Symmetry of zeros in the critical strip for Riemann Hypothesis
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If it can be proven that there are no zeros for real values greater than $0.5$ in the critical strip, does this prove that there are no zeros in the critical strip having a real value of less than $0.5$?
riemann-zeta riemann-hypothesis
edited Jan 17 at 12:00
idriskameni
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
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add a comment |
$begingroup$
If it can be proven that there are no zeros for real values greater than $0.5$ in the critical strip, does this prove that there are no zeros in the critical strip having a real value of less than $0.5$?
riemann-zeta riemann-hypothesis
edited Jan 17 at 12:00
idriskameni
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
$endgroup$
add a comment |
$begingroup$
If it can be proven that there are no zeros for real values greater than $0.5$ in the critical strip, does this prove that there are no zeros in the critical strip having a real value of less than $0.5$?
riemann-zeta riemann-hypothesis
edited Jan 17 at 12:00
idriskameni
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
$endgroup$
If it can be proven that there are no zeros for real values greater than $0.5$ in the critical strip, does this prove that there are no zeros in the critical strip having a real value of less than $0.5$?
riemann-zeta riemann-hypothesis
riemann-zeta riemann-hypothesis
edited Jan 17 at 12:00
idriskameni
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
edited Jan 17 at 12:00
idriskameni
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
edited Jan 17 at 12:00
idriskameni
583319
edited Jan 17 at 12:00
idriskameni
583319
edited Jan 17 at 12:00
idriskameni
583319
583319
asked Jan 17 at 11:41
Archie CallArchie Call
1
asked Jan 17 at 11:41
Archie CallArchie Call
1
asked Jan 17 at 11:41
Archie CallArchie Call
1
1
add a comment |
add a comment |
1 Answer
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Yes, it would prove that.
The Riemann $zeta$ function is known to satisfy the following functional equation:
$$
zeta(s) = 2^spi^{s-1}sinleft(frac{pi s}2right)Gamma(1-s)zeta(1-s)
$$
Assuming $z_0$ is a root in the critical strip which is not on the critical line, then $1-z_0$ is also a root on the critical strip which is on the opposite side of the critical line. So if $z_0$ has real part less than $0.5$, then $1-z_0$ has real part greater than $0.5$, and vice versa. Of course this also means that the absence of roots to the right of the critical strip imples the absence of roots on the left side.
This phenomenon of roots off the critical line appearing in pairs is actually why the Riemann hypothesis is so important. One manifestation is the following: If $pi:Bbb R^+to Bbb N$ is the prime counting function (i.e. $pi(x)$ is the number of primes less than or equal to $x$), then we have
$$
pi(x) = int_2^xfrac{dx}{ln x} + text{correction terms}
$$
The corrections include an infinite sum with one term for each non-trivial root of the Riemann $zeta$ function. If there are roots not on the critical line, then this symmetry means there are actually two roots of that magnitude, meaining we have more, larger correction terms.
answered Jan 17 at 11:54
ArthurArthur
115k7116198
$endgroup$
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
add a comment |
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$begingroup$
Yes, it would prove that.
The Riemann $zeta$ function is known to satisfy the following functional equation:
$$
zeta(s) = 2^spi^{s-1}sinleft(frac{pi s}2right)Gamma(1-s)zeta(1-s)
$$
Assuming $z_0$ is a root in the critical strip which is not on the critical line, then $1-z_0$ is also a root on the critical strip which is on the opposite side of the critical line. So if $z_0$ has real part less than $0.5$, then $1-z_0$ has real part greater than $0.5$, and vice versa. Of course this also means that the absence of roots to the right of the critical strip imples the absence of roots on the left side.
This phenomenon of roots off the critical line appearing in pairs is actually why the Riemann hypothesis is so important. One manifestation is the following: If $pi:Bbb R^+to Bbb N$ is the prime counting function (i.e. $pi(x)$ is the number of primes less than or equal to $x$), then we have
$$
pi(x) = int_2^xfrac{dx}{ln x} + text{correction terms}
$$
The corrections include an infinite sum with one term for each non-trivial root of the Riemann $zeta$ function. If there are roots not on the critical line, then this symmetry means there are actually two roots of that magnitude, meaining we have more, larger correction terms.
answered Jan 17 at 11:54
ArthurArthur
115k7116198
$endgroup$
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
add a comment |
$begingroup$
Yes, it would prove that.
