What is the value of the Dirichlet Eta Function at s=1/2?
$begingroup$
Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac12);$
$$etaleft(frac12right) = sum_n=1^inftyfrac(-1)^(n+1)sqrtn$$
A web calculator gives the value to be 0.6, which seems to be right.
number-theory
edited Nov 13 '18 at 22:59
User525412790
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
$endgroup$
add a comment |
$begingroup$
Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac12);$
$$etaleft(frac12right) = sum_n=1^inftyfrac(-1)^(n+1)sqrtn$$
A web calculator gives the value to be 0.6, which seems to be right.
number-theory
edited Nov 13 '18 at 22:59
User525412790
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
$endgroup$
add a comment |
$begingroup$
Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac12);$
$$etaleft(frac12right) = sum_n=1^inftyfrac(-1)^(n+1)sqrtn$$
A web calculator gives the value to be 0.6, which seems to be right.
number-theory
edited Nov 13 '18 at 22:59
User525412790
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
$endgroup$
Although $eta(1)$ is known to be $ln(2)$, I have not seen an analytically calculated value for $eta(frac12);$
$$etaleft(frac12right) = sum_n=1^inftyfrac(-1)^(n+1)sqrtn$$
A web calculator gives the value to be 0.6, which seems to be right.
number-theory
number-theory
edited Nov 13 '18 at 22:59
User525412790
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
edited Nov 13 '18 at 22:59
User525412790
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
edited Nov 13 '18 at 22:59
User525412790
313114
edited Nov 13 '18 at 22:59
User525412790
313114
edited Nov 13 '18 at 22:59
User525412790
313114
313114
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
asked Nov 13 '18 at 10:08
Akira BergmanAkira Bergman
214
214
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Isn't just
$$etaleft(frac12right)=sum_n=1^inftyfrac(-1)^(n+1)sqrtn=left(1-sqrt2right) zeta left(frac12right)approx 0.6048986434$$
Edit
Remember the general relation
$$etaleft(sright)=left(1-2^1-sright) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
$$etaleft(sright)=frac 12+left( log (2)-frac12right), s^0.895$$
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
A careful computation shows that the numerical value is
$$0.6048986434216303702472...$$
which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt2)log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
$endgroup$
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
|
show 3 more comments
$begingroup$
The numerical value of 0.604898... is provided in http://oeis.org/A113024 .
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Isn't just
$$etaleft(frac12right)=sum_n=1^inftyfrac(-1)^(n+1)sqrtn=left(1-sqrt2right) zeta left(frac12right)approx 0.6048986434$$
Edit
Remember the general relation
$$etaleft(sright)=left(1-2^1-sright) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
$$etaleft(sright)=frac 12+left( log (2)-frac12right), s^0.895$$
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Isn't just
$$etaleft(frac12right)=sum_n=1^inftyfrac(-1)^(n+1)sqrtn=left(1-sqrt2right) zeta left(frac12right)approx 0.6048986434$$
Edit
Remember the general relation
$$etaleft(sright)=left(1-2^1-sright) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
$$etaleft(sright)=frac 12+left( log (2)-frac12right), s^0.895$$
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Isn't just
$$etaleft(frac12right)=sum_n=1^inftyfrac(-1)^(n+1)sqrtn=left(1-sqrt2right) zeta left(frac12right)approx 0.6048986434$$
Edit
Remember the general relation
$$etaleft(sright)=left(1-2^1-sright) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
$$etaleft(sright)=frac 12+left( log (2)-frac12right), s^0.895$$
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Isn't just
$$etaleft(frac12right)=sum_n=1^inftyfrac(-1)^(n+1)sqrtn=left(1-sqrt2right) zeta left(frac12right)approx 0.6048986434$$
Edit
Remember the general relation
$$etaleft(sright)=left(1-2^1-sright) zeta (s)$$ If you want a quick and dirty shortcut evaluation, for $0 leq s leq 1$, you could use
$$etaleft(sright)=frac 12+left( log (2)-frac12right), s^0.895$$
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
edited Nov 14 '18 at 4:53
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
answered Nov 13 '18 at 10:43
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
A careful computation shows that the numerical value is
$$0.6048986434216303702472...$$
which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt2)log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
$endgroup$
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
|
show 3 more comments
$begingroup$
A careful computation shows that the numerical value is
$$0.6048986434216303702472...$$
which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt2)log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
$endgroup$
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
|
show 3 more comments
$begingroup$
A careful computation shows that the numerical value is
$$0.6048986434216303702472...$$
which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt2)log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
$endgroup$
A careful computation shows that the numerical value is
$$0.6048986434216303702472...$$
which is not $0.6$. One should be aware that the above series converge really slowly.
As Claude Leibovici indicates, one can relate its value to the Riemann's Zeta function value at $1/2$. However, as far as I know, there is no analytic formula of $zeta(1/2)$, so this is why you haven't seen an "analytically calculated value for $eta(1/2)$".
EDIT2: As pointed again in the comments by leftaroundabout, I missread the OEIS link given in the answer of R. J. Mathar. What equals $$gamma/2 + pi/4 - (1/2 + sqrt2)log(2) + log(pi)/2,$$where $gamma$ is the Euler-Mascheroni constant, is $eta'(1/2)/eta(1/2)$ not $eta(1/2)$.
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
edited Nov 14 '18 at 21:30
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
answered Nov 13 '18 at 10:48
Josué Tonelli-CuetoJosué Tonelli-Cueto
3,7121127
3,7121127
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
|
show 3 more comments
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
$begingroup$
You are very correct ! We start a no-end loop. By the way $to +1$
$endgroup$
– Claude Leibovici
Nov 13 '18 at 10:51
4
4
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
It does seem to be expressible exactly in terms of the Euler-Mascheroni constant though.
$endgroup$
– leftaroundabout
Nov 13 '18 at 16:01
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
$begingroup$
@leftaroundabout Do you have a reference, I searched, but I didn't find it.
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 18:47
1
1
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
$begingroup$
@user3059799 Corrected, editing from the phone is hard
$endgroup$
– Josué Tonelli-Cueto
Nov 13 '18 at 20:11
1
1
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
$begingroup$
The formula seems incorrect - I get the wrong answer when calculating it. OEIS seems to say that if $c$ is the OP's constant, and $d$ is the constant described here, then your formula gives $d/c$. The formula on OEIS for $d$ includes $zeta(1/2)$, so I suspect all they are giving is Claude Leibovici's formula in disguise.
$endgroup$
– user98602
Nov 14 '18 at 1:18
|
show 3 more comments
$begingroup$
The numerical value of 0.604898... is provided in http://oeis.org/A113024 .
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
$endgroup$
add a comment |
$begingroup$
The numerical value of 0.604898... is provided in http://oeis.org/A113024 .
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
$endgroup$
add a comment |
$begingroup$
The numerical value of 0.604898... is provided in http://oeis.org/A113024 .
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
$endgroup$
The numerical value of 0.604898... is provided in http://oeis.org/A113024 .
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
answered Nov 13 '18 at 14:29
R. J. MatharR. J. Mathar
511
511
add a comment |
add a comment |
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