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Probability density function for Riemann-zeta zeros

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$begingroup$

Curious about the expected probability distribution for the spacing between Riemann zeta zeros, of the form $s_n=sigma+it_n$, where $sigma=0.5$ and $t_n$ is the imaginary part of the $n$-th zero.

The mean spacing between zeros decreases slowly as the height $t_n$ goes up, somewhat confounding the issue, but taking a narrow slice of 100,000 zeros starting at the billionth zero ($t_{1000000000}$) should make that variation negligible. Here's a histogram of the spacings in that region:

The longer upper tail precludes a normal distribution. The red curve is a best fit gamma distribution, which doesn't quite do it either. The mean spacing at the beginning of the range covered is $0.351087$ compared to $0.351073$ at the end, so that variation is small.

What is the closest distribution to model the spacing?

probability-theory riemann-zeta

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edited Mar 13 at 0:26

Joe Knapp

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

$endgroup$

add a comment | 

0

1

$begingroup$

Curious about the expected probability distribution for the spacing between Riemann zeta zeros, of the form $s_n=sigma+it_n$, where $sigma=0.5$ and $t_n$ is the imaginary part of the $n$-th zero.

The mean spacing between zeros decreases slowly as the height $t_n$ goes up, somewhat confounding the issue, but taking a narrow slice of 100,000 zeros starting at the billionth zero ($t_{1000000000}$) should make that variation negligible. Here's a histogram of the spacings in that region:

The longer upper tail precludes a normal distribution. The red curve is a best fit gamma distribution, which doesn't quite do it either. The mean spacing at the beginning of the range covered is $0.351087$ compared to $0.351073$ at the end, so that variation is small.

What is the closest distribution to model the spacing?

probability-theory riemann-zeta

share|cite|improve this question

edited Mar 13 at 0:26

Joe Knapp

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

$endgroup$

add a comment | 

$begingroup$

Curious about the expected probability distribution for the spacing between Riemann zeta zeros, of the form $s_n=sigma+it_n$, where $sigma=0.5$ and $t_n$ is the imaginary part of the $n$-th zero.

The mean spacing between zeros decreases slowly as the height $t_n$ goes up, somewhat confounding the issue, but taking a narrow slice of 100,000 zeros starting at the billionth zero ($t_{1000000000}$) should make that variation negligible. Here's a histogram of the spacings in that region:

The longer upper tail precludes a normal distribution. The red curve is a best fit gamma distribution, which doesn't quite do it either. The mean spacing at the beginning of the range covered is $0.351087$ compared to $0.351073$ at the end, so that variation is small.

What is the closest distribution to model the spacing?

probability-theory riemann-zeta

share|cite|improve this question

edited Mar 13 at 0:26

Joe Knapp

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

$endgroup$

share|cite|improve this question

edited Mar 13 at 0:26

Joe Knapp

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

share|cite|improve this question

edited Mar 13 at 0:26

Joe Knapp

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

asked Mar 10 at 13:16

Joe KnappJoe Knapp

733312

1 Answer
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$begingroup$

Just looking at other distributions, the three-parameter Burr type XII distribution fits pretty well. Here's the result as above around $t_{1e9}$:

The Burr type XII pdf is:

$f(x|alpha,c,k)={{kcover alpha}({xover alpha})^{c-1}over{(1+({xover a})^c)^{k+1}}}$

For the above fit,

$alpha = 0.837947, c=2.76292$ and $k=8.68889 $

However, the best fit parameters change at other heights, e.g. at $t_{10e9}$:

$alpha = 0.776292, c=2.73405$ and $k=9.38153 $

Note the mean has shifted down slightly. The mean is given by:

$mu = alpha kGamma(k-1/c)Gamma(1+1/c)/Gamma(k+1)$

The Burr type XII distribution is used most often to study mortality, survival, failure rates and the like. I guess the interval between a zeta zero and the next could be thought of as its lifespan.

share|cite|improve this answer

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

$endgroup$

add a comment | 

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0

$begingroup$

Just looking at other distributions, the three-parameter Burr type XII distribution fits pretty well. Here's the result as above around $t_{1e9}$:

The Burr type XII pdf is:

$f(x|alpha,c,k)={{kcover alpha}({xover alpha})^{c-1}over{(1+({xover a})^c)^{k+1}}}$

For the above fit,

$alpha = 0.837947, c=2.76292$ and $k=8.68889 $

However, the best fit parameters change at other heights, e.g. at $t_{10e9}$:

$alpha = 0.776292, c=2.73405$ and $k=9.38153 $

Note the mean has shifted down slightly. The mean is given by:

$mu = alpha kGamma(k-1/c)Gamma(1+1/c)/Gamma(k+1)$

The Burr type XII distribution is used most often to study mortality, survival, failure rates and the like. I guess the interval between a zeta zero and the next could be thought of as its lifespan.

share|cite|improve this answer

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

$endgroup$

add a comment | 

0

$begingroup$

Just looking at other distributions, the three-parameter Burr type XII distribution fits pretty well. Here's the result as above around $t_{1e9}$:

The Burr type XII pdf is:

$f(x|alpha,c,k)={{kcover alpha}({xover alpha})^{c-1}over{(1+({xover a})^c)^{k+1}}}$

For the above fit,

$alpha = 0.837947, c=2.76292$ and $k=8.68889 $

However, the best fit parameters change at other heights, e.g. at $t_{10e9}$:

$alpha = 0.776292, c=2.73405$ and $k=9.38153 $

Note the mean has shifted down slightly. The mean is given by:

$mu = alpha kGamma(k-1/c)Gamma(1+1/c)/Gamma(k+1)$

The Burr type XII distribution is used most often to study mortality, survival, failure rates and the like. I guess the interval between a zeta zero and the next could be thought of as its lifespan.

share|cite|improve this answer

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

$endgroup$

add a comment | 

$begingroup$

Just looking at other distributions, the three-parameter Burr type XII distribution fits pretty well. Here's the result as above around $t_{1e9}$:

The Burr type XII pdf is:

$f(x|alpha,c,k)={{kcover alpha}({xover alpha})^{c-1}over{(1+({xover a})^c)^{k+1}}}$

For the above fit,

$alpha = 0.837947, c=2.76292$ and $k=8.68889 $

However, the best fit parameters change at other heights, e.g. at $t_{10e9}$:

$alpha = 0.776292, c=2.73405$ and $k=9.38153 $

Note the mean has shifted down slightly. The mean is given by:

$mu = alpha kGamma(k-1/c)Gamma(1+1/c)/Gamma(k+1)$

The Burr type XII distribution is used most often to study mortality, survival, failure rates and the like. I guess the interval between a zeta zero and the next could be thought of as its lifespan.

share|cite|improve this answer

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

$endgroup$

share|cite|improve this answer

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312

answered Mar 12 at 14:33

Joe KnappJoe Knapp

733312