The appearance of subharmonic eigenvalue orbits, recently discovered in the periodic spectrum of pulsating Poiseuille flow, can be traced back to the coalescence of eigenvalues at exceptional points. While the resonances associated with these points give rise to algebraic growth, the development of non-modal stability theory exploiting specific perturbation structures with much larger potential for transient energy growth has led to waning interest in spectral degeneracies. $$\rho=\prod_(s) > 0$.Spectral degeneracies where eigenvalues and eigenvectors simultaneously coalesce, also known as exceptional points, are a natural consequence of the strong non-normality of the Orr–Sommerfeld operator describing the evolution of infinitesimal disturbances in parallel shear flows. It will work (still have to go through a round of double-checking to confirm this) if So you need to look at cases where $P$ is infinite but sparse enough.
I am pretty sure there are no such examples. I could not find "standard" examples (characters modulo $m$) with a hole if $P$ contains sufficiently many primes. Also, I assume that if there is a "thick" hole, there is only one. This does not contradict the universality property: this property is true if $\sigma1$). That's what it does for the standard RH case.įinal update: non-trivial examples with a hole It will shrink as $t$ increases, in the end to a single point (or worse, an empty set if there are some unexpected roots) but incredibly slowly. But if you make a video of the orbit starting at $t=0$, no matter how fast your computer is, for a very, very long time the hole will be visible to the naked eye. There is one thing where I am most certainly wrong: the hole is reduced to to a single point (the origin), contrarily to what I thought initially (a hole of radius $\rho>0$). My guess is that for the product to be zero if $\sigma\rho$ for all $z=\sigma+it$ in the complex plane, for a fixed value of $\sigma$, say $0.75$. Does this also hold for the Euler product? If yes (and I assume that the answer is yes), then can the product still be equal to zero? The norm of the product can be very close to zero. I'm working on a number theory tutorial, and I'd like to introduce the students to a situation where the orbit not only never crosses the origin (as for the standard Riemann Hypothesis if $0.50$.Īssuming this is correct, and since convergence for a specific $\sigma_0=0.75$ at $t=0$ implies convergence for all $t$ for that particular $\sigma_0$, it implies that the abscissa of convergence is well below $1$. See example here and here for a truncated (finite) product when $\chi$ is constant equal to $1$.The hole in the orbit shrinks to a single point (the origin) if you add all the primes. imaginary part), for a fixed value of $\sigma$, and $t$ varying between $0$ and (say) $1000$. By orbit, I mean a scatterplot of the product in the complex plane (real vs. Is this the case, or is this still a conjecture?Īlso, I'd love to see a plot of the orbit for a fixed value of $\sigma$, say $\sigma=0.75$. Since the densities, for both set of primes, are identical, the signs will alternate nicely on average, and one would guess that the product converges not only for $\sigma>1$, but actually for $\sigma>0$. Also $s=\sigma+ it$ to use the standard notation. Here $\chi(2)=\pm 1$, the sign does not matter. The product is over all primes (in increasing order), with $\chi(p)=+1$ if $p \bmod 4 =3$ and I am interested in the convergence of the following Euler product:
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