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Solutions to functional equational of Riemann zeta function
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Is it possible to determine all the meromorphic functions $f : Bbb C to Bbb C$ which have a pole only at $s=1$, of order $1$, and such that
$$f(s) = 2^s pi^s-1 sin(pi s / 2) Gamma(1-s) f(1-s)$$
for all $s in Bbb C$?
Are they other solutions than multiples of $zeta$ ?
functional-equations riemann-zeta
edited Jul 23 at 14:08
Nosrati
19.4k41544
asked Jul 23 at 11:53
Alphonse
1,767622
add a comment |Â
up vote
1
down vote
favorite
Is it possible to determine all the meromorphic functions $f : Bbb C to Bbb C$ which have a pole only at $s=1$, of order $1$, and such that
$$f(s) = 2^s pi^s-1 sin(pi s / 2) Gamma(1-s) f(1-s)$$
for all $s in Bbb C$?
Are they other solutions than multiples of $zeta$ ?
functional-equations riemann-zeta
edited Jul 23 at 14:08
Nosrati
19.4k41544
asked Jul 23 at 11:53
Alphonse
1,767622
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is it possible to determine all the meromorphic functions $f : Bbb C to Bbb C$ which have a pole only at $s=1$, of order $1$, and such that
$$f(s) = 2^s pi^s-1 sin(pi s / 2) Gamma(1-s) f(1-s)$$
for all $s in Bbb C$?
Are they other solutions than multiples of $zeta$ ?
functional-equations riemann-zeta
edited Jul 23 at 14:08
Nosrati
19.4k41544
asked Jul 23 at 11:53
Alphonse
1,767622
Is it possible to determine all the meromorphic functions $f : Bbb C to Bbb C$ which have a pole only at $s=1$, of order $1$, and such that
$$f(s) = 2^s pi^s-1 sin(pi s / 2) Gamma(1-s) f(1-s)$$
for all $s in Bbb C$?
Are they other solutions than multiples of $zeta$ ?
functional-equations riemann-zeta
edited Jul 23 at 14:08
Nosrati
19.4k41544
asked Jul 23 at 11:53
Alphonse
1,767622
edited Jul 23 at 14:08
Nosrati
19.4k41544
edited Jul 23 at 14:08
Nosrati
19.4k41544
edited Jul 23 at 14:08
Nosrati
19.4k41544
19.4k41544
asked Jul 23 at 11:53
Alphonse
1,767622
asked Jul 23 at 11:53
Alphonse
1,767622
asked Jul 23 at 11:53
Alphonse
1,767622
1,767622
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
-1
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Hint: Consider
$$xi(s)=dfrac12s(s-1)pi^-fracs2Gamma(fracs2)zeta(s)$$
which is entire on $mathbbC$, let $xi(s)$ be every entire function so it's enough that
$$f(s)=dfracxi(s)dfrac12s(s-1)pi^-fracs2Gamma(fracs2)$$
answered Jul 23 at 12:07
Nosrati
19.4k41544
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
Hint: Consider
$$xi(s)=dfrac12s(s-1)pi^-fracs2Gamma(fracs2)zeta(s)$$
which is entire on $mathbbC$, let $xi(s)$ be every entire function so it's enough that
$$f(s)=dfracxi(s)dfrac12s(s-1)pi^-fracs2Gamma(fracs2)$$
answered Jul 23 at 12:07
Nosrati
19.4k41544
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
add a comment |Â
up vote
-1
down vote
Hint: Consider
$$xi(s)=dfrac12s(s-1)pi^-fracs2Gamma(fracs2)zeta(s)$$
which is entire on $mathbbC$, let $xi(s)$ be every entire function so it's enough that
$$f(s)=dfracxi(s)dfrac12s(s-1)pi^-fracs2Gamma(fracs2)$$
answered Jul 23 at 12:07
Nosrati
19.4k41544
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Hint: Consider
$$xi(s)=dfrac12s(s-1)pi^-fracs2Gamma(fracs2)zeta(s)$$
which is entire on $mathbbC$, let $xi(s)$ be every entire function so it's enough that
$$f(s)=dfracxi(s)dfrac12s(s-1)pi^-fracs2Gamma(fracs2)$$
answered Jul 23 at 12:07
Nosrati
19.4k41544
Hint: Consider
$$xi(s)=dfrac12s(s-1)pi^-fracs2Gamma(fracs2)zeta(s)$$
which is entire on $mathbbC$, let $xi(s)$ be every entire function so it's enough that
$$f(s)=dfracxi(s)dfrac12s(s-1)pi^-fracs2Gamma(fracs2)$$
answered Jul 23 at 12:07
Nosrati
19.4k41544
answered Jul 23 at 12:07
Nosrati
19.4k41544
answered Jul 23 at 12:07
Nosrati
19.4k41544
answered Jul 23 at 12:07
Nosrati
19.4k41544
19.4k41544
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
add a comment |Â
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
What do you mean by "let $xi(s)$ be every entire function", since you just defined $xi(s)$ above?!
– Alphonse
Jul 23 at 14:31
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
With arbitrary $xi(s)$ in formula, we obtain such $f$. for instance $xi(s)=e^s$.
– Nosrati
Jul 23 at 14:58
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
But if $xi$ is arbitrary, then the functional equation between $f(s)$ and $f(1-s)$ won't be satisfied in general. I really don't get the point of your answer, sorry.
– Alphonse
Jul 23 at 15:00
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
I think we indeed get $f=$ multiple of $zeta$ by Hamburger theorem (see here).
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
So your answer seems to contradict Hamburger theorem?!
– Alphonse
Jul 24 at 9:59
add a comment |Â
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