8
$\begingroup$

How to visualize in a nice geometric way (e.g., like in this YT video) the Riemann zeta function, $\zeta (z)$?

$\endgroup$
  • 3
    $\begingroup$ FWIW, here's a barebones version of one of the pictures shown in that video: ParametricPlot[ReIm[Zeta[r Exp[I t]]], {r, 0, 4}, {t, 0, 2 π}, Mesh -> True, PlotStyle -> None]. You might also be interested in this. $\endgroup$ – J. M. is away Dec 12 '16 at 13:45
  • $\begingroup$ @J.M. Nice ParametricPlot. Why not post an answer? $\endgroup$ – corey979 Dec 12 '16 at 13:51
  • $\begingroup$ It's "barebones", as I said. Somebody else might want to make something more stylish out of it. :) If you're looking for other visualization ideas, you might be able to adapt the ones used at the Wolfram Functions site. $\endgroup$ – J. M. is away Dec 12 '16 at 14:02
  • $\begingroup$ Plot the real function $\xi(1/2+it), t \in \mathbb{R}$ where $\xi(s) = s(s-1)\pi^{-s/2} \Gamma(s/2) \zeta(s)$, then do the same with the Hurwitz zeta functions, and see how the RH fails for those $\endgroup$ – user1952009 Dec 12 '16 at 18:57
  • $\begingroup$ @user1952009, and RiemannXi[] is built-in, too! $\endgroup$ – J. M. is away Dec 13 '16 at 3:02
6
$\begingroup$

One of the visualisations in the linked video is similar to the following. Parameter t is the imaginary value (height) above the real axis.

Manipulate[
   Module[{p, c, z},
      p = Table[RiemannSiegelZ[t]*{Cos[t], Sin[t]}, {t, 0, tmax - dt, dt}];
      c = Table[ColorData["DarkRainbow", t/tmax], {t, 0, tmax - dt, dt}];
      z = Im[N[ZetaZero[Range[tmax]]]];
      Graphics[{
         Red, Line[{{-4, 0}, {4, 0}}], Line[{{0, -4}, {0, 4}}],
         Thick, Line[Take[p,Floor[t/dt]], VertexColors->Take[c,Floor[t/dt]]]},
         PlotLabel -> ToString[Select[z, # <= t &]], Background -> Black,
         PlotRange -> 4 {{-1, 1}, {-1, 1}}, BaseStyle -> {FontSize -> 12},
         ImageSize -> 600]
      ],
{{tmax, 60., "Maximum t"}, 3, 100, Appearance -> "Labeled"},
{{dt, 0.1, "Delta t"}, 0.03, 0.3, Appearance -> "Labeled"},
{{t, 0.0, "Parameter t"}, 0.0, tmax, 1, Appearance -> "Labeled"}]

zeta manipulation

Or try an animation:

Module[{tmax = 60, dt = 0.1, p, c, z},
   p = Table[RiemannSiegelZ[t]*{Cos[t], Sin[t]}, {t, 0, tmax - dt, dt}];
   c = Table[ColorData["DarkRainbow", t/tmax], {t, 0, tmax - dt, dt}];
   z = Im[N[ZetaZero[Range[tmax]]]];
   Animate[
      Graphics[{
      Red, Line[{{-4, 0}, {4, 0}}], Line[{{0, -4}, {0, 4}}],
      Thick, Line[Take[p, Floor[t/dt]], VertexColors->Take[c,Floor[t/dt]]]},
      PlotLabel -> ToString[Select[z, # <= t &]], Background -> Black,
      PlotRange -> 4 {{-1, 1}, {-1, 1}}, BaseStyle -> {FontSize -> 12},
      ImageSize -> 500],
  {t, 0.0, tmax}, AnimationDirection -> ForwardBackward]
]

The curving coloured line plots the complex value of w=Zeta[1/2+ I t] as real t increases from zero. The red axes are the real part of w (horizontal) and imaginary part of w (vertical). As each non-trivial zeta-function root is encountered on this critical line x=1/2, the curve passes through the origin and the plot label appends its t value to a list.

$\endgroup$
  • $\begingroup$ I'd add that this way of visualizing $\zeta$ is especially interesting for demonstrating Lehmer's phenomenon. $\endgroup$ – J. M. is away Dec 13 '16 at 14:28
7
$\begingroup$

One way is to show the convergence of the defining sum.

Background

The zeta function is defined as

$$\zeta (s)=\sum\limits_{k=1}^{\infty}\frac{1}{k^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\ldots$$

which is:

  • divergent for $s=1$;
  • gives nice values for other positive integers, e.g. Zeta[2]$=\frac{\pi^2}{6}$;
  • is well defined on the complex plane if $\Re (s)>1$;
  • when $\Re (s)\leq 1$, the sum is divergent, e.g. $\zeta (-1)=1+2+3+4+\ldots$. However, one can perform a unique analytic continuation that ascribes a value of $-\frac{1}{12}$ to $\zeta (-1)$ (not to be covered here).

For $s\in\mathbb{C}$, e.g. $s=2+i$, the sum will be

$$\frac{1}{1^2 1^i}+\frac{1}{2^2 2^i}+\frac{1}{3^2 3^i}+\frac{1}{4^2 4^i}+\ldots$$

Each of the terms consists of a purely real and a purely imaginary exponent (referred to as part in short). The real part describes the length of the number (i.e., $\frac{1}{1^2}=1$ says the number is 1 unit from $(0,0)$; $\frac{1}{2^2}=1/4$ says that the number is $1/4$ units from the previous number, etc.); the imaginary part describes an angle. Given this, each term describes a vector starting from the end of the previous one.

