Sum of Stieltjes constants
$begingroup$
Does anyone know of any papers or resources dealing with the following question: For which values of $s=sigma+it$ does the following sum of Stieltjes constants hold,
$$sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n=0,$$
where $gamma_n$ is the $n$-th Stieltjes constant. See here for information on the Stieltjes constants. I'm investigating when we have $zeta(s)=1/(s-1)$ for $sneq 1$ and thought this might help me given that
$$zeta(s) = frac{1}{s-1}+sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n.$$
analytic-number-theory riemann-zeta
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
$endgroup$
add a comment |
$begingroup$
Does anyone know of any papers or resources dealing with the following question: For which values of $s=sigma+it$ does the following sum of Stieltjes constants hold,
$$sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n=0,$$
where $gamma_n$ is the $n$-th Stieltjes constant. See here for information on the Stieltjes constants. I'm investigating when we have $zeta(s)=1/(s-1)$ for $sneq 1$ and thought this might help me given that
$$zeta(s) = frac{1}{s-1}+sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n.$$
analytic-number-theory riemann-zeta
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
$endgroup$
add a comment |
$begingroup$
Does anyone know of any papers or resources dealing with the following question: For which values of $s=sigma+it$ does the following sum of Stieltjes constants hold,
$$sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n=0,$$
where $gamma_n$ is the $n$-th Stieltjes constant. See here for information on the Stieltjes constants. I'm investigating when we have $zeta(s)=1/(s-1)$ for $sneq 1$ and thought this might help me given that
$$zeta(s) = frac{1}{s-1}+sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n.$$
analytic-number-theory riemann-zeta
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
$endgroup$
Does anyone know of any papers or resources dealing with the following question: For which values of $s=sigma+it$ does the following sum of Stieltjes constants hold,
$$sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n=0,$$
where $gamma_n$ is the $n$-th Stieltjes constant. See here for information on the Stieltjes constants. I'm investigating when we have $zeta(s)=1/(s-1)$ for $sneq 1$ and thought this might help me given that
$$zeta(s) = frac{1}{s-1}+sum_{n=0}^inftyfrac{(-1)^n}{n!}gamma_n(s-1)^n.$$
analytic-number-theory riemann-zeta
analytic-number-theory riemann-zeta
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
asked Dec 18 '12 at 9:13
AntinousAntinous
5,83842453
5,83842453
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Building on the answer of Manzoni, the Mathematica command
s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]
Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]},
{theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red],
Directive[Thick, Green]}]
Generates this plot
This shows that
$$
frac{1}{|s-1|}<|zeta(s)|
$$
for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $rho=1/2+i14.13ldots$.
By Rouche's Theorem, $zeta(s)$ and $zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.
One could presumably do this for higher zeros as well.
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
$endgroup$
add a comment |
$begingroup$
I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $zeta'$ near $zeta$ zeros) :
begin{array} {llll}
&0.5+14.1347251417i&mapsto &0.47940300622352cdots&+14.04751638282159cdots i\
&0.5+21.0220396387 i&mapsto &0.507180820934280248966cdots&+20.980648953392265333cdots i\
&0.5+25.0108575801i&mapsto &0.4893602291391196345cdots&+24.9838122121001409936cdots i\
end{array}
To find the approximative value of these zeros let's consider the first order expansion of $zeta$ near one of its zeros $s_i$ :
$$zeta(s_i+epsilon_i)=zeta(s_i)+epsilon_izeta'(s_i)+O(epsilon_i^2)$$
Since $zeta(s_i)=0$ and since you want $ zeta(s_i+epsilon_i)=frac 1{s_i+epsilon_i-1} $ you will have :
$$frac 1{s_i+epsilon_i-1}approx epsilon_izeta'(s_i)$$
and, to first order, the zeros will be nearly :
$$s_i+epsilon_i approx s_i+frac 1{zeta'(s_i)(s_i-1)}$$
With the values of $zeta$ and $zeta'$ for the three first zeros we get the approximate table :
begin{array} {ll}
&0.482882 &+ 14.0472i\
&0.508179 &+ 20.9810i\
&0.489891 &+ 24.9835i\
end{array}
Not sure it will help much...
