17
$\begingroup$

Is it true that

$$\lim_{y\rightarrow\infty}\dfrac{\sum_{n=1}^{y}n^{-1/2-iy}}{\zeta(1/2+iy)}=1$$

? Below is a plot of $$\sum_{n=1}^{y}\dfrac{1}{n^{s}}\text{for }s=\dfrac{1}{2}+iy$$

enter image description here

set against its smooth analytic continuation. Is this less expensive computationally for large $y$?

Notes

x = 1/2; Plot[{Re[\!\( \*SubsuperscriptBox[\(\[Sum]\), \(n = 1\), \(y\)] 
\*FractionBox[\(1\), SuperscriptBox[\(n\), \(x + I\ y\)]]\)], 
Re[Zeta[x + I y]], Im[\!\( \*SubsuperscriptBox[\(\[Sum]\), \(n = 1\), \(y\)] 
\*FractionBox[\(1\), SuperscriptBox[\(n\), \(x + I\ y\)]]\)], 
Im[Zeta[x + I y]]}, {y, 0, 30}] 

Included with subsuperscript boxes for ease of reading when pasted into Mathematica.

Update

To address Mercio's point:

enter image description here

The plot of the quotients (as posed in the original question) is very messy (includes grid-line at $1$), and clearly doesn't "tend to $1$" in the strict sense of the meaning. Perhaps if zeros of each function were excepted, it would make more sense. Suggestions of re-phrasings of the question are welcomed!

Update 2

Question now rephrased - improvements suggested by Raymond Manzoni:

Is it true that $$\lim_{y\rightarrow\infty}\zeta(\dfrac{1}{2}+iy)+\dfrac {1} {2}\dfrac {1} {[y/\pi]^{1/2 + iy}} - \sum_ {n = 1}^{[y/ \pi]}\dfrac {1} {n^{1/2 + iy}}=0?$$

which John M has more or less answered below.

$\endgroup$
  • 2
    $\begingroup$ $\sum n^{-s}$ diverges when $\operatorname{Re}s<1$ (in particular, when $s=1/2+iy$). So, no, you have to use reflection formula or something like that to define $\zeta(s)$ in the 'interesting' area. $\endgroup$ – Grigory M May 22 '14 at 13:59
  • 2
    $\begingroup$ @GrigoryM: The upper bound for $n$ in the summation depends on $y$ in the given formula. It is not therefore (to me) a priori impossible that this makes some sense for some reason. $\endgroup$ – J. J. May 22 '14 at 14:03
  • 3
    $\begingroup$ I guess another to phrase it is: Is it true that $$\lim_{y \to \infty}\frac{\sum_{n=1}^y\frac1{n^{\frac12+iy}}}{\zeta(\frac12+iy)} = 1$$ $\endgroup$ – Daniel R May 22 '14 at 14:10
  • 5
    $\begingroup$ Your denominator $\zeta(1/2+iy)$ may be approximated using the Euler Maclaurin expansion as exposed here with the finite sums $S_y(N):=\sum_{k=1}^Nk^{-1/2-iy}$ as long as $\;2\,\pi\,N >\ |y|$. The next 'error terms' contribute nearly $\dfrac 1{\sqrt{y}}$ in your specific case $N=y$ (which satisfies the previous inequality!). The two links at the end should provide, I hope, interesting additional information. $\endgroup$ – Raymond Manzoni May 25 '14 at 23:32
  • 6
    $\begingroup$ uh ... guys ? this is false for a stupid reason : unless the zeroes of the approximation are exactly the same as the zeroes of the zeta function (which would be very unexpected), the quotient of the two functions takes values as close to 0 and infinity as you want, so it can't have a limit. A valid question would be if the series is convergent for nonzero $y$ maybe ? $\endgroup$ – mercio May 28 '14 at 17:42
12
$\begingroup$

Please allow me to write what is true:

We have the estimate in the critical strip ($s = \sigma + it$): $$\zeta(s) = \sum_{n < N} n^{-s} + \frac{N^{1-s}}{s-1} + O(N^{-\sigma}).$$ Therefore, $$\zeta\left(\frac{1}{2}+it\right) = \sum_{n < t} \frac{1}{n^{\frac{1}{2}+it}} + O(t^{-1/2}).$$ So $$\lim_{t \rightarrow \infty} \left | \zeta\left(\frac{1}{2}+it\right) - \sum_{n < t} \frac{1}{n^{\frac{1}{2}+it}} \right| = 0.$$

You may also be interested in this recent preprint.

