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Question Number 3250 by Filup last updated on 08/Dec/15

It is known that :

ζ (s) = \underset{i = 1}{\sum^{\infty}} i^{- s}

Prove that :

ζ (s) = \underset{i = 1}{\prod^{\infty}} (1 - \frac{1}{π {(i)}^{s}})

where π (n) = n th prime

π (1) = 2, π (2) = 3, π (3) = 5 \dots

Commented by 123456 last updated on 08/Dec/15

\sum_{n ⩾ 1} \frac{1}{n^{s}} = \prod_{p \in P} \frac{1}{1 - p^{- s}}

ζ (s) = \prod_{p \in P} \frac{1}{1 - p^{- s}}

\frac{1}{ζ (s)} = \prod_{p \in P} 1 - p^{- s}

Answered by prakash jain last updated on 08/Dec/15

ζ (s) = \frac{1}{1^{s}} + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + . .

\frac{1}{2^{s}} ζ (s) = \frac{1}{2^{s}} + \frac{1}{4^{s}} + \frac{1}{6^{s}} + . .

(1 - \frac{1}{2^{s}}) ζ (s) = 1 + \frac{1}{3^{s}} + \frac{1}{5^{s}} + \dots (a l l f a c t o r s o f \frac{1}{2^{s}} r e m o v e d)

(1 - \frac{1}{3^{s}}) (1 - \frac{1}{2^{s}}) ζ (s) = 1 + \frac{1}{5^{s}} + \dots (a l l m u l t i p l e s o f 3^{s} r e m o v e d)

r e p e a t i n g t h e p r o c e s s f o r a l l p r i m e s n u m b e r

e a c h s t e p r e m o v e m u l t i p l e s o f a p r i m e n u m b e r

ζ (s) \underset{i = 1}{\prod^{\infty}} (1 - \frac{1}{π {(i)}^{s}}) = 1

\frac{1}{ζ (s)} = \underset{i = 1}{\prod^{\infty}} (1 - \frac{1}{π {(i)}^{s}})

\frac{1}{ζ (s)} = \underset{i = 1}{\prod^{\infty}} (1 - π {(i)}^{- s})

As mentioned in the question π (i) is i th prime .

Usual convention is π is prime density .

Commented by RasheedAhmad last updated on 08/Dec/15

ζ (s) = \frac{1}{2^{s}} + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + . .

\overset{?}{\Rightarrow} \frac{1}{2^{s}} ζ (s) = \frac{1}{2^{s}} + \frac{1}{4^{s}} + \frac{1}{6^{s}} + . .

Commented by prakash jain last updated on 08/Dec/15

t y p o . c o r r e c t e d .

Commented by Filup last updated on 08/Dec/15

A m a z i n g!

Commented by Filup last updated on 08/Dec/15

I am aware π (i) is the p r i m e d e n s i t y .

But I have seen π (i) be used on several

occasions as the n th prime .

But If it is more aesthetic, i^{'} ll stick to

P (i) = i th prime :)

Thank you regardess!

Answered by 123456 last updated on 08/Dec/15

you can go by other way too

note that

\frac{1}{1 - q} = 1 + q + q^{2} + \dots (∣ q ∣< 1)

so

\frac{1}{1 - p^{- s}} = 1 + \frac{1}{p^{s}} + \frac{1}{p^{2 s}} + \dots (s > 1 and since p > 1 \Rightarrow 1 / p < 1)

\frac{1}{1 - p^{- s}} = \underset{n = 0}{\sum^{+ \infty}} p^{- n s}

\prod_{p \in P} \frac{1}{1 - p^{- s}} = \prod_{p \in P} \underset{n = 0}{\sum^{+ \infty}} p^{- n s}

by unique factorization we have

\prod_{p \in P} \frac{1}{1 - p^{- s}} = \underset{n = 0}{\sum^{+ \infty}} n^{- s} = ζ (s)

\frac{1}{ζ (s)} = \prod_{p \in P} 1 - p^{- s}

Commented by Filup last updated on 08/Dec/15

A w e s o m e!

Commented by prakash jain last updated on 20/May/17

123456, are u still using app ?