The Riemann $zeta$ function is known to satisfy the following functional equation:
$$
zeta(s) = 2^spi^{s-1}sinleft(frac{pi s}2right)Gamma(1-s)zeta(1-s)
$$
Assuming $z_0$ is a root in the critical strip which is not on the critical line, then $1-z_0$ is also a root on the critical strip which is on the opposite side of the critical line. So if $z_0$ has real part less than $0.5$, then $1-z_0$ has real part greater than $0.5$, and vice versa. Of course this also means that the absence of roots to the right of the critical strip imples the absence of roots on the left side.
This phenomenon of roots off the critical line appearing in pairs is actually why the Riemann hypothesis is so important. One manifestation is the following: If $pi:Bbb R^+to Bbb N$ is the prime counting function (i.e. $pi(x)$ is the number of primes less than or equal to $x$), then we have
$$
pi(x) = int_2^xfrac{dx}{ln x} + text{correction terms}
$$
The corrections include an infinite sum with one term for each non-trivial root of the Riemann $zeta$ function. If there are roots not on the critical line, then this symmetry means there are actually two roots of that magnitude, meaining we have more, larger correction terms.
answered Jan 17 at 11:54
ArthurArthur
115k7116198
$endgroup$
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
add a comment |
$begingroup$
Yes, it would prove that.
The Riemann $zeta$ function is known to satisfy the following functional equation:
$$
zeta(s) = 2^spi^{s-1}sinleft(frac{pi s}2right)Gamma(1-s)zeta(1-s)
$$
Assuming $z_0$ is a root in the critical strip which is not on the critical line, then $1-z_0$ is also a root on the critical strip which is on the opposite side of the critical line. So if $z_0$ has real part less than $0.5$, then $1-z_0$ has real part greater than $0.5$, and vice versa. Of course this also means that the absence of roots to the right of the critical strip imples the absence of roots on the left side.
This phenomenon of roots off the critical line appearing in pairs is actually why the Riemann hypothesis is so important. One manifestation is the following: If $pi:Bbb R^+to Bbb N$ is the prime counting function (i.e. $pi(x)$ is the number of primes less than or equal to $x$), then we have
$$
pi(x) = int_2^xfrac{dx}{ln x} + text{correction terms}
$$
The corrections include an infinite sum with one term for each non-trivial root of the Riemann $zeta$ function. If there are roots not on the critical line, then this symmetry means there are actually two roots of that magnitude, meaining we have more, larger correction terms.
answered Jan 17 at 11:54
ArthurArthur
115k7116198
$endgroup$
Yes, it would prove that.
The Riemann $zeta$ function is known to satisfy the following functional equation:
$$
zeta(s) = 2^spi^{s-1}sinleft(frac{pi s}2right)Gamma(1-s)zeta(1-s)
$$
Assuming $z_0$ is a root in the critical strip which is not on the critical line, then $1-z_0$ is also a root on the critical strip which is on the opposite side of the critical line. So if $z_0$ has real part less than $0.5$, then $1-z_0$ has real part greater than $0.5$, and vice versa. Of course this also means that the absence of roots to the right of the critical strip imples the absence of roots on the left side.
This phenomenon of roots off the critical line appearing in pairs is actually why the Riemann hypothesis is so important. One manifestation is the following: If $pi:Bbb R^+to Bbb N$ is the prime counting function (i.e. $pi(x)$ is the number of primes less than or equal to $x$), then we have
$$
pi(x) = int_2^xfrac{dx}{ln x} + text{correction terms}
$$
The corrections include an infinite sum with one term for each non-trivial root of the Riemann $zeta$ function. If there are roots not on the critical line, then this symmetry means there are actually two roots of that magnitude, meaining we have more, larger correction terms.
answered Jan 17 at 11:54
ArthurArthur
115k7116198
edited Jan 17 at 12:11
answered Jan 17 at 11:54
ArthurArthur
115k7116198
answered Jan 17 at 11:54
ArthurArthur
115k7116198
answered Jan 17 at 11:54
ArthurArthur
115k7116198
115k7116198
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
add a comment |
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
$begingroup$
$sum_{p^k le x} log p = x- sum_rho frac{x^rho}{rho}+C+O(x^{-1})$. If $Re(rho) > 1/2$ then $x^{1-rho}$ is much smaller than $x^rho$ so we don't really care of it. That's part of the problem : the RH fails for most Euler products without functional equation, and the functional equation has only very hardly sensible consequences on the distribution of the primes. Together with the explicit formula, most of the information we have is the density of zeros, and that "exponentiating the primes" gives the integers, whose distribution is well-understood.
$endgroup$
– reuns
Jan 17 at 16:29
add a comment |
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