Approximate functional equation for the Riemann zeta function
$begingroup$
The Riemann zeta function admits the approximation $$zeta(s)simsum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}},$$ in the critical strip, which is known as the approximate functional equation for the Riemann zeta function. Here $gamma$ is the multiplier from the functional equation $zeta(s)=gamma(1-s)zeta(1-s)$. However, the both sums seem to tend to $zeta(s)$ as $N, Mtoinfty$. I would like to get an explanation why we do not have a duplication and the sum of the two series equals $zeta(s)$ and not $2zeta(s)$.
riemann-zeta
asked Dec 12 '18 at 12:44
DuracDurac
113
$endgroup$
add a comment |
$begingroup$
The Riemann zeta function admits the approximation $$zeta(s)simsum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}},$$ in the critical strip, which is known as the approximate functional equation for the Riemann zeta function. Here $gamma$ is the multiplier from the functional equation $zeta(s)=gamma(1-s)zeta(1-s)$. However, the both sums seem to tend to $zeta(s)$ as $N, Mtoinfty$. I would like to get an explanation why we do not have a duplication and the sum of the two series equals $zeta(s)$ and not $2zeta(s)$.
riemann-zeta
asked Dec 12 '18 at 12:44
DuracDurac
113
$endgroup$
add a comment |
$begingroup$
The Riemann zeta function admits the approximation $$zeta(s)simsum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}},$$ in the critical strip, which is known as the approximate functional equation for the Riemann zeta function. Here $gamma$ is the multiplier from the functional equation $zeta(s)=gamma(1-s)zeta(1-s)$. However, the both sums seem to tend to $zeta(s)$ as $N, Mtoinfty$. I would like to get an explanation why we do not have a duplication and the sum of the two series equals $zeta(s)$ and not $2zeta(s)$.
riemann-zeta
asked Dec 12 '18 at 12:44
DuracDurac
113
$endgroup$
The Riemann zeta function admits the approximation $$zeta(s)simsum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}},$$ in the critical strip, which is known as the approximate functional equation for the Riemann zeta function. Here $gamma$ is the multiplier from the functional equation $zeta(s)=gamma(1-s)zeta(1-s)$. However, the both sums seem to tend to $zeta(s)$ as $N, Mtoinfty$. I would like to get an explanation why we do not have a duplication and the sum of the two series equals $zeta(s)$ and not $2zeta(s)$.
riemann-zeta
riemann-zeta
asked Dec 12 '18 at 12:44
DuracDurac
113
asked Dec 12 '18 at 12:44
DuracDurac
113
asked Dec 12 '18 at 12:44
DuracDurac
113
asked Dec 12 '18 at 12:44
DuracDurac
113
asked Dec 12 '18 at 12:44
DuracDurac
113
113
add a comment |
add a comment |
1 Answer
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For $;s:=sigma+it;$ with $,sigmain(0,1);$ Hardy and Littlewood's approximate functional equation states that :
$$tag{1}zeta(s)=sum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}}+Oleft(N^{-sigma}right)+Oleft(|t|^{frac 12-sigma},M^{sigma-1}right)$$
$$tag{2}gamma(s):=pi^{1/2-s}frac{Gamma(s/2)}{Gamma((1-s)/2)};$$
with the hypothesis $;|t|approx 2pi,N,M;$ (see page $4$ of Gourdon and Sebah's paper "Numerical evaluation of the Riemann Zeta-function" for details).
But this implies that nor $N$ nor $M$ are going to $,+infty;$ ... especially since both series would be divergent at the limit!.
In fact we usually suppose $;N=M=leftlfloorsqrt{dfrac {|t|}{2pi}}rightrfloor;$ and with the precise remainder in $(1)$ considered as an asymptotic expansion in $N$, obtain the Riemann-Siegel formula.
This remainder is not easy to evaluate and we will follow up with $,sigma=dfrac 12$.
For $;s=frac 12+it;$ the $;gamma(1-s),$ factor verifies $;|gamma(1-s)|=1,$ but with a phase factor that we can't neglect while the sum at the right will simply be the complex conjugate of the first sum.
The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $,zeta(s),$ using the finite sum $;displaystylesum_{n=1}^Xfrac{1}{n^s};$ like Euler-Maclaurin seem to impose $,X,$ to be larger than $dfrac{|t|}{2pi}$ to be precise (as illustrated in this answer).
The Riemann-Siegel formula allows to replace this sum of $X=left[dfrac{|t|}{2pi}right]$ terms with the sum restrained to the $left[sqrt{X}right]$ first terms added to the sum of the $left[sqrt{X}right]$ first terms terms of $zeta(1-s)$ multiplied by $,gamma(1-s)$ : computing $[2times],10^5$ terms instead of $,10^{10}$ makes a difference when we search large zeros! (Riemann-Siegel versus Euler-Maclaurin is described here).
