MatsGranvik

John Baez question mathoverflow my answer

Sep 23rd, 2018
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  1. Clear[t, m, s, M, r, b, M, v, d, x, k, q, z, nn, H, T, TT, g, t, L];
  2. (*b=N[Sum[Sum[(BernoulliB[2*r]/((2*r)!))*(-d k^(1-2 r) E^(I x) \
  3. Abs[StirlingS1[2*r-1,m]] Gamma[1+m,s Log[d*k]] \
  4. Log[d*k]^(-1-m)),{m,1,2*r-1}],{r,1,q-1}],30];*)
  5. b = 4;
  6. M = 7;
  7. v = 2;
  8. k = 20;
  9. q = 8;
  10. c = 1;
  11. dd = 2;
  12. nn = 1;
  13.  
  14. z[t_] = Sum[1/n^(1/2 + I*t), {n, 1, k}] +
  15. k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2;
  16. z1 = ListLinePlot[Table[N[Re[z[t]]], {t, 0, 60, 1/10}],
  17. DataRange -> {0, 60}, ImageSize -> Large,
  18. PlotStyle -> {Thickness[0.004]}];
  19. z2 = Table[
  20. Graphics[{Arrowheads[0.025],
  21. Arrow[{{Im[ZetaZero[n]], -1/2}, {Im[ZetaZero[n]], 0}}]}], {n, 1,
  22. 13}];
  23. z3 = Table[
  24. Graphics[
  25. Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -1}],
  26. 90 Degree], Medium]], {n, 1, 13}];
  27. Show[z1, z2, z3]
  28.  
  29. h[t_] = Sum[(I d^(1/2 - I t) E^(I t x) n^(-(1/2) - I t))/(2 (-x +
  30. Log[d] +
  31. Log[n])) - (I d^(1/2 + I t) E^(I t x) n^(-(1/2) +
  32. I t))/(2 (x + Log[d] + Log[n])), {n, 1,
  33. k}] + -(1/2) I E^(x/2) ExpIntegralEi[
  34. 1/2 I (I + 2 t) (x - Log[d k])] +
  35. 1/2 I E^(-x/2) ExpIntegralEi[
  36. 1/2 (1 + 2 I t) (x +
  37. Log[d k])] - (I d^(1/2 - I t) E^(I t x) k^(-(1/2) -
  38. I t))/(4 (-x + Log[d] +
  39. Log[k])) + (I d^(1/2 + I t) E^(I t N[x]) k^(-(1/2) +
  40. I t))/(4 (x + Log[d] + Log[k]));
  41. t1 = -100;
  42. t2 = 100;
  43. L = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
  44. l1 = ListLinePlot[
  45. Re[Sum[Sum[L*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]],
  46. DataRange -> {1/200, dd}, PlotRange -> {-80, 80},
  47. ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
  48. l2 = Table[
  49. Graphics[{Arrowheads[0.025],
  50. Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
  51. l3 = Table[
  52. Graphics[
  53. Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
  54. 1, 7}];
  55. Show[l1, l2, l3]
  56.  
  57. nn = k;
  58. m[t_] = Sum[
  59. Sum[(Sum[1/n^(1/2 + I*t), {n, 1, k}] +
  60. k^(1 - (1/2 + I*t))/((1/2 + I*t) - 1) - (k^(-(1/2 + I*t)))/2)*
  61. MoebiusMu[d]/N[d]^(1/2 + I*t - 1), {d, Divisors[z]}]/z^c, {z, 1,
  62. nn}]; m1 =
  63. ListLinePlot[Table[N[Re[m[t]]], {t, 1/10, 60, 1/10}],
  64. DataRange -> {0, 60}, PlotRange -> {-3, 7 + 1/2},
  65. ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
  66. m2 = Table[
  67. Graphics[{Arrowheads[0.025],
  68. Arrow[{{Im[ZetaZero[n]], -1}, {Im[ZetaZero[n]], 0}}]}], {n, 1,
  69. 13}];
  70. m3 = Table[
  71. Graphics[
  72. Style[Rotate[Text[N[Im[ZetaZero[n]]], {Im[ZetaZero[n]], -2}],
  73. 90 Degree], Medium]], {n, 1, 13}];
  74. Show[m1, m2, m3]
  75.  
  76. nn = k;
  77. t1 = -100;
  78. t2 = 100;
  79. T = Table[N[(h[t2] - h[t1])], {x, 1/200, dd, 1/200}];
  80. f1 = ListLinePlot[
  81. Re[Sum[Sum[T*MoebiusMu[d], {d, Divisors[z]}]/z^c, {z, 1, nn}]],
  82. DataRange -> {1/200, dd}, PlotRange -> {-80, 80},
  83. ImageSize -> Large, PlotStyle -> {Thickness[0.004]}];
  84. f2 = Table[
  85. Graphics[{Arrowheads[0.025],
  86. Arrow[{{Log[n], -35}, {Log[n], 0}}]}], {n, 1, 7}];
  87. f3 = Table[
  88. Graphics[
  89. Style[Rotate[Text[Log[n], {Log[n], -52}], 90 Degree], Large]], {n,
  90. 1, 7}];
  91. Show[f1, f2, f3]
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