Does this normalization of the Riemann zeta function make sense?












0












$begingroup$


For $c=0$ the following should be true for the $n$-th Gram point:
$$frac{x}{2 pi e}logleft(frac{x}{2 pi e}right) = frac{x}{2 pi e}log left(frac{x}{2 pi e}right) + frac{-c+n}{e}-frac{vartheta (x)}{pi e}$$
and for $c=frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $vartheta (x)$ is the Riemann-Siegel theta function.



That equation can be solved for $x$ like this:



$$x=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



where $W$ is the Lambert W function.



Then name the right hand side the function $f(x)$:
$$f(x)=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:



$$g(n,x)=f(...f(f(f(f(x))))...)$$



This should allow us to cut the real part of the Riemann zeta function:



$$Releft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



at complementary Gram points $c=frac{1}{2}$ into a family of functions with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$.
In case I did not say this correctly, this how it is written in Mathematica:



Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


Plot of the superimposed real parts:



real parts superimposed



where the end points at $0$ and $1$ are complementary Gram points.



The average of the 101 curves looks like this:



real parts average



This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:



cosine plus one



Mathematica code for the normalized real parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]


We can do the same for the imaginary part of Riemann zeta:



$$Imleft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



Again we look at the family of functions between complementary Gram points $c=frac{1}{2}$ with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$.
In Mathematica:



Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


The plot of the superimposed imaginary parts looks like this:



imaginary parts superimposed



where the middle common point at $frac{1}{2}$ is at a complementary Gram point.



The average of the 101 curves is:



average of imaginary parts



This again looks like $sin (x)$ on the interval $left[0,2piright]$:



sine function




Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?




Mathematica code for the normalized imaginary parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]









share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 30 '17 at 21:25










  • $begingroup$
    I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 11:50






  • 1




    $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 31 '17 at 12:04










  • $begingroup$
    @reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 12:28








  • 1




    $begingroup$
    The more precise version is $vartheta^{-1}(t)$.
    $endgroup$
    – reuns
    Jul 31 '17 at 12:36
















0












$begingroup$


For $c=0$ the following should be true for the $n$-th Gram point:
$$frac{x}{2 pi e}logleft(frac{x}{2 pi e}right) = frac{x}{2 pi e}log left(frac{x}{2 pi e}right) + frac{-c+n}{e}-frac{vartheta (x)}{pi e}$$
and for $c=frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $vartheta (x)$ is the Riemann-Siegel theta function.



That equation can be solved for $x$ like this:



$$x=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



where $W$ is the Lambert W function.



Then name the right hand side the function $f(x)$:
$$f(x)=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:



$$g(n,x)=f(...f(f(f(f(x))))...)$$



This should allow us to cut the real part of the Riemann zeta function:



$$Releft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



at complementary Gram points $c=frac{1}{2}$ into a family of functions with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$.
In case I did not say this correctly, this how it is written in Mathematica:



Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


Plot of the superimposed real parts:



real parts superimposed



where the end points at $0$ and $1$ are complementary Gram points.



The average of the 101 curves looks like this:



real parts average



This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:



cosine plus one



Mathematica code for the normalized real parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]


We can do the same for the imaginary part of Riemann zeta:



$$Imleft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



Again we look at the family of functions between complementary Gram points $c=frac{1}{2}$ with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$.
In Mathematica:



Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


The plot of the superimposed imaginary parts looks like this:



imaginary parts superimposed



where the middle common point at $frac{1}{2}$ is at a complementary Gram point.



The average of the 101 curves is:



average of imaginary parts



This again looks like $sin (x)$ on the interval $left[0,2piright]$:



sine function




Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?




Mathematica code for the normalized imaginary parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]









share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 30 '17 at 21:25










  • $begingroup$
    I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 11:50






  • 1




    $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 31 '17 at 12:04










  • $begingroup$
    @reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 12:28








  • 1




    $begingroup$
    The more precise version is $vartheta^{-1}(t)$.
    $endgroup$
    – reuns
    Jul 31 '17 at 12:36














0












0








0


2



$begingroup$


For $c=0$ the following should be true for the $n$-th Gram point:
$$frac{x}{2 pi e}logleft(frac{x}{2 pi e}right) = frac{x}{2 pi e}log left(frac{x}{2 pi e}right) + frac{-c+n}{e}-frac{vartheta (x)}{pi e}$$
and for $c=frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $vartheta (x)$ is the Riemann-Siegel theta function.