The links should help you more as well as this nice paper by Carl Erickson's "A Geometric Perspective on the Riemann Zeta Function's Partial Sums".
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
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add a comment |
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1 Answer
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$begingroup$
For $;s:=sigma+it;$ with $,sigmain(0,1);$ Hardy and Littlewood's approximate functional equation states that :
$$tag{1}zeta(s)=sum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}}+Oleft(N^{-sigma}right)+Oleft(|t|^{frac 12-sigma},M^{sigma-1}right)$$
$$tag{2}gamma(s):=pi^{1/2-s}frac{Gamma(s/2)}{Gamma((1-s)/2)};$$
with the hypothesis $;|t|approx 2pi,N,M;$ (see page $4$ of Gourdon and Sebah's paper "Numerical evaluation of the Riemann Zeta-function" for details).
But this implies that nor $N$ nor $M$ are going to $,+infty;$ ... especially since both series would be divergent at the limit!.
In fact we usually suppose $;N=M=leftlfloorsqrt{dfrac {|t|}{2pi}}rightrfloor;$ and with the precise remainder in $(1)$ considered as an asymptotic expansion in $N$, obtain the Riemann-Siegel formula.
This remainder is not easy to evaluate and we will follow up with $,sigma=dfrac 12$.
For $;s=frac 12+it;$ the $;gamma(1-s),$ factor verifies $;|gamma(1-s)|=1,$ but with a phase factor that we can't neglect while the sum at the right will simply be the complex conjugate of the first sum.
The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $,zeta(s),$ using the finite sum $;displaystylesum_{n=1}^Xfrac{1}{n^s};$ like Euler-Maclaurin seem to impose $,X,$ to be larger than $dfrac{|t|}{2pi}$ to be precise (as illustrated in this answer).
The Riemann-Siegel formula allows to replace this sum of $X=left[dfrac{|t|}{2pi}right]$ terms with the sum restrained to the $left[sqrt{X}right]$ first terms added to the sum of the $left[sqrt{X}right]$ first terms terms of $zeta(1-s)$ multiplied by $,gamma(1-s)$ : computing $[2times],10^5$ terms instead of $,10^{10}$ makes a difference when we search large zeros! (Riemann-Siegel versus Euler-Maclaurin is described here).
The links should help you more as well as this nice paper by Carl Erickson's "A Geometric Perspective on the Riemann Zeta Function's Partial Sums".
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
add a comment |
$begingroup$
For $;s:=sigma+it;$ with $,sigmain(0,1);$ Hardy and Littlewood's approximate functional equation states that :
$$tag{1}zeta(s)=sum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}}+Oleft(N^{-sigma}right)+Oleft(|t|^{frac 12-sigma},M^{sigma-1}right)$$
$$tag{2}gamma(s):=pi^{1/2-s}frac{Gamma(s/2)}{Gamma((1-s)/2)};$$
with the hypothesis $;|t|approx 2pi,N,M;$ (see page $4$ of Gourdon and Sebah's paper "Numerical evaluation of the Riemann Zeta-function" for details).
But this implies that nor $N$ nor $M$ are going to $,+infty;$ ... especially since both series would be divergent at the limit!.
In fact we usually suppose $;N=M=leftlfloorsqrt{dfrac {|t|}{2pi}}rightrfloor;$ and with the precise remainder in $(1)$ considered as an asymptotic expansion in $N$, obtain the Riemann-Siegel formula.
This remainder is not easy to evaluate and we will follow up with $,sigma=dfrac 12$.
For $;s=frac 12+it;$ the $;gamma(1-s),$ factor verifies $;|gamma(1-s)|=1,$ but with a phase factor that we can't neglect while the sum at the right will simply be the complex conjugate of the first sum.
The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $,zeta(s),$ using the finite sum $;displaystylesum_{n=1}^Xfrac{1}{n^s};$ like Euler-Maclaurin seem to impose $,X,$ to be larger than $dfrac{|t|}{2pi}$ to be precise (as illustrated in this answer).
The Riemann-Siegel formula allows to replace this sum of $X=left[dfrac{|t|}{2pi}right]$ terms with the sum restrained to the $left[sqrt{X}right]$ first terms added to the sum of the $left[sqrt{X}right]$ first terms terms of $zeta(1-s)$ multiplied by $,gamma(1-s)$ : computing $[2times],10^5$ terms instead of $,10^{10}$ makes a difference when we search large zeros! (Riemann-Siegel versus Euler-Maclaurin is described here).
The links should help you more as well as this nice paper by Carl Erickson's "A Geometric Perspective on the Riemann Zeta Function's Partial Sums".
