argument of the Riemann zeta function
$begingroup$
what does it mean that the function $ F(t) $
$$ F(t)= frac{argzeta (frac{1}{2}+it)}{sqrt{loglog(t)}} $$
is distributed as a 'Gaussian Random variable ?? in the limit $ t to infty $
a) $$ argzeta (frac{1}{2}+it)=(1+o(1))sqrt{loglog(t)}$$
b) the Argument of the Zeta function on the critical line is almost $ sqrt{loglog(t)} $
riemann-zeta
$endgroup$
add a comment |
$begingroup$
what does it mean that the function $ F(t) $
$$ F(t)= frac{argzeta (frac{1}{2}+it)}{sqrt{loglog(t)}} $$
is distributed as a 'Gaussian Random variable ?? in the limit $ t to infty $
a) $$ argzeta (frac{1}{2}+it)=(1+o(1))sqrt{loglog(t)}$$
b) the Argument of the Zeta function on the critical line is almost $ sqrt{loglog(t)} $
riemann-zeta
$endgroup$
$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47
add a comment |
$begingroup$
what does it mean that the function $ F(t) $
$$ F(t)= frac{argzeta (frac{1}{2}+it)}{sqrt{loglog(t)}} $$
is distributed as a 'Gaussian Random variable ?? in the limit $ t to infty $
a) $$ argzeta (frac{1}{2}+it)=(1+o(1))sqrt{loglog(t)}$$
b) the Argument of the Zeta function on the critical line is almost $ sqrt{loglog(t)} $
riemann-zeta
$endgroup$
what does it mean that the function $ F(t) $
$$ F(t)= frac{argzeta (frac{1}{2}+it)}{sqrt{loglog(t)}} $$
is distributed as a 'Gaussian Random variable ?? in the limit $ t to infty $
a) $$ argzeta (frac{1}{2}+it)=(1+o(1))sqrt{loglog(t)}$$
b) the Argument of the Zeta function on the critical line is almost $ sqrt{loglog(t)} $
riemann-zeta
riemann-zeta
edited Dec 31 '12 at 17:12
amWhy
1
1
asked Dec 31 '12 at 17:10
Jose GarciaJose Garcia
4,31611637
4,31611637
$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47
add a comment |
$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47
$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47
$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider a measurable real-valued function $G$ defined on $(0,+infty)$. For every $Tgt0$ and real number $x$, define $ell_T(x)$ as the Lebesgue measure of the set ${tleqslant Tmid G(t)leqslant x}$.
One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$,
$$
limlimits_{Ttoinfty}T^{-1}ell_T(x)=frac1{sqrt{2pi}}int_{-infty}^xmathrm e^{-z^2/2}mathrm dz,
$$
that is,
$$
limlimits_{Ttoinfty}T^{-1}int_0^Tmathbf 1_{G(t)leqslant x},mathrm dt=mathbb P(Zleqslant x),
$$
where $Z$ is a standard normal random variable.
There exists some variants of this definition but the idea remains the same.
$endgroup$
add a comment |
$begingroup$
What Selberg proved is that for $Esubset mathbb R$, we have that the limit as $Ttoinfty$ of
$$
frac{1}{T}mu(Tle tle 2T,|,arg(zeta(1/2+i t)/sqrt{1/2loglog t}in E)
$$
where $mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
frac{1}{sqrt{2pi}}int_E exp(-x^2/2), dx.
$$
Edit: Here's an example that may help clarify, using the harmonic conjugate $log|zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $log|zeta(1/2+i t)|/sqrt{1/2loglog(t))}$, for $50le tle 100$. Note it looks nothing like a Gaussian.
Here's a histogram of $50000$ equally spaced values values taken by this function:
Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.
$endgroup$
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider a measurable real-valued function $G$ defined on $(0,+infty)$. For every $Tgt0$ and real number $x$, define $ell_T(x)$ as the Lebesgue measure of the set ${tleqslant Tmid G(t)leqslant x}$.