ADDED LATER (at Daniel's request):

First, we assume that $\sigma = \operatorname{Re}{z} > 1$, so that the sum $\sum n^{-s}$ converges absolutely. We can write the sum $\sum_{n \geq N} n^{-s}$ in terms as a Stieltjes integral, $$\sum_{n = N}^\infty \frac{1}{n^s} = N^{-s} + \int_N^\infty x^{-s} \;d(\lfloor x \rfloor),$$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Now we can integrate by parts to get $$\int_N^\infty x^{-s} \;d(\lfloor x \rfloor) = -N^{-s+1} + s \int_N^\infty x^{-s-1} \lfloor x \rfloor \; dx.$$ Let $\{x\}$ denote the fractional part of $x$, i.e. $\{x\} = x - \lfloor x \rfloor$. We get $$\zeta(s) = \sum_{n < N} n^{-s} + N^{-s} + \frac{N^{1-s}}{s-1} - s \int_N^\infty x^{-s-1} \{x\} \;dx.$$ Since $\{x\} < 1$, the integral on the right actually converges for all $\sigma = \operatorname{Re}(s) > 0$, i.e. we get an analytic continuation of the Riemann zeta function from $\sigma > 1$ to $\sigma > 0$. In fact, for $\sigma>0$, we can bound the integral, $$\left| \int_N^\infty x^{-s-1} \{x\} \;dx \right| < \frac{N^{-\sigma}}{\sigma}.$$

Anyway, this is all standard material available in any book on analytic number theory.

$\endgroup$
  • $\begingroup$ Could you elaborate a bit, especially on the estimate in the critical strip? $\endgroup$ – Daniel R May 30 '14 at 14:02
  • $\begingroup$ @DanielR - Hi I'll add that to my answer. $\endgroup$ – John M May 30 '14 at 16:46
  • $\begingroup$ Thank you for providing an explanation for the proof - to be honest, I am not sure that I fully understand each step (though I understand in it in a broad conceptual sense), but I shall persevere with the links you provided, and hopefully gain a deeper understanding as a result! $\endgroup$ – martin Jun 1 '14 at 21:30
  • $\begingroup$ @martin - If there are particular steps that are confusing, then I can point you somewhere that explains better. $\endgroup$ – John M Jun 1 '14 at 21:44
  • $\begingroup$ Thank you - I will have a good look and focus my questions carefully ! :) $\endgroup$ – martin Jun 1 '14 at 22:25
15
+50
$\begingroup$

(Updated)
As a complement to John M's proof (+1) I'll add some 'visual clarifications' (whatever this may mean...) about the behavior of the finite sum of $\zeta\;$ for a large but fixed ordinate $y\in\mathbb{R}^+$ : $$\tag{1}S_y(N)=\sum_{k=1}^N\frac 1{k^{\,1/2+iy}}$$ Let's suppose that $y$ is the first ordinate of a nontrivial zero larger than $10000$ and represent all the partial sums $\,S_y(N)$ for $\;N=1\cdots 3183=[y/\pi]\;$ in the complex plane :

N=3183

The cross at the middle is at the origin ($z=0$) while '1' represents $z=S_y(1)=1$ and '2' $\;S_y(2)=1+\dfrac {2^{-iy}}{\sqrt{2}}$ and so on up to '3183' representing $\;S_y(3183)$ very near to $0$ again. $$-$$ The picture shows figures looking like Cornu (or Euler) spirals. Let's justify this :
the sum $S_y(N)$ is obtained by addition of terms $\,\displaystyle \frac 1{k^{1/2+iy}}=\frac {e^{-iy\log(k)}}{\sqrt{k}}$.
The next term will thus be $\,\displaystyle \frac {e^{-iy\log(k+1)}}{\sqrt{k+1}}$.
Now for $k\gg 1$ the denominator will change only slightly while the change of phase of the numerator will be $\;\delta=-y\;(\log(k+1)-\log(k))=-y\,\log(1+1/k)\approx -\dfrac yk$.

