An inequality involving the Dirichlet eta function
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I would be interested in proving the following inequality involving Dirichlet's eta function $eta(s)$ at different values which, after some numerical investigations, I am sure is true
$|chi(1-s)eta(s+2)-eta(3-s)|>0$ $qquad$ $1/2<Re(s)<1$
Here $chi(1-s)=frac{Gamma(s)}{(2pi)^s}2cosleft(frac{pi s}{2} right)frac{1-2^s}{1-2^{1-s}}$ is the factor entering in the functional equation of the Dirichlet eta function
$eta(1-s)=chi(1-s)eta(s)$
Do you have any suggestion on how to attack this problem and/or references that could be helpful? Thank you very much in advance!
functional-equations zeta-functions
asked Nov 22 at 18:06
Frobenius
613
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up vote
0
down vote
favorite
I would be interested in proving the following inequality involving Dirichlet's eta function $eta(s)$ at different values which, after some numerical investigations, I am sure is true
$|chi(1-s)eta(s+2)-eta(3-s)|>0$ $qquad$ $1/2<Re(s)<1$
Here $chi(1-s)=frac{Gamma(s)}{(2pi)^s}2cosleft(frac{pi s}{2} right)frac{1-2^s}{1-2^{1-s}}$ is the factor entering in the functional equation of the Dirichlet eta function
$eta(1-s)=chi(1-s)eta(s)$
Do you have any suggestion on how to attack this problem and/or references that could be helpful? Thank you very much in advance!
functional-equations zeta-functions
asked Nov 22 at 18:06
Frobenius
613
You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would be interested in proving the following inequality involving Dirichlet's eta function $eta(s)$ at different values which, after some numerical investigations, I am sure is true
$|chi(1-s)eta(s+2)-eta(3-s)|>0$ $qquad$ $1/2<Re(s)<1$
Here $chi(1-s)=frac{Gamma(s)}{(2pi)^s}2cosleft(frac{pi s}{2} right)frac{1-2^s}{1-2^{1-s}}$ is the factor entering in the functional equation of the Dirichlet eta function
$eta(1-s)=chi(1-s)eta(s)$
Do you have any suggestion on how to attack this problem and/or references that could be helpful? Thank you very much in advance!
functional-equations zeta-functions
asked Nov 22 at 18:06
Frobenius
613
I would be interested in proving the following inequality involving Dirichlet's eta function $eta(s)$ at different values which, after some numerical investigations, I am sure is true
$|chi(1-s)eta(s+2)-eta(3-s)|>0$ $qquad$ $1/2<Re(s)<1$
Here $chi(1-s)=frac{Gamma(s)}{(2pi)^s}2cosleft(frac{pi s}{2} right)frac{1-2^s}{1-2^{1-s}}$ is the factor entering in the functional equation of the Dirichlet eta function
$eta(1-s)=chi(1-s)eta(s)$
Do you have any suggestion on how to attack this problem and/or references that could be helpful? Thank you very much in advance!
functional-equations zeta-functions
functional-equations zeta-functions
asked Nov 22 at 18:06
Frobenius
613
asked Nov 22 at 18:06
Frobenius
613
edited Nov 22 at 19:05
asked Nov 22 at 18:06
Frobenius
613
asked Nov 22 at 18:06
Frobenius
613
asked Nov 22 at 18:06
Frobenius
613
613
You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00
add a comment |
You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00
You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00
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You mean: there are no zeros for $Re s > 1/2?$
– gammatester
Nov 22 at 18:58
Yes. This is exactly what I mean. I have, however, corrected the question, because the region that I have investigated is $1/2<Re{s} < 1$.
– Frobenius
Nov 22 at 19:03
Ok, I think I have established the result by using that, letting $s=sigma +it$, we have $1/7<|zeta(2+s)/zeta(3-s)|<7$ and $chi(1-s)=O(t^{sigma})$ for sufficiently large $t$. Thanks anyway.
– Frobenius
Nov 23 at 14:00