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Am I right that the Franca-Leclair approximation is a better approximation to the points on the critical line where the real part of the Riemann zeta function is zero and the imaginary part of log zeta is equal to $+-\frac{\pi}{2}$ rather than the actual zeta zeros?

Compare the Franca-Leclair approximation:

$$\Im(\rho _n) \approx 2 \pi \exp (1) \exp \left(W\left(\frac{n-\frac{11}{8}}{\exp (1)}\right)\right)$$

(*Mathematica 8*)
a = N[Table[
   2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]], {n, 1, 42}]]

{14.5213, 20.6557, 25.4927, 29.7394, 33.6245, 37.2574, 40.7006, \
43.994, 47.1651, 50.2337, 53.2144, 56.1189, 58.9563, 61.7338, \
64.4577, 67.133, 69.764, 72.3544, 74.9073, 77.4257, 79.9118, 82.3678, \
84.7957, 87.1972, 89.5737, 91.9268, 94.2576, 96.5674, 98.8571, \
101.128, 103.38, 105.615, 107.833, 110.036, 112.223, 114.395, \
116.554, 118.698, 120.83, 122.949, 125.056, 127.151}

to the roots found for $\Re\{\zeta(1/2+I*b_n)\}=0$ when using the Franca-Leclair approximation above as a starting point:

b = x /. Table[
   FindRoot[
    Re[Zeta[1/2 + I*x]] == 0, {x, 
     2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]]}], {n, 1, 42}]


{14.5179, 20.654, 25.4915, 29.7385, 33.6238, 37.2567, 40.7, 43.9935, \
47.1647, 50.2333, 53.2141, 56.1186, 58.956, 61.7335, 64.4574, \
67.1327, 69.7638, 72.3542, 74.9071, 77.4254, 79.9116, 82.3676, \
84.7955, 87.197, 89.5736, 91.9266, 94.2575, 96.5672, 98.8569, \
101.127, 103.38, 105.615, 107.833, 110.036, 112.223, 114.395, \
116.554, 118.698, 120.83, 122.949, 125.056, 127.151}

And to verify that $\Im\{\log(\zeta(1/2+I*b_n))\} = +- \frac{\pi}{2}$ for those latter values $b_n$:

Table[Im[Log[Zeta[1/2 + I*b[[n]]]]], {n, 1, 42}]

{1.5708, -1.5708, 1.5708, -1.5708, 1.5708, -1.5708, -1.5708, 1.5708, \
-1.5708, 1.5708, 1.5708, -1.5708, -1.5708, 1.5708, -1.5708, 1.5708, \
1.5708, 1.5708, -1.5708, 1.5708, 1.5708, -1.5708, 1.5708, -1.5708, \
1.5708, -1.5708, -1.5708, 1.5708, 1.5708, -1.5708, -1.5708, 1.5708, \
1.5708, -1.5708, 1.5708, 1.5708, 1.5708, -1.5708, -1.5708, 1.5708, \
1.5708, -1.5708}

Print["Pi/2"]
N[Pi/2]
1.5708

And to remind you what the Riemann zeta zeros are for comparison:

N[Im[ZetaZero[Range[42]]]]

{14.1347, 21.022, 25.0109, 30.4249, 32.9351, 37.5862, 40.9187, \
43.3271, 48.0052, 49.7738, 52.9703, 56.4462, 59.347, 60.8318, \
65.1125, 67.0798, 69.5464, 72.0672, 75.7047, 77.1448, 79.3374, \
82.9104, 84.7355, 87.4253, 88.8091, 92.4919, 94.6513, 95.8706, \
98.8312, 101.318, 103.726, 105.447, 107.169, 111.03, 111.875, 114.32, \
116.227, 118.791, 121.37, 122.947, 124.257, 127.517}
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  • 1
    $\begingroup$ The question title could perhaps be phrased better. $\endgroup$ – Mats Granvik May 12 '16 at 20:39
  • $\begingroup$ and $\frac{x}{\ln x}$ is an approximation to $\pi(x)$, so what ? $\endgroup$ – reuns Jun 5 '16 at 1:31

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