$\delta\approx -\dfrac {y}k$ gives a special role to the values of $k$ such that $\dfrac yk\approx f\pi$ with $f$ integer :

  • for $f=2m$ (i.e. $k=\left[ \frac y{2m\pi}\right]$) we have $\,\delta\approx -2m\pi\,$ : for values of $k$ near of $\left[ \frac y{2m\pi}\right]$ the terms have nearly the same phase and their addition will nearly give a straight line (see f=2, f=4, and so on on the picture)
  • for $f=2m+1$ (i.e. $k=\left[ \frac y{(2m+1)\pi}\right]$) we have $\,\delta\approx -(2m+1)\pi\,$ : for values of $k$ near of $\left[ \frac y{(2m+1)\pi}\right]$ two consecutive terms nearly cancel each other and that's what is happening in the middle of the 'nodes' f=1, f=3, f=5 and so on.

Let's zoom the center of the final node $f=1$ :

zoom

The line nearly crossing the origin is obtained with the term $k=3183=\left[ y/\pi\right]$ while the line at its left and right came from $k=3182$ and $k=3181$ respectively. The value $S_y(3183)$ itself is far out of the picture (say $12$ or more times higher) while $S_y(3182)$ is far at the bottom on the other side. Taking the middle of these two values should bring us not too far from our target even if the different partial sums $S_y(N)$ don't really 'go down to $0$' ! (they merely turn around it)

All this explains (but doesn't prove) that an excellent approximation for $\zeta$ may be obtained (near the zeros at least) with the formula : $$\zeta\left(\frac 12+iy\right)\approx -\frac 1{2\,[y/\pi]^{1/2+iy}}+\sum_{n=1}^{[y/ \pi]}\frac{1}{n^{1/2+iy}}$$

(numerically the absolute error appears majored by $\dfrac{2.5}{y^{3/2}}$ in the range $(10,10000)$, and probably for larger values, while the initial sum from the question was majored by $\dfrac{0.9}{y^{1/2}}$)

Now what happens when we continue adding terms after $\dfrac y{\pi}$ ? Well the difference of phase $\,\delta\,$ will become smaller than $\pi$ and we will turn round and round and obtain an ever growing ball of black wool :-) (illustration for $N=10^7$)

black ball

This discussion was rather qualitative and may be followed by these more general and precise expansions provided not only for $\,N=\left[\dfrac y{\pi}\right]\,$ terms but also for $\left[\dfrac y{e}\right]$ terms with $0<e<2\pi$.

But the story doesn't end here and Riemann himself found that you didn't need to compute the sum of $[y]$ terms (or $\left[\dfrac y{\pi}\right]$ or $\left[\dfrac y{2\pi}\right]$ or whatever) but that $\left[\sqrt{\dfrac y{2\pi}}\right]$ terms were enough, at least if you accept some correction terms! (Riemann-Siegel formula)
and basic facts about Riemann $\zeta$ and the Euler Maclaurin formula.

As a fun alternative you may examine with care my first picture : notice that the distance from f=1 to f=3 is $1$, that the distance from f=3 to f=5 is $\dfrac 1{\sqrt{2}}$ and think 'Symmetry'!

All the pictures presented here were produced interactively using the (old) CSE applet that I created for Matthew R. Watkins (see the comments in case of problems).
Vitaliy Kaurov kindly provided nicer pictures using Mathematica.

$\endgroup$
  • 2
    $\begingroup$ $\Large\text{Yay!}$ $\endgroup$ – Daniel R May 31 '14 at 3:41
  • 1
    $\begingroup$ Thanks, that's an interesting perspective, and thank you for the links. $\endgroup$ – John M May 31 '14 at 8:11
  • 2
    $\begingroup$ @martin Mathematica or Wolfram Language version $\endgroup$ – Vitaliy Kaurov Jun 2 '14 at 6:40
  • 1
    $\begingroup$ Excellent work @VitaliyKaurov ! (antialiasing makes things nicer) $\endgroup$ – Raymond Manzoni Jun 2 '14 at 7:31
  • 1
    $\begingroup$ @martin: Concerning the CSE there is indeed some security 'blocking' in recent versions of Windows (7 and more and possibly Vista) or java (that I'll suppose installed and activated; btw you should notice the same problem with the linked applet by Pugh). To make it work I modified /configuration panel/java (32 bits)/security panel/add url and added the CSE link (a better way may exist...). Anyway Vitaly proposed you a useful alternative so enjoy! Cheers, $\endgroup$ – Raymond Manzoni Jun 2 '14 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.