Back to Riemann, Part 12

One of the issues I hadn’t covered in much detail in the last post was what happens if Riemann’s hypothesis fails somewhere. That is, what if one of the zeta zeros is not on the critical line x = 1/2.

Remember that in the explicit formula for the prime counting function, we’re summing up the effects of ALL of the zeros for a specific value of x (x being the range of counting numbers that contains the prime numbers we want). To make that easier to do, we pair up each zero with its mirror version: u = v + wi with mirror(u) = v – wi.

The equation for that zero-pair then becomes:

-(2/|u|) * (x^v) * cos(w*ln(x) – a)

If the zero lies on the critical line, we get:
-(2/sqrt((1/2)^2 + w^2) * sqrt(x) * cos(w * ln(x) – atan(w))

If it doesn’t, then we end up with two pairs of zeros:
v + wi, v – wi, 1 – v + wi and 1 – v – wi

And we get two equations:

-(2/sqrt((v)^2 + w^2) * x^v * cos(w * ln(x) – atan(w/v))
-(2/sqrt((1-v)^2 + w^2) * x^(1-v) * cos(w * ln(x) – atan(w/(1-v)))

Right away, it should be obvious that if v is really close to 0.5, that we’re going to have two results that are very similar, in effect doubling the impact of this zero on the total for the psi(x) summation. And, although size of w for the zero is going to be much larger than v is (say we use the 410th zero: 693.1769701, and v is 0.51), meaning that the variation in v won’t affect the absolute value of v + wi very much, the variations will become much more noticeable in the x^v and x^(1-v) parts as we calculate psi(x) for larger values of x (i.e. -> x = 100,000^(0.1) or 100,000^(0.9)).

Then, what would that look like? Well, first, it depends on how far from v = 1/2 the zeros are. We know that there are no zeros for v < 0 or v > 0, and it turns out that there are no zeros directly on the boundaries v = 0 or v = 1. In fact, the zeros are going to be closer to v = 1/2, anyway. But let’s assume that the rogue zero is somewhere between 0.1 and 0.9. If we have one zero at 0.9 + w, then we’ll get matching zeros at 0.9 – w, and 0.1 + w and 0.1 + w.

Now, let’s arbitrarily pick one of the zeros for w, say the 13th one, at 59.347044003.

Then, the plots for 2/sqrt((v)^2 + w^2) and 2/sqrt((1-v)^2 + w^2), with v running from 0.1 to 0.9 are above.
This is pretty flat.
If we add them together, remembering that we’re not counting v = 0.5 twice…

They are flat, except that, as mentioned before, having 4 zeros instead of 2 is doubling the magnitude of these pairs.

Now, the place where we really get a lot more contribution from these zeros is in the parts x^v and x^(1-v). If v is 0.9 and 1 – v is 0.1, then if x is 1000 and v is 0.9, (2/(|u|)) * x^v = 8.4, compared with 0.533 when v=1/2.

The reason for using x=1000 is simply that the contribution for (2/(|u|)) * x^0.5 is so small if x is small. Besides, looking at psi(1000) starts getting us up into somewhat larger values of x, and the entire point is what happens when x tends towards infinity.

Adding the results from the above graph together, we get:

where the v-notch represents the smallest value for the contribution for v = 0.5 and w = 59.347044003. If the zeros here are off the critical line, they really throw off the results of the psi(x) approximation.

(Correct approximation for psi(100.)

Compare the correct graph above, for psi(100), with the 11th zero set so that v=0.9 and 1-v=0.1, below.

(Summations of the first 11 zeros, comparing the correct values to v=0.1/v=0.9 for the 11th zero.)

der Veen and de Craats talk about the “music of the primes.” We saw this a little bit in the last blog entry. The cosine component for each non-trivial zero of the explicit formula creates a growing, slowing cosine result as x gets bigger. These ringing bell waveshapes get added together for ALL zeros, spanning the entire range of x, in the psi(x) equation. This is why the Riemann hypothesis is so important. If even one zero is off the critical line, you get the 2 pairs of zeros adding together instead of the one pair, throwing off the step-shape of the approximation. And that invalidates the prime number theorem.

The way der Veen and de Craats put it, each non-trivial zero, in the explicit formula, acts like the next octave up on the “musical scale”. These are smooth, tonal increments from one zero to the next. If any of the non-trivial zeros is not on the critical line, the amplitude of that doubled-up zero pair becomes an ugly discordant spike.

I’m showing the first 10 zero pairs from the explicit formula to show what I mean. I introduced the first and tenth zeros in the last blog entry. Adding them all together gives the approximation for psi(100) in the above graph.

Also, I just want to show the magnitude contributions for the first ten zeroes, as x goes to 6,250. The relative spacing between one zero and the next remains constant, independent of x.

But it’s not a linear, smooth or consistent spacing.

(Left: Relative spacing of the contributions of the first 10 zeros for the explicit formula. Right: For comparison’s sake, the relative spacing for the first 10 prime numbers. Note that spacing of the zeros doesn’t reflect the spacing of the primes themselves.)

Anyway, the objective for this blog entry was just to consider the impact of having one of the non-trivial zeros not being on the critical line. The answer is that it really depends on how far off from the line the two matching zero pairs (v and 1 – v) are. If they’re very close to 0.5, then they just double the magnitude of the -(2/|u|) * (x^v) * cos(w*ln(x) – a) contribution for that zero.

If they’re farther apart, such as being at v = 0.9 and 1 – v = 0.1, then the results can be up to 8 times greater. It all depends on how big x is where you’re looking.

Having a proof showing that the Riemann hypothesis is true, or even that it is false, would remove the uncertainty from the question. After 158 years, we’re still waiting for someone to figure out what that proof is.

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