That equation can be solved for $x$ like this:



$$x=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



where $W$ is the Lambert W function.



Then name the right hand side the function $f(x)$:
$$f(x)=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:



$$g(n,x)=f(...f(f(f(f(x))))...)$$



This should allow us to cut the real part of the Riemann zeta function:



$$Releft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



at complementary Gram points $c=frac{1}{2}$ into a family of functions with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$.
In case I did not say this correctly, this how it is written in Mathematica:



Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


Plot of the superimposed real parts:



real parts superimposed



where the end points at $0$ and $1$ are complementary Gram points.



The average of the 101 curves looks like this:



real parts average



This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:



cosine plus one



Mathematica code for the normalized real parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]


We can do the same for the imaginary part of Riemann zeta:



$$Imleft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



Again we look at the family of functions between complementary Gram points $c=frac{1}{2}$ with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$.
In Mathematica:



Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


The plot of the superimposed imaginary parts looks like this:



imaginary parts superimposed



where the middle common point at $frac{1}{2}$ is at a complementary Gram point.



The average of the 101 curves is:



average of imaginary parts



This again looks like $sin (x)$ on the interval $left[0,2piright]$:



sine function




Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?




Mathematica code for the normalized imaginary parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]









share|cite|improve this question











$endgroup$




For $c=0$ the following should be true for the $n$-th Gram point:
$$frac{x}{2 pi e}logleft(frac{x}{2 pi e}right) = frac{x}{2 pi e}log left(frac{x}{2 pi e}right) + frac{-c+n}{e}-frac{vartheta (x)}{pi e}$$
and for $c=frac{1}{2}$ it should be true for the $n$-th complementary Gram point. $vartheta (x)$ is the Riemann-Siegel theta function.



That equation can be solved for $x$ like this:



$$x=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



where $W$ is the Lambert W function.



Then name the right hand side the function $f(x)$:
$$f(x)=2 pi e exp left(Wleft(frac{-c+n-frac{vartheta (x)}{pi }+frac{x log left(frac{x}{2 pi e}right)}{2 pi }}{e}right)right)$$



and iterate it as many times as computer power allows. This iteration is the normalizing function that we call $g(n,x)$:



$$g(n,x)=f(...f(f(f(f(x))))...)$$



This should allow us to cut the real part of the Riemann zeta function:



$$Releft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



at complementary Gram points $c=frac{1}{2}$ into a family of functions with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $x=1$.
In case I did not say this correctly, this how it is written in Mathematica:



Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


Plot of the superimposed real parts:



real parts superimposed



where the end points at $0$ and $1$ are complementary Gram points.



The average of the 101 curves looks like this:



real parts average



This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:



cosine plus one



Mathematica code for the normalized real parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-1.3, 6.4}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Re[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]


We can do the same for the imaginary part of Riemann zeta:



$$Imleft(zeta left(frac{1}{2}+i g(n,1)right)right)$$



Again we look at the family of functions between complementary Gram points $c=frac{1}{2}$ with intervals $n=left[k,k+1right]$ where $k=0,1,2,3,4,...,100$. The seed point is $y=1$.
In Mathematica:



Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}]


The plot of the superimposed imaginary parts looks like this:



imaginary parts superimposed



where the middle common point at $frac{1}{2}$ is at a complementary Gram point.



The average of the 101 curves is:



average of imaginary parts



This again looks like $sin (x)$ on the interval $left[0,2piright]$:



sine function




Would such a normalization with the function $g(n,x)$, be any easier to study, than the Riemann zeta function directly?




Mathematica code for the normalized imaginary parts:



Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E*
E^LambertW[((N[x]/(2*Pi))*Log[N[x]/(2*Pi*E)] - c + n -
RiemannSiegelTheta[N[x]]/Pi)/E];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
Table[ListLinePlot[
Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}],
DataRange -> {0, 1}, PlotRange -> {-5, 5}], {k, 0, 100}], k]]
Clear[x, n, c, f, g]
c = 1/2;
f[x_] = 2*Pi*E^1*
E^LambertW[((x/(2*Pi))*Log[x/(2*Pi*E)] - c + n -
RiemannSiegelTheta[x]/Pi)/E^1];
g[y_] = f[f[f[f[y]]]];
Show[Monitor[
ListLinePlot[
Sum[Table[Im[Zeta[1/2 + I*g[1]]], {n, k, k + 1, 1/80}], {k, 0,
100}]/101, DataRange -> {0, 1}], k]]






analytic-number-theory riemann-zeta fixed-point-theorems experimental-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 4 '18 at 13:25







