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Question Number 3250 by Filup last updated on 08/Dec/15
It is known that: ζ(s)=Σ_(i=1) ^∞ i^(−s) Prove that: ζ(s)=Π_(i=1) ^∞ (1−(1/(π(i)^s ))) where π(n)=nth prime π(1)=2, π(2)=3, π(3)=5...
$$\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}: \\ $$$$\zeta\left({s}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{−{s}} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\zeta\left({s}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\pi\left({i}\right)^{{s}} }\right) \\ $$$$\mathrm{where}\:\pi\left({n}\right)={n}\mathrm{th}\:\mathrm{prime} \\ $$$$\pi\left(\mathrm{1}\right)=\mathrm{2},\:\:\pi\left(\mathrm{2}\right)=\mathrm{3},\:\:\pi\left(\mathrm{3}\right)=\mathrm{5}… \\ $$
Commented by 123456 last updated on 08/Dec/15
Σ_(n≥1) (1/n^s )=Π_(p∈P) (1/(1−p^(−s) )) ζ(s)=Π_(p∈P) (1/(1−p^(−s) )) (1/(ζ(s)))=Π_(p∈P) 1−p^(−s)
$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{{s}} }=\underset{{p}\in\mathbb{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} } \\ $$$$\zeta\left({s}\right)=\underset{{p}\in\mathbb{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} } \\ $$$$\frac{\mathrm{1}}{\zeta\left({s}\right)}=\underset{{p}\in\mathbb{P}} {\prod}\mathrm{1}−{p}^{−{s}} \\ $$
Answered by prakash jain last updated on 08/Dec/15
ζ(s)=(1/1^s )+(1/2^s )+(1/3^s )+(1/4^s )+.. (1/2^s )ζ(s)=(1/2^s )+(1/4^s )+(1/6^s )+.. (1−(1/2^s ))ζ(s)=1+(1/3^s )+(1/5^s )+... (all factors of (1/2^s ) removed) (1−(1/3^s ))(1−(1/2^s ))ζ(s)=1+(1/5^s )+... (all multiples of 3^s removed) repeating the process for all primes number each step remove multiples of a prime number ζ(s)Π_(i=1) ^∞ (1−(1/(π(i)^s )))=1 (1/(ζ(s)))=Π_(i=1) ^∞ (1−(1/(π(i)^s ))) (1/(ζ(s)))=Π_(i=1) ^∞ (1−π(i)^(−s) ) As mentioned in the question π(i) is ith prime. Usual convention is π is prime density.
$$\zeta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+.. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{s}} }\zeta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+\frac{\mathrm{1}}{\mathrm{6}^{{s}} }+.. \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}} }\right)\zeta\left({s}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\frac{\mathrm{1}}{\mathrm{5}^{{s}} }+…\:\left({all}\:{factors}\:{of}\:\frac{\mathrm{1}}{\mathrm{2}^{{s}} }\:{removed}\right) \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{{s}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{s}} }\right)\zeta\left({s}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}^{{s}} }+…\:\left({all}\:{multiples}\:{of}\:\mathrm{3}^{{s}} \:{removed}\right) \\ $$$${repeating}\:{the}\:{process}\:{for}\:{all}\:{primes}\:{number} \\ $$$${each}\:{step}\:{remove}\:{multiples}\:{of}\:{a}\:{prime}\:{number} \\ $$$$\zeta\left({s}\right)\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\pi\left({i}\right)^{{s}} }\right)=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\zeta\left({s}\right)}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\pi\left({i}\right)^{{s}} }\right) \\ $$$$\frac{\mathrm{1}}{\zeta\left({s}\right)}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\pi\left({i}\right)^{−{s}} \right) \\ $$$$\mathrm{As}\:\mathrm{mentioned}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\pi\left({i}\right)\:\mathrm{is}\:{i}\mathrm{th}\:\mathrm{prime}. \\ $$$$\mathrm{Usual}\:\mathrm{convention}\:\mathrm{is}\:\pi\:\mathrm{is}\:\mathrm{prime}\:\mathrm{density}. \\ $$
Commented by RasheedAhmad last updated on 08/Dec/15
ζ(s)=(1/2^s )+(1/2^s )+(1/3^s )+(1/4^s )+.. ⇒^? (1/2^s )ζ(s)=(1/2^s )+(1/4^s )+(1/6^s )+..
