Why is it impossible to invert the analytic continuation of a Dirichlet series?
$begingroup$
By Mathematica (and the truncated Euler MacLaurin formula) I know that:
$$zeta(s)=lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right) tag{1}$$
when the real part of $s$ is greater than $0$.
Question:
I would like to understand better why it is impossible to find a
Dirichlet series for the reciprocal of the Right Hand Side (RHS) of $(1)$.
In other words I am looking for to invert the reciprocal of the Riemann zeta function:
$$frac{1}{zeta(s)}=frac{1}{lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right)} tag{2}$$
when the real part of $s$ is greater than $0$.
For $Re(s)>1$ there is the well known relation:
$$frac{1}{zeta(s)}=sum _{n=1}^{infty} frac{mu(n)}{n^s} tag{3}$$
and the Riemann hypothesis is that $(3)$ converges for $Re(s)>frac{1}{2}$.
In a sense I already know the answer to the question. It is because $frac{1}{(s-1) k^{s-1}}$ is like a constant and $sum _{n=1}^k frac{1}{n^s}$ is the Dirichlet series. But can you explain it better to me?
elementary-number-theory analytic-number-theory riemann-zeta dirichlet-series
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
$endgroup$
add a comment |
$begingroup$
By Mathematica (and the truncated Euler MacLaurin formula) I know that:
$$zeta(s)=lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right) tag{1}$$
when the real part of $s$ is greater than $0$.
Question:
I would like to understand better why it is impossible to find a
Dirichlet series for the reciprocal of the Right Hand Side (RHS) of $(1)$.
In other words I am looking for to invert the reciprocal of the Riemann zeta function:
$$frac{1}{zeta(s)}=frac{1}{lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right)} tag{2}$$
when the real part of $s$ is greater than $0$.
For $Re(s)>1$ there is the well known relation:
$$frac{1}{zeta(s)}=sum _{n=1}^{infty} frac{mu(n)}{n^s} tag{3}$$
and the Riemann hypothesis is that $(3)$ converges for $Re(s)>frac{1}{2}$.
In a sense I already know the answer to the question. It is because $frac{1}{(s-1) k^{s-1}}$ is like a constant and $sum _{n=1}^k frac{1}{n^s}$ is the Dirichlet series. But can you explain it better to me?
elementary-number-theory analytic-number-theory riemann-zeta dirichlet-series
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
$endgroup$
$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
1
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30
add a comment |
$begingroup$
By Mathematica (and the truncated Euler MacLaurin formula) I know that:
$$zeta(s)=lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right) tag{1}$$
when the real part of $s$ is greater than $0$.
Question:
I would like to understand better why it is impossible to find a
Dirichlet series for the reciprocal of the Right Hand Side (RHS) of $(1)$.
In other words I am looking for to invert the reciprocal of the Riemann zeta function:
$$frac{1}{zeta(s)}=frac{1}{lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right)} tag{2}$$
when the real part of $s$ is greater than $0$.
For $Re(s)>1$ there is the well known relation:
$$frac{1}{zeta(s)}=sum _{n=1}^{infty} frac{mu(n)}{n^s} tag{3}$$
and the Riemann hypothesis is that $(3)$ converges for $Re(s)>frac{1}{2}$.
In a sense I already know the answer to the question. It is because $frac{1}{(s-1) k^{s-1}}$ is like a constant and $sum _{n=1}^k frac{1}{n^s}$ is the Dirichlet series. But can you explain it better to me?
elementary-number-theory analytic-number-theory riemann-zeta dirichlet-series
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
$endgroup$
By Mathematica (and the truncated Euler MacLaurin formula) I know that:
$$zeta(s)=lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right) tag{1}$$
when the real part of $s$ is greater than $0$.
Question:
I would like to understand better why it is impossible to find a
Dirichlet series for the reciprocal of the Right Hand Side (RHS) of $(1)$.
In other words I am looking for to invert the reciprocal of the Riemann zeta function:
$$frac{1}{zeta(s)}=frac{1}{lim_{kto infty } , left(sum _{n=1}^k frac{1}{n^s}+frac{1}{(s-1) k^{s-1}}right)} tag{2}$$
when the real part of $s$ is greater than $0$.
For $Re(s)>1$ there is the well known relation:
$$frac{1}{zeta(s)}=sum _{n=1}^{infty} frac{mu(n)}{n^s} tag{3}$$
and the Riemann hypothesis is that $(3)$ converges for $Re(s)>frac{1}{2}$.
In a sense I already know the answer to the question. It is because $frac{1}{(s-1) k^{s-1}}$ is like a constant and $sum _{n=1}^k frac{1}{n^s}$ is the Dirichlet series. But can you explain it better to me?
elementary-number-theory analytic-number-theory riemann-zeta dirichlet-series
elementary-number-theory analytic-number-theory riemann-zeta dirichlet-series
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
edited Jan 20 at 15:23
J. W. Tanner
2,1371117
2,1371117
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
asked Jan 20 at 15:14
Mats GranvikMats Granvik
3,29132250
3,29132250
$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
1
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30
add a comment |
$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
1
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30
$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
1
1
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30
add a comment |
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$begingroup$
If there was such a Dirichlet series for the reciprocal of the RHS of $(1)$ then a formula for the Riemann zeta zeros would follow after some effort. One could perhaps say that there is no formula for the zeta zeros because the RHS of $(3)$ is an epsilon away from converging on the critical line.
$endgroup$
– Mats Granvik
Jan 20 at 15:23
$begingroup$
You should look at the Dirichlet L-function $beta(s) = lim_{N to infty}sum_{n=0}^N (-1)^n (2n+1)^{-s}$. It converges for $Re(s) > 0$ and iff it has no zeros for $Re(s) > sigma$ then $frac{1}{beta(s)}= sum_{n=0}^infty (-1)^n (2n+1)^{-s} mu(n)$ for $Re(s) >sigma$. For $Re(s)$ large enough $frac{1}{sum_{n=0}^N (-1)^n (2n+1)^{-s}} = sum_{k=0}^infty (1-sum_{n=0}^N (-1)^n (2n+1)^{-s})^k$, once reordered it is a Dirichlet series. It shouldn't converge beyond $Re(s) > 1$. For $zeta$ those things won't be Dirichlet series anymore due to the additional term $frac{N^{1-s}}{s-1}$.
$endgroup$
– reuns
Jan 21 at 12:23
1
$begingroup$
Also the explicit formula for $sum_{n le x} mu(n)$ tell us how to regularize/continue $sum_{n=1}^infty mu(n)n^{-s}$ to any vertical strip with no zeros.
$endgroup$
– reuns
Jan 21 at 12:30