On the Taylor series of the logarithm of the Riemann zeta function.
0
$begingroup$
For $Re(s)>1$, the Riemann zeta function $zeta(s)$ is defined by
$$zeta(s) = sum_{n=1}^infty n^{-s},$$
or equivalently by the Euler product
$$zeta(s) = prod_{p} (1-p^{-s})^{-1}, $$
where $p$ runs over the entire set of primes. My question is, what is the Taylor series of $log zeta(s)$ ?
real-analysis number-theory riemann-zeta
edited Dec 17 '18 at 10:49
amWhy
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
$endgroup$
add a comment |
0
$begingroup$
For $Re(s)>1$, the Riemann zeta function $zeta(s)$ is defined by
$$zeta(s) = sum_{n=1}^infty n^{-s},$$
or equivalently by the Euler product
$$zeta(s) = prod_{p} (1-p^{-s})^{-1}, $$
where $p$ runs over the entire set of primes. My question is, what is the Taylor series of $log zeta(s)$ ?
real-analysis number-theory riemann-zeta
edited Dec 17 '18 at 10:49
amWhy
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
$endgroup$
2
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
2
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50
add a comment |
0
0
0
$begingroup$
For $Re(s)>1$, the Riemann zeta function $zeta(s)$ is defined by
$$zeta(s) = sum_{n=1}^infty n^{-s},$$
or equivalently by the Euler product
$$zeta(s) = prod_{p} (1-p^{-s})^{-1}, $$
where $p$ runs over the entire set of primes. My question is, what is the Taylor series of $log zeta(s)$ ?
real-analysis number-theory riemann-zeta
edited Dec 17 '18 at 10:49
amWhy
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
$endgroup$
For $Re(s)>1$, the Riemann zeta function $zeta(s)$ is defined by
$$zeta(s) = sum_{n=1}^infty n^{-s},$$
or equivalently by the Euler product
$$zeta(s) = prod_{p} (1-p^{-s})^{-1}, $$
where $p$ runs over the entire set of primes. My question is, what is the Taylor series of $log zeta(s)$ ?
real-analysis number-theory riemann-zeta
real-analysis number-theory riemann-zeta
edited Dec 17 '18 at 10:49
amWhy
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
edited Dec 17 '18 at 10:49
amWhy
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
edited Dec 17 '18 at 10:49
amWhy
1
edited Dec 17 '18 at 10:49
amWhy
1
edited Dec 17 '18 at 10:49
amWhy
1
1
asked Oct 8 '16 at 15:54
Calc.Calc.
12
asked Oct 8 '16 at 15:54
Calc.Calc.
12
asked Oct 8 '16 at 15:54
Calc.Calc.
12
12
2
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
2
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50
add a comment |
2
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
2
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50
2
2
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
2
2
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50
add a comment |
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2
$begingroup$
What's the Taylor Series of $logleft[1 + left(zeta(s) - 1 right) right]$?
$endgroup$
– ÍgjøgnumMeg
Oct 8 '16 at 15:57
$begingroup$
Perhaps you could try to take the Taylor series of $log(x)$ and put the Taylor series of $zeta(s)$ inside it followed by lots of expansions.
$endgroup$
– Simply Beautiful Art
Oct 8 '16 at 16:03
2
$begingroup$
The Taylor series of $log zeta(s)$ around what $s$ ? if it is around $s_0, Re(s_0) > 1$ then $$log zeta(s_0+s) = sum_{k=0}^infty s^k frac{1}{k!}sum_{p^m} frac{p^{-s_0 m} (-ln p^m)^k}{m}$$ with radius of convergence at least $Re(s_0) - 1$
$endgroup$
– reuns
Oct 8 '16 at 16:50