Mats Granvik

















asked Jul 30 '17 at 10:46









Mats GranvikMats Granvik

3,29932252




3,29932252












  • $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 30 '17 at 21:25










  • $begingroup$
    I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 11:50






  • 1




    $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 31 '17 at 12:04










  • $begingroup$
    @reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 12:28








  • 1




    $begingroup$
    The more precise version is $vartheta^{-1}(t)$.
    $endgroup$
    – reuns
    Jul 31 '17 at 12:36


















  • $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 30 '17 at 21:25










  • $begingroup$
    I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 11:50






  • 1




    $begingroup$
    I don't understand what you wrote. What is supposed to be this iteration ?
    $endgroup$
    – reuns
    Jul 31 '17 at 12:04










  • $begingroup$
    @reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
    $endgroup$
    – Mats Granvik
    Jul 31 '17 at 12:28








  • 1




    $begingroup$
    The more precise version is $vartheta^{-1}(t)$.
    $endgroup$
    – reuns
    Jul 31 '17 at 12:36
















$begingroup$
I don't understand what you wrote. What is supposed to be this iteration ?
$endgroup$
– reuns
Jul 30 '17 at 21:25




$begingroup$
I don't understand what you wrote. What is supposed to be this iteration ?
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– reuns
Jul 30 '17 at 21:25












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I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
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– Mats Granvik
Jul 31 '17 at 11:50




$begingroup$
I have a missing variable $n$. The function $g(y)$ should be $g(n,y)$. $n$ is usually in the Riemann zeta function called $t$ for the quantity of the imaginary variable.
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– Mats Granvik
Jul 31 '17 at 11:50




1




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$begingroup$
I don't understand what you wrote. What is supposed to be this iteration ?
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– reuns
Jul 31 '17 at 12:04




$begingroup$
I don't understand what you wrote. What is supposed to be this iteration ?
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– reuns
Jul 31 '17 at 12:04












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@reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
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– Mats Granvik
Jul 31 '17 at 12:28






$begingroup$
@reuns I tried to write $g(x)=f(...f(f(f(f(x))))...)$ in Mathematica but it would parse, so I wrote $g(y)=f(...f(f(f(f(y))))...)$ instead. It is a function composition. $y$ is the seed point for $x$. $n$ is the variable that I feed into the function,.
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– Mats Granvik
Jul 31 '17 at 12:28






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The more precise version is $vartheta^{-1}(t)$.
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– reuns
Jul 31 '17 at 12:36




$begingroup$
The more precise version is $vartheta^{-1}(t)$.
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– reuns
Jul 31 '17 at 12:36










1 Answer
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$begingroup$

I don't know what you want to say with this iteration.





See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.



Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$






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    $begingroup$

    I don't know what you want to say with this iteration.





    See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.



    Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      I don't know what you want to say with this iteration.





      See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.



      Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$






      share|cite|improve this answer











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        3












        3








        3





        $begingroup$

        I don't know what you want to say with this iteration.





        See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.



        Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$






        share|cite|improve this answer











        $endgroup$



        I don't know what you want to say with this iteration.





        See this. Defining the arguments correctly (zero on $s > 1$, continuous on $Re(s) > 1$, and extended by continuous variation on horizontal lines to $Re(s) le 1$) we can express the number of non-trivial zeros $zeta(rho) = 0, Im(rho) in (0,t]$ with $$ N(t) =frac{text{arg }(Z(t))}{pi} = frac{vartheta(t)}{pi}+ frac{text{arg }(zeta(1/2+it))}{pi} = frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t) + mathcal{O}(log t)$$ where $frac{t}{2pi} log(frac{t}{2pi e}) + 7/8+mathcal{O}(1/t)$ comes from the Stirling approximation applied to $vartheta(t)$, and $ mathcal{O}(log t)$ comes from a bound for $text{arg }(zeta(1/2+it))$.



        Assuming the Riemann hypothesis and that $text{arg }(zeta(1/2+it))$ has zero-mean, the best approximation of the $n$th non-trivial zero is $$rho_n approx 1/2+i vartheta^{-1}(n)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 7:05









        Kevin O'Bryant

        1885




        1885










        answered Jul 31 '17 at 12:49









        reunsreuns

        21.1k21250




        21.1k21250






























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