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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...
Itisknownthat:ζ(s)=i=1isProvethat:ζ(s)=i=1(11π(i)s)whereπ(n)=nthprimeπ(1)=2,π(2)=3,π(3)=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)
n11ns=pP11psζ(s)=pP11ps1ζ(s)=pP1ps
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.
ζ(s)=11s+12s+13s+14s+..12sζ(s)=12s+14s+16s+..(112s)ζ(s)=1+13s+15s+(allfactorsof12sremoved)(113s)(112s)ζ(s)=1+15s+(allmultiplesof3sremoved)repeatingtheprocessforallprimesnumbereachstepremovemultiplesofaprimenumberζ(s)i=1(11π(i)s)=11ζ(s)=i=1(11π(i)s)1ζ(s)=i=1(1π(i)s)Asmentionedinthequestionπ(i)isithprime.Usualconventionisπisprimedensity.
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 )+..
ζ(s)=12s+12s+13s+14s+..?12sζ(s)=12s+14s+16s+..
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!
Iamawareπ(i)istheprimedensity.ButIhaveseenπ(i)beusedonseveraloccasionsasthenthprime.ButIfitismoreaesthetic,illsticktoP(i)=ithprime:)Thankyouregardess!
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)
youcangobyotherwaytoonotethat11q=1+q+q2+(q∣<1)so11ps=1+1ps+1p2s+(s>1andsincep>11/p<1)11ps=+n=0pnspP11ps=pP+n=0pnsbyuniquefactorizationwehavepP11ps=+n=0ns=ζ(s)1ζ(s)=pP1ps
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?
123456,areustillusingapp?

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