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For example, to compute the zeroes of the Riemann zeta function using the Euler-Maclaurin summation method one has to do O(T) work. The Euler-Maclaurin summation method for zeta is given by $ \zeta(s)= \sum^{N-1}_{n=1} \frac{1}{n^{s}} + \frac{N^{1-s}}{s-1}+\frac{N^{-s}}{2}+\frac{B_2}{2}sN^{-s-1} +\ldots + \\ \frac{B_{2v}}{(2v)!} \frac{(s+2v-2)!}{(s-1)!}N^{-s-2v+1} + R_{2v}$

where

$R_{2v}=-\frac{s(s+1) \ldots (s+2v-1)}{(2v+1)!} \int^\infty_N B_{2v+1}(x-[x]-\frac{1}{2}x^{-s-2v-1} \mathrm{d} x.$

may be computed in O(t) time. The Riemann-Siegel formula counterpart may be computed in $O\sqrt(t)$ time. I haven't the first clue of why this is true and this concept of time taken for computations. Anybody care to direct me somewhere which would give an explanation? Thanks.

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(this is a kind of follow up to a discussion about partial sums starting with an introduction to $\zeta\,$)

Let's compare the two proposed methods to evaluate $\zeta\left(s\right)$ with $s=\frac 12+i\,t\ $ (Euler-Maclaurin may be applied everywhere, Riemann-Siegel may be used too if $\,\Re(s)\not =\frac 12$ but this is less documented) :

  • Using Euler-Maclaurin you will need to compute the sum of at least the $\,N\approx \dfrac t{2\pi}\ $ first terms to get an accurate result (ignoring the 'cost' of a few Bernoulli terms and neglecting the remainder $R_{2v}$) from your expression : $$\zeta(s)= \sum^{N}_{n=1} \frac{1}{n^{s}} + \frac{N^{1-s}}{s-1}-\frac 1{2\,N^s}+\frac{B_2}{2}sN^{-s-1} +\ldots + R_{2v}$$ To illustrate this let's add $3000$ terms ($\displaystyle 1,\,2^{-1/2-it},\,3^{-1/2-it},\cdots$) in the complex plane with the first term at the right at $1+0\,i$ and finishing near the middle at $0+0\,i\ $ because I choose $t$ to be the first value larger than $10000$ such that $\zeta\left(\frac 12+i\,t\right)=0\,$. This gave the picture :

method 1

At this point let's apply the functional equation to the $N$ first terms of the series for $\zeta$ (consider every power term at $1-s$ instead of $s$ and multiply by $\gamma(1-s)$ as exposed at Wikipedia or at the bottom of this text) then we will get :

functional equation

(in the simple case $\,\Re(s)=\frac 12\,$ this is merely a 'mirror effect' applied to the first picture : complex numbers are replaced by their conjugates and a global rotation depending smoothly of $t$ is applied)

Now let's superpose the second picture turned $180^\circ$ to the first one : result

and observe the superposition of the first values (at the right on the first picture) to the final 'nodes' (at the left of the first picture).

  • This is the whole point of the Riemann-Siegel formula :
    from the $\,N\approx \dfrac {\Im(s)}{2\pi}\,$ terms needed to evaluate $\zeta(s)$ the first $[\sqrt{N}]$ terms provide half of the contribution while the remainder may be obtained with the $[\sqrt{N}]$ first terms terms of $\zeta(1-s)$ multiplied by $\,\gamma(1-s)$.

Replacing a sum over $N$ terms for Euler-Maclaurin by a sum over $[\sqrt{N}]$ terms is not uninteresting!

This may be done because of the functional equation : $$\zeta(s)=\gamma(1-s)\zeta(1-s)$$ with $\;\displaystyle\gamma(s):=\pi^{1/2-s}\frac{\Gamma(s/2)}{\Gamma((1-s)/2)}\;$ to give the "approximate functional equation" : $$\zeta(s)=\sum_{n=1}^m\frac 1{n^s}+\gamma(1-s)\sum_{n=1}^m\frac 1{n^{1-s}} + R(s)$$ the price is a not so easy to evaluate remainder : $$R(s)=\frac{\Gamma(1-s)}{2\pi i}\int_{C_m}\frac{(-x)^{s-1}e^{-mx}}{e^x-1}dx$$ (an approximation of this integral is needed for a precise evaluation)

The $\gamma$ notation used in the approximate functional equation is from Carl Erickson's very interesting and recommended article " A Geometric Perspective on the Riemann Zeta Function's Partial Sums".