Code

Defining a partial sum of $n$ terms:

zeta[s_, n_] := Table[1/k^s, {k, 1, n}]

one can visualize the addition of partial vectors in the complex plane with

Manipulate[
 DynamicModule[{loc = {2., 1.}}, 
  LocatorPane[Dynamic[loc], 
   Dynamic@ListLinePlot[
     Accumulate@
      Partition[
       Join[{{0, 0}}, ReIm /@ zeta[loc[[1]] + loc[[2]] I, n]], 2, 1], 
     PlotRange -> {{-1, 3}, {-2, 2}}, Frame -> True, 
     AxesOrigin -> {1, 0}, AspectRatio -> 1, 
     PlotLabel -> If[loc[[1]] <= 1, "Divergent", "Convergent"], 
     Epilog -> {Red, PointSize[Large], 
       Point@ReIm@Zeta[loc[[1]] + loc[[2]] I]}]
   ]
  ]
 , {n, {100, 500, 1000}, ControlType -> PopupMenu}]

enter image description here

enter image description here

enter image description here

The Locator shows the input value $s$; the red point shows the exact value Zeta[s]; the spiral with colorful segments shows the partial vectors in the complex plane. n can be varied. One can see that when $\Re (s)$ is close to $1$, the convergence is very slow.

$\endgroup$
  • $\begingroup$ You could use Accumulate[Partition[ReIm[Prepend[1/Range[n]^({1, I}.loc), 0]], 2, 1]] instead for the expression within ListLinePlot[]. $\endgroup$ – J. M. is away Dec 12 '16 at 13:36
  • 1
    $\begingroup$ $\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^\infty (n^{-s}-\int_n^{n+1} x^{-s}dx)$ for $Re(s) > 0$ is more interesting, and the convergence is slow when $t$ gets large (see the Lindelöf hypothesis) $\endgroup$ – user1952009 Dec 12 '16 at 18:59
5
$\begingroup$
rz[u_, v_] := With[{i = Complex[u, v]}, ReIm[Zeta[i]]]
szf[{u_, v_}, n_Integer] := {{0, 0}}~Join~
  N@Table[ReIm[Sum[1/j^(u + I v), {j, 1, k}]], {k, 1, n}]
Manipulate[
 Row[{ParametricPlot[{u, v}, {u, -r, r}, {v, -r, r}, 
    MeshFunctions -> {#3 &, #4 &}, Mesh -> {10, 10}, 
    Exclusions -> None, ImageSize -> 400, 
    Epilog -> {Red, PointSize[0.02], Point[p]}],
   ParametricPlot[rz[u, v], {u, -r, r}, {v, -r, r}, 
    MeshFunctions -> {#3 &, #4 &}, Mesh -> {10, 10}, 
    Exclusions -> None, ImageSize -> 400, 
    Epilog -> {Red, PointSize[0.02], Point[rz @@ p], Green, 
      Line[szf[p, number]], Black, Point[szf[p, number]]
      }, PlotRange -> {{-r, r}, {-r, r}}]}], {{p, {1, 
    1}}, {-r, -r}, {r, r}, 
  Slider2D}, {{r, 2}, {2, 5, 10, 20}}, {number, Range[2, 100]}]

enter image description here enter image description here enter image description here

If you want to see behaviour along Re[z]=1/2:

tab1 = Table[{1/2, j}, {j, -30, 30, 0.1}];
tab2 = Table[ReIm[Zeta[1/2 + j I]], {j, -30, 30, 0.1}];
lp1 = ListPlot[tab1, Joined -> True, PlotStyle -> Dashed, 
     Epilog -> {Red, PointSize[0.02], Point[#]}, Frame -> True, 
     ImageSize -> 400] & /@ tab1;
lp2 = ListPlot[tab2, Joined -> True, PlotStyle -> Dashed, 
     Epilog -> {Red, PointSize[0.02], Point[#]}, Frame -> True, 
     ImageSize -> 400] & /@ tab2;
an = Row /@ Thread[{lp1, lp2}];

Animated gif is here

$\endgroup$
2
$\begingroup$

I improved Corey's answer by animating the partial sum graph.

I feel this gives good insight into what the zeta function is computing as Im(s) increases.

I added an adjustment for Re(s), to show how zeros only happen along Re(s) = ½.

zeta[s_, n_] := Table[1/k^s, {k, 1, n}]

Animate[
 Manipulate[
  LocatorPane[Dynamic[{re, im}],
   Dynamic@ListLinePlot[
     Accumulate@Partition[
       Join[{{0, 0}}, ReIm /@ zeta[re + im I, n]], 2, 1],
     PlotRange -> {{-4, 4}, {-4, 4}}, Frame -> True, AspectRatio -> 1,
      Epilog -> {Blue, PointSize[Large], Point@ReIm@Zeta[re + im I]}]],
  {{re, 0.5, "Re(s)"}, 0, 1, Appearance -> "Labeled"},
  {{n, 50, "Iterations"}, 10, 200, 1, Appearance -> "Labeled"}],
 {{im, 0, "Im(s)"}, 0, 1000, AnimationRate -> 2, Appearance -> "Labeled"}]

I am fascinated by the asymmetric-symmetry.

Zeta Partial Sum

$\endgroup$

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.