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Building on this question: Zeros of $\zeta(s) \pm \zeta(1-s)$, I experimented further with:

$$\zeta(s) \pm \zeta(\overline s)$$

Assuming $s=\sigma + ti$, I observed that this function also has many "semi-trivial" as well as "non-trivial" zeros for each $\sigma$. Furthermore these "non-trivial" zeros all seem to reside very close to the Riemann non trivial zeros at $\sigma=\frac12$.

However, what I found curious is that only when $\frac12 < \sigma < 1$ the function:

$$\zeta(s) + \zeta(\overline s)$$

suddenly loses all of its "non-trivial" zeros (i.e. the ones near the Riemann zeros), whilst still retaining all of its "semi-trivial" zeros (they disappear when $\sigma >1$). Is there a logical explanation or even proof for this?

P.S.:

In an attempt to find out more, I used the alternating zeta-function $\eta(s)$ and rewrote it as:

$$\eta(s) - \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \sin(t \ln(n))}{n^\sigma}$$

and

$$\eta(s) + \eta(\overline s) =\displaystyle 2i \sum _{n=1}^{\infty } \frac{e^{\pi i n} \cos(t \ln(n))}{n^\sigma}$$

These functions look very symmetrical, but it seems that the denominator $n^\sigma$ drives the infinite alternating sum of the cosines to always positive near the non-trivial zeros whilst keeping the semi-trivial ones intact, when $\sigma > \frac12$. Could these functions be rewritten as an integral?

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Your functions are just the real and imaginary parts of the zeta function. –  John Pardon Apr 2 '12 at 4:48
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As unknown points out your functions are related to the real and imaginary parts of zeta: $ \zeta(s)=\overline{\zeta(\overline{s})}$. Certainly there are a lot of zeros with Re(s) in (1/2,1), how do you define "close to a zeta zero"? –  joro Apr 2 '12 at 6:59
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1 Answer

As "unknown" points out your functions are related to the real and imaginary parts of zeta: $ \zeta(s)=\overline{\zeta(\overline{s})}$.

Not sure your statement that $\zeta(s) + \zeta(\overline{s})$ "suddenly loses all of its "non-trivial" zeros (i.e. the ones near the Riemann zeros)" is correct.

There are a lot of zeros with $\frac12 < \sigma < 1$.

Here are the first few zeros in the above interval found with sage:

0.50462947732965045372 + 21.0i
0.52448011283041710992 + 30.379693112598828167i
0.58339455491020710687 + 30.0i
0.53642761519952422834 + 33.0i
0.65407326676423894749 + 47.747059280547112743i
0.50578106968814753741 + 48.0i
0.5282879698726187585 + 49.840960886611267987i
0.55323613245493607226 + 50.0i
0.50688116551268832104 + 53.0i
0.53003774683011668413 + 60.851647312217356793i
0.51632607137165855174 + 59.0i
0.696668107848621598 + 61.381499982325861301i
0.65871381704750571336 + 61.0i
0.54078375897341852744 + 65.06334913779052855i
0.57819848053572496195 + 65.0i
0.51405235699573598995 + 69.642351828461338856i
0.57314472284372395716 + 75.0i
0.65581349324544466194 + 89.0i
0.54628477636795993949 + 92.0i
0.52175065892232430153 + 94.608107654221020801i
0.61139715426268535099 + 96.0i
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@Joro: I have tried to illustrate my point in the attached graph s13.postimage.org/pp9j4zw1z/non_trivial_test.jpg I believe that a true non trivial zero $t_1 =14.134725...$ at $\sigma=\frac12$ always has a counterpart $t_1'$ that moves closer or further away depending on the $\sigma$. However, when $\sigma > \frac12$, both zeros disappear and become f.i. point B when $\sigma = \frac23$. There are also "semi-trivial zeros" on the right that are not connected to the non-trivial zeros and remain zero (e,g, point A on the yellow curve) and only when $\sigma > 1$ these do also disappear. –  Agno Apr 2 '12 at 18:01
    
However... ...after more carefully looking at the precise point where both zeros disappear, I found that the tilting value for $\sigma$ (for specifically both $t_1$ and $t_1'$ to disappear as zeros) does actually lie slightly above $\frac12$. So, back to the drawing board. –  Agno Apr 2 '12 at 18:13
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