Approximate functional equation for the Riemann zeta function












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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







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    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





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      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 12 '18 at 12:44









      DuracDurac

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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".






          share|cite|improve this 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".






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              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".






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                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".






                share|cite|improve this answer











                $endgroup$



                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".







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 22:40

























                answered Dec 12 '18 at 14:02









                Raymond ManzoniRaymond Manzoni

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                37.1k563117






























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