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
add a comment |
$begingroup$
For $;s:=sigma+it;$ with $,sigmain(0,1);$ Hardy and Littlewood's approximate functional equation states that :
$$tag{1}zeta(s)=sum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}}+Oleft(N^{-sigma}right)+Oleft(|t|^{frac 12-sigma},M^{sigma-1}right)$$
$$tag{2}gamma(s):=pi^{1/2-s}frac{Gamma(s/2)}{Gamma((1-s)/2)};$$
with the hypothesis $;|t|approx 2pi,N,M;$ (see page $4$ of Gourdon and Sebah's paper "Numerical evaluation of the Riemann Zeta-function" for details).
But this implies that nor $N$ nor $M$ are going to $,+infty;$ ... especially since both series would be divergent at the limit!.
In fact we usually suppose $;N=M=leftlfloorsqrt{dfrac {|t|}{2pi}}rightrfloor;$ and with the precise remainder in $(1)$ considered as an asymptotic expansion in $N$, obtain the Riemann-Siegel formula.
This remainder is not easy to evaluate and we will follow up with $,sigma=dfrac 12$.
For $;s=frac 12+it;$ the $;gamma(1-s),$ factor verifies $;|gamma(1-s)|=1,$ but with a phase factor that we can't neglect while the sum at the right will simply be the complex conjugate of the first sum.
The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $,zeta(s),$ using the finite sum $;displaystylesum_{n=1}^Xfrac{1}{n^s};$ like Euler-Maclaurin seem to impose $,X,$ to be larger than $dfrac{|t|}{2pi}$ to be precise (as illustrated in this answer).
The Riemann-Siegel formula allows to replace this sum of $X=left[dfrac{|t|}{2pi}right]$ terms with the sum restrained to the $left[sqrt{X}right]$ first terms added to the sum of the $left[sqrt{X}right]$ first terms terms of $zeta(1-s)$ multiplied by $,gamma(1-s)$ : computing $[2times],10^5$ terms instead of $,10^{10}$ makes a difference when we search large zeros! (Riemann-Siegel versus Euler-Maclaurin is described here).
The links should help you more as well as this nice paper by Carl Erickson's "A Geometric Perspective on the Riemann Zeta Function's Partial Sums".
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
For $;s:=sigma+it;$ with $,sigmain(0,1);$ Hardy and Littlewood's approximate functional equation states that :
$$tag{1}zeta(s)=sum_{n=1}^Nfrac{1}{n^s}+gamma(1-s)sum_{n=1}^Mfrac{1}{n^{1-s}}+Oleft(N^{-sigma}right)+Oleft(|t|^{frac 12-sigma},M^{sigma-1}right)$$
$$tag{2}gamma(s):=pi^{1/2-s}frac{Gamma(s/2)}{Gamma((1-s)/2)};$$
with the hypothesis $;|t|approx 2pi,N,M;$ (see page $4$ of Gourdon and Sebah's paper "Numerical evaluation of the Riemann Zeta-function" for details).
But this implies that nor $N$ nor $M$ are going to $,+infty;$ ... especially since both series would be divergent at the limit!.
In fact we usually suppose $;N=M=leftlfloorsqrt{dfrac {|t|}{2pi}}rightrfloor;$ and with the precise remainder in $(1)$ considered as an asymptotic expansion in $N$, obtain the Riemann-Siegel formula.
This remainder is not easy to evaluate and we will follow up with $,sigma=dfrac 12$.
For $;s=frac 12+it;$ the $;gamma(1-s),$ factor verifies $;|gamma(1-s)|=1,$ but with a phase factor that we can't neglect while the sum at the right will simply be the complex conjugate of the first sum.
The interest of the Riemann-Siegel formula (and the approximate functional equation) is that alternative evaluations of $,zeta(s),$ using the finite sum $;displaystylesum_{n=1}^Xfrac{1}{n^s};$ like Euler-Maclaurin seem to impose $,X,$ to be larger than $dfrac{|t|}{2pi}$ to be precise (as illustrated in this answer).
The Riemann-Siegel formula allows to replace this sum of $X=left[dfrac{|t|}{2pi}right]$ terms with the sum restrained to the $left[sqrt{X}right]$ first terms added to the sum of the $left[sqrt{X}right]$ first terms terms of $zeta(1-s)$ multiplied by $,gamma(1-s)$ : computing $[2times],10^5$ terms instead of $,10^{10}$ makes a difference when we search large zeros! (Riemann-Siegel versus Euler-Maclaurin is described here).
The links should help you more as well as this nice paper by Carl Erickson's "A Geometric Perspective on the Riemann Zeta Function's Partial Sums".
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
edited Dec 13 '18 at 22:40
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Dec 12 '18 at 14:02
Raymond ManzoniRaymond Manzoni
37.2k563117
37.2k563117
add a comment |
add a comment |
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