(*Mathematica 8 start*)
Clear[n, k, t, z, FL, NZ]
NZ[t_] = Arg[Zeta[1/2 + I*t]]/Pi;
Plot[NZ[t], {t, 280, 284}]
Plot[NZ[t], {t, 282.3, 282.6}]

Look at these two graphs of $$\arg(\zeta(1/2+I\cdot t))$$ around the Riemann zeta function zero 0.5 + 282.465 I:

Failure zoom out

Failure zoom in

Is it correct that close to $t=282.465$ the graph should make a jump increase of size $2$, and then should continue only slightly downwards, and then should suddenly decrease a step of almost equal to $1$? Should it be that way?


The argument jumps by $\pi$ at each zero because the curve $\zeta(\frac12+it)$ passes right through the origin. (With the principal argument the jump will be upwards or downwards according to whether the curve passes from the lower half-plane to the upper one, or vice versa).

The large jump by $2\pi$ just before the zero is just caused by normalization of the argument to fall within the standard interval $(-\pi,\pi]$. It shows that the curve crosses the negative real axis shortly before the zero at 282.465.

Note that the almost stable downwards slope of the graph between the jumps is because what you're really plotting is minus the the Riemann-Siegel theta function, plus $\pi$ times the number of critical zeroes below $t$, modulo $2\pi$, and the theta function is quite smooth for such large $t$.

  • $\begingroup$ What about the sudden downward fall of $1/2$ is that as it should be too? $\endgroup$ – Mats Granvik Nov 6 '15 at 15:30
  • $\begingroup$ @MatsGranvik: I don't see any jumps of size $1/2$ -- only ones of size $1$ and $2$. $\endgroup$ – hmakholm left over Monica Nov 6 '15 at 15:30
  • $\begingroup$ Well it is not exactly $1/2$ but only close to $1/2$. $\endgroup$ – Mats Granvik Nov 6 '15 at 15:31
  • $\begingroup$ @MatsGranvik: I don't even see any jump close to 1/2. The graph jumps from near $-1$ to near $1$, a distance of $2$, and then from near $1$ to near $0$, a distance of $1$. Are you misunderstanding the vertical scale on the graph you show? $\endgroup$ – hmakholm left over Monica Nov 6 '15 at 15:32
  • 1
    $\begingroup$ @MatsGranvik: I can't vouch for the correctness, but it doesn't look wrong to me. $\endgroup$ – hmakholm left over Monica Nov 6 '15 at 15:33

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