$$\zeta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+.. \\ $$$$\overset{?} {\Rightarrow}\frac{\mathrm{1}}{\mathrm{2}^{{s}} }\zeta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+\frac{\mathrm{1}}{\mathrm{6}^{{s}} }+.. \\ $$
Commented by prakash jain last updated on 08/Dec/15
typo. corrected.
$${typo}.\:{corrected}. \\ $$
Commented by Filup last updated on 08/Dec/15
Amazing!
$${Amazing}! \\ $$
Commented by Filup last updated on 08/Dec/15
I am aware π(i) is the prime density. But I have seen π(i) be used on several occasions as the nth prime. But If it is more aesthetic, i′ll stick to P(i)=ith prime :) Thank you regardess!
$$\mathrm{I}\:\mathrm{am}\:\mathrm{aware}\:\pi\left({i}\right)\:\mathrm{is}\:\mathrm{the}\:{prime}\:{density}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\pi\left({i}\right)\:\mathrm{be}\:\mathrm{used}\:\mathrm{on}\:\mathrm{several} \\ $$$$\mathrm{occasions}\:\mathrm{as}\:\mathrm{the}\:{n}\mathrm{th}\:\mathrm{prime}. \\ $$$$\mathrm{But}\:\mathrm{If}\:\mathrm{it}\:\mathrm{is}\:\mathrm{more}\:\mathrm{aesthetic},\:\mathrm{i}'\mathrm{ll}\:\mathrm{stick}\:\mathrm{to} \\ $$$$\left.{P}\left({i}\right)={i}\mathrm{th}\:\mathrm{prime}\::\right) \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{regardess}! \\ $$
Answered by 123456 last updated on 08/Dec/15
you can go by other way too note that (1/(1−q))=1+q+q^2 +... (∣q∣<1) so (1/(1−p^(−s) ))=1+(1/p^s )+(1/p^(2s) )+... (s>1 and since p>1⇒1/p<1) (1/(1−p^(−s) ))=Σ_(n=0) ^(+∞) p^(−ns) Π_(p∈P) (1/(1−p^(−s) ))=Π_(p∈P) Σ_(n=0) ^(+∞) p^(−ns) by unique factorization we have Π_(p∈P) (1/(1−p^(−s) ))=Σ_(n=0) ^(+∞) n^(−s) =ζ(s) (1/(ζ(s)))=Π_(p∈P) 1−p^(−s)
$$\mathrm{you}\:\mathrm{can}\:\mathrm{go}\:\mathrm{by}\:\mathrm{other}\:\mathrm{way}\:\mathrm{too} \\ $$$$\mathrm{note}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{q}}=\mathrm{1}+{q}+{q}^{\mathrm{2}} +…\:\:\:\left(\mid{q}\mid<\mathrm{1}\right) \\ $$$$\mathrm{so} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} }=\mathrm{1}+\frac{\mathrm{1}}{{p}^{{s}} }+\frac{\mathrm{1}}{{p}^{\mathrm{2}{s}} }+…\:\left({s}>\mathrm{1}\:\mathrm{and}\:\mathrm{since}\:{p}>\mathrm{1}\Rightarrow\mathrm{1}/{p}<\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} }=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{p}^{−{ns}} \\ $$$$\underset{{p}\in\mathbb{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} }=\underset{{p}\in\mathbb{P}} {\prod}\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{p}^{−{ns}} \\ $$$$\mathrm{by}\:\mathrm{unique}\:\mathrm{factorization}\:\mathrm{we}\:\mathrm{have} \\ $$$$\underset{{p}\in\mathbb{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} }=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{n}^{−{s}} =\zeta\left({s}\right) \\ $$$$\frac{\mathrm{1}}{\zeta\left({s}\right)}=\underset{{p}\in\mathbb{P}} {\prod}\mathrm{1}−{p}^{−{s}} \\ $$
Commented by Filup last updated on 08/Dec/15
Awesome!
$${Awesome}! \\ $$
Commented by prakash jain last updated on 20/May/17
123456, are u still using app?
$$\mathrm{123456},\:\mathrm{are}\:\mathrm{u}\:\mathrm{still}\:\mathrm{using}\:\mathrm{app}? \\ $$

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