i alone.

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  • $\begingroup$ Nice, Gottfried. Thanks very much. However, Coffey gives $\xi(1/2)^{(2)}$ as approx. 0.022972 from an integral calculation and this should be equal to $Tr_2$ from manipulating the log derivative of the Hadamard product formula for xi. Since your partial sums exceed this, it suggests the series is not convergent. So there is an inconsistency somewhere between my analysis, Conrad's comment, and your partial sums. (I had initially erroneously assumed gaps of at least one between consecutive large $z_k$, ensuring convergence, but this gap assumption is not true.) $\endgroup$ – Tom Copeland May 27 '20 at 15:29
  • $\begingroup$ P. 529 of "Relations and positivity results for the derivatives of the Riemann $\xi$ function" by Coffey. $\endgroup$ – Tom Copeland May 27 '20 at 15:33
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    $\begingroup$ Okay, being a little more careful with my old notes: $$Tr_2 = \xi^{(2)}(1/2) / (2\xi(1/2)) \simeq .022972/(2 \cdot .497) \simeq .02311$$, so everything is consistent. $\endgroup$ – Tom Copeland May 27 '20 at 17:39
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    $\begingroup$ Second sanity check: $$12(Tr_2^2-Tr_4) \simeq 12((.02311)^2-.0000372) \simeq.005962$$ equal to $$\xi^{(4)}(1/2)/\xi(1/2) \simeq .0.002963/.497 \simeq .005962.$$ $\endgroup$ – Tom Copeland May 27 '20 at 21:18
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    $\begingroup$ Conversely, using the Faber polynomials, we can tease out the traces from the coefficients of $\Omega$; e.g., recalling that $$\Omega^{(n)}(0)=(-1)^n \xi^{(n)}(1/2)/\xi(1/2),$$ then $$2 Tr_4 = -4 \Omega^{(4)}(0)/4! + 2 (\Omega^{(2)}(0)/2)^2$$ $$ \simeq -4 \cdot .005962/4! + 2 (-.04622/2)^2 \simeq .00007448,$$ and from Gottfried $$2Tr_4 \simeq 2 \cdot .000037173 = .00007436.$$ $\endgroup$ – Tom Copeland Jun 2 '20 at 18:37

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