For a formal derivation of these expressions your may consult too Titchmarsh's book about zeta or the chap.7 of Edwards' excellent book with a direct link to similar formulas here.

The details of actual computations are clearly exposed by Takusagawa and Gourdon&Sebah (for more see Edwards' book or Pugh's thesis.

Raymond Manzoni
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  • Thanks @J.M. I have a generator for that ! – Raymond Manzoni Apr 12 '13 at 13:15
  • Only very slightly related: have you by any chance seen Berry's papers on Riemann-Siegel? I have them, but haven't had the time to read them. My understanding was that he made a few tweaks to the formula to make it more efficient, but I haven't had the chance to test out that claim. – J. M. ain't a mathematician Apr 12 '13 at 13:22
  • @J.M. I have many papers of Berry (most of them unread... the guy is really very productive !) and appreciate very much his geometrical view on many topics. The two papers that seem to correspond to your indications may be the 1995 paper "The Riemann–Siegel expansion for the zeta function: high orders and remainders" (following Odlyzko and Schönhage's 'parallel evaluation') and the 1999 "The Riemann Zeros and Eigenvalue Asymptotics". Berry seems to handle with care the remainder $R(s)$ (justifying the notoriously difficult error terms...). – Raymond Manzoni Apr 12 '13 at 14:25
  • The method I used in the applet was simply the method used in Edwards' book (around 2000 obtaining the coefficients of the Taylor expansion took about an hour computer time) and I handled only the case $\Re(s)=\frac 12$ (Euler-Maclaurin is used elsewhere so that for large $t$ it is not recommanded to change the real value... :-)). Perhaps that the next version of the applet... (ok it didn't change since 200x...). Anyway thanks to point out Berry's excellent work ! – Raymond Manzoni Apr 12 '13 at 14:28
  • Yes, those two. :) I've used CRVZ extrapolation on Dirichlet $\eta$ outside the critical strip instead of Euler-Maclaurin in my own experiments. Ah, 2000... simpler times and simpler computers. Thanks for the response! – J. M. ain't a mathematician Apr 12 '13 at 14:33
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    Nice wall-paper. Look familiar? http://www.math.harvard.edu/~elkies/M229.09/index.html –  Sep 11 '13 at 22:00
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    @Andrew: The basic motif is deeper than you think. See this picture from Feynman's Masterpiece 'QED: The Strange Theory of Light and Matter. Quantum Electodynamics behaves this way as well as the other fundamental interactions QCD and Weak interactions ! (concerning gravity I don't know...) – Raymond Manzoni Sep 11 '13 at 22:16
  • Well, I never overlooked the depth. Wallpaper was from algebra http://en.wikipedia.org/wiki/Wallpaper_group not the pre-Raphaelites https://www.google.com/search?q=pre-raphaelite+wallpaper&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a –  Sep 11 '13 at 22:43
  • I think there's a small typo in the text, should be "More exactly the m first terms of the sum for ζ(s) will be added to the m first terms of the sum for ζ(1 - s)...". – user258279 Mar 15 '17 at 06:59
  • Your links only seem to give an approximation for the remainder R(s) for the Riemann-Siegel Z function. Is this the same R(s) as you show for the Riemann-Siegel formula? – user258279 Mar 15 '17 at 07:07
  • @user258279: You are right the formulation is somewhat incorrect (I noticed other mistakes and will update this answer). Numerical expressions for the remainder of the Riemann-Siegel formula (derived from my $R(s)$ here) are rather complicated compared say to Euler Maclaurin and this explains that most implementations are proposed only for $,\Re(s)=\dfrac 12$. In this specific case we may obtain quickly the real $Z$ function (very practical to get zeros of $\zeta$) – Raymond Manzoni Mar 15 '17 at 23:35
  • because the Riemann-Siegel $\theta$-function is easy to evaluate... Excellent continuation, – Raymond Manzoni Mar 15 '17 at 23:38
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    For your R(s), I calculated the remainder R(s) for the Riemann-Siegel Z function using Ken Takusagawa's code and just divided it by exp(iθ(t)). It worked! – user258279 Mar 16 '17 at 11:40