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
Building on the answer of Manzoni, the Mathematica command
s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]
Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]},
{theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red],
Directive[Thick, Green]}]
Generates this plot
This shows that
$$
frac{1}{|s-1|}<|zeta(s)|
$$
for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $rho=1/2+i14.13ldots$.
By Rouche's Theorem, $zeta(s)$ and $zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.
One could presumably do this for higher zeros as well.
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
$endgroup$
add a comment |
$begingroup$
Building on the answer of Manzoni, the Mathematica command
s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]
Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]},
{theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red],
Directive[Thick, Green]}]
Generates this plot
This shows that
$$
frac{1}{|s-1|}<|zeta(s)|
$$
for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $rho=1/2+i14.13ldots$.
By Rouche's Theorem, $zeta(s)$ and $zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.
One could presumably do this for higher zeros as well.
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
$endgroup$
add a comment |
$begingroup$
Building on the answer of Manzoni, the Mathematica command
s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]
Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]},
{theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red],
Directive[Thick, Green]}]
Generates this plot
This shows that
$$
frac{1}{|s-1|}<|zeta(s)|
$$
for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $rho=1/2+i14.13ldots$.
By Rouche's Theorem, $zeta(s)$ and $zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.
One could presumably do this for higher zeros as well.
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
$endgroup$
Building on the answer of Manzoni, the Mathematica command
s[theta_, eps_, k_] := ZetaZero[k] + eps Exp[I theta]
Plot[{Abs[1/(s[theta, .1, 1] - 1)], Abs[Zeta[s[theta, .1, 1]]]},
{theta, 0, 2 Pi}, PlotStyle -> {Directive[Thick, Red],
Directive[Thick, Green]}]
Generates this plot
This shows that
$$
frac{1}{|s-1|}<|zeta(s)|
$$
for $s$ on the circle of radius $.1$ around the lowest nontrivial Riemann zero $rho=1/2+i14.13ldots$.
By Rouche's Theorem, $zeta(s)$ and $zeta(s)-1/(s-1)$ have the same number of zeros inside the circle, namely $1$.
One could presumably do this for higher zeros as well.
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
edited Dec 23 '18 at 20:29
Glorfindel
3,41581830
3,41581830
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
answered Jan 11 '13 at 23:40
stopplestopple
1,504817
1,504817
add a comment |
add a comment |
$begingroup$
I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $zeta'$ near $zeta$ zeros) :
begin{array} {llll}
&0.5+14.1347251417i&mapsto &0.47940300622352cdots&+14.04751638282159cdots i\
&0.5+21.0220396387 i&mapsto &0.507180820934280248966cdots&+20.980648953392265333cdots i\
&0.5+25.0108575801i&mapsto &0.4893602291391196345cdots&+24.9838122121001409936cdots i\
end{array}
To find the approximative value of these zeros let's consider the first order expansion of $zeta$ near one of its zeros $s_i$ :
$$zeta(s_i+epsilon_i)=zeta(s_i)+epsilon_izeta'(s_i)+O(epsilon_i^2)$$
Since $zeta(s_i)=0$ and since you want $ zeta(s_i+epsilon_i)=frac 1{s_i+epsilon_i-1} $ you will have :
$$frac 1{s_i+epsilon_i-1}approx epsilon_izeta'(s_i)$$
and, to first order, the zeros will be nearly :
$$s_i+epsilon_i approx s_i+frac 1{zeta'(s_i)(s_i-1)}$$
With the values of $zeta$ and $zeta'$ for the three first zeros we get the approximate table :
begin{array} {ll}
&0.482882 &+ 14.0472i\
&0.508179 &+ 20.9810i\
&0.489891 &+ 24.9835i\
end{array}
Not sure it will help much...
