Partial Euler productEuler product on the critical lineThe partial sum and partial product of $zeta$functionEuler product for Riemann zeta and analytic continuationEuler product of Dirichlet seriesWhy do you need to prove the error term goes to zero for the complete derivation of the Euler Product Formula?Convergence of the Euler productConvergence of Euler Product for Leibniz Pi FormulaDoes the Euler product stand for $a(n)=rad(n)$?Is there an Euler product for the Prime Zeta function?How does one compute the Euler product for the Dirichlet Beta function?
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Partial Euler product
Euler product on the critical lineThe partial sum and partial product of $zeta$functionEuler product for Riemann zeta and analytic continuationEuler product of Dirichlet seriesWhy do you need to prove the error term goes to zero for the complete derivation of the Euler Product Formula?Convergence of the Euler productConvergence of Euler Product for Leibniz Pi FormulaDoes the Euler product stand for $a(n)=rad(n)$?Is there an Euler product for the Prime Zeta function?How does one compute the Euler product for the Dirichlet Beta function?
$begingroup$
The Riemann Zeta function defined as $$zeta(s) = sum_n=1^infty n^-s$$ For $Re(s)>1$ is convergent and admits the Euler product representation $$zeta(s) = prod_p (1-p^-s)^-1$$ For partial Euler product $ prod_p<x(1-p^-s)^-1$ we obviously will have $$zeta_x(s) = prod_p<x (1-p^-s)^-1$$where $zeta_x(s)$ is a $zeta(s)$ with "thrown out" summands with $n$ having in fuctorisation $pgeq x$ My question: How can I write with correct math notation this function as a Dirichlet series, something like: $$zeta_x(s) = sum_n=1^infty frac a(n)n^s $$ where $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$
sequences-and-series analytic-number-theory riemann-zeta euler-product
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
$endgroup$
add a comment |
$begingroup$
The Riemann Zeta function defined as $$zeta(s) = sum_n=1^infty n^-s$$ For $Re(s)>1$ is convergent and admits the Euler product representation $$zeta(s) = prod_p (1-p^-s)^-1$$ For partial Euler product $ prod_p<x(1-p^-s)^-1$ we obviously will have $$zeta_x(s) = prod_p<x (1-p^-s)^-1$$where $zeta_x(s)$ is a $zeta(s)$ with "thrown out" summands with $n$ having in fuctorisation $pgeq x$ My question: How can I write with correct math notation this function as a Dirichlet series, something like: $$zeta_x(s) = sum_n=1^infty frac a(n)n^s $$ where $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$
sequences-and-series analytic-number-theory riemann-zeta euler-product
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
$endgroup$
add a comment |
$begingroup$
The Riemann Zeta function defined as $$zeta(s) = sum_n=1^infty n^-s$$ For $Re(s)>1$ is convergent and admits the Euler product representation $$zeta(s) = prod_p (1-p^-s)^-1$$ For partial Euler product $ prod_p<x(1-p^-s)^-1$ we obviously will have $$zeta_x(s) = prod_p<x (1-p^-s)^-1$$where $zeta_x(s)$ is a $zeta(s)$ with "thrown out" summands with $n$ having in fuctorisation $pgeq x$ My question: How can I write with correct math notation this function as a Dirichlet series, something like: $$zeta_x(s) = sum_n=1^infty frac a(n)n^s $$ where $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$
sequences-and-series analytic-number-theory riemann-zeta euler-product
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
$endgroup$
The Riemann Zeta function defined as $$zeta(s) = sum_n=1^infty n^-s$$ For $Re(s)>1$ is convergent and admits the Euler product representation $$zeta(s) = prod_p (1-p^-s)^-1$$ For partial Euler product $ prod_p<x(1-p^-s)^-1$ we obviously will have $$zeta_x(s) = prod_p<x (1-p^-s)^-1$$where $zeta_x(s)$ is a $zeta(s)$ with "thrown out" summands with $n$ having in fuctorisation $pgeq x$ My question: How can I write with correct math notation this function as a Dirichlet series, something like: $$zeta_x(s) = sum_n=1^infty frac a(n)n^s $$ where $$a(n)=1, n= .... ?$$ $$a(n)=0, n= .... ?$$
sequences-and-series analytic-number-theory riemann-zeta euler-product
sequences-and-series analytic-number-theory riemann-zeta euler-product
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
asked Mar 16 at 10:27
Aleksey DruggistAleksey Druggist
14219
14219
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
$endgroup$
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
$endgroup$
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
add a comment |
$begingroup$
I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
$endgroup$
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
add a comment |
$begingroup$
I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
$endgroup$
I would say that $a(n) = 1$ if $n$ is $x$-smooth. See https://en.wikipedia.org/wiki/Smooth_number
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
answered Mar 16 at 10:33
Esteban CrespiEsteban Crespi
2,8761621
2,8761621
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
add a comment |
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
So $ zeta_x (s)= sum_n: x-smooth n^-s$ ?
$endgroup$
– Aleksey Druggist
Mar 16 at 11:15
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
Yes that's right
$endgroup$
– Esteban Crespi
Mar 16 at 11:57
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
$begingroup$
@AlekseyDruggist This is indeed the way we prove $sum_n=1^infty n^-s = prod_p frac11-p^-s$ for $Re(s) > 1$
$endgroup$
– reuns
Mar 16 at 12:58
add a comment |
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