Need someone to show me how the Zeta function is equal to Euler's product formula?












1












$begingroup$


$zeta(x)$=$1+frac{1}{2^x}+frac{1}{3^x}+frac{1}{4^x}+frac{1}{5^x}+frac{1}{6^x}+cdots$
Euler's product formula states, for all $x$ greater than $1$ we have:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$
The right hand side of the equation is an infinite product of the form $1over 1-frac{1}{p^x}$ where $p$ runs through all prime numbers. Also the author states that if we think of the infinite product as a limit, it will get closer to the limit value and that is the Zeta function itself.
Can someone show me a proof of this:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$






riemann-zeta infinite-product euler-product







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  • 1




    $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – angryavian
    Nov 29 '18 at 0:29










  • $begingroup$
    Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 0:35










  • $begingroup$
    Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
    $endgroup$
    – reuns
    Nov 29 '18 at 4:40












  • $begingroup$
    @reuns not quite but I can see there is some connection though.
    $endgroup$
    – DDP
    Nov 29 '18 at 17:53






  • 1




    $begingroup$
    Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
    $endgroup$
    – reuns
    Nov 29 '18 at 19:22


















1












$begingroup$


$zeta(x)$=$1+frac{1}{2^x}+frac{1}{3^x}+frac{1}{4^x}+frac{1}{5^x}+frac{1}{6^x}+cdots$
Euler's product formula states, for all $x$ greater than $1$ we have:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$
The right hand side of the equation is an infinite product of the form $1over 1-frac{1}{p^x}$ where $p$ runs through all prime numbers. Also the author states that if we think of the infinite product as a limit, it will get closer to the limit value and that is the Zeta function itself.
Can someone show me a proof of this:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$






riemann-zeta infinite-product euler-product







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – angryavian
    Nov 29 '18 at 0:29










  • $begingroup$
    Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 0:35










  • $begingroup$
    Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
    $endgroup$
    – reuns
    Nov 29 '18 at 4:40












  • $begingroup$
    @reuns not quite but I can see there is some connection though.
    $endgroup$
    – DDP
    Nov 29 '18 at 17:53






  • 1




    $begingroup$
    Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
    $endgroup$
    – reuns
    Nov 29 '18 at 19:22
















1












1








1





$begingroup$


$zeta(x)$=$1+frac{1}{2^x}+frac{1}{3^x}+frac{1}{4^x}+frac{1}{5^x}+frac{1}{6^x}+cdots$
Euler's product formula states, for all $x$ greater than $1$ we have:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$
The right hand side of the equation is an infinite product of the form $1over 1-frac{1}{p^x}$ where $p$ runs through all prime numbers. Also the author states that if we think of the infinite product as a limit, it will get closer to the limit value and that is the Zeta function itself.
Can someone show me a proof of this:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$






riemann-zeta infinite-product euler-product







share|cite|improve this question









$endgroup$




$zeta(x)$=$1+frac{1}{2^x}+frac{1}{3^x}+frac{1}{4^x}+frac{1}{5^x}+frac{1}{6^x}+cdots$
Euler's product formula states, for all $x$ greater than $1$ we have:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$
The right hand side of the equation is an infinite product of the form $1over 1-frac{1}{p^x}$ where $p$ runs through all prime numbers. Also the author states that if we think of the infinite product as a limit, it will get closer to the limit value and that is the Zeta function itself.
Can someone show me a proof of this:
$zeta(x)$=$1over 1-frac{1}{2^x}$$times$$1over 1-frac{1}{3^x}$$times$$1over 1-frac{1}{5^x}$$times$$1over 1-frac{1}{7^x}$$times$$1over 1-frac{1}{11^x}$$times cdots$






riemann-zeta infinite-product euler-product




riemann-zeta infinite-product euler-product






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 0:26









DDPDDP

164




164








  • 1




    $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – angryavian
    Nov 29 '18 at 0:29










  • $begingroup$
    Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 0:35










  • $begingroup$
    Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
    $endgroup$
    – reuns
    Nov 29 '18 at 4:40












  • $begingroup$
    @reuns not quite but I can see there is some connection though.
    $endgroup$
    – DDP
    Nov 29 '18 at 17:53






  • 1




    $begingroup$
    Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
    $endgroup$
    – reuns
    Nov 29 '18 at 19:22
















  • 1




    $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – angryavian
    Nov 29 '18 at 0:29










  • $begingroup$
    Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
    $endgroup$
    – Eevee Trainer
    Nov 29 '18 at 0:35










  • $begingroup$
    Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
    $endgroup$
    – reuns
    Nov 29 '18 at 4:40












  • $begingroup$
    @reuns not quite but I can see there is some connection though.
    $endgroup$
    – DDP
    Nov 29 '18 at 17:53






  • 1




    $begingroup$
    Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
    $endgroup$
    – reuns
    Nov 29 '18 at 19:22










1




1




$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– angryavian
Nov 29 '18 at 0:29




$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– angryavian
Nov 29 '18 at 0:29












$begingroup$
Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
$endgroup$
– Eevee Trainer
Nov 29 '18 at 0:35




$begingroup$
Mathologer also has done his own communication of it on YouTube - youtube.com/watch?v=LFwSIdLSosI
$endgroup$
– Eevee Trainer
Nov 29 '18 at 0:35












$begingroup$
Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
$endgroup$
– reuns
Nov 29 '18 at 4:40






$begingroup$
Let $A_k$ be the set of integers ($ge 1$) whose prime factors are $le k$. Then $$sum_{n in A_k} n^{-s} = prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})= prod_{p le k} frac{1}{1-p^{-s}}$$ Is it clear to you ? Therefore $|zeta(s)- prod_{p le k} frac{1}{1-p^{-s}}| = ??$
$endgroup$
– reuns
Nov 29 '18 at 4:40














$begingroup$
@reuns not quite but I can see there is some connection though.
$endgroup$
– DDP
Nov 29 '18 at 17:53




$begingroup$
@reuns not quite but I can see there is some connection though.
$endgroup$
– DDP
Nov 29 '18 at 17:53




1




1




$begingroup$
Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
$endgroup$
– reuns
Nov 29 '18 at 19:22






$begingroup$
Expand the finite product $prod_{p le k} (1+sum_{r=1}^infty (p^r)^{-s})$. If $n = prod_i p_i^{r_i}$ then what will be the coefficient of $n^{-s}$ ?
$endgroup$
– reuns
Nov 29 '18 at 19:22












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