argument of the Riemann zeta function












3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    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.



    plot of log(abs(zeta))



    Here's a histogram of $50000$ equally spaced values values taken by this function:



    histogram of log(abs(zeta))



    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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











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    2 Answers
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    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '12 at 19:03









        DidDid

        248k23224463




        248k23224463























            4












            $begingroup$

            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.



            plot of log(abs(zeta))



            Here's a histogram of $50000$ equally spaced values values taken by this function:



            histogram of log(abs(zeta))



            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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
















            4












            $begingroup$

            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.



            plot of log(abs(zeta))



            Here's a histogram of $50000$ equally spaced values values taken by this function:



            histogram of log(abs(zeta))



            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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














            4












            4








            4





            $begingroup$

            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.



            plot of log(abs(zeta))



            Here's a histogram of $50000$ equally spaced values values taken by this function:



            histogram of log(abs(zeta))



            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.






            share|cite|improve this answer











            $endgroup$



            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.



            plot of log(abs(zeta))



            Here's a histogram of $50000$ equally spaced values values taken by this function:



            histogram of log(abs(zeta))



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            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


















            • $begingroup$
              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


















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