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# Euler - Cornu like spirals ( spiros, clothoids ) of simple zeta zeros

Posted 11 years ago

## Side note

This is a good place to note that Wolfram Language was used and mentioned in a recent preprint by Yuri Matiyasevich (known for his negative solution of Hilbert's tenth problem) and Gleb Beliakov:

Approximation of Riemann's zeta function by finite Dirichlet series: multiprecision numerical approach

where authors also give thanks to our own Oleksandr Pavlyk in the Acknowledgements section.

## Visualizations

Recently there were series of forum discussions about interesting visualizations related to simple zeta zeros that are easily done with Wolfram Language. I will give a short summary here, with explanations and code almost word by word from:

All started from a question whether the following is true:

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

A simple logic goes along these lines. 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 $$\lim_ {t \rightarrow \infty} \left | \zeta\left(\frac{1}{2}+it\right) - \sum_{n < t} \frac{1}{n^{\frac{1}{2}+it}} \right| = 0.$$

As a complement to the proof there are some "visual clarifications" (whatever this may mean...) about the behavior of the finite sum of $\zeta$ for a large but fixed ordinate $y\in\mathbb{R}^+$ :

$$S_ y(N)=\sum_ {k=1}^N\frac 1{k^{1/2+iy}}$$

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 :

Manzoni[n_, y_] := Transpose@{ Re@#, Im@# }&@Accumulate[Range[n]^-(.5 + y I)]

ListLinePlot[Manzoni[3183, Im[ZetaZero[1, 10000]]], Frame -> True,
GridLines -> Automatic, PlotStyle -> Directive[Red, Opacity[.5]],
ImageSize -> 700, PlotRange -> All, AspectRatio -> Automatic]


1st point represents $z=S_ y(1)=1$ and 2nd $S_ y(2)=1+\dfrac {2^{-iy}}{\sqrt{2}}$ and so on up to 3183 representing $S_ y(3183)$ very near to $0$.

The picture shows figures looking like Cornu (or Euler) spirals. Let's justify this :
the sum $S_ y(N)$ is obtained by addition of $\,\displaystyle \frac 1{k^{1/2+iy}}=\frac {e^{-iy\log(k)}}{\sqrt{k}}\,$ terms.
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:

ListLinePlot[Manzoni[3183, Im[ZetaZero[1, 10000]]], Frame -> True,
GridLines -> Automatic, PlotStyle -> Directive[Red, Opacity[.4]],
PlotRange -> .0008 {{-1, 1}, {-1, 1}}, AspectRatio -> Automatic]


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). All this explains (but doesn't prove) that an excellent approximation for $\zeta$ may be obtained (near the zeros at least) with the formula :

$$\zeta\left(\frac 12+iy\right)\approx \sum_ {n=1}^{[y/ \pi]}'\frac{1}{n^{1/2+iy}}=-\frac 1{2\,[y/\pi]^{1/2+iy}}+\sum_ {n=1}^{[y/ \pi]}\frac{1}{n^{1/2+iy}}$$

(numerically the absolute error appears majored by $\dfrac{4.5}{y^{3/2}}$ in the range $(10,10000)$, and probably for larger values, while the initial sum from the question was majored by $\dfrac{0.9}{y^{1/2}}$)

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Posted 1 month ago
 A Geometric Perspective on the Riemann Zeta Function's Partial Sums by Carl Erickson visualizes the convergence of the alternating sum—which I reference here—gives similar visualisations.
Posted 3 years ago
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