Mixing
Expression for Prime Zeta Function in terms of Mobius and Multiplicity
up vote
0
down vote
favorite
A few years ago I had an idea but was a little unsure, so I asked a question and got a fantastic answer. See An algebraic manipulation of the Zeta function
The answer to the above-mentioned question reveals an interesting expression involving the Prime zeta function:
$sinmathbb{R}$, such that $s > 1$
$P(s) = sum_p frac{1}{p^{s}}$
The interesting expression is:
$P(s)zeta(s) = sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Where $omega(n)$ is the number of distinct prime divisors of n.
The zeta function can be expressed in terms of the Mobius function (See eqn 69: http://mathworld.wolfram.com/RiemannZetaFunction.html):
$frac{1}{zeta(s)} = sum_{n = 1}^infty frac{mu(n)}{n^{s}}$
So, with this in mind, the interesting expression becomes:
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Now this looks very similar, to me at least, to the more well known expression for the prime zeta function (See eqn 10: http://mathworld.wolfram.com/PrimeZetaFunction.html):
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)]$
So my question is can we derive any "interesting" formulas from this:
$sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)] = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
I understand that "interesting" is open to interpretation, I am just interested to see what results people can come to from this (if any). Thank you in advance.
riemann-zeta
asked 3 hours ago
KnowsNothing
204
add a comment |
up vote
0
down vote
favorite
A few years ago I had an idea but was a little unsure, so I asked a question and got a fantastic answer. See An algebraic manipulation of the Zeta function
The answer to the above-mentioned question reveals an interesting expression involving the Prime zeta function:
$sinmathbb{R}$, such that $s > 1$
$P(s) = sum_p frac{1}{p^{s}}$
The interesting expression is:
$P(s)zeta(s) = sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Where $omega(n)$ is the number of distinct prime divisors of n.
The zeta function can be expressed in terms of the Mobius function (See eqn 69: http://mathworld.wolfram.com/RiemannZetaFunction.html):
$frac{1}{zeta(s)} = sum_{n = 1}^infty frac{mu(n)}{n^{s}}$
So, with this in mind, the interesting expression becomes:
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Now this looks very similar, to me at least, to the more well known expression for the prime zeta function (See eqn 10: http://mathworld.wolfram.com/PrimeZetaFunction.html):
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)]$
So my question is can we derive any "interesting" formulas from this:
$sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)] = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
I understand that "interesting" is open to interpretation, I am just interested to see what results people can come to from this (if any). Thank you in advance.
riemann-zeta
asked 3 hours ago
KnowsNothing
204
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A few years ago I had an idea but was a little unsure, so I asked a question and got a fantastic answer. See An algebraic manipulation of the Zeta function
The answer to the above-mentioned question reveals an interesting expression involving the Prime zeta function:
$sinmathbb{R}$, such that $s > 1$
$P(s) = sum_p frac{1}{p^{s}}$
The interesting expression is:
$P(s)zeta(s) = sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Where $omega(n)$ is the number of distinct prime divisors of n.
The zeta function can be expressed in terms of the Mobius function (See eqn 69: http://mathworld.wolfram.com/RiemannZetaFunction.html):
$frac{1}{zeta(s)} = sum_{n = 1}^infty frac{mu(n)}{n^{s}}$
So, with this in mind, the interesting expression becomes:
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Now this looks very similar, to me at least, to the more well known expression for the prime zeta function (See eqn 10: http://mathworld.wolfram.com/PrimeZetaFunction.html):
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)]$
So my question is can we derive any "interesting" formulas from this:
$sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)] = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
I understand that "interesting" is open to interpretation, I am just interested to see what results people can come to from this (if any). Thank you in advance.
riemann-zeta
asked 3 hours ago
KnowsNothing
204
A few years ago I had an idea but was a little unsure, so I asked a question and got a fantastic answer. See An algebraic manipulation of the Zeta function
The answer to the above-mentioned question reveals an interesting expression involving the Prime zeta function:
$sinmathbb{R}$, such that $s > 1$
$P(s) = sum_p frac{1}{p^{s}}$
The interesting expression is:
$P(s)zeta(s) = sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Where $omega(n)$ is the number of distinct prime divisors of n.
The zeta function can be expressed in terms of the Mobius function (See eqn 69: http://mathworld.wolfram.com/RiemannZetaFunction.html):
$frac{1}{zeta(s)} = sum_{n = 1}^infty frac{mu(n)}{n^{s}}$
So, with this in mind, the interesting expression becomes:
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
Now this looks very similar, to me at least, to the more well known expression for the prime zeta function (See eqn 10: http://mathworld.wolfram.com/PrimeZetaFunction.html):
$P(s) = sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)]$
So my question is can we derive any "interesting" formulas from this:
$sum_{k = 1}^infty frac{mu(k)}{k} ln[zeta(ks)] = sum_{k = 1}^infty frac{mu(k)}{k^{s}} sum_{n = 2}^infty frac{omega(n)}{n^{s}}$
I understand that "interesting" is open to interpretation, I am just interested to see what results people can come to from this (if any). Thank you in advance.
riemann-zeta
riemann-zeta
asked 3 hours ago
KnowsNothing
204
asked 3 hours ago
KnowsNothing
204
asked 3 hours ago
KnowsNothing
204
asked 3 hours ago
KnowsNothing
204
asked 3 hours ago
KnowsNothing
204
204
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004664%2fexpression-for-prime-zeta-function-in-terms-of-mobius-and-multiplicity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
