Mixing
Does this normalization of the Riemann zeta function make sense?
$begingroup$
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:
where the end points at $0$ and $1$ are complementary Gram points.
The average of the 101 curves looks like this:
This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:
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:
where the middle common point at $frac{1}{2}$ is at a complementary Gram point.
The average of the 101 curves is:
This again looks like $sin (x)$ on the interval $left[0,2piright]$:
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
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
$endgroup$
|
show 3 more comments
$begingroup$
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:
where the end points at $0$ and $1$ are complementary Gram points.
The average of the 101 curves looks like this:
This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:
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:
where the middle common point at $frac{1}{2}$ is at a complementary Gram point.
The average of the 101 curves is:
This again looks like $sin (x)$ on the interval $left[0,2piright]$:
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
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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:
where the end points at $0$ and $1$ are complementary Gram points.
The average of the 101 curves looks like this:
This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:
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:
where the middle common point at $frac{1}{2}$ is at a complementary Gram point.
The average of the 101 curves is:
This again looks like $sin (x)$ on the interval $left[0,2piright]$:
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
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
$endgroup$
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:
where the end points at $0$ and $1$ are complementary Gram points.
The average of the 101 curves looks like this:
This looks very much like the function $1+cos (x+pi )$ on the interval $left[0,2piright]$:
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:
where the middle common point at $frac{1}{2}$ is at a complementary Gram point.
The average of the 101 curves is:
This again looks like $sin (x)$ on the interval $left[0,2piright]$:
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
analytic-number-theory riemann-zeta fixed-point-theorems experimental-mathematics
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
edited Feb 4 '18 at 13:25
Mats Granvik
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
asked Jul 30 '17 at 10:46
Mats GranvikMats Granvik
3,31432250
3,31432250
$begingroup$
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
|
show 3 more comments
$begingroup$
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 ?
$endgroup$
– 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
$begingroup$
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
1
$begingroup$
I don't understand what you wrote. What is supposed to be this iteration ?
$endgroup$
– reuns
Jul 31 '17 at 12:04
$begingroup$
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
$begingroup$
@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
1
1
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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 $vartheta^{-1}(t)$.
$endgroup$
– reuns
Jul 31 '17 at 12:36
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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 $$rho_n approx 1/2+i vartheta^{-1}(n)$$
edited Jan 3 at 7:05
Kevin O'Bryant
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
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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 $$rho_n approx 1/2+i vartheta^{-1}(n)$$
edited Jan 3 at 7:05
Kevin O'Bryant
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
$endgroup$
add a comment |
$begingroup$
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 $$rho_n approx 1/2+i vartheta^{-1}(n)$$
edited Jan 3 at 7:05
Kevin O'Bryant
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
$endgroup$
add a comment |
$begingroup$
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 $$rho_n approx 1/2+i vartheta^{-1}(n)$$
edited Jan 3 at 7:05
Kevin O'Bryant
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
$endgroup$
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 $$rho_n approx 1/2+i vartheta^{-1}(n)$$
edited Jan 3 at 7:05
Kevin O'Bryant
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
edited Jan 3 at 7:05
Kevin O'Bryant
1885
edited Jan 3 at 7:05
Kevin O'Bryant
1885
edited Jan 3 at 7:05
Kevin O'Bryant
1885
1885
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
answered Jul 31 '17 at 12:49
reunsreuns
19.8k21147
19.8k21147
add a comment |
add a comment |
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$begingroup$
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