One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$,
$$
limlimits_{Ttoinfty}T^{-1}ell_T(x)=frac1{sqrt{2pi}}int_{-infty}^xmathrm e^{-z^2/2}mathrm dz,
$$
that is,
$$
limlimits_{Ttoinfty}T^{-1}int_0^Tmathbf 1_{G(t)leqslant x},mathrm dt=mathbb P(Zleqslant x),
$$
where $Z$ is a standard normal random variable.
There exists some variants of this definition but the idea remains the same.
$endgroup$
add a comment |
$begingroup$
Consider a measurable real-valued function $G$ defined on $(0,+infty)$. For every $Tgt0$ and real number $x$, define $ell_T(x)$ as the Lebesgue measure of the set ${tleqslant Tmid G(t)leqslant x}$.
One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$,
$$
limlimits_{Ttoinfty}T^{-1}ell_T(x)=frac1{sqrt{2pi}}int_{-infty}^xmathrm e^{-z^2/2}mathrm dz,
$$
that is,
$$
limlimits_{Ttoinfty}T^{-1}int_0^Tmathbf 1_{G(t)leqslant x},mathrm dt=mathbb P(Zleqslant x),
$$
where $Z$ is a standard normal random variable.
There exists some variants of this definition but the idea remains the same.
$endgroup$
add a comment |
$begingroup$
Consider a measurable real-valued function $G$ defined on $(0,+infty)$. For every $Tgt0$ and real number $x$, define $ell_T(x)$ as the Lebesgue measure of the set ${tleqslant Tmid G(t)leqslant x}$.
One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$,
$$
limlimits_{Ttoinfty}T^{-1}ell_T(x)=frac1{sqrt{2pi}}int_{-infty}^xmathrm e^{-z^2/2}mathrm dz,
$$
that is,
$$
limlimits_{Ttoinfty}T^{-1}int_0^Tmathbf 1_{G(t)leqslant x},mathrm dt=mathbb P(Zleqslant x),
$$
where $Z$ is a standard normal random variable.
There exists some variants of this definition but the idea remains the same.
$endgroup$
Consider a measurable real-valued function $G$ defined on $(0,+infty)$. For every $Tgt0$ and real number $x$, define $ell_T(x)$ as the Lebesgue measure of the set ${tleqslant Tmid G(t)leqslant x}$.
One says that the function $G$ is asymptotically distributed as a standard normal random variable if, for every real number $x$,
$$
limlimits_{Ttoinfty}T^{-1}ell_T(x)=frac1{sqrt{2pi}}int_{-infty}^xmathrm e^{-z^2/2}mathrm dz,
$$
that is,
$$
limlimits_{Ttoinfty}T^{-1}int_0^Tmathbf 1_{G(t)leqslant x},mathrm dt=mathbb P(Zleqslant x),
$$
where $Z$ is a standard normal random variable.
There exists some variants of this definition but the idea remains the same.
answered Dec 31 '12 at 19:03
DidDid
248k23224463
248k23224463
add a comment |
add a comment |
$begingroup$
What Selberg proved is that for $Esubset mathbb R$, we have that the limit as $Ttoinfty$ of
$$
frac{1}{T}mu(Tle tle 2T,|,arg(zeta(1/2+i t)/sqrt{1/2loglog t}in E)
$$
where $mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
frac{1}{sqrt{2pi}}int_E exp(-x^2/2), dx.
$$
Edit: Here's an example that may help clarify, using the harmonic conjugate $log|zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $log|zeta(1/2+i t)|/sqrt{1/2loglog(t))}$, for $50le tle 100$. Note it looks nothing like a Gaussian.
Here's a histogram of $50000$ equally spaced values values taken by this function:
Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.
