Is there an analytic function with zeroes only at $-2n$, and zeroes at $frac12pm it$, and further, symmetric zeroes within the critical strip?
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Is there an analytic function with zeroes only at:
- every $-2n$,
- $frac12pm it$, and
- at least one at $frac12pmepsilonpm it$ where $0<epsilon<frac12, tneq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?
A answer assuming the Riemann hypothesis true would be fine.
To my mind the uniqueness of any analytic continuation of $zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.
If not, uniqueness of $zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.
complex-analysis riemann-zeta riemann-hypothesis
asked Sep 5 at 9:06
Robert Frost
3,9511036
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up vote
0
down vote
favorite
1
Is there an analytic function with zeroes only at:
- every $-2n$,
- $frac12pm it$, and
- at least one at $frac12pmepsilonpm it$ where $0<epsilon<frac12, tneq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?
A answer assuming the Riemann hypothesis true would be fine.
To my mind the uniqueness of any analytic continuation of $zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.
If not, uniqueness of $zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.
complex-analysis riemann-zeta riemann-hypothesis
asked Sep 5 at 9:06
Robert Frost
3,9511036
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
Is there an analytic function with zeroes only at:
- every $-2n$,
- $frac12pm it$, and
- at least one at $frac12pmepsilonpm it$ where $0<epsilon<frac12, tneq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?
A answer assuming the Riemann hypothesis true would be fine.
To my mind the uniqueness of any analytic continuation of $zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.
If not, uniqueness of $zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.
complex-analysis riemann-zeta riemann-hypothesis
asked Sep 5 at 9:06
Robert Frost
3,9511036
Is there an analytic function with zeroes only at:
- every $-2n$,
- $frac12pm it$, and
- at least one at $frac12pmepsilonpm it$ where $0<epsilon<frac12, tneq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?
A answer assuming the Riemann hypothesis true would be fine.
To my mind the uniqueness of any analytic continuation of $zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.
If not, uniqueness of $zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.
complex-analysis riemann-zeta riemann-hypothesis
complex-analysis riemann-zeta riemann-hypothesis
asked Sep 5 at 9:06
Robert Frost
3,9511036
asked Sep 5 at 9:06
Robert Frost
3,9511036
edited Sep 5 at 11:32
asked Sep 5 at 9:06
Robert Frost
3,9511036
asked Sep 5 at 9:06
Robert Frost
3,9511036
asked Sep 5 at 9:06
Robert Frost
3,9511036
3,9511036
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
down vote
accepted
Take $zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)$.
Every condition that you mentioned holds.
edited Sep 5 at 11:31
Robert Frost
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
 |Â
show 2 more comments
up vote
-1
down vote
I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.
answered Sep 5 at 9:35
Uri George Peterzil
578
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Take $zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)$.
Every condition that you mentioned holds.
edited Sep 5 at 11:31
Robert Frost
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
 |Â
show 2 more comments
up vote
3
down vote
accepted
Take $zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)$.
Every condition that you mentioned holds.
edited Sep 5 at 11:31
Robert Frost
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Take $zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)$.
Every condition that you mentioned holds.
edited Sep 5 at 11:31
Robert Frost
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
Take $zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)$.
Every condition that you mentioned holds.
edited Sep 5 at 11:31
Robert Frost
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
edited Sep 5 at 11:31
Robert Frost
3,9511036
edited Sep 5 at 11:31
Robert Frost
3,9511036
edited Sep 5 at 11:31
Robert Frost
3,9511036
3,9511036
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
answered Sep 5 at 9:17
José Carlos Santos
122k16101186
122k16101186
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
 |Â
show 2 more comments
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $zeta$ functions plus $frac14$.
– José Carlos Santos
Sep 5 at 9:34
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost I've edited my answer. What do you think now?
– José Carlos Santos
Sep 5 at 10:28
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
@RobertFrost What about$$zeta(z)left(z-frac14-iright)left(z-frac14+iright)left(z-frac34-iright)left(z-frac34+iright)?$$But note that the extra zeros of my previous answer ($frac14$ and $frac34$) were already symmetric with respect to the real axis.
– José Carlos Santos
Sep 5 at 10:36
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $tneq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $zeta(z)g(a+bi)$ and $zeta(z)/g(a+bi)$ look like?
– Robert Frost
Sep 5 at 11:05
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
@RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one.
– José Carlos Santos
Sep 5 at 11:24
 |Â
show 2 more comments
up vote
-1
down vote
I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.
answered Sep 5 at 9:35
Uri George Peterzil
578
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
add a comment |Â
up vote
-1
down vote
I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.
answered Sep 5 at 9:35
Uri George Peterzil
578
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.
answered Sep 5 at 9:35
Uri George Peterzil
578
I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.
answered Sep 5 at 9:35
Uri George Peterzil
578
answered Sep 5 at 9:35
Uri George Peterzil
578
answered Sep 5 at 9:35
Uri George Peterzil
578
answered Sep 5 at 9:35
Uri George Peterzil
578
578
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
add a comment |Â
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH.
– Robert Frost
Sep 5 at 9:39
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
That doesn't sound right.
– Uri George Peterzil
Sep 5 at 15:34
add a comment |Â
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