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
add a comment |
$begingroup$
I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $zeta'$ near $zeta$ zeros) :
begin{array} {llll}
&0.5+14.1347251417i&mapsto &0.47940300622352cdots&+14.04751638282159cdots i\
&0.5+21.0220396387 i&mapsto &0.507180820934280248966cdots&+20.980648953392265333cdots i\
&0.5+25.0108575801i&mapsto &0.4893602291391196345cdots&+24.9838122121001409936cdots i\
end{array}
To find the approximative value of these zeros let's consider the first order expansion of $zeta$ near one of its zeros $s_i$ :
$$zeta(s_i+epsilon_i)=zeta(s_i)+epsilon_izeta'(s_i)+O(epsilon_i^2)$$
Since $zeta(s_i)=0$ and since you want $ zeta(s_i+epsilon_i)=frac 1{s_i+epsilon_i-1} $ you will have :
$$frac 1{s_i+epsilon_i-1}approx epsilon_izeta'(s_i)$$
and, to first order, the zeros will be nearly :
$$s_i+epsilon_i approx s_i+frac 1{zeta'(s_i)(s_i-1)}$$
With the values of $zeta$ and $zeta'$ for the three first zeros we get the approximate table :
begin{array} {ll}
&0.482882 &+ 14.0472i\
&0.508179 &+ 20.9810i\
&0.489891 &+ 24.9835i\
end{array}
Not sure it will help much...
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
add a comment |
$begingroup$
I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $zeta'$ near $zeta$ zeros) :
begin{array} {llll}
&0.5+14.1347251417i&mapsto &0.47940300622352cdots&+14.04751638282159cdots i\
&0.5+21.0220396387 i&mapsto &0.507180820934280248966cdots&+20.980648953392265333cdots i\
&0.5+25.0108575801i&mapsto &0.4893602291391196345cdots&+24.9838122121001409936cdots i\
end{array}
To find the approximative value of these zeros let's consider the first order expansion of $zeta$ near one of its zeros $s_i$ :
$$zeta(s_i+epsilon_i)=zeta(s_i)+epsilon_izeta'(s_i)+O(epsilon_i^2)$$
Since $zeta(s_i)=0$ and since you want $ zeta(s_i+epsilon_i)=frac 1{s_i+epsilon_i-1} $ you will have :
$$frac 1{s_i+epsilon_i-1}approx epsilon_izeta'(s_i)$$
and, to first order, the zeros will be nearly :
$$s_i+epsilon_i approx s_i+frac 1{zeta'(s_i)(s_i-1)}$$
With the values of $zeta$ and $zeta'$ for the three first zeros we get the approximate table :
begin{array} {ll}
&0.482882 &+ 14.0472i\
&0.508179 &+ 20.9810i\
&0.489891 &+ 24.9835i\
end{array}
Not sure it will help much...
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
$endgroup$
I don't know papers about your question but some numerical investigations return zeros near the Riemann zeros (because $frac 1{s-1}$ is small for Riemann's zeros and of the smooth character of $zeta'$ near $zeta$ zeros) :
begin{array} {llll}
&0.5+14.1347251417i&mapsto &0.47940300622352cdots&+14.04751638282159cdots i\
&0.5+21.0220396387 i&mapsto &0.507180820934280248966cdots&+20.980648953392265333cdots i\
&0.5+25.0108575801i&mapsto &0.4893602291391196345cdots&+24.9838122121001409936cdots i\
end{array}
To find the approximative value of these zeros let's consider the first order expansion of $zeta$ near one of its zeros $s_i$ :
$$zeta(s_i+epsilon_i)=zeta(s_i)+epsilon_izeta'(s_i)+O(epsilon_i^2)$$
Since $zeta(s_i)=0$ and since you want $ zeta(s_i+epsilon_i)=frac 1{s_i+epsilon_i-1} $ you will have :
$$frac 1{s_i+epsilon_i-1}approx epsilon_izeta'(s_i)$$
and, to first order, the zeros will be nearly :
$$s_i+epsilon_i approx s_i+frac 1{zeta'(s_i)(s_i-1)}$$
With the values of $zeta$ and $zeta'$ for the three first zeros we get the approximate table :
begin{array} {ll}
&0.482882 &+ 14.0472i\
&0.508179 &+ 20.9810i\
&0.489891 &+ 24.9835i\
end{array}
Not sure it will help much...
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
edited Dec 18 '12 at 13:26
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
answered Dec 18 '12 at 11:05
Raymond ManzoniRaymond Manzoni
37.3k563117
37.3k563117
add a comment |
add a comment |
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