$endgroup$
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
add a comment |
$begingroup$
What Selberg proved is that for $Esubset mathbb R$, we have that the limit as $Ttoinfty$ of
$$
frac{1}{T}mu(Tle tle 2T,|,arg(zeta(1/2+i t)/sqrt{1/2loglog t}in E)
$$
where $mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
frac{1}{sqrt{2pi}}int_E exp(-x^2/2), dx.
$$
Edit: Here's an example that may help clarify, using the harmonic conjugate $log|zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $log|zeta(1/2+i t)|/sqrt{1/2loglog(t))}$, for $50le tle 100$. Note it looks nothing like a Gaussian.
Here's a histogram of $50000$ equally spaced values values taken by this function:
Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.
$endgroup$
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
add a comment |
$begingroup$
What Selberg proved is that for $Esubset mathbb R$, we have that the limit as $Ttoinfty$ of
$$
frac{1}{T}mu(Tle tle 2T,|,arg(zeta(1/2+i t)/sqrt{1/2loglog t}in E)
$$
where $mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
frac{1}{sqrt{2pi}}int_E exp(-x^2/2), dx.
$$
Edit: Here's an example that may help clarify, using the harmonic conjugate $log|zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $log|zeta(1/2+i t)|/sqrt{1/2loglog(t))}$, for $50le tle 100$. Note it looks nothing like a Gaussian.
Here's a histogram of $50000$ equally spaced values values taken by this function:
Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.
$endgroup$
What Selberg proved is that for $Esubset mathbb R$, we have that the limit as $Ttoinfty$ of
$$
frac{1}{T}mu(Tle tle 2T,|,arg(zeta(1/2+i t)/sqrt{1/2loglog t}in E)
$$
where $mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
frac{1}{sqrt{2pi}}int_E exp(-x^2/2), dx.
$$
Edit: Here's an example that may help clarify, using the harmonic conjugate $log|zeta(1/2+i t)|$ (which is implemented in Mathematica). The analog of Selberg's theorem is true for this function as well. The plot is of $log|zeta(1/2+i t)|/sqrt{1/2loglog(t))}$, for $50le tle 100$. Note it looks nothing like a Gaussian.
Here's a histogram of $50000$ equally spaced values values taken by this function:
Extreme negative values (near the Riemann zeros) are extremely scarce, as are large positive values. The fit to the bell curve is not yet good, but the Riemann zeta function approaches its asymptotic behavior very slowly.
edited Dec 23 '18 at 20:29
Glorfindel
3,41981830
3,41981830
answered Dec 31 '12 at 19:03
stopplestopple
1,494817
1,494817
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
add a comment |
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
then stopple what you mean is that $$ frac{1}{T}int_{T}^{2T}dt frac{arg xi(1/2+it)}{sqrt{(1/2)loglog(t)}}= frac{1}{sqrt{2pi}}int_{E}exp(-x^{2}/2) $$ is this true
$endgroup$
– Jose Garcia
Jan 2 '13 at 11:31
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Well, no. Your right side depends on $E$; your left side does not.
$endgroup$
– stopple
Jan 2 '13 at 15:29
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
@Jose: Very roughly what Selberg's theorem says is that if you make a histogram of the values of $arg(zeta(1/2+it))/sqrt{1/2loglog(t)}$, it will look like the Gaussian, i.e. a bell curve.
$endgroup$
– stopple
Jan 2 '13 at 17:01
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
sorry stopple i am a physicit so i do not understand much about probability (only for quantum mechanics) so if your reressent the function $$ argzeta(1/2+it)/sqrt{1/2loglog(t)} $$ will have the form of a Gaussian ?? thanks for your advise :) and your patience
$endgroup$
– Jose Garcia
Jan 2 '13 at 20:05
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
$begingroup$
@Jose: see edit above.
$endgroup$
– stopple
Jan 2 '13 at 21:11
add a comment |
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$begingroup$
A deterministic function of a deterministic variable cannot be distributed in any fashion as a random variable. On the other hand, it may be interesting to analyze the result when the independent variable $t$ is normally distributed. As $t rightarrow infty$ however, it may be that the phase of $F$ changes so rapidly that it may be better to treat it statistically.
$endgroup$
– Ron Gordon
Dec